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HEAT AND MASS TRANSFER LABORATORY MANUAL

V Semester (17BT5DLHMT)

DAYANANDA SAGAR COLLEGE OF ENGINEERING

Accredited by National Assessment & Accreditation Council (NAAC) with ’A’ Grade (An Autonomous Institution affiliated to Visvesvaraya Technological University, Belagavi

& ISO 9001:2008 Certified)

DEPARTMENT OF BIOTECHNOLOGY SHAVIGE MALLESWARA HILLS , KUMARASWAMY LAYOUT

BENGALURU-560078

Name of the Student : Semester /Section : USN : Batch :

Heat and Mass Transfer Lab: [2017]

Dept. of Biotechnology, Dayananda Sagar College of Engineering Bengaluru

Vision of the Institute

To impart quality technical education with a focus on Research and Innovation emphasizing on Development of Sustainable and Inclusive Technology for the benefit of society.

Mission of the Institute

To provide an environment that enhances creativity and Innovation in pursuit of Excellence.

To nurture teamwork in order to transform individuals as responsible leaders and entrepreneurs.

To train the students to the changing technical scenario and make them to understand the importance of Sustainable and Inclusive technologies.



HEAT AND MASS TRANSFER LABORATORY MANUAL

V Semester (17BT5DLHMT)

DAYANANDA SAGAR COLLEGE OF ENGINEERING (An Autonomous Institution affiliated to Visvesvaraya Technological University, Belagavi)

DEPARTMENT OF BIOTECHNOLOGY SHAVIGE MALLESWARA HILLS

KUMARASWAMY LAYOUT BENGALURU-560078

Name of the Student : Semester /Section : USN : Batch :




DEPARTMENT OF BIOTECHNOLOGY, BENGALURU-560078

VISION OF THE DEPARTMENT

To impart quality education, training and research in multidisciplinary domains of biotechnology for the benefit of society and environment.

MISSION OF THE DEPARTMENT

To provide globally acceptable technical education in the field of biotechnology by encouraging innovative thinking with practical insights.

To promote team work and nurture students to serve society with ethical and environmental responsibilities.

To foster students for higher studies, R&D activities and professional career in emerging trends of biotechnology.

PROGRAMME EDUCATIONAL OBJECTIVES [PEOs]

PEO 1: Skill - Enable our graduates to identify, analyse and solve industrial and environmental problems by implementing acquired skills. PEO 2: Career - Encourage our graduates to apply their engineering knowledge as an individual or in a team to excel in higher studies, research, teaching and industry. PEO 3: Lifelong learning - Instil in our graduates a desire to engage in lifelong learning that will foster their career with an impact on society.

PEO 4: Ethics - Inculcate in our graduates to develop high level of professionalism and ethical

attitude with awareness of current issues in relation to safety, health and environment.

PROGRAMME SPECIFIC OUTCOMES [PSOs]

Graduates will be able to

Apply engineering principles to biological systems.

Analyse the genome and proteome by advanced molecular techniques and computational tools to address major challenges in Pharma and Health.

Develop eco-friendly solutions to address complex environmental problems.




DEPARTMENT OF BIOTECHNOLOGY, BENGALURU-560078

HEAT AND MASS TRANSFER LABORATORY (SYLLABUS) V SEMESTER B. E (BT)

Sub. Code: 17BT5DLHMT IA Marks : 25

Hrs/Week : 3 Exam Hrs : 3

Total Hrs: 36 Exam Marks : 50

Course Objectives: 1. To calculate Cold fluid, hot fluid and overall Heat transfer co-efficient for Double pipe and Shell &

Tube Heat Exchanger.

2. To determine the value of diffusion co-efficients, prove Rayliegh’s equation and efficiency of

extraction using the fundamental knowledge of Mass transfer.

3. To analyse and determine the design parameters in condensation equipment by using heat transfer

concepts.

4. To apply the basic principles of Heat & Mass Transfer to develop solutions for complex problems

engaging themselves in lifelong learning.

Syllabus: 1. Double pipe heat exchanger

2. Shell and tube exchanger

3. Vertical condenser

4. Horizontal condenser

5. Transient heat conduction

6. Diffusivity

7. Liquid – Liquid extraction

8. Simple distillation

9. Steam distillation

10. Centrifugation

11. Ethanol estimation

12. Homogeniser

Course outcomes: After completion of the course, the graduates will be able to

1. Compute the Heat transfer co-efficient for Heat Exchanger.

2. Analyze and compute the Boiling Heat and overall transfer co-efficient using fundamentals of Heat

transfer.

3. Compute and Estimate the protein content using Homogeniser.

4. Determine the value of diffusion coefficients and efficiency of extraction using the fundamental

knowledge of Mass transfer.

5. Compute the design parameters in condensation equipment by using heat transfer concepts.

6. Apply the basic principles of Heat & Mass Transfer to develop solutions for complex problems

engaging themselves in lifelong learning.




DEPARTMENT OF BIOTECHNOLOGY

BIOPROCESS CONTROL & AUTOMATION LABORATORY (10BTL67)

I - CYCLE

1. Determination of the cold fluid side heat transfer co – efficient, hot fluid side heat transfer co –

efficient and overall heat transfer co – efficient for Double Pipe Heat Exchanger.

2. Determination of the cold fluid side heat transfer co – efficient, hot fluid side heat transfer

co – efficient and overall heat transfer co – efficient for Shell and Tube Heat Exchanger


co – efficient and overall heat transfer co – efficient for Vertical Condenser.


co – efficient and overall heat transfer co – efficient for Horizontal Condenser.

5. Verification of experimental predicted value of temperature with theoretically predicted

values and to plot the graph of temperature Vs. time.

6. Estimation of the amount of ethanol present in an unknown sample colorometrically.

II – CYCLE

1. Determination of the diffusion co – efficient (diffusivity) of organic vapor in air.

2. Verification of Rayleigh’s equation by differentially distilling the given binary mixture.

(System: Organic sample - Water).

3. Determination of Vaporization efficiency and Thermal efficiency and Studying the

characteristics of steam distillation.

4. Determination of the liquid – liquid equilibrium for ternary system (Benzene - Acetic Acid).

5. Disruption of yeast cells by mechanical method using Homogeniser and estimation of total

protein content by Biuret method.

6. Determination of the percentage of recovery of yeast cells by centrifugation for the

following RPM.



DAYANANDA SAGAR COLLEGE OF ENGINEERING DEPARTMENT OF BIOTECHNOLOGY

BENGALURU – 560078

DO’s

Adhere and follow timings, proper dress code with appropriate foot wear. Bags and other personal items must be stored in designated place. Come prepared with the viva, procedure, and other details of the experiment. Secure long hair, loose clothing & know safety and emergency procedures. Do check for the correct ranges/rating and carry one meter/instrument at a time

Inspect all equipment/meters for damage prior to use Conduct the experiments accurately as directed by the teacher. Immediately report any sparks/ accidents/ injuries/ any other untoward incident to the

faculty /instructor. Handle the apparatus/meters/computers gently and with care In case of an emergency or accident, follow the safety procedure. Switch OFF the power supply after completion of experiment

DONT’s The use of mobile/ any other personal electronic gadgets is prohibited in the laboratory. Do not make noise in the Laboratory & do not sit on experiment table. Do not make loose connections and avoid overlapping of wires Don’t switch on power supply without prior permission from the concerned staff. Never point/touch the CRO/Monitor screen with the tip of the open pen/pencil/any other

sharp object. Never leave the experiments while in progress. Do not insert/use pen drive/any other storage devices into the CPU. Do not leave the Laboratory without the signature of the concerned staff in observation book.



Experiment No: ___1_______ Date: ____________

DOUBLE PIPE HEAT EXCHANGER Aim:

To determine -

1) The cold fluid side heat transfer coefficient

2) The hot fluid side heat transfer coefficient &

3) The overall heat transfer coefficient

Apparatus/Components required:

Thermometers, bucket, stop watch and spring balance

Theory: TRANSFER FUNCTION OF MERCURY THERMOMETER:

It is the simplest type of heat exchanger and is used when the required heat transfer area is

relatively small. The DPHE consists of concentric pipes, connecting tees, return heads and return

bends. The packing glands support the inner pipe within the outer pipe. A double pipe heat

exchanger arranged in two legs is known as single hair-pin. Tees are provided with nozzles or

screwed connections for permitting the entry and exit of the annulus fluid which crosses from one

leg to another through return head .The return bend connects two legs of inner pipe to each other.

Double pipe heat exchangers are usually employed for decreasing the temperature of a hot fluid

with the help of a cold fluid when flow rates are low. These are commonly used in refrigeration

service. These exchangers are usually assembled in effective lengths of 3.65, 4.57,6 m. the distance

in each leg over which the heat transfer takes place is termed as the effective length.

But this type of heat exchanger requires large floor area for small heat transfer and it has maximum

leakage points.

DPHE are used for heat transfer areas ranging between 9 to 14 m2. It is simple in construction and

easy to clean.

Temperature can be defined as the amount of energy that a substance has. Heat exchangers are

used to transfer that energy from one substance to another. In process units it is necessary to

control the temperature of incoming and outgoing streams. These streams can either be gases or

liquids.

Heat exchangers raise or lower the temperature of these streams by transferring heat to or from

the stream. Heat exchangers are a device that exchange the heat between two fluids of different

temperatures that are separated by a solid wall. The temperature gradient, or the differences in

temperature facilitate this transfer of heat. Transfer of heat happens by three principle means:

radiation, conduction and convection. In the use of heat exchangers radiation does take place.



However, in comparison to conduction and convection, radiation does not play a major role.

Conduction occurs as the heat from the higher temperature fluid passes through the solid wall.

To maximize the heat transfer, the wall should be thin and made of a very conductive material.

The biggest contribution to heat transfer in a heat exchanger is made through convection.

In a heat exchanger forced convection allows for the transfer of heat of one moving stream to

another moving stream. With convection as heat is transferred through the pipe wall it is mixed

into the stream and the flow of the stream removes the transferred heat. This maintains a

temperature gradient between the two fluids.

The double-pipe heat exchanger is one of the simplest types of heat exchangers. It is called a

double-pipe exchanger because one fluid flows inside a pipe and the other fluid flows between

that pipe and another pipe that surrounds the first. This is a concentric tube construction. Flow

in a double-pipe heat exchanger can be co-current or counter-current. There are two flow

configurations: co-current is when the flow of the two streams is in the same direction, counter

current is when the flow of the streams is in opposite directions.

As conditions in the pipes change: inlet temperatures, flow rates, fluid properties, fluid

composition, etc., the amount of heat transferred also changes. This transient behavior leads to

change in process temperatures, which will lead to a point where the temperature distribution

becomes steady. When heat is beginning to be transferred, this changes the temperature of the

fluids. Until these temperatures reach a steady state their behavior is dependent on time.

In this double-pipe heat exchanger a hot process fluid flowing through the inner pipe transfers

its heat to cooling water flowing in the outer pipe. The system is in steady state until conditions

change, such as flow rate or inlet temperature. These changes in conditions cause the

temperature distribution to change with time until a new steady state is reached. The new steady

state will be observed once the inlet and outlet temperatures for the process and coolant fluid

become stable. In reality, the temperatures will never be completely stable, but with large

enough changes in inlet temperatures or flow rates a relative steady state can be experimentally

observed.



As with any process the analysis of a heat exchanger begins with an energy and material

balance. Before doing a complete energy balance a few assumptions can be made. The first

assumption is that the energy lost to the surroundings from the cooling water or from the U-

bends in the inner pipe to the surroundings is negligible. We also assume negligible potential or

kinetic energy changes and constant physical properties such as specific heats and density.

These assumptions also simplify the basic heat-exchanger equations.

The determination of the overall heat-transfer coefficient is necessary in order to determine the

heat transferred from the inner pipe to the outer pipe. This coefficient takes into account all of

the conductive and convective resistances (k and h, respectively) between fluids separated by

the inner pipe, and also takes into account thermal resistances caused by fouling (rust, scaling,

i.e.) on both sides of the inner pipe. For a double-pipe heat exchanger the overall heat transfer

coefficient, U, can be expressed as

In a heat exchanger the log-mean temperature difference is the appropriate average temperature

difference to use in heat transfer calculations. The equation for the log-mean temperature

difference is

Fluid properties such as density, viscosity and heat capacity are evaluated at the average

temperatures. The average is found by using the inlet and outlet values

Thermal conductivity, k, can be evaluated at the average of the average temperatures. In a

double-pipe heat exchanger the inner pipe is made of a conductive metal and is thin. The

problem can be further simplified if the equipment is assumed to be clean, which means that no

scaling exists. This is a poor simplification with the double-pipe heat exchanger in the

laboratory, because it is many years old. The fouling factors Rfo and Rfi can be looked up from

various sources, including Standards of the Tubular Exchange Manufacturers Association, or

lumped together and determined experimentally.

The only part of the overall heat-transfer coefficient that needs to be determined is the

convective heat-transfer coefficients. Correlations are used to relate the Reynolds number to the

heat-transfer coefficient. The Reynolds number is a dimensionless ratio of the inertial and

viscous forces in flow.



In the inner pipe if the Reynolds is less than 2000 this is considered to be laminar flow and the

Nusselt number is equal to 4.36. If the Reynolds number is greater than 10,000, the Nusselt

number is given by

This gives a Nusselt number that can then be use to find hi

The convective heat transfer coefficient in the annulus is more difficult to determine. The

hydraulic diameter is used to find the Reynolds number. The hydraulic diameter is defined as

the cross-sectional area perpendicular to flow divided by the wetted perimeter. With the

Reynolds number calculated the same correlations apply and with these ho can be determined.

Once all the separate heat-transfer coefficients are calculated an overall heat transfer coefficient

is calculated. Now everything that was necessary for an energy balance is available. With the

previous assumptions made earlier the dynamic equations would be

With the transient data taken experimentally an overall heat-transfer coefficient can be

determined at each time step. This can be solved numerically.

Procedure:

1. Arrange the experimental setup for counter current flow.

2. Open the steam inlet valve & keep steam pressure constant at 2 kg/cm2.

3. Adjust the hot fluid flow rate at medium level.



4. Adjust the cold fluid rate at minimum.

5. After steady state is reached, not down the inlet temperature, outlet temperature and mass

flow rate of cold side fluid and hot side fluid respectively.

6. Keeping hot side fluid flow rate constant, 6 readings are taken at different cold fluid flow

rate from minimum to maximum.

OBSERVATIONS:

1. inner pipe inner diameter di = ------- cm

2. inner pipe outer diameter do = ------ cm

3. outer pipe inner diameter Di = ------- cm

4. length of the pipe L = ------- cm

5. no of hairpins = -----------

6. length of each hairpin = -------- cm

7. thickness of the pipe xw = ------cm= [do-di]/2

8. thermal conductivity of the material of pipe k=------W/mK

CALCULATIONS:

Tube side calculations (hot water)

1.) Qh = mhcph∆t , where ∆t= t h1- t h2

2.) NRe = di vhρh/µ, where vh= mh/ρhA & A= Πdi2/4

3.) For NRe >10,000, Use Dittus – boelter eqn of cooling hi= 0.023 (NRe)0.8

(NPr)0.4

k/ di where

NPr (prandtl number) = (cph µh / kh)

Annular side calculations

4.) QC = mccpc∆t , where ∆t= tc2-tc1

5.) NRe = devcρc /µc , where de= (Di2

- do2)/do, vc=mc/ρcA & A= Π(Di

2- do

2) /4

6.) For NRe >10,000,Use Dittus – boelter eqn of heating,ho=0.023 (NRe)0.8

(NPr)0.3

k / de,

where NPr (prandtl number) = (cpc µc / kc)

7.) Clean overall heat transfer coefficient UC = 1/[(1/ hi)+(xw do/kwdl)+(do/hodi )]

where dl = [do – di ]/ ln(do/di) – log mean diameter

TABULAR COLUMN: 1

Sl.no tc1

tc2

t h1 t h2 Cold water flow rate (mc) Hot water flow rate

kg/sec (mh) Water

collected

in kg

Time

in sec

Kg/sec

1

2

3



4

5

6

Calculations

Note: All the physical properties should be evaluated at avg. temperatures and units must be in SI

system, using Perry’s chemical engineer’s Hand book (7th

addition).

Annular side fluid

Sl.

no

Qc

W

Avg

temp cpc

kJ/kgK µc

kg/msec

kc

W/mK

ρc

kg/m3

NRe

NPr Vc

m/sec

hi

W/m2K

1

2

3

4

5

6

Tube side fluid

sl.

no

Qh W

Avg.

temp cph

,kJ/kgK µh kg/msec

kh

W/mK

ρh

kg/m3

NRe NPr Vh

m/sec

ho

W/m2K

1

2

3

4

5

6

Theoretical Overall heat transfer coefficient

Sl.no ho , W/m2K hi , W/m

2K UC , W/m

2K

Actual heat Transfer coefficient

Ua= Q c / (Ai∆TLMTD) where Ai = ΠdoL and dirt factor Rd = 1/Ua - 1/UC



Sl.no Q c , W ∆TLMTD Ua , W/m

2K dirt factor Rd

m2K/W

Results: 1. Cold fluid side heat transfer coefficient hi = --------- W/m

2K

2. Hot fluid side heat transfer coefficient ho = -------- W/m2K

3. Theoretical Clean overall heat transfer coefficient UC = ---------- W/m2K

4. Dirt factor Rd= --------- m2K/W (compare it with data hand book) & give reasons for

deviation if there is a variation.

Applications:

Remarks:

Signature of Staff Incharge with date:

Probable viva questions:

1. What are the modes of heat transfer?

2. When you say heat is conducted?

3. What is the difference between conduction and convection?

4. Give practical example of natural convection?

5. What do you mean by forced convection?



Experiment No: _____2_____ Date: ____________

SHELL AND TUBE HEAT EXCHANGER

Aim: To determine -

1) The cold fluid side heat transfer coefficient

2) The hot fluid side heat transfer coefficient &

3) The overall heat transfer coefficient

Apparatus/Components required: Thermometers, bucket, stop watch and spring balance

Theory:

The simple double pipe heat exchanger is inadequate for flow rates that cannot readily be handled

in a few tubes. If several double pipes are used in parallel, the weight of metal required for the outer

tubes becomes so large that the shell and tube construction, where one shell serves for many tubes,

will be more economical. In STHE based on the no. of times the fluid in both shell and tube side

goes through ( w.r.t heat transfer) we have 1-1, 1-2,2-4 shell and tube heat exchangers with first no.

indicating the no. of shell fluid passes and the second no. indicating that of tube fluid passes.

Tubular heat exchangers are so important and so widely used in the process industries that their

design has been widely developed. The Standards devised and accepted by TEMA (Tubular

Exchangers Manufacturers Association) are available covering in detail the materials, methods of

construction, technique of design and dimensions of exchangers.

Principle types of constructions of STHE are fixed tube sheet heat exchangers, U-tube heat

exchanger, packed –lantern-ring exchanger, outside –packed floating-head exchanger, internal

floating – head exchanger , pull-through floating – head exchanger and falling film exchangers.

In this heat exchanger the shell side and tube side heat transfer coefficients are of comparable

importance and both must be large if a large satisfactory overall coefficient is to be attained. The

velocity and turbulence of the shell-side liquid are important. To have this, to promote cross-flow

and to raise the average velocity of the shell-side fluid, baffles are installed in the shell. In the

construction the baffles consist of circular disks of sheet metal with one side cut away (one fourth

of shell inner diameter). Such baffles are called 25 percent baffles. The baffles are perforated to

receive the tubes. To minimize leakage the clearances between baffles and shell and tubes should be

small. The center to center distance between baffles is called baffle spacing and it should not be less

than one fifth of shell diameter.

Another important part of STHE is tube and tube sheet. Tubes are drawn to definite wall thickness

in terms of BWG and true out side diameter (OD), and they are available in all common metals.

Standard lengths of tubes for STHE are 8, 12, 16, 20ft.

The tubes are arranged in a triangular or square layout known as triangular or square pitch (pitch

means center to center distance between adjacent tubes). Triangular pitch is used unless the shell

side tends to foul badly, because more heat transfer area can be packed into a shell of a given

diameter than with the square pitch. TEMA standards specify a minimum pitch of 1.25 times the

out side diameter of the tubes for triangular pitch.



In multi pass exchangers the flow of fluid is not truly countercurrent and there exists some parallel

flow also and hence the usual LMTD value can not be used as such and to account for this we

introduce a correction factor FG to have actual LMTD. The value of FG is less than unity.

A shell and tube heat exchanger is a class of heat exchanger designs. It is the most common type of

heat exchanger in oil refineries and other large chemical processes, and is suited for higher-pressure

applications. As its name implies, this type of heat exchanger consists of a shell (a large pressure

vessel) with a bundle of tubes inside it. One fluid runs through the tubes, and another fluid flows

over the tubes (through the shell) to transfer heat between the two fluids. The set of tubes is called a

tube bundle, and may be composed of several types of tubes: plain, longitudinally finned, etc.

Theory and Application:

Two fluids, of different starting temperatures, flow through the heat exchanger. One flows through

the tubes (the tube side) and the other flows outside the tubes but inside the shell (the shell side).

Heat is transferred from one fluid to the other through the tube walls, either from tube side to shell

side or vice versa. The fluids can be either liquids or gases on either the shell or the tube side. In

order to transfer heat efficiently, a large heat transfer area should be used, leading to the use of

many tubes. In this way, waste heat can be put to use. This is an efficient way to conserve energy.

Heat exchangers with only one phase (liquid or gas) on each side can be called one-phase or single-

phase heat exchangers. Two-phase heat exchangers can be used to heat a liquid to boil it into a gas

(vapor), sometimes called boilers, or cool a vapor to condense it into a liquid (called condensers),

with the phase change usually occurring on the shell side. Boilers in steam engine locomotives are

typically large, usually cylindrically-shaped shell-and-tube heat exchangers. In large power plants

with steam-driven turbines, shell-and-tube surface condensers are used to condense the

exhaust steam exiting the turbine into condensate water which is recycled back to be turned into

steam in the steam generator.

Shell and tube heat exchanger design There can be many variations on the shell and tube design. Typically, the ends of each tube are

connected to plenums (sometimes called water boxes) through holes in tube sheets. The tubes may

be straight or bent in the shape of a U, called U-tubes.

https://en.wikipedia.org/wiki/Phase_(matter)

https://en.wikipedia.org/wiki/Boiler

https://en.wikipedia.org/wiki/Condenser_(heat_transfer)

https://en.wikipedia.org/wiki/Locomotive

https://en.wikipedia.org/wiki/Power_plant

https://en.wikipedia.org/wiki/Turbine

https://en.wikipedia.org/wiki/Surface_condenser

https://en.wikipedia.org/wiki/Steam

https://en.wikipedia.org/wiki/Water



In nuclear power plants called pressurized water reactors, large heat exchangers called steam

generators are two-phase, shell-and-tube heat exchangers which typically have U-tubes. They are

used to boil water recycled from a surface condenser into steam to drive a turbine to produce power.

Most shell-and-tube heat exchangers are either 1, 2, or 4 pass designs on the tube side. This refers

to the number of times the fluid in the tubes passes through the fluid in the shell. In a single pass

heat exchanger, the fluid goes in one end of each tube and out the other.

Surface condensers in power plants are often 1-pass straight-tube heat exchangers. Two and four

pass designs are common because the fluid can enter and exit on the same side. This makes

construction much simpler.

https://en.wikipedia.org/wiki/File:U-tube_heat_exchanger.PNG

https://en.wikipedia.org/wiki/Pressurized_water_reactor

https://en.wikipedia.org/wiki/Steam_generator_(nuclear_power)

https://en.wikipedia.org/wiki/Steam_generator_(nuclear_power)

https://en.wikipedia.org/wiki/Steam_turbine

https://en.wikipedia.org/wiki/File:Straight-tube_heat_exchanger_1-pass.PNG



There are often baffles directing flow through the shell side so the fluid does not take a short cut

through the shell side leaving ineffective low flow volumes. These are generally attached to the

tube bundle rather than the shell in order that the bundle is still removable for maintenance.

Counter current heat exchangers are most efficient because they allow the highest log mean

temperature difference between the hot and cold streams. Many companies however do not use two

pass heat exchangers with a u-tube because they can break easily in addition to being more

expensive to build. Often multiple heat exchangers can be used to simulate the counter current flow

of a single large exchanger.

Selection of tube material: To be able to transfer heat well, the tube material should have good thermal conductivity. Because

heat is transferred from a hot to a cold side through the tubes, there is a temperature difference

through the width of the tubes. Because of the tendency of the tube material to thermally expand

differently at various temperatures, thermal stresses occur during operation. This is in addition to

any stress from high pressures from the fluids themselves. The tube material also should be

compatible with both the shell and tube side fluids for long periods under the operating conditions

(temperatures, pressures, pH, etc.) to minimize deterioration such as corrosion. All of these

requirements call for careful selection of strong, thermally-conductive, corrosion-resistant, high

quality tube materials, typically metals, including aluminium, copper alloy, stainless steel, carbon

steel, non-ferrous copper alloy, Inconel, nickel, Hastelloy and titanium. Fluoropolymers such

as Perfluoroalkoxy alkane (PFA) and Fluorinated ethylene propylene (FEP) are also used to

produce the tubing material due to their high resistance to extreme temperatures. Poor choice of

tube material could result in a leak through a tube between the shell and tube sides causing fluid

cross-contamination and possibly loss of pressure.

Design and construction standards:

https://en.wikipedia.org/wiki/File:Straight-tube_heat_exchanger_2-pass.PNG

https://en.wikipedia.org/wiki/Stress_(physics)

https://en.wikipedia.org/wiki/PH

https://en.wikipedia.org/wiki/Corrosion

https://en.wikipedia.org/wiki/Metal

https://en.wikipedia.org/wiki/Aluminium

https://en.wikipedia.org/wiki/Copper_alloy

https://en.wikipedia.org/wiki/Stainless_steel

https://en.wikipedia.org/wiki/Carbon_steel

https://en.wikipedia.org/wiki/Carbon_steel

https://en.wikipedia.org/wiki/Non-ferrous

https://en.wikipedia.org/wiki/Inconel

https://en.wikipedia.org/wiki/Nickel

https://en.wikipedia.org/wiki/Hastelloy

https://en.wikipedia.org/wiki/Titanium

https://en.wikipedia.org/wiki/Fluoropolymers

https://en.wikipedia.org/wiki/Perfluoroalkoxy_alkane

https://en.wikipedia.org/wiki/Fluorinated_ethylene_propylene

https://en.wikipedia.org/wiki/Leak



Standards of the Tubular Exchange Manufacturers Association (TEMA), 9th edition, 2009

EN 13445-3 "Unfired Pressure Vessels - Part 3: Design", Section 13 (2012)

ASME Boiler and Pressure Vessel Code, Section VIII, Division 1, Part UHX

Procedure: 1. Arrange the experimental setup for counter current flow.

2. Open the steam inlet valve & keep steam pressure constant at 0.2 kg/cm2.

3. Adjust the cold fluid rate at minimum.

4. After steady state is reached, not down the inlet temperature, outlet temperature and mass

floe rate of cold side fluid.

5. Keeping hot side fluid flow rate constant, 6 readings are taken at different cold fluid flow

rate from minimum to maximum, for three different steam pressures.

OBSERVATIONS AND CALCULATIONS:

1. inner pipe inner diameter di = ------- cm

2. inner pipe outer diameter Do = ------ cm

3. Shell diameter DS = ------- cm

4. Fraction of cross sectional area of shell occupied by baffle window fb= 0.1955

5. length of the tube L = ------- cm

6. Total no. of tubes N = --------

7. No of tubes in the baffle window Nb = fbx N -----------

8. P = baffle spacing = ------- m

9. p = tube pitch=-------m

10. thickness of the tube xw = ------cm

11. thermal conductivity of the material of tube k=------W/m2K

Shell side calculations (hot water)

1.) Qh=Qc

2.) Where λ = latent heat of vaporization ( at TS )

(Find TS at Absolute pressure = gauge pressure + atmospheric pressure)

3.) hi = 0.729[k3ρ

2 g λ ]

1/4 [do Nt µ ∆t]

-(1/4) ∆t = TS – TW and TW = [TS+tCavg] / 2

Tube side calculations

Note: For tube side calculations use DPHE formulae

4) Clean overall heat transfer coefficient UC = 1/[(1/ hi)+(xw do/kdl)+(do/hodi )]

where dl = [do – di ]/ ln(do/di) – log mean diameter

https://en.wikipedia.org/wiki/ASME_Boiler_and_Pressure_Vessel_Code_(BPVC)



Readings

Calculations

Note: All the physical properties should be evaluated at avg. temperatures and units must be in SI

system, using Perry’s chemical engineer’s Hand book (7th

addition).

TABULAR COLUMN:

Tube side fluid (find properties at average temperature)

Sl.

no

Qc

W

Avg

temp

TS+TW/2

λ

kJ/kg µ

kg/m.s

k

W/mK

ρ

kg/m3

NRe NPr hi

W/m2K

1

2

3

4

5

6

Shell side fluid(find properties at average temperature)

sl.no QS W

Avg.tem

tc1+ tc2/2

cpc,

kJ/kgK µc

kg/ms

kc

W/mK

ρc ,

kg/m3

NPr ho W/m2K

1

2

3

4

5

6

Theoretical Overall heat transfer coefficient

Sl.no ho , W/m2K hi , W/m

2K UC , W/m

2K



Actual heat Transfer coefficient

Ua= Q c / (NAi∆TLMTD) where Ai = ΠdoL and dirt factor Rd = 1/Ua - 1/UC

LMTD correction factor FG =

To find FG , calculate R= [t h1-t h2]/ [tc2- tc1], S= [tc2- tc1]/ [t h1-tc1] then refer perry’s hand book

page no. 11-6.

Sl.no Q c , W ∆TLMTD Corrected

LMTD

Ua , W/m2K Dirt factor

Rd m2K/W

Results: 1. Tube side heat transfer coefficient hi = --------- W/m

2K

2. Shell side heat transfer coefficient ho = -------- W/m2K

3. Theoretical Clean overall heat transfer coefficient UC = ---------- W/m2K

4. Dirt factor Rd= --------- m2K/W (compare it with data hand book) & give reasons for

deviation if there is a variation.

Applications: The simple design of a shell and tube heat exchanger makes it an ideal cooling solution for a wide variety of applications. One of the most common applications is the cooling of hydraulic fluid and oil in engines, transmissions and hydraulic power packs. With the right choice of materials they can also be used to cool or heat other mediums, such as swimming pool water or charge air. One of the big advantages of using a shell and tube heat exchanger is that they are often easy to service, particularly with models where a floating tube bundle (where the tube plates are not welded to the outer shell) is available.

Remarks:


Probable viva questions: 1. What are the advantages and disadvantages of DPHE?

2. Mention some industrial applications of heat exchangers?

3. Define radiation?

4. Define the heat law governing the radiation mode of heat transfer?

5. Why tubular heat exchangers (STHE) are preferred over DPHE?

https://en.wikipedia.org/wiki/Hydraulic_fluid

https://en.wikipedia.org/wiki/Hydraulic_fluid

https://en.wikipedia.org/w/index.php?title=Hydraulic_power_pack&action=edit&redlink=1




VERTICAL CONDENSER

Aim:

To determine -

1. Overall heat transfer co – efficient, U

2. Cold fluid side heat transfer co – efficient, hi.

3. Steam side film co – efficient, hs.

4. To draw Wilson’s plot and hence calculate the value of hi theoretically from the graph.

Apparatus/Components required: Thermometer, stop watch, bucket, spring balance etc.

Theory:

CONDENSATION:

The change of state of a substance from vapor to liquid is known as condensation. The process

of condensation of a pure material is isothermal process as the condensing temperature of a

single pure substance depends only on pressure. The condensation of vapors on the surface of

tubes cooler than the condensing temperature of the vapor is important when vapors such as

those of water, hydrocarbons and other volatile substances are processed.

Condenser is heat exchange equipment employed to condense vapor or mixture of vapors. It

involves removal of latent heat with the help of a suitable cooling media. Ex: Cooling water.

The condensing vapors may consist of a single substance, a mixture of condensable and non

condensable substances or a mixture of two or more condensable vapors. Friction loses in a

condenser are normally small, so that condensation is essentially a constant pressure process.

The condensing temperature of a single pure substance depends only on the pressure and

therefore the process of condensation of a pure substance is isothermal. Also, the condensate is

a pure liquid. Mixed vapors, condensing at a constant pressure, condense over a temperature

range and yield a condensate of variable composition until the entire vapor stream is condensed.

Vertical condensers are excellent for use as condenser tube coolers whether they may be of the

1 -2 type with condensation in the shell or 1 – 1 type with condensation in the vertical tubes

follows essentially the same mechanism as condensation outside the vertical tubes if the

interference of the shell baffles is neglected to grow continuously in its descent down the inside

or outside of the tubes, it may change from streamline to turbulent flow at some point. The local

condensing co – efficient decreases continually from the top downward until at some point the

film changes to turbulent flow.



The value of the condensing film co – efficient for a given quantity of vapor on a given surface

is significantly affected by the position of the condenser. In the vertical condenser about 60% of

the vapor condenses in the upper half of the tube. The ratio of the theoretical horizontal and

vertical condensing co – efficient is given by (0.725)/0.294 (L/D)1/4.

When the condenser is employed for a distillation column, vertical condenser is not suitable

because of the gravity return of the condensate, since it must be elevated considerably above the

column, which in many cases is tall by itself. Maintenance and structural support for the vertical

condenser may be costly and considerably more difficult. On the other hand, it is desired not

only to condense the overhead vapor but also to cool the condensate the vertical condenser is

admirably suited.

A common example of the condensation of one constituent from its mixture with a second non -

condensable substance is the condensation of water from a mixture of steam and air.

Condensation occurs by two distinct mechanisms:

DROP-WISE CONDENSATION:

When a saturated vapor comes in contact with a cold surface, it condenses and if the

condensate does not wet the surface, the droplets are formed on the surface. These droplets

grow and ultimately fall from the surface on which further condensation takes place. The

condensation occurring by this mechanism is known as drop-wise condensation. In drop-wise

condensation, the condensate begins to form at microscopic nucleation sites. Typical sites are

tiny pits, scratches and dust specks. The drops grow and coalesce with their neighbours to

form visible fine drops like those often seen on the outside of a cold – water pitcher in a

humid room. The fine drops, in turn, coalesce into rivulets, which flow down the tube under

the action of gravity, sweep away condensate and clear the surface of more droplets. [

During drop-wise condensation, large areas of cold tube are bare and directly exposed to the

vapor. Because of the absence of liquid the resistance to heat flow at these bare areas is very

low and the heat transfer co-efficient is correspondingly high.

The average heat transfer co-efficient for drop-wise condensation may be 5-8 times that for

film-type condensation. On long tubes, condensation on some of the surface may be film-wise

condensation and the reminder is drop-wise condensation.

The appearance of drop-wise condensation depends upon the wetting or non wetting of the

surface by the liquid, and fundamentally, the phenomenon lies in the field of surface

chemistry. Drop-wise condensation is obtainable only when the cooling surface is

contaminated. It is more easily maintained on a smooth contaminated surface than on a rough

contaminated surface.

The quantity of contaminant or promoter required to cause drop-wise condensation is minute

and apparently only a monomolecular film is necessary.

Effective drop promoters are strongly adsorbed by the surface and the substances that merely

prevent wetting are ineffective. Some promoters are especially effective on certain metals.



Ex: Mercaptans on copper alloys, oleic acid etc.

The average co-efficient obtainable in pure drop-wise condensation is as high as

20,000 Btu/ft2 h F.

FILM-WISE CONDENSATION:

When a saturated vapor comes in contact with a cold surface, it condenses and if the

condensate does not wet the surface, it forms a continuous film of the condensate through

which it must be transferred. The additional vapour is then required to condense in to the

liquid film rather than directly on the surface. The condensate ultimately flows down the

surface under the influence of gravity. The condensation occurring by this mechanism is

known as film-wise condensation.

In film-wise condensation, which is more common than drop-wise condensation, the

liquid condensate forms a film or continuous layer of liquid that flows over the surface of the

tube under the action of gravity. It is the layer of liquid interposed between the vapor and the

wall of the tube which provides the resistance to heat flow and therefore which fixes the

magnitude of the heat transfer co-efficient.

Film-wise condensation occurs on tubes of the common metals if both the steam and the tube

of the common metals if both the steam and the tube are clean, in the presence or absence of

air, on rough or on polished surfaces.

NUSSELT EQUATION

Nusselt equations are based on the following ASSUMPTIONS:

1. The vapor and liquid at the outside boundary of the liquid layer are in thermodynamic

equilibrium, so that the only resistance to the flow of heat is that offered by the layer of

condensable flowing downward in laminar flow under the action of gravity.

2. The velocity of the liquid at the wall is zero.

3. The velocity of the liquid at the outside of the film is not influenced by the velocity of the

vapor.

4. The temperatures of the wall and the vapor are constant.

5. Superheat in vapor is neglected, the condensate is assumed to leave the tube at the

condensing temperature and the physical properties of the liquid are taken at the mean

film temperature.

VERTICAL TUBES:

In film-wise condensation, the Nusselt theory shows that the condensate film starts to form at

the top of the tube and that the thickness of the film increases rapidly in the first few inches at

the top and then more and more slowly in the remaining length.

The heat is assumed to flow through the condensate film solely by conduction and the local co-

efficient is therefore given by



hx = kf _______ (1)

δ Where δ = local film thickness.

The local film thickness changes inversely with the film thickness since there is a temperature

gradient in the film; the properties of the liquid are evaluated at the average film temperature,

Tf given by

Tf = Th – 3(Th-Tw) = Th – 3ΔTo

4 4 For condensation on a vertical surface, for which cosβ = 1

δ = 3μT' 1/3

_______ (2)

ρ2gcosβ

Where δ = thickness of the layer

δ = 3μfT' 1/3

_______ (3)

ρf2g

Substituting for δ in equation (1) gives the local heat transfer co-efficient, at a distance, L from

the top of the vertical surface

hx = kf ρf2g

1/3 _______ (4)

3μfT'

For condensation on the outside of a vertical tube the local co-efficient is given by

hx = dq = λdm _______ (5)

ΔTodAo ΔToπDodL

Where λ = heat of vaporization.

m = local flow rate of condensate.

Since m = T' πDo Equation (5) may be written as

hx = λdT' _______ (6)

ΔTodL The average co-efficient, h for the entire tube is defined by

h = qT = mT λ = T'b λ _______ (7)

ΔToAo πDoLTΔTo LTΔTo Where qT = total rate of heat transfer.

mT = total rate of condensation.

LT = total tube length.

T'b = condensate loading at the bottom of the tube.

Eliminating hx from equations (4) and (6) and then solving for ΔTo gives



ΔTo = 3μfT' 1/3

= λdT'

ρf2g kfdL

_______ (8)

Substituting ΔTo from (8) into equation (7) gives

h = T'bkf ρf2g

1/3 dL

_______ (9)

LT 3μf T'1/3

dT' Rearranging equation (9) and integrating between the limits gives

T'b LT

h ∫ T'1/3

d T’ = T'b kf ρf

2g

1/3 ∫ dL

0 LT 3μf 0

h = 4kf ρf2g

1/3 _______ (10)

3 3μfTb'

Equation (10) can be rearranged as

h μf2 1/3

= 1.47 4Tb' - 1/3

_______ (11)

kf3 ρf

2g μf

The reference temperature for evaluating μf , kf and ρf is defined by the relation

Tf = Th – 3(Th-Tw) = Th – 3ΔTo _______ (12)

4 4 Where Tf = reference temperature.

Th = temperature of condensing vapor.

Tw = temperature of outside surface of tube wall.

Equation (11) is used in an equivalent form, in which the term Tb' has been eliminated by

combining equation (7) and (11) to give

h = 0.943 kf3 ρf

2g λ

1/4 _______ (13)

LΔToμf



NUSSELT THEORY:

CONDENSATION ON A VERTICAL PLATE:

In the process of condensation on a vertical surface, a film of condensate is formed as

shown.

Consider a vertical plate having unit width perpendicular to the plane maintained at a

constant temperature of Tw on which a pure saturated vapor at temperature Ts is

condensing. The film condensation is assumed so that a condensate film growing from the

zero thickness at the top edge of the plate will continuously flow down.

Let δ be the thickness of the condensate film.

ASSUMPTIONS:

1. The drainage of the condensate film from the surface is given by laminar flow

only.

2. The heat is transferred through the film by conduction.

3. The physical properties of the condensate are constant and heat delivered by the

vapor is latent heat only.

4. The liquid-vapor interface is at the saturation temperature, Ts.

5. The vapor exerts no shear stress at the liquid-vapor interface and the temperature of

the surface of the plate is constant.

Consider the infinite thin section at a distance, Z of thickness, dz inside the moving

condensate.

Consider a differential control volume defined by dx.dz at a distance X from the plate.

Ts

Tw

Z

s

dZ

X d

X

μ du + d2u dx dZ

dx dx2

μ du dZ

dy

ρgdxdz

dZ

μ du dZ

dy

dx

μ du dZ

dy



Let u be the velocity of liquid in the Z-direction.

Tangential shear force + Tangential shear force + gravitational = 0

acting upward acting downward force

- μ du dZ + μ du + d2u dx dZ + ρgdxdz = 0

dx dx dx2

μ d2u dx dZ = - ρgdxdz

dx2

d2u = - ρg

dx2

μ

(Or) the differential tangential force must be offset by the gravitational force acting

downward.

μ du dZ _ μ du + d2u dx dZ = ρgdxdz

dx dx dx2

_ μ d2u dx dZ = ρgdxdz

dx2

d2u = - ρg _______ (1)

dx2

μ

On integrating we get

du = - ρg x + C1 _______ (2)

dx μ

u = - ρg x

2+ C1x + C2 _______ (3)

2μ

Boundary conditions:

1. Since the liquid adheres to the wall of the plate, u=0 at x=0, C2=0.

2. At the outer boundary of the film there is no tangential shear stress.

du = 0 at x = δ

dx

C1 = ρg δ

μ

Putting the values of C1 and C2 in equation (3) we get,



u = - ρg x2+ ρg δ x

2μ μ

u = - ρg x δ - x2

μ 2

The mass flow rate of the condensate at a distance Z from the top edge is given by δ

m = ρ ∫ u dx

0 δ

= ρ2g ∫ x δ - x

2 dx

μ 0 2

δ = ρ

2g x

2 δ –

x

3

μ 2 6 0

= ρ2g δ

3 –

δ

3

μ 2 6 = ρ

2g δ

3

μ 3

dm = ρ2g δ

2 dδ _______ (4)

μ

Equation (4) indicates that we move from a section z to z+dz and the condensate film

thickness increases from δ to δ+dδ, the mass flow rate of condensate increases by

ρ2g δ

2dδ i.e. condensate rate from Z to Z+dZ is ρ

2g δ

2dδ.

μ μ



Applying energy balance over a control volume of dimensions δ, dz we have

Let λ be the latent heat of condensation of vapor

Rate of heat released due to condensation at the = Rate of heat conducted through the

Liquid-vapor interface. film.

ρ2g δ

2dδ λ = K (Ts-Tw) dz _______ (5)

μ δ

δ 3

dδ = K (Ts-Tw) μ dz

λρ2g

On integrating

δ 4

= K (Ts-Tw) μ z

4 λρ2g

1/4

δ = K (Ts-Tw) μ z _______ (6)

λρ2g

The local heat transfer co-efficient is given by

K (Ts-Tw) = hz (Ts-Tw)

δ

hz = K (Ts-Tw) = K _______ (7)

δ (Ts-Tw) δ

Substituting equation (6) in (7) we get

1/4

hz = λρ2g K

3 _______ (8)

4(Ts-Tw) μ z

dz dm

k

δ

K (Ts-Tw) dz

δ



The mean heat transfer co-efficient over the length L of the plate will be

L

h = 1 ∫ hz dz

L 0

1/4

L

= 1 λρ2g K

3 ∫ dz

L 4(Ts-Tw)μ 0 z1/4

1/4

= 4 λρ

2g K

3

3 4(Ts-Tw) μL

h = 4 hz=L _______ (9)

3

Thus, the average heat transfer co-efficient over the length L of the plate is 4/3 times the local

co-efficient at z=L.

1/4

h = 4 λρ2g K

3

3*41/4

(Ts-Tw)μL

1/4

h = 0.943 λρ2g K

3 (Ts-Tw)μL

1/4

h = 0.943 K 3ρ

2g

λ

_______ (10)

ΔTμL



The Reynold’s number for condensate film is given by

NRe = DeG'

μ

= 4 Af G

P Af

μ

= 4 G

P μ

= 4 M

μ

Where G’ = mass flow rate per unit area = G

Af

M = mass flow rate of condensate per unit length = G

πDo

If 4 m < 1800, the condensate fluid flow is laminar.

μ

If 4 m > 1800, the condensate fluid flow is turbulent.

μ Where m = mass flow rate of vapor per unit length of vertical plate (kg/m.s).

Procedure:

1. The inlet valve is opened and the cold fluid is allowed to flow through the condenser.

2. The flow rate of the cold fluid is adjusted to minimum.

3. The steam inlet valve is opened and the steam pressure is kept constant throughout

the experiment.

4. After the cold fluid temperature becomes steady, the inlet temperature, outlet

temperature and flow rate of the cold fluid are noted down and also the time required

for 1 cm rise in the condensate level in the condensate tank is noted.

5. Keeping the steam pressure constant, 4 readings are taken for different flow rates of

the cold fluid from minimum to maximum.

6. The experiment is repeated for another constant steam pressure.



OBSERVATION:

1. Inner pipe outer dia, do = 16 mm.

2. Inner pipe inner dia, di = 13 mm.

3. Outer pipe inner dia, Di = 28 mm.

4. Length of the pipe, L = 1500 mm.

5. Condensate vessel dia, D = 140 mm.

6. Weight of empty bucket = ________ kg.

TABULAR COLUMN:

Sl.

No:

Steam

pressure

Guage

(kg/cm2)

Cold fluid

temperature

Mass collected

In bucket +

Mass of

Empty

Bucket

(kg)

Mass of water

Collected

(kg)

Time of

collection

(sec)

Condensate flow

rate.

Time required

For 1 cm rise in

Height

(sec)

Inlet

t1°C

Outlet

t2°C

P1

P2

mh

kg/s

mc

kg/s

Vc

m/s

1

( Vc )0.8

NRe NPr hi

W/m2

ºK

ΔTLMTD

ºK

Qc

W

Uoexp

W/m2

ºK

1

Uoexp

W/m2

ºK

ho

W/m2

ºK

Uotheo

W/m2

ºK



Tf ºK

FORMULAS:

1. Qc = mc Cpc (ΔT)c = mc Cpc (t2 - t1).

2. Ts = temperature of steam at 1.2 bar pressure.

3. λ = from steam table for that pressure and temperature in (kJ/kg°K).

4. Mass flow rate of condensate = ________.

5. Condensate vessel area = A' = πD2

i. 4

6. Height of condensate water collected = h.

7. Time required to collect the condensate = t.

8. Mass flow rate of condensate, mh = A'h ρ (kg/sec).

a. T

9. (ΔT)LMTD = (Ts-Tc2) – (Ts – Tc1)

a. ln (Ts-Tc2)

b. (Ts–Tc1)

10. Twall = Tw = Ts+Tavg

2

11. Δt = Ts – Twall. 12.

1/4

13. ho = 0.725 Kf

3ρf

2 g

λf

ΔTf μf do

1 = 1 + Xw do + 1 do

14. (Uo)th ho Km dw hi dI

1/4

ho = 0.725 Kf3ρf

2g λf

μf ΔTf do



CALCULATION:

15. t1 = _____ °K. t2 = _____ °K.

16. tavg = t1°K + t2°K = = _______°K.

i. 2

17. Operating pressure = 0.2 kg/cm2 = 0.2 bar.

18. Barometric pressure = 1.0 kg/cm2 = 1.0 bar.

19. Total pressure = 1+0.2 = 1.2 bar.

20. Temperature of steam at 1.2 Bar = _________°K.

21. Cold fluid (water) properties at tavg are:

22. K = ________(W/m°K).

23. ρ = 1/volume = ________(kg/m3) = 992.875kg/ m3

24. μ = _________kg/ms.

25. λ = _________ KJ/kg°K.

26. Cpc = 4.178 kJ/kg °K.

27. A' = πd2

4

28. Mass flow rate of condensate = A'hρ

T

29. = __________ kg/s.

30. MASS = mass of water in bucket – mass of bucke

a. time

31. Qc = mc

ρ

32. Ac = area of annulus in which cold fluid is flowing

(Di2 – do2) * π

4

33. Ac = ________ m2

34. Vc = Qc

Ac

35. De = (Di2 – do2)

Do



36. NRe = DeVc ρ

μ

If NRe > 2100 and NRe < 1000, JH factor is used.

37. JH = (hi/CpG)(Cpμ/K)2/3(μW/μ)0.14

hi = ________ W/m2°K.

Where G = mc/Ac =

38. Qc = mc Cpc Δt

39. Uo exp = Qc

(ΔT LMTD)(area of heat transfer)

Where (ΔT)LMTD =(Ts-Tc2) – (Ts – Tc1)

ln (Ts-Tc2)

(Ts–Tc1)

Where Ts = _________ temperature of steam at 1.2 bar pressure. 44. Area of heat transfer = πdoL

45. Uo exp =

46. Tw = Tf = Ts + T avg

2

47. ΔTf = Ts – Tavg

48. Properties at Tf :

i. Kf = ________(W/m°K).

ii. ρf = 1/volume = ________(kg/m3) = __________kg/ m3

iii. μf = _________kg/ms.

iv. λf = _________ J/kg°K.



49. Cpc = 4.178 kJ/kg °K. 1/4

50. ho = 0.943 Kf

3ρf

2 g

λf

ΔTf μf L

51. 1 = 1 + Xw do + 1 do

(Uo)th ho Km dw hi dI

Where Xw = do - dI

2

dw = do - dI

ln do

dI

45. (Uo)theo = _________ W/m2°K.

If NRe > 10,000

We use NNu = 0.023 [(NRe) 0.8 ( Npr)

0.33]

46. Npr = Cp μ

47. K

48. Where Cp = specific heat capacity of water at mean temperature.

49. μ = viscosity of water at mean cold temperature.

50. K = thermal conductivity of water at mean temperature.

51. NNu = hi * de = 0.023 [(NRe) 0.8 ( Npr)

0.33]

52. K

53. hi = ________ W/m2°K.

54. Qc = mc Cpc ΔTc

55. (ΔT)ln = _______°K.

56. Area of heat transfer = πdoL

57. (Uo)Experimental = Qc_____________________

58. (ΔT)ln * area of heat transfer



HEAT TRANSFER CO-EFFICIENT OF STEAM SIDE FLUID:

1. Average temperature of condensate =

2. Tf = Ts+Tavg

2


PROPERTIES OF CONDENSATE WATER AT Tf:

1. ρf = kg/ m3

2. Cpf = J/kg °K.

3. Kf = W/m°K. 4. μf = kg/ms.

5. λf = J/kg°K.

By Nusselt equation

1/4

ho = 0.943 Kf3ρf

2g λf

μf ΔT do

1 = 1 + Xw do + 1 do


Where Xw = do - dI

2

dw = do - dI

ln do

dI

(Uo)theo = _________ W/m2°K.



WILSON PLOT

Results:

SL.

NO:

Working

pressure

(Bar)

NRe hi

W/m2ºK

ho

W/m2ºK

Uo exp

W/m2ºK

Uo theo9

W/m2ºK

Applications:

1

UoExp

INTERCEPT

1

Vc0.8



Remarks:



1. What is overall heat transfer coefficient?

2. When extended surface type of heat exchangers is used?

3. Why LMTD is required?

4. Justify the countercurrent flow over co-current flow?

5. What are condensers?




HORIZONTAL CONDENSER

Aim: To determine

1. Overall heat transfer co – efficient, U

2. Cold fluid side heat transfer co – efficient, hi.

3. Steam side film co – efficient, hs.

4. To draw Wilson’s plot and hence calculate the value of hi theoretically from the graph.


Thermometer, stop watch, bucket, spring balance etc.

Theory:

CONDENSATION:

The change of state of a substance from vapor to liquid is known as condensation. The process

of condensation of a pure material is isothermal process as the condensing temperature of a

single pure substance depends only on pressure. The condensation of vapors on the surface of

tubes cooler than the condensing temperature of the vapor is important when vapors such as

those of water, hydrocarbons and other volatile substances are processed. Condensation can be

defined as heat transfer accompanied by phase change, which is obviously more intricate than simple

heat exchange between fluids. Condensation is essentially a constant pressure process in that friction

losses are small. Hence, a process of condensation of pure substances is isothermal.

A vapor may condense on a cold surface in one of the two ways, which are well described by the

two terms – Drop wise and Film wise. The latter being more common than the former involves the

formation of a film of liquid that flows over by gravity. But in drop wise condensation, the

condensate begins to form at microscopic nucleation sites. The typical sites are tiny pits, scratches

and dust specks. The drops grow and coalesce with thin neighbours to form visible fine drops like

those often seen on the outside of a cold water pitcher in a humid room.

In general, the co – efficient of a horizontal tube is considerably larger than that on a vertical tube

under otherwise similar condition unless the tubes are very short or there are many horizontal tubes

in each stack. Vertical condenser is preferred when the condensate must be appreciably cooled.

Mixtures of vapor are usually cooled and condensed inside vertical tubes, so that the inert gas is

continually swept away from the heat transfer surface by the incoming stream.

Condenser is heat exchange equipment employed to condense vapor or mixture of vapors. It

involves removal of latent heat with the help of a suitable cooling media. Ex: Cooling water.

The condensing vapors may consist of a single substance, a mixture of condensable and non

condensable substances or a mixture of two or more condensable vapors. Friction loses in a

condenser are normally small, so that condensation is essentially a constant pressure process.

The condensing temperature of a single pure substance depends only on the pressure and



therefore the process of condensation of a pure substance is isothermal. Also, the condensate is

a pure liquid. Mixed vapors, condensing at a constant pressure, condense over a temperature

range and yield a condensate of variable composition until the entire vapor stream is condensed.

A common example of the condensation of one constituent from its mixture with a second non -

condensable substance is the condensation of water fro a mixture of steam and air.

Condensation occurs by two distinct mechanisms:

1. DROP-WISE CONDENSATION:

When a saturated vapor comes in contact with a cold surface, it condenses and if the condensate

does not wet the surface, the droplets are formed on the surface. These droplets grow and

ultimately fall from the surface on which further condensation takes place. The condensation

occurring by this mechanism is known as drop-wise condensation. In drop-wise condensation,

the condensate begins to form at microscopic nucleation sites. Typical sites are tiny pits,

scratches and dust specks. The drops grow and coalesce with their neighbours to form visible

fine drops like those often seen on the outside of a cold – water pitcher in a humid room. The

fine drops, in turn, coalesce into rivulets, which flow down the tube under the action of gravity,

sweep away condensate and clear the surface of more droplets.

During drop-wise condensation, large areas of cold tube are bare and directly exposed to the

vapor. Because of the absence of liquid the resistance to heat flow at these bare areas is very

low and the heat transfer co-efficient is correspondingly high.

The average heat transfer co-efficient for drop-wise condensation may be 5-8 times that for

film-type condensation. On long tubes, condensation on some of the surface may be film-wise

condensation and the reminder is drop-wise condensation.

The appearance of drop-wise condensation depends upon the wetting or non wetting of the

surface by the liquid, and fundamentally, the phenomenon lies in the field of surface chemistry.

Drop-wise condensation is obtainable only when the cooling surface is contaminated. It is more

easily maintained on a smooth contaminated surface than on a rough contaminated surface.

The quantity of contaminant or promoter required to cause drop-wise condensation is minute

and apparently only a monomolecular film is necessary.

Effective drop promoters are strongly adsorbed by the surface and the substances that merely

prevent wetting are ineffective. Some promoters are especially effective on certain metals.

Ex: Mercaptans on copper alloys, oleic acid etc.

The average co-efficient obtainable in pure drop-wise condensation is as high as

20,000 Btu/ft2 h F.

2. FILM-WISE CONDENSATION:

When a saturated vapor comes in contact with a cold surface, it condenses and if the condensate

does not wet the surface, it forms a continuous film of the condensate through which it must be

transferred. The additional vapour is then required to condense in to the liquid film rather than

directly on the surface. The condensate ultimately flows down the surface under the influence

of gravity. The condensation occurring by this mechanism is known as film-wise condensation.



In film-wise condensation, which is more common than drop-wise condensation, the liquid

condensate forms a film or continuous layer of liquid that flows over the surface of the tube

under the action of gravity. It is the layer of liquid interposed between the vapor and the wall of

the tube which provides the resistance to heat flow and therefore which fixes the magnitude of

the heat transfer co-efficient.

Film-wise condensation occurs on tubes of the common metals if both the steam and the tube of

the common metals if both the steam and the tube are clean, in the presence or absence of air,

on rough or on polished surfaces.

NUSSELT EQUATION

Nusselt equations are based on the following ASSUMPTIONS:

1. The vapor and liquid at the outside boundary of the liquid layer are in thermodynamic

equilibrium, so that the only resistance to the flow of heat is that offered by the layer of

condensable flowing downward in laminar flow under the action of gravity.

2. The velocity of the liquid at the wall is zero.

3. The velocity of the liquid at the outside of the film is not influenced by the velocity of

the vapor.

4. The temperatures of the wall and the vapor are constant.

5. Superheat in vapor is neglected, the condensate is assumed to leave the tube at the

condensing temperature and the physical properties of the liquid are taken at the mean

film temperature.

VERTICAL TUBES:

In film-wise condensation, the Nusselt theory shows that the condensate film starts to form at

the top of the tube and that the thickness of the film increases rapidly in the first few inches at

the top and then more and more slowly in the remaining length.

The heat is assumed to flow through the condensate film solely by conduction and the local co-

efficient is therefore given by

hx = kf _______ (1)

δ Where δ = local film thickness.

The local film thickness changes inversely with the film thickness since there is a temperature

gradient in the film; the properties of the liquid are evaluated at the average film temperature,

Tf given by

Tf = Th – 3(Th-Tw) = Th – 3ΔTo

4 4 For condensation on a vertical surface, for which cosβ = 1

δ = 3μT' 1/3

_______ (2)

ρ2gcosβ



Where δ = thickness of the layer

δ = 3μfT' 1/3

_______ (3)

ρf2g

Substituting for δ in equation (1) gives the local heat transfer co-efficient, at a distance, L from

the top of the vertical surface

hx = kf ρf2g

1/3 _______ (4)

3μfT'

For condensation on the outside of a vertical tube the local co-efficient is given by

hx = dq = λdm _______ (5)

ΔTodAo ΔToπDodL

Where λ = heat of vaporization.

m = local flow rate of condensate.

Since m = T' πDo Equation (5) may be written as

hx = λdT' _______ (6)

ΔTodL The average co-efficient, h for the entire tube is defined by

h = qT = mT λ = T'b λ _______ (7)

ΔToAo πDoLTΔTo LTΔTo Where qT = total rate of heat transfer.

mT = total rate of condensation.

LT = total tube length.

T'b = condensate loading at the bottom of the tube.

Eliminating hx from equations (4) and (6) and then solving for ΔTo gives

ΔTo = 3μfT' 1/3

= λdT'

ρf2g kfdL

_______ (8)

Substituting ΔTo from (8) into equation (7) gives

h = T'bkf ρf2g

1/3 dL

_______ (9)

LT 3μf T'1/3

dT'



Rearranging equation (9) and integrating between the limits gives

T'b LT

h ∫ T'1/3

d T’ = T'b kf ρf

2g

1/3 ∫ dL

0 LT 3μf 0

h = 4kf ρf2g

1/3 _______ (10)

3 3μfTb'

Equation (10) can be rearranged as

h μf2 1/3

= 1.47 4Tb' - 1/3

_______ (11)

kf3 ρf

2g μf

The reference temperature for evaluating μf , kf and ρf is defined by the relation

Tf = Th – 3(Th-Tw) = Th – 3ΔTo _______ (12)

4 4 Where Tf = reference temperature.

Th = temperature of condensing vapor.

Tw = temperature of outside surface of tube wall.

Equation (11) is used in an equivalent form, in which the term Tb' has been eliminated by

combining equation (7) and (11) to give

h = 0.943 kf3 ρf

2g λ

1/4 _______ (13)

LΔToμf

HORIZONTAL TUBES

Corresponding to equation (11) and (13) for vertical tubes, the following equations apply to

single horizontal tubes,

h = μf 1/3

kf3 ρf

2g



1/3

h = 1.51 4T'' _______ (14)

μf

h = 0.725 kf3 ρf

2g λ

1/4 _______ (15)

ΔToDoμf

Where T'' = condensate loading per unit length of tube (pounds/foot/hour).

Ts'' = mT _______ (16)

LN

Where mT = total condensate flow rate for entire stack.

L = length of one tube.

N = number of tubes in stack.

Practically, because of the fact that some condensate splashes away from each individual tube

instead of dripping completely to the tube below, it is more accurate to use a value of Ts''

calculated by

Ts'' = mT _______ (17)

LN 2/3

Equation (15) becomes

h = 0.725 kf3 ρf

2g λ

1/4 _______ (18)

N 2/3

ΔToDoμf

NUSSELT THEORY:

CONDENSATION ON A VERTICAL PLATE:

In the process of condensation on a vertical surface, a film of condensate is formed as shown.

Consider a vertical plate having unit width perpendicular to the plane maintained at a

constant temperature of Tw on which a pure saturated vapor at temperature Ts is

Ts

Tw

Z

s

dZ

X d

X



condensing. The film condensation is assumed so that a condensate film growing from the

zero thickness at the top edge of the plate will continuously flow down.

Let δ be the thickness of the condensate film.

ASSUMPTIONS:

1. The drainage of the condensate film from the surface is given by laminar flow

only.

2. The heat is transferred through the film by conduction.

3. The physical properties of the condensate are constant and heat delivered by the

vapor is latent heat only.

4. The liquid-vapor interface is at the saturation temperature, Ts.

5. The vapor exerts no shear stress at the liquid-vapor interface and the temperature of

the surface of the plate is constant.

Consider the infinite thin section at a distance, Z of thickness, dz inside the moving

condensate.

Consider a differential control volume defined by dx.dz at a distance X from the plate.

Let u be the velocity of liquid in the Z-direction.

Tangential shear force + Tangential shear force + gravitational = 0

acting upward acting downward force

- μ du dZ + μ du + d2u dx dZ + ρgdxdz = 0

dx dx dx2

μ d2u dx dZ = - ρgdxdz

dx2

d2u = - ρg

dx2

μ

(Or) the differential tangential force must be offset by the gravitational force acting downward.

μ du + d2u dx dZ

dx dx2

μ du dZ

dy

ρgdxdz

dZ

μ du dZ

dy

dx

μ du dZ

dy



μ du dZ _ μ du + d2u dx dZ = ρgdxdz

dx dx dx2

_ μ d2u dx dZ = ρgdxdz

dx2

d2u = - ρg _______ (1)

dx2

μ

On integrating we get

du = - ρg x + C1 _______ (2)

dx μ

u = - ρg x

2+ C1x + C2 _______ (3)

2μ

Boundary conditions: 1. Since the liquid adheres to the wall of the plate, u=0 at x=0, C2=0.

2. At the outer boundary of the film there is no tangential shear stress.

du = 0 at x = δ

dx

C1 = ρg δ

μ

Putting the values of C1 and C2 in equation (3) we get,

u = - ρg x2+ ρg δ x

2μ μ

u = - ρg x δ - x2

μ 2

The mass flow rate of the condensate at a distance Z from the top edge is given by δ

m = ρ ∫ u dx

0

δ

= ρ2g ∫ x δ - x

2 dx

μ 0 2



δ

= ρ2g ∫ x2

δ – x

3

μ 0 2 6

= ρ2g δ

3 –

δ

3

μ 2 6 = ρ

2g δ

3

μ 3

dm = ρ2g δ

2 dδ _______ (4)

μ

Equation (4) indicates that we move from a section z to z+dz and the condensate film

thickness increases from δ to δ+dδ, the mass flow rate of condensate increases by

ρ2g δ

2dδ i.e. condensate rate from Z to Z+dZ is ρ

2g δ

2dδ.

μ μ

Applying energy balance over a control volume of dimensions δ, dz we have

Let λ be the latent heat of condensation of vapor

Rate of heat released due to condensation at the = Rate of heat conducted through the

Liquid-vapor interface. film.

ρ2g δ

2dδ λ = K (Ts-Tw) dz _______ (5)

μ δ

δ 3

dδ = K (Ts-Tw) μ dz

λρ2g

On integrating

δ 4

= K (Ts-Tw) μ z

4 λρ2g

dz dm

k

δ

K (Ts-Tw) dz

δ



1/4

δ = K (Ts-Tw) μ z _______ (6)

λρ2g

The local heat transfer co-efficient is given by

K (Ts-Tw) = hz (Ts-Tw)

δ

hz = K (Ts-Tw) = K _______ (7)

δ (Ts-Tw) δ

Substituting equation (6) in (7) we get

1/4

hz = λρ2g K

_______ (8)

4(Ts-Tw) μ z

The mean heat transfer co-efficient over the length L of the plate will be

L

h = 1 ∫ hz dz

L 0

1/4

L

= 1 λρ2g K

3 ∫ dz

L 4(Ts-Tw)μ 0 z1/4

1/4

= 4 λρ

2g K

3

3 4(Ts-Tw) μL

h = 4 hz=L _______ (9)

3

Thus, the average heat transfer co-efficient over the length L of the plate is 4/3 times the local co-

efficient at z=L.



1/4

h = 4 λρ2g K

3

3*41/4

(Ts-Tw)μL

1/4

h = 0.943 λρ2g K

3 (Ts-Tw)μL

1/4

h = 0.943 K 3ρ

2g

λ

_______ (10)

ΔTμL Similarly, for laminar flow condensation on a HORIZONTALTUBE Nusselt gives

h = 0.725 kf3 ρf

2g λ

1/4

Doμ(Ts-Tw)

h = 0.725 kf3 ρf

2g λ

1/4

ΔTDoμ

The Reynold’s number for condensate film is given by

NRe = DeG'

μ

= 4 Af G

P Af

μ

= 4 G

P μ

= 4 M

μ

Where G’ = mass flow rate per unit area = G

Af



M = mass flow rate of condensate per unit length = G

ΠDo

If 4 m < 1800, the condensate fluid flow is laminar.

μ

If 4 m > 1800, the condensate fluid flow is turbulent.

μ Where m = mass flow rate of vapor per unit length of vertical plate (kg/m.s).

Effect of non – condensable gases:

If a non – condensable gas is also present, for example, air in the condensing vapors in a

condenser because of leakage of hinders the process of heat transfer. The non – condensable

gas collects in the vicinity of the condensate surface and the condensing vapor must have to

diffuse through the gas film. The presence of diffusion resistance into the process of

condensation decreases the rate of condensation far below that for a pure metal. Presence of

air about 1% by volume can reduce the heat transfer co –efficient by 60% of its value for no

air. Air vent is always provided on almost all the condensers to eliminate air in the system.

PROCEDURE:

1. The inlet valve is opened and the cold fluid is allowed to flow through the condenser.

2. The flow rate of the cold fluid is adjusted to minimum.

3. The steam inlet valve is opened and the steam pressure is kept constant throughout the

experiment.

4. After the cold fluid temperature becomes steady, the inlet temperature, outlet temperature

and flow rate of the cold fluid are noted down and also the time required for 1 cm rise in

the condensate level in the condensate tank is noted.

5. Keeping the steam pressure constant, 4 readings are taken for different flow rates of the

cold fluid from minimum to maximum.

6. The experiment is repeated for another constant steam pressure.

OBSERVATION:

1. Inner pipe outer dia, do = 16 mm.

2. Inner pipe inner dia, dI = 13 mm.

3. Outer pipe inner dia, DI = 28 mm.

4. Length of the pipe, L = 1500 mm.

5. Condensate vessel dia, D = 140 mm.

6. Weight of empty bucket = ________ kg.



TABULAR COLUMN:

Sl.

No:

Steam pressure

Guage

(kg/cm2)

Cold fluid

temperature

Mass collected

In bucket +

Bucket wt.

(kg)

Mass of water

Collected

(kg)

Time of

collection

(sec)

Condensate flow

rate.

Time required

For 1 cm rise in

Height

(sec)

t1°C

t2°C

P1

P2

FORMULAS:

1. Qc = mc Cpc (ΔT)c = mc Cpc (t2 - t1).

2. Ts = temperature of steam at 1.2 bar pressure.

3. λ = from steam table for that pressure and temperature in (kJ/kg°K).

4. Mass flow rate of condensate = ________.

5. Condensate vessel area = A' = πD2

4

6. Height of condensate water collected = h.

7. Time required to collect the condensate = t.

8. Mass flow rate of condensate, mh = A'h ρ (kg/sec).

T



9. (ΔT)LMTD = (Ts-Tc2) – (Ts – Tc1)

ln (Ts-Tc2)

(Ts–Tc1)

10. Twall = Tw = Ts+Tavg

2

11. Δt = Ts – Twall.

1/4

ho = 0.725 Kf

3ρf

2 g

λf

ΔTf μf do

12. 1 = 1 + Xw do + 1 do


1/4

ho = 0.725 Kf3ρf

2g λf

μf ΔTf do

CALCULATION:

1. t1 = _____ °K. t2 = _____ °K.

2. tavg = t1°K + t2°K = = _______°K.

2

3. Operating pressure = 0.2 kg/cm2 = 0.2 bar.

4. Barometric pressure = 1.0 kg/cm2 = 1.0 bar.

5. Total pressure = 1+0.2 = 1.2 bar.

6. Temperature of steam at 1.2 Bar = _________°K.

Cold fluid (water) properties at tavg are:

7. K = ________(W/m°K).

8. ρ = 1/volume = ________(kg/m3) = 992.875kg/ m3

9. μ = _________kg/ms.

10. λ = _________ KJ/kg°K.

11. Cpc = 4.178 kJ/kg °K.



12. A' = πd2

4 m2

13. Mass flow rate of condensate = A'hρ

T kg/s.

14. mc = mass of water in bucket – mass of bucket

time

15. Qc = mc

ρ

16. Ac = area of annulus in which cold fluid is flowing

= (Di2 – do2) * π

4

17. Vc = Qc

Ac

18. De = (Di2 – do2)

Do

19. NRe = DeVc ρ

μ

If NRe > 2100 and NRe < 10000, JH factor (from the graph) is used.

20. JH = (hi/CpG)(Cpμ/K)2/3(μW/μ)0.14

hi = ________ W/m2°K.

Where G = mc/Ac =

21. Qc = mc Cpc Δt



22. Uo exp = Qc

(ΔT LMTD)(Area of heat transfer)

Where (ΔT) LMTD = (Ts-Tc2) – (Ts – Tc1)

ln (Ts-Tc2)

(Ts–Tc1)

Where Ts = _________ temperature of steam at 1.2 bar pressure.

23. Area of heat transfer = πdoL m2.

24. Uo exp

25. Tw = Tf = Ts + T avg

2


Properties at Tf :

Kf = ________(W/m°K).

ρf = 1/volume = ________(kg/m3) = __________kg/ m3

μf = _________kg/ms.

λf = _________ J/kg°K.

Cpc = 4.178 kJ/kg °K.

1/4

27. ho = 0.725 Kf 3ρf

2 g

λf

ΔTf μf do

28. 1 = 1 + Xw do + 1 do




Where Xw = do - dI

2

dw = do - dI

ln do

dI

29. 1 =

(Uo)theo

If NRe > 10,000

We use NNu = hi * de = 0.023 [(NRe) 0.8 ( Npr)

0.14]

K

1. Npr = Cp μ

K

Where Cp = specific heat capacity of water at mean temperature.

μ = viscosity of water at mean cold temperature.

K = thermal conductivity of water at mean temperature.

2. NNu = hi * de = 0.023 [(NRe) 0.8 ( Npr)

0.14]

K

3. Qc = mc Cpc ΔTc

4. (ΔT)ln = (Ts-Tc2) – (Ts – Tc1)

ln (Ts-Tc2)

(Ts–Tc1)

5. Area of heat transfer , Ah= πdoL

6. (Uo)Experimental = Qc_____________________

(ΔT)ln * Ah



HEAT TRANSFER CO-EFFICIENT OF STEAM SIDE FLUID:

1. Average temperature of condensate =

2. Tf = Ts+Tavg

2


PROPERTIES OF CONDENSATE WATER AT Tf:

1. ρf = kg/ m3

2. Cpf = J/kg °K.

3. Kf = W/m°K. 4. μf = kg/ms.

5. λf = J/kg°K.

By Nusselt equation

1/4

ho = 0.725 Kf3ρf

2g λf

μf ΔTf do

1 = 1 + Xw do + 1 do


Where Xw = do - dI

2

dw = do - dI

ln do

dI

1 =

(Uo)theo



WILSON PLOT

Results:

SL.

NO:

Working

pressure

(Bar)

NRe hi

W/m2ºK

ho

W/m2ºK

Uo exp

W/m2ºK

Uo theo

W/m2ºK

1

UoExp

INTERCEPT

1

Vc0.8



Applications: Remarks:



1. What is the importance of Heat transfer in chemical engineering?

2. What is Fourier law of heat conduction?

3. What is the significance of thermal conductivity?

4. Define insulation?

5. What is the significance of individual heat transfer coefficient?



Experiment No: __5________ Date: ____________

TRANSIENT HEAT CONDUCTION UNDER CONSTANT TEMPERATURE

Aim:

1) To verify the experimentally predicted value of temperature with theoretically

predicted value.

2) To plot a graph of temperature vs time

3) To plot a graph of temperature vs distance

Apparatus/Components required: Thermometer, stop watch, metal rod etc

Theory:

A solid is said to be in steady state if its temperature does not change with time. If there is an

abrupt change in its surface temperature or environment, it takes some time for the body to attain

equilibrium temperature or steady state temperature. During this interim period, the temperature

varies with time and the body is said to be in an unsteady state or transient state.

Unsteady state heat flow is very common in all heating or cooling problems. Hardening by

quenching, cooling of I.C. engine cylinders, and heating of boiler tubes are common examples of

unsteady state heat flow. Transient heat conduction problem can be periodic heat flow problem or

non periodic heat flow problem. Periodic heat flow problems are those in which the temperature

varies on a regular basis. In the non periodic type , the temperature at any point within the system

varies non linearly with time.

Generally temperature of the body T is given as the function of x,y,z and t.

T = f (x,y,z,t)

By heat balance it is found that

The rate of change of temperature with respect to time is proportional to the second derivative of

the temperature gradient with respect to distance. The proportionality constant ‘’ is called the

‘thermal diffusivity’ of the solid and is a property of the material. It is the ratio of the thermal

conductivity ‘k’ to the product of the density and heat capacity. By determining the thermal

conductivity of the material it is possible to identify the material by comparing the thermal

conductivity of the material from the literature.

T t

=

2T

x2



Classical examples of transient heat conduction are quenching of fillets, burning of bricks,

annealing of solids, manufacture of glass vulcanization of rubber, steaming of wood. As the heating

is continued, the temperature at a given point asymptotically approaches the temperature of the

heating medium where as point near the surface quickly approaches the temperature of the

surroundings, those in the interior lag far behind.

The primary law that describes conduction heat transfer is Fourier's Law of Heat Conduction.

Fourier's Law is based upon the observation that the conductive heat flux is directly proportional to

the negative of the temperature gradient or

--- (1)

Introducing the thermal conductivity as the constant of proportionality, we then have the equation

--- (2)

In this experiment we will utilize Fourier's Law to study the problem of transient, one dimensional

heat conduction in a cylinder and to use the law in determining the thermal conductivity of a solid.

A cylindrical element which is embedded with several thermocouples is heated at one end by an

electric hot plate and cooled at the other end by flowing water. The side of the cylinder is very well

insulated so that the heat conduction is assumed to be one dimensional. A schematic of the

apparatus is shown below.

We consider the transient problem in which the cylinder begins at some constant, uniform

temperature and then suddenly the hot plate is turned on so that a heat flux is imposed at the lower

boundary. Our describing equation for conservation of energy balances the internal energy change

with the axial heat conduction. Hence, in differential form we write



--- (3)

where α is the thermal diffusivity of the cylinder material. One way to solve this equation is to

discretize the space domain, write it in finite difference form, and solve the subsequent system of

algebraic equations for the discritized temperatures. If we use a second order correct approximation,

the finite difference form of Eq. above becomes

--- (4)

The subscript on the T represents the space (or z) discetization while the superscript represents the

time discetization. Note that in Eq. (4) we have not specified the time discretization (j or j-1) for the

spatial derivative. There are two choices for the time discretization for the temperatures in the

spatial derivative. They could be evaluated at the previous time step, j-1, or at the current time step,

j. The algorithm is called explicit if the temperatures are evaluated at the previous time step, and

clearly this makes the algebraic system very easy to solve since Eq. (4) can be solved directly forTi

(j) with no coupling to the other spatial nodes (i+1 and i-1). However, the explicit approach does

not always lead to a stable solution (can you say blow-up) and in fact stability is only guaranteed

when

Since for most materials α is of the order of 10-5, this stability criteria often leads to unacceptable

time steps or spatial grids. The implicit algorithm, when the spatial derivative is evaluated at the

current time step, does not have this stability problem, but does require simultaneous solution of the

spatial node equations. Microsoft Excel is a powerful spreadsheet tool that can carry out these

simultaneous calculations. Hence, for a spatial domain with N spatial nodes, we would have N

simultaneous equations of the form

---(5)

to be solved for the Ti (j) 's.

In steady state the axial temperature profile should be linear which confirms Fourier's Law. The

steady state heat transfer is determined by measuring the mass flow rate and temperature change of

a coolant stream which passes over one end of the element, or q = m c ( ) T - T p out in coolant & &

(6) Then the thermal conductivity can be calculated by



PROCEDURE:

1. the thermometer is immersed into the respective ( thermo well) openings

2. The cold water is allowed at one end of the pipe and steam at constant temperature is

allowed at the other end.

3. Initial temperature of the system is noted down and steam is supplied simultaneously.

(steam at 0.5 kg/cm2)

4. the temperature at different points along the length of the rod are noted down at 5 min

interval

5. The procedure is continued till temperature reaches steady value.

Observation and calculation

1. diameter of the mild steel rod = -------- mm

2. length of mild steel rod=------- mm

3. Thermowell points

0-50, 0-150, 0-250, 0-350, 0-450, 0-550 (all dimensions in mm)

Readings

Experimentally

Time in

minutes

T1oC T2

oC T3

oC T4

oC T5

oC T6

oC T7

oC

Calculations

{[T – Ta] / [TS –Ta] } =1-[X /L] – [2/ Π ]{∑ [(1/n) sin (n ΠX/L) e(-n Π α ө)

}

Where

T = temperature at any location x at time ө

Ta = room temperature

TS = steam temperature

α = thermal diffusivity = thermal conductivity / (density x specific heat)= k/ρCP

X = distance, L = total length of metal rod

Calculations

Theoretically from above equation

T1oC T2

oC T3

oC T4

oC T5

oC



Results: The theoretical and experimental values of variation of temperature with time and distance

are as tabulated.

Applications: Remarks:



1. Explain the process of condensation?

2. What do you mean by film type and drop wise condensation? Which one is preferred

industrially?

3. Give few examples for film type and drop wise condensation?

4. What do you say about the type of condensation in your condenser experiment?

5. What is equivalent diameter? How do you find the same mathematically?

6. When we go for equivalent diameter?




DIFFUSIVITY CO-EFFICIENT OF VAPOUR IN AIR

Aim: To determine the diffusivity co-efficient for Acetone – Air system.

Apparatus/Components required: Arnold cell, magnetic stirrer, heater, beaker with hot water, Thermometer, Stop clock.

Theory: DIFFUSION is the movement of individual component through a mixture under the

influence of the physical stimulus.

The most common cause of diffusion is the concentration gradient of the diffusing

component in such a direction so as to equalize concentration and destroy the gradient.

In a two - phase system, which is not at equilibrium, spontaneous alteration occurs

throughout the Molecular diffusion bringing the entire system to a state of equilibrium

where vapor alteration stops.

At the end we observe that the concentration varies in both the phases.

Molecular diffusion is with movement of individual molecular diffusion ultimately leads to

completely uniform concentration of substance through out the solution, which may initially

have been non uniform. The rate at which a substance (solute) moves in a solution at any

point in any direction depends on the concentration gradient at that point and in that

direction. In order to express quantitatively, an approximate rate is needed. The rates are

described in terms of molar flux, moles per area time. The area being measured normal to

the direction of diffusion. Two types of fluxes are often used to describe the molecular

diffusion. J & N types. J is defined as flux of a constituent relative to the molar velocities of

all other constituents. While, N may be defined as the flux of a constituent relative to a fixed

point or location in space.

If the diffusion is purely by molecular motion, then J may be given by the equation

J = -DAB ∂CA / ∂Z or J = -CDAB ∂XA / ∂Z

Where DAB = diffusivity of diffusion coefficient of constituent A in solution B.



This equation is known as Fick’s law of diffusion, which is analogous to Newton’s Law of

viscosity and Fourier law of heat conduction.

Two types of situations are often encountered in mass transfer operations:

1) Constituent A diffusing in stagnant non-diffusing constituent B.

Ex: Gas absorption where a solute gas like NH3 in liquid water (solvent) with no diffusion of

water vapor into the stream of gaseous mixture of Air-NH3 where water diffuses into the air

stream with no Air diffusion into the water stream.

2) Diffusion of both A and B in Counter-Current direction (equimolar).

Ex: Separation of Benzene for Toluene by Distillation, where Benzene and Toluene diffuses

across the interface i.e. Benzene (A) diffuses from vapor phase to liquid phase while

Toluene (B) diffuses from liquid to vapor phase.

The true driving force for this diffusion is actually the chemical potential and not the

concentration. The rate at which the solute moves at any point in any direction must

therefore depend on the Concentration gradient at that point and in that direction the rate

will be expressed in terms of molar flux. Mass transfer takes place from the surface from the

molecular diffusion alone. The rate of fall of the liquid surface since the diffusivity co-

efficient can be calculated by the equation

NA = DAB PT (PA1-PA2)/RTZPBM Where PBM = log mean difference of partial pressure of B.

PBM = [(PB2-PB1)/ (ln PB1/PB2)]

This equation may be further modified for DAB as

DAB = [RT (Z1²-Z2²) ρ1/PT (ln PB1/PB2) 2t M] And for equimolar counter current diffusion,

NA = DAB PT (PA1-PA2)/RTZPBM

Diffusion can occur by two different mechanisms: interstitial diffusion and substitutional diffusion.

Picture an impurity atom in an otherwise perfect lattice. The atom can sit either on the lattice itself,

substituted for one of the atoms of the bulk material, or it can sit in an interstice (interstitial). These

two positions give rise to the two different diffusion mechanisms.

Substitutional Diffusion Substitutional diffusion occurs by the movement of atoms from one atomic site to another. In a

perfect lattice, this would require the atoms to “swap places” within the lattice. A straight-forward

swapping of atoms would require a great deal of energy, as the swapping atoms would need to



physically push other atoms out of the way in order to swap places. In practice, therefore, this is not

the mechanism by which substitutional diffusion occurs.

Substitutional diffusion occurs only if a vacancy is present. A vacancy is a “missing atom” in the

lattice. If a vacancy is present, one of the adjacent atoms can move into the vacancy, creating a

vacancy on the site that the atom has just left. In the same way that there is an equal probability of

an atom moving into any adjacent atomic site, there is an equal probability that any of the adjacent

atoms will move into the vacancy. It is often useful to think of this mechanism as the diffusion of

vacancies, rather than the diffusion of atoms.

The diffusion of an atom is therefore dependent upon the presence of a vacancy on an adjacent site,

and the rate of diffusion is therefore dependent upon two factors: how easily vacancies can form in

the lattice, and how easy it is for an atom to move into a vacancy. The dependence upon the

presence of vacancies makes substitutional diffusion slower than interstitial diffusion, which we

will look at now.

Interstitial Diffusion In this case, the diffusing atom is not on a lattice site but on an interstice. The diffusing atom is free

to move to any adjacent interstice, unless it is already occupied. The rate of diffusion is therefore

controlled only by the ease with which a diffusing atom can move into an interstice. Theoretically,

at very high impurity concentrations movement may be restricted by the presence of atoms in the

adjacent interstices. In practice, however, it is very likely that a new phase would be formed before

this had an effect.

DERIVATION: DIFFUSION THROUGH A STAGNANT GAS FILM Consider a narrow tube of uniform cross-section, which is partially filled with pure liquid

A, is maintained at a constant temperature and pressure. Gas B which flows across the open

end of the tube, has a negligible solubility in liquid A vaporizes and diffuses into the gas

phase.

FLOW OF GAS B→

∆Z

PURE LIQUID A

↑NAZIZ+∆Z

---------------

---------------

↑ NAZIZ



Consider the control volume S∆Z.

Where S= Cross-Sectional area of the tube.

Mass Balance on A over the control volume for a steady state operation yields,

[Moles of A leaving at Z+∆Z] - [Moles of A entering at Z] = 0 __ (1)

S NAZIZ+∆Z - S NAZIZ = 0

Dividing Equation (1) by the volume, S∆Z and evaluating in the limit ∆Z→0 we get the

differential equation,

For component A, dNA = 0 __ (2)

dZ

For component B, dNB = 0 __ (3)

dZ

Considering only at plane Z1, and since the gas B is insoluble in liquid A, we realize that

NB, the net flux of B, is zero throughout the diffusion path, accordingly B is a stagnant gas.

We have, in a binary system, containing A and B; the molar flux is given by

NA = - CDAB dyA + yA (NA+NB)

dZ

NA = - CDAB dyA __ (4)

1- yA dZ

The equation is integrated between two boundary conditions

i. At Z=Z1, yA=yA1

ii. At Z=Z2, yA=yA2

Assuming the diffusivity is to be independent of concentration and NA is constant along the

diffusion path

By integrating equation (4) we get

Z2 yA2

NA ʃ dZ = CDAB ʃ - dyA

Z1 yA1 1- yA



NA = CDAB ln ( 1- yA1 ) _ (5)

Z2-Z1 1- yA2

The log mean average concentration of component B is defined as –

yB, lm = yB2-yB1

Ln (yA2/yA1)

Since yB = 1- yA

yB, lm = (1- yA2) – (1- yA1)

Ln (yA2/yA1)

= yA1 – yA2

Ln (yA2/yA1) ___ (6)

Substituting equation (6) in equation (5) we get

NA = CDAB (yA1 – yA2) _ (7)

Z2-Z1 yB, lm

For an ideal gas C = n = P

V RT

For a mixture of ideal gases, yA = PA

P

For an ideal gas mixture equation (7) becomes

NA = DABPT (PA1 –PA2) _ (8)

RT (Z2-Z1) PB, lm

DAB = NA RT (Z2-Z1) PB, lm

A PT (PA1 –PA2)



Procedure:

1. The diffusion cell is cleaned and rinsed thoroughly with acetone. The vessel containing water

is heated on the heater which acts as a constant temperature bath. The bath temperature can

be adjusted by heater and the regulator. The temperature is noted down with the help of the

thermometer which is dipped in the water bath. The uniformity of the temperature is ensured

by constant stirring with the magnetic stirrer.

2. The diffusivity cell is filled with acetone up to a known level and the blower is started and

adjusted for steady state of air across the diffusivity cell.

3. After the steady state is achieved the stop watch is immediately started and the time required

for 0.5mm decrease in the level of acetone is noted down. The corresponding temperature,

initial and final levels of acetone is also noted.

4. The above procedure is repeated for different temperatures.

5. The diffusion co-efficient at different temperatures is calculated.


Room temperature = ___________ ºC.

1. Diameter of diffusivity cell = 0.7 cm.

2. Cross – section area of the cell = Пd² m2.

4

3. Molecular weight of acetone = 58.04g.

4. Density of acetone = 789g/cc.

5. Universal gas constant = 63.187 mm Hg m³.

Kg mol

TABULAR COLUMN:

SL

NO

Initial

reading

Z1(cm)

Final

reading

Z2 (cm)

Zavg

(cm)

Temp

(°c)

Time

(sec)

NA/A

gm.mol/cm²sec

PA1

mmHg

PB1

mmHg

PBm

mmHg

DAB

m²/s

1.

2.

3.

4.

5.

6.



FORMULAS USED:

1. NA = Height * Density of Acetone

A (molecular wt.)Acetone * Time

Where Height = (Z2-Z1) * 10ˉ² m.

2. Zavg = Z1+ Z2 = cm.

2

3. PB1 = PT - PA1

PT = 760 mmHg.

4. PB2 = PT – PA2

5. PBm = PB2- PB1

ln (PB2/ PB1) 6. R = 62.86 ton cm³

g mol ºK

= 62.86 * 10ˉ6 * 750

10ˉ³ 760

R= 63.187 mm Hg.

7. DAB = NA RT Zavg PBm

A PT (PA1 –PA2)

7.

8.

9.

10.



VAPOR PRESSURE DATA OF ACETONE

NATURE OF GRAPH:

Graph of vapour pressure Vs temperature

Vapour pressure

mm Hg

TEMPERATURE ºC

Temperature

(ºC)

Vapor pressure

(mm Hg)

7.7 100

22.7 200

39.4 400

56.5 760



Graph of DAB Vs temperature

DAB

m2/s

Temperature ºC

Results:

Diffusivity co-efficient is found to be DAB = __________________ m²/s.



Applications:

1. Diffusion Bonding: Diffusion bonding is a method of joining metals similar to welding, but relies only on the surfaces diffusing into one another as a means of 'welding'. This process is aided by high pressure and/or temperature.

2. Heat treatment of metals.

3. Sintering of a Powder Compact: The processes governing powder metallurgy and one type

of ceramic material processing are greatly dependent on diffusion processes that combine

distinct powdered grains into one cohesive material. This process, also known as sintering.

4. Carburization: Carburization is the process by which carbon is diffused into the surface of

steel in order to increase its hardness.

Remarks:


Probable viva questions: 1. What is meant by diffusivity? 2. What are different types of diffusion? 3. Mention the examples of diffusivity process. 4. Explain the principle of diffusion. 5. Explain the laws involved in the diffusion process.




SIMPLE DISTILLATION

Aim:

To verify Rayliegh’s equation ln (F/W) =dXA/y-XA by differentially distilling the given binary

mixture (SYSTEM: methanol-water).

Apparatus/Components required: Beakers, thermometers, measuring jar, pipette, round bottom flask, balance, specific gravity

bottle etc.

Theory: Distillation is a chemical engineering unit operation where two liquids are separated from each

other. You do this by simply heating the liquids until they boil. The liquids will have different

vapor pressures, and so one of them will evaporate before the other does. The vapor is then led

into a condenser, which takes off its heat and condenses it back to liquid. A fractionating

column can be used to improve the separation. Distillation is a mass transfer unit operation.

Distillation is probably the most common technique for purifying organic liquids. The

vaporized and condensed component is called distillate and the other component as bottom

product this has been used for a long time, to distill alcohol, and produced distilled beverages.

Distillation is a traditional technique used to separate and purify liquids in the chemistry lab and

throughout industry. It uses the difference in boiling points of liquids to separate these liquids

from other substances in solution. A liquid has a specific boiling temperature at standard

pressure conditions. If there are two liquids in a solution, the liquid with the lowest boiling

point can be boiled out of the solution without removing a significant amount of the other

liquid. By vaporizing the liquid and condensing it in another container, one can obtain a pure

sample of the liquid. Any impurities, and liquids with higher boiling points, remain in the

original container. Distillation is a method of separating mixtures based on differences in their

volatilities in a boiling liquid mixture. Distillation is a unit operation, or a physical separation

process, and not a chemical reaction. If during an infinite number of successive flash

vaporizations of a liquid only an infinitesimal portion of a liquid were flashed each time, the net

result would be differential (or) simple distillation.

A batch of liquid is charged to a kettle with a heating device. The charge is boiled slowly and

the vapors are withdrawn as rapidly as they form to condenser, where they are liquefied and the

condensate (distillate0 is collected in the receiver. The first portion of the distillate will be

richest bin the more volatile substance and as in the more volatile substance and as the

distillation proceeds, the vaporized product becomes linear. The distillate can therefore be

collected in several separate batches called “CUT OFF” to give a series of distilled products of



various purities. The vapor issuing from a true differential distillation is at any time in

equilibrium with the liquid from which it rises but changes continuously in composition.

Commercially, distillation has a number of uses. It is used to separate crude oil into more

fractions for specific uses such as transport, power generation, and heating. Water is distilled to

remove impurities, such as salt from seawater. Air is distilled to separate its components—

notably oxygen, nitrogen, and argon—for industrial use. Distillation of fermented solutions has

been used since ancient times to produce distilled beverages with higher alcohol content. The

premises where distillation is carried out, especially distillation of alcohol are known as a

distillery.

IDEALIZED DISTILLATION MODEL

The boiling point of a liquid is the temperature at which the vapor pressure of the liquid equals

the pressure in the liquid, enabling bubbles to form without being crushed. A special case is the

normal boiling point, where the vapor pressure of the liquid equals the ambient pressure. It is a

common misconception that in a liquid mixture at a given pressure, each component boils at the

boiling point corresponding to the given pressure and the vapors of each component will collect

separately and purely. This, however, does not occur even in an idealized system. Idealized

models of distillation are essentially governed by Raoult's law and Dalton's law, and assume

that vapor-liquid equilibria are attained.

Raoult's law assumes that a component contributes to the total vapor pressure of the mixture in

proportion to its percentage of the mixture and its vapor pressure when pure, or succinctly:

partial pressure equals mole fraction multiplied by vapor pressure when pure. If one component

changes another component's vapor pressure, or if the volatility of a component is dependent on

its percentage in the mixture, the law will fail.

Dalton's law states that the total vapor pressure is the sum of the vapor pressures of each

individual component in the mixture. When a multi-component liquid is heated, the vapor

pressure of each component will rise, thus causing the total vapor pressure to rise. When the

total vapor pressure reaches the pressure surrounding the liquid, boiling occurs and liquid turns

to gas throughout the bulk of the liquid. Note that a mixture with a given composition has one

boiling point at a given pressure, when the components are mutually soluble. An implication of

one boiling point is that lighter components never cleanly "boil first". At boiling point, all

volatile components boil, but for a component, its percentage in the vapor is the same as its

percentage of the total vapor pressure. Lighter components have a higher partial pressure and

thus are concentrated in the vapor, but heavier volatile components also have a (smaller) partial

pressure and necessarily evaporate also, albeit being less concentrated in the vapor. Indeed,

batch distillation and fractionation succeed by varying the composition of the mixture. In batch

distillation, the batch evaporates, which changes its composition; in fractionation, liquid higher

in the fractionation column contains more lights and boils at lower temperatures.

http://en.wikipedia.org/wiki/Normal_boiling_point

http://en.wikipedia.org/wiki/Raoult%27s_law

http://en.wikipedia.org/wiki/Vapor_pressure



The idealized model is accurate in the case of chemically similar liquids, such as benzene and

toluene. In other cases, severe deviations from Raoult's law and Dalton's law are observed, most

famously in the mixture of ethanol and water. These compounds, when heated together, form an

azeotrope, which is a composition with a boiling point higher or lower than the boiling point of

each separate liquid. Virtually all liquids, when mixed and heated, will display azeotropic

behavior. Although there are computational methods that can be used to estimate the behavior

of a mixture of arbitrary components, the only way to obtain accurate vapor-liquid equilibrium

data is by measurement.

It is not possible to completely purify a mixture of components by distillation, as this would

require each component in the mixture to have a zero partial pressure. If ultra-pure products are

the goal, then further chemical separation must be applied. When a binary mixture is evaporated

and the other component, e.g. a salt, has zero partial pressure for practical purposes, the process

is simpler and is called evaporation in engineering.

Principles of Distillation

Thermodynamics play an important role in the distillation of essential oils. In order to vaporize

any liquid, energy in the form of heat must be applied. This energy is termed latent heat. When

a vapor is converted back to liquid, there will be a reduction and release of that latent heat.

In order for a liquid to be vaporized, latent heat must be applied to this liquid until its

temperature attains the point where its vapor pressure becomes equal to the surrounding

atmospheric pressure. When it reaches this point the temperature will rise no further. However,

if heat continues to be applied, the liquid will take up this latent heat and then vaporize at the

appropriate rate. At this point the liquid has said to have reached its “boiling point” under the

prevailing pressure.

When mutually insoluble compounds such as water and essential oil are present, the total

pressure exerted by the mixed vapor then becomes the sum of the partial pressures exerted by

each constituent present. This liquid mixture will boil when its temperature is raised to the point

where the combined vapor pressure of its components becomes equal to the surrounding

pressure.

Simple distillation

Simple distillation is designed to evaporate a volatile liquid from a solution of non-volatile

substances; the vapor is then condensed in the water condenser and collected in the receiver. In

simple distillation, a liquid is boiled and the vapors work through the apparatus until they reach

the condenser where they are cooled and reliquify. Liquids are separated based upon their

differences in boiling point. This operation is an improvement over flash vaporization. Simple

distillation is a combination of multiple flash vaporizations. In flash vaporization, ratio of

(W/D) is very high. To decrease ratio of (W/D), multiple flash vaporization will be carried out.

In practice, heating of feed is continuous, the vapor formation and condensation is also

http://en.wikipedia.org/wiki/Benzene

http://en.wikipedia.org/wiki/Toluene

http://en.wikipedia.org/wiki/Azeotrope



continuous. Simple distillation is a distillation without reflux. In this type of distillation, the

mixture is subjected to separation by vaporization by heating at very slow heating rate, so that a

small differential quantity of liquid mixture is vaporized. The vapor formed which is in

equilibrium, is immediately removed and condensed.

In simple distillation, all the hot vapors produced are immediately channeled into a condenser

which cools and condenses the vapors. Therefore, the distillate will not be pure - its

composition will be identical to the composition of the vapors at the given temperature and

pressure, and can be computed from Raoult's law. As a result, simple distillation is usually used

only to separate liquids whose boiling points differ greatly (rule of thumb is 25 °C), or to

separate liquids from involatile solids or oils. For these cases, the vapor pressures of the

components are usually sufficiently different that Raoult's law may be neglected due to the

insignificant contribution of the less volatile component. In this case, the distillate may be

sufficiently pure for its intended purpose. The boiling point of a pure organic liquid is a

physical property of that liquid. It is defined as the temperature at which the vapor pressure of

the liquid exactly equals the pressure exerted on it. Boiling points can be determined using the

technique of simple distillation. Distillation is a technique that is used to purify a mixture of

liquids or to obtain a boiling point of a pure liquid (in the case of this course).

Essentially, the liquid is heated to boiling and the vapors condensed above the boiling liquid.

Simple distillation is a procedure by which two liquids with different boiling points can be

separated. Simple distillation (the procedure outlined below) can be used effectively to separate

liquids that have at least fifty degrees difference in their boiling points degree of separation. As

the liquid being distilled is heated, the vapors that form will be richest in the component of the

mixture that boils at the lowest temperature. Purified compounds will boil, and thus turn into

vapors, over a relatively small temperature range (2 or 3°C); by carefully watching the

temperature in the distillation flask, it is possible to affect a reasonably good separation. As

distillation progresses, the concentration of the lowest boiling component will steadily decrease.

Eventually the temperature within the apparatus will begin to change; a pure compound is no

longer being distilled. The temperature will continue to increase until the boiling point of the

next-lowest-boiling compound is approached. When the temperature again stabilizes, another

pure fraction of the distillate can be collected. This fraction of distillate will be primarily the

compound that boils at the second lowest temperature. This process can be repeated until all the

fractions of the original mixture have been separated. An example of a simple distillation is the

separation of a solution of salt and water into two separate pure substances.

RAYLEIGH’S EQUATION:

This is an important equation applicable for simple distillation process. The usefulness of this

equation is, if, three of the four quantities are known then the unknown can be calculated.

Usually F and XF are known and unknown is either W or XW. To derive this equation

following assumption is made:

The vapors coming out of the distillation column is in equilibrium with liquid and in small

quantity but changes continuously in composition.



Consider ‘F’ moles of feed with concentration XF in a vessel. The vessel is then subjected to

simple distillation, until W moles of residue with concentration XW is left over in the vessel.

During distillation process at any time, let ‘L’ be the moles of liquid present in vessel with

concentration X. when this liquid is subjected to flash vaporization, let dL moles of liquid is

vaporized with concentration y.

Let L-dL be the liquid left over in the vessel with the concentration X-dX.

Writing the component balance:

Initial amount in vessel = amount left in vessel + amount vaporized.

L.X = (X-dX)(L-dL) + y.dL

L.X = L.X-L.dX-X.dL+dL.dX + y.dL

L.dX = dL(y-X)

dX = dL

y-X L ___ (1)

Integrating equation (1) between the following limits

When X=XF, L=F

X=XW, L=W

W XW

∫ dL = ∫ dX

L y-X

F XF

F XF

∫ dL = ∫ dX

L y-X

W XW

XW

ln (F/W) = ∫ dX

y-X XF

XW

ln (F/W) = ∫ dX

y*-X

XF

Where y* = equilibrium quantity.



ADVANTAGES:

1. Simpler setup.

2. Faster distillation times.

3. Consumes less energy.

DISADVANTAGES:

1. Requires the liquids to have large boiling point difference (>70°c).

2. Gives poorer separation.

3. Only works well with relatively pure liquids.

4. Simple distillation is uneconomical for small units and purity requirements are not very

stringent and rigid.

Procedure:

1. 200ml of methanol/ethanol and 100ml of water is taken in a round bottom flask of

the simple distillation setup.

2. The round bottom flask is then connected to a horizontal condenser and then to a

flask.

3. The solution in the flask is heated and the vapors are condensed. The distillate is then

collected.

4. The distillation flask is cooled then the volume and specific gravity of the distillate

and the residue are found.

PROCEDURE FOR CALIBRATION CHART: 1. The weight of the empty specific gravity bottle is found out.

2. 10 beakers of 100 ml is taken and to these methanol/ethanol -water mixture is added

in increasing order of the methanol and decreasing order of water.

3. Then the mixture is placed in specific gravity bottle and each weight is noted and

their specific gravity, mole and mole fraction of methanol are calculated.

4. The calibration chart of specific gravity Vs. mole fraction of methanol is plotted.

From this mole fraction of distillate and residue are obtained.

5. Another graph of equilibrium data of liquid Vs. vapor for methanol-water is plotted

and from this the area under the curve is calculated.



OBSERVATION AND CALCULATION:

1. Weight of empty specific gravity bottle = __________ g

2. Weight of empty specific gravity bottle + water = __________ g

3. Weight of empty specific gravity bottle + distillate = __________ g

4. Weight of empty specific gravity bottle + residue =__________ g

5. Weight of residue = ___________ g

6. Weight of distillate =____________ g

7. Volume of distillate collected = _________ ml

8. Volume of residue collected = _________ml

9. Specific gravity of residue = weight of residue

Weight of water

10. Specific gravity of distillate = weight of distillate

Weight of water

11. Molecular weight of methanol = 32.04g

12. Molecular weight of water = 18g

TABULAR COLUMN 1:

SL.

NO:

Vol. of

methan

ol/etha

nol

Vol.

of

water

Wt. of

mixture

Specific

gravity

Wt. of

methanol/

ethanol

Wt.

of

water

Moles of

methanol/

ethanol

Moles

of

water

mole

fraction of

methanol/

ethanol

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.



TABULAR COLUMN 2:

SL.NO: LIQUID

XA

VAPOUR

y*

__1___

y* - XA

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

CALCULATION:

1. Specific gravity = weight of mixture

Weight of water

2. Weight of methanol/ethanol = volume of methanol/ethanol * Specific gravity

3. Weight of water = volume of water * Specific gravity

4. Moles of methanol/ethanol = Mass of methanol /ethanol

Molecular weight of methanol

5. Moles of methanol/ethanol = Mass of methanol/ethanol


6. Moles of methanol/ethanol = Mass of methanol/ethanol


7. Moles of water= Mass of water

Molecular weight of water

8. Mole fraction of methanol/ethanol = Moles of methanol/ethanol

Moles of methanol/ethanol + Moles of water

9. F = (vol. * density) + (vol.* density)

(Molecular weight) water (Molecular weight) methanol

10. W = vol. of residue * density of residue

Average molecular weight



11. Average molecular weight = MA(XW) + MB(1-XW)

12. XF = [(vol. * density)/ (mol. Wt.)] methanol

[(vol. * density)/ (mol. Wt.)] methanol + [(vol. * density)/ (mol. Wt.)] water

13. L.H.S of Rayliegh’s equation ln (F/W) =

ln (F/W) = ____________

14. R.H.S of Rayliegh’s equation XW

∫ dX = _________

y*-X

XF

From graph of X(x-axis) Vs 1/( y*-X) (on y-axis)

Graph of X(x-axis) Vs 1/( y*-X) (on y-axis)

1

y*-X Area

Xw XF

x



Graph of density Vs mole fraction

Density

g/cc

Xw XD

Mole fraction

Results:

From the graphs and calculations the Rayliegh’s equation is verified.

Applications: 1. Simple distillation is in use for separating a dissolution formed by two liquids depending on

their different boiling-points.

2. Used in separating relatively pure liquids with large boiling differences (or) liquids with

impurities.

Remarks:




Probable viva questions: 1. What is meant by distillation? 2. What are different types of distillation? 3. Mention the applications of distillation. 4. Explain the principle of distillation. 5. Explain the laws involved in the distillation process.




STEAM DISTILLATION

Aim:

To determine the vaporization and thermal efficiencies in steam distillation of the given sample.ie.

Nitrobenzene.


Condensers, beakers, conical flask, thermometer, measuring jar, separating funnel, specific gravity

bottle.

Theory:

Distillation is a method of separating mixtures based on differences in their volatilities in a boiling

liquid mixture. Distillation is a unit operation, or a physical separation process, and not a chemical

reaction. Commercially, distillation has a number of uses. It is used to separate crude oil into more

fractions for specific uses such as transport, power generation, and heating. Water is distilled to

remove impurities, such as salt from seawater. Air is distilled to separate its components—notably

oxygen, nitrogen, and argon—for industrial use. Distillation of fermented solutions has been used

since ancient times to produce distilled beverages with higher alcohol content. The premises where

distillation is carried out, especially distillation of alcohol are known as a distillery.

Idealized distillation model

The boiling point of a liquid is the temperature at which the vapor pressure of the liquid equals the

pressure in the liquid, enabling bubbles to form without being crushed. A special case is the normal

boiling point, where the vapor pressure of the liquid equals the ambient atmospheric pressure.It is a

common misconception that in a liquid mixture at a given pressure, each component boils at the

boiling point corresponding to the given pressure and the vapors of each component will collect

separately and purely. This, however, does not occur even in an idealized system. Idealized models

of distillation are essentially governed by Raoult's law and Dalton's law, and assume that vapor-

liquid equilibria are attained.

Raoult's law assumes that a component contributes to the total vapor pressure of the mixture in

proportion to its percentage of the mixture and its vapor pressure when pure, or succinctly: partial

pressure equals mole fraction multiplied by vapor pressure when pure. If one component changes

another component's vapor pressure, or if the volatility of a component is dependent on its

percentage in the mixture, the law will fail.

Dalton's law states that the total vapor pressure is the sum of the vapor pressures of each individual

component in the mixture. When a multi-component liquid is heated, the vapor pressure of each



http://en.wikipedia.org/wiki/Atmospheric_pressure

http://en.wikipedia.org/wiki/Raoult%27s_law

http://en.wikipedia.org/wiki/Vapor_pressure



component will rise, thus causing the total vapor pressure to rise. When the total vapor pressure

reaches the pressure surrounding the liquid, boiling occurs and liquid turns to gas throughout the

bulk of the liquid. Note that a mixture with a given composition has one boiling point at a given

pressure, when the components are mutually soluble.

An implication of one boiling point is that lighter components never cleanly "boil first". At boiling

point, all volatile components boil, but for a component, its percentage in the vapor is the same as

its percentage of the total vapor pressure. Lighter components have a higher partial pressure and

thus are concentrated in the vapor, but heavier volatile components also have a (smaller) partial

pressure and necessarily evaporate also, albeit being less concentrated in the vapor. Indeed, batch

distillation and fractionation succeed by varying the composition of the mixture. In batch

distillation, the batch evaporates, which changes its composition; in fractionation, liquid higher in

the fractionation column contains more lights and boils at lower temperatures.

The idealized model is accurate in the case of chemically similar liquids, such as benzene and

toluene. In other cases, severe deviations from Raoult's law and Dalton's law are observed, most

famously in the mixture of ethanol and water. These compounds, when heated together, form an

azeotrope, which is a composition with a boiling point higher or lower than the boiling point of

each separate liquid. Virtually all liquids, when mixed and heated, will display azeotropic behavior.

Although there are computational methods that can be used to estimate the behavior of a mixture of

arbitrary components, the only way to obtain accurate vapor-liquid equilibrium data is by

measurement.

It is not possible to completely purify a mixture of components by distillation, as this would require

each component in the mixture to have a zero partial pressure. If ultra-pure products are the goal,

then further chemical separation must be applied. When a binary mixture is evaporated and the

other component, e.g. a salt, has zero partial pressure for practical purposes, the process is simpler

and is called evaporation in engineering.

PrinciplesofDistillation

Thermodynamics play an important role in the distillation of essential oils. In order to vaporize any

liquid, energy in the form of heat must be applied. This energy is termed latent heat. When a vapor

is converted back to liquid, there will be a reduction and release of that latent heat.

In order for a liquid to be vaporized, latent heat must be applied to this liquid until its temperature

attains the point where its vapor pressure becomes equal to the surrounding atmospheric pressure.

When it reaches this point the temperature will rise no further. However, if heat continues to be

applied, the liquid will take up this latent heat and then vaporize at the appropriate rate. At this

point the liquid has said to have reached its “boiling point” under the prevailing pressure.

When mutually insoluble compounds such as water and essential oil are present, the total pressure

exerted by the mixed vapor then becomes the sum of the partial pressures exerted by each

constituent present. This liquid mixture will boil when its temperature is raised to the point where

the combined vapor pressure of its components becomes equal to the surrounding pressure.

http://en.wikipedia.org/wiki/Benzene

http://en.wikipedia.org/wiki/Toluene

http://en.wikipedia.org/wiki/Azeotrope



Steam distillation is a purifying technique used to distill a mixture of immiscible substances. The

solution boils at lower temperature than either of the separate components (below 100 oC), a factor

that causes substances to co distill and prevent decomposition of unstable or high boiling point

substances; furthermore, because the substances are immiscible, the distillate will be divided into

layers; the desired product can then be isolated using a separatory funnel. Steam distillation is often

used in isolating liquids from natural sources such as Eugenol extraction from cloves.

STEAM DISTILLATION:

Steam distillation is a method for distilling compounds which are heat-sensitive. This process

involves using bubbling steam through a heated mixture of the raw material. By Raoult's law, some

of the target compound will vaporize (in accordance with its partial pressure). The vapor mixture is

cooled and condensed, usually yielding a layer of oil and a layer of water. Certain compounds

namely organic compounds, which have high boiling points can be distilled by using steam

distillation. Steam distillation is a process of purifying a substance through the application of steam.

Steam distillation is a special type of distillation for temperature sensitive materials like natural

aromatic compounds.

Steam distillation of various aromatic herbs and flowers can result in two products; an essential oil

as well as a watery herbal distillate. The essential oils are often used in perfumery and

aromatherapy while the watery distillates have many applications in aromatherapy, food processing

and skin care.

Open steam distillation is restricted to systems where water is insoluble for the component

involved. In inert distillation, an inert gas is used instead of steam. The conditions to be fulfilled for

steam distillation are:

i. Component must be insoluble with water.

ii. Component tube distilled are thermally unstable, are of the components present in the feed at high

temperature reacts with another component present and hence normal distillation cannot be

carried out.

PRINCIPLE:

Steam distillation works on the principle that immiscible objects when mixed together can

lower the boiling point of each other.

When a mixture of two practically immiscible liquid is heated while being agitated to expose

the surfaces of both the liquid to vapor phase, each constituent independently exerts its own

vapor pressure as a function of temperature. Consequently, the vapor pressure of the whole

system increases. Boiling begins when the sum of the partial pressure of two immiscible liquids

just exceeds the atmospheric pressure.



ADVANTAGES:

1. Low cost.

2. Simple.

DISADVANTAGES:

1. Time consuming.

2. High temperatures used may alter the compounds.

PROCEDURE:

1. About 200 ml of nitrobenzene and 100 ml water is measured using a measuring jar and it is

transferred into the round bottom flask and the round bottom flask is then connected to a water

cooled condenser.

2. Steam is passed at constant temperature from one end into the round bottom flask, as a result

nitrobenzene inside the round bottom flask starts boiling.

3. At particular temperature vapors of nitrobenzene are formed which passes through the water

cooled condenser get condensed and gets collected as distillate in a beaker on the other hand.

4. The distillation process is continued till about 60 ml of the initial liquid taken distillated out.

5. This collected sample is poured into a separating funnel. As the density of nitrobenzene is less

than water, nitrobenzene gets collected at the bottom while water on the top of the separating

funnel.

6. Nitrobenzene solution is drawn from the bottom of separating funnel until the total volume of

nitrobenzene collected is 60 ml.

7. The steam supply is stopped and the volume of water in distillate and volume of water and

nitrobenzene in residue is measured.

8. The specific gravity of nitrobenzene and water in distillate and in residue is noted down using a

specific gravity bottle.

9. Vaporization and thermal efficiencies are calculated.

GRAPH:

1. A graph of ln PS on Y-axis and ln PW on X-axis is plotted. From this graph the slope d ln (PS)

is found out.

ln (PW)

2. A graph of temperature on X – axis and vapor pressure of nitrobenzene, PS on Y- axis is

plotted.

3. A graph of temperature on X – axis and vapor pressure of water, PW on Y – axis is plotted.




1. Room temperature = ________ ºC.

2. Distillation temperature = _________ ºC.

3. Volume of water in residue = _________ ml.

4. Volume of nitrobenzene in residue = _________ ml.

5. Volume of water in distillate = _________ ml.

6. Volume of nitrobenzene in distillate = ________ ml.

7. Weight of empty specific gravity bottle = ________ g.

8. Weight of empty specific gravity bottle + water =________ g.

9. Weight of empty specific gravity bottle + water in residue = _______ g.

10. Weight of empty specific gravity bottle + water in distillate = ________ g.

11. Weight of empty specific gravity bottle + nitrobenzene in residue = _______ g.

12. Weight of empty specific gravity bottle + nitrobenzene in distillate =_________ g.

13. Density of water in distillate =[(weight of specific gravity bottle + water in

distillate)-(weight of empty specific gravity bottle)]

[(weight of specific gravity bottle + water)-(weight

Of empty specific gravity bottle)]

= ______________ g/cc.

14. Density of water in residue=[(weight of specific gravity bottle + water in

residue)-(weight of empty specific gravity bottle)]



= ______________ g/cc.

15. Density of nitrobenzene in residue=[(weight of specific gravity bottle +

nitrobenzene in residue)-(weight of empty specific gravity bottle)]



= _______________ g/cc.



16. Density of nitrobenzene in distillate =[(weight of specific gravity bottle +

nitrobenzene in distillate)-(weight of empty specific gravity bottle)]



= _______________ g/cc.

17. vapor pressure of nitrobenzene at distillation temperature,

PS (from graph) = ________ mm Hg. 18. vapor pressure of water at distillation temperature,

PW (from graph) =________mm Hg.

19. Molecular weight of nitrobenzene, MS = 123.11g.

20. Molecular weight of water, MW = 18.02g.

21. Latent heat of water, λW = 540 cal/gm ºC.

22. Specific heat of nitrobenzene, CPS = 0.41 cal/gm ºC.

23. Specific heat of water, CPW = 1.1 cal/gm ºC.

CALCULATIONS:

VAPORIZATION EFFICIENCY:

1. ( WS ) = (volume * density) Nitrobenzene in distillate

WW Actual (volume * density) water in distillate

2. ( WS ) = PSMS

WW Ideal PWMW

3. vaporization efficiency, EV = ( WS )

WW Actual * 100

( WS )

WW Ideal

THERMAL EFFICIENCY:

1. St = thermal requirement of steam

Unit weight of sample distilled

2. St = (1+R) CPS (Td-Tr) + λS + PSMS

λW + (Tb-Td) CPW PWMW



3. R = weight of sample in residue

Weight of sample in distillate

= [(volume of Nitrobenzene in residue)*(density of

Nitrobenzene in residue)]

[(volume of Nitrobenzene in distillate)*(density of

Nitrobenzene in distillate)]

4. Boiling point of water, Tb = 150ºC.

5. λS = λW * d ln (PS) * + MS

ln (PW) MW

Where slope = d ln (PS)

ln (PW)

6. Sa = Actual requirement of steam

Unit weight of sample distilled

= (volume * density) water in residue + (volume*density) water In distillate

(Volume * density) nitrobenzene in distillate

7. Thermal efficiency, Et = St * 100

Sa

VAPOR PRESSURE DATA OF NITROBENZENE AND WATER FROM PERRY 7TH

EDITION (PAGE 2-71)

sl

no:

Temperature

TºC

vapor pressure

of nitrobenzene

PS (mm Hg)

Vapor pressure

of water

PW(mm Hg)

ln PS ln PW

1 44.4 1 69.69

2 71.6 5 243.9

3 84.9 10 416.0

4 99.3 20 714.18

5 115.4 40 801.66



NATURE OF GRAPH:

ln Ps

ln Pw

Vapor pressure

Of nitro benzene

mm Hg

Temperature



Vapor pressure of water

mm Hg

Temperature ºC

Results:

1. Vaporization efficiency = ________

2. Thermal efficiency = ________

Applications:

Steam distillation is employed in the following areas –

1. Manufacture of essential oils. Ex: perfumes.

2. Synthetic procedures of complex organic compounds.Ex: eucalyptus oil, orange oil.

3. Petroleum refineries and petrochemical plants.

4. Production of consumer food products.

Remarks:




Probable viva questions: 1. What is meant by simple distillation? 2. What are different types of distillation? 3. Mention the applications of distillation. 4. Explain the principle of distillation. 5. Explain the laws involved in the distillation process.




LIQUID – LIQUID EXTRACTION

Aim: To determine the percentage recovery of solute from a solution using a solvent.

Apparatus/Components required: Flask shaker, volumetric flask 250 ml, separating funnel, volumetric flask 500 ml, measuring

jar 100 ml, burette, pipette 10 ml and beakers.

Theory:

Extraction is a terminology used in the mass transfer operations when a desired solute is

selectively removed from rest of inerts or non desired materials by use of a liquid solvent. If the

solute desired is present in another liquid mixture then the operation is termed as Liquid-liquid

extraction and if the solute is present in a solid phase the operation is termed as leaching.

Liquid-liquid extraction is a mass transfer operation in which a solute dispersed or dissolved in a

liquid media is separated in by use of a solvent which has differential solubiliies with media and

solute and has a phase characteristics. The substance i.e. solute when dissolved in a medium can be

taken into another liquid which is insoluble with the medium but has a higher solubility with solute.

Due to the insolubility there will be two separate phases, one in which the solvent with a higher

percentage of solute called as Extract and the other being the original medium in which the solute

of lower percentage called as Raffinate.

SOLVENT C

B+C Rich in B



EXRACT ΔA %

Let F = kgs of feed.

E = kgs of Extract.

R = kgs of Raffinate.

B = solvent.

C = solute.

A = the media in which C is dissolved.

Then if Y is the mole fraction in the Extract

Y = kg C __ (1)

Kg A+kg B+kg C

If X is the weight of C in the raffinate then,

X = kg C __ (2)

Kg A+kg B+kg C

We can define X' and Y' which will be the ratios of mass of C to mass of A + B in raffinate and

extract.

i.e. X' = X , Y' = Y

1 – X 1-Y

We can also define X = weight fraction of C on a B free basis.

i.e. X = Mass C in raffinate Y = Mass C in Extract

Mass A + B Mass A + C

And

N = Mass C

Mass A + C

Then NF, NR, NE will be the values o B/A+C in feed. Raffinate and Extract.

Liquid-liquid extraction is also known as solvent extraction and partitioning, is a method to

separate compounds based on their relative solubilities in two different immiscible liquids,

usually water and an organic solvent. It is an extraction of a substance from one liquid phase into

another liquid phase. Liquid-liquid extraction is a basic technique in chemical laboratories,

where it is performed using a separatory funnel. This type of process is commonly performed

after a chemical reaction as a part of the work-up.

In other words, this is the separation of a substance from a mixture by preferentially dissolving

that substance in a suitable solvent. By this process a soluble compound is usually separated

from an insoluble compound. Solvent extraction in an industrial application, the process is done

continuously by pumping an organic and aqueous stream into a mixer. This mixes the organic

component with the aqueous component and allows ion transfer between them. The mixing

continues until equilibrium is reached. Once the ion transfer is complete (equilibrium is

reached), the mixture flows into a vessel, where the organic and aqueous are allowed to separate,

similar to the way oil and water would separate after mixing them. Fresh material is



continuously fed into the mixer, and a two continuous streams is removed from the settler (one

organic and one aqueous). The process is commonly used to process copper and uranium.

Liquid-liquid extraction is possible in aqueous systems: in a system consisting of a molten metal

in contact with molten salt, metals can be extracted from one phase to the other. This is related to

a mercury electrode where a metal can be reduced, the metal will often then dissolve in the

mercury to form an amalgam which modifies its electrochemistry greatly. For example it is

possible for sodium cations to be reduced at a mercury cathode to form sodium amalgam, while

at an inert electrode (such as platinum) the sodium cations are not reduced. Instead water is

reduced to hydrogen. If a detergent or fine solid can stabilize an emulsion which in the solvent

extraction is known as a third phase.

The water-benzene and IPA constitutes a system of immiscible liquids i.e. water and benzene are

insoluble in each other for all concentrations. For such a system if solvent (B) is added to a

binary homogeneous liquid mixture (A-C) containing a diluent (A) and a solute (C), then C

separates (distributes itself) between A and B with no mixing of A and B. such mixing and then

separating two heterogeneous liquid constitutes a stage and if the exit streams are in equilibrium,

then a combination of mixing and separating operations is known as an equilibrium stage. The

two liquid phases are accordingly called as ‘Extract’ (solvent rich phase) and ‘Raffinate’ (diluent

rich phase). If the operations are carried out in such a manner that if the fresh solvent is added to

‘raffinate phase’ then the operation is termed as cross-current operations.

ADVANTAGES:

1. Low cost.

2. Low energy consumption.

3. Variety of solvents is available.

4. Easy to scale-up.

DISADVANTAGES:

1. Emulsification may occur.

2. Effluent must be treated.

Procedure:

1. About 2 ml of Acetic acid is taken in 250 ml volumetric flask and is made to 250 ml by adding

benzene, which forms a homogeneous solution acts as a feed solution to the experiment.

2. Two separating funnels are taken to which 50 ml of feed solution is added to each followed by 20

ml and 40 ml of water added to the two separating funnels respectively.

3. The separating funnels are then shaken for about 30 minutes and then allowed to settle.

4. On settling the mixture is going to form two separate phases. The bottom aqueous layer (Extract)

is removed from the bottom and the volume is measured.



5. About 10 ml of the aqueous layer (Extract) is pipetted out into a conical flask and titrated against

standardized NaOH solution using phenolphthalein as the indicator till the end point of pale pink

is obtained, the volume of NaOH run down is recorded.

6. The volume of NaOH run down is used to determine the concentration of the acetic acid in the

aqueous layer.

7. From the concentration of acetic acid in different extract layers the percentage recovery is

calculated.

PREPARATION OF SOLUTIONS:

1. 0.1N oxalic acid solution:

About 0.63gm of oxalic acid crystals is weighed and transferred into 100 ml volumetric flask and

it is made up to the mark by adding water.

2. 0.5N NaOH solution:

About 5 gm of NaOH pellets is weighed and transferred into 250 ml volumetric flask and it is

made up to the mark by adding water.

3. Feed solution:

4 ml of Acetic acid is measured using a measuring jar and it is added to a 500 ml volumetric

flask and it is made up to the mark by adding benzene.

4. Standardization of NaOH solution:

About 10 ml of oxalic acid is pipetted out into a conical flask and it is titrated against NaOH

solution using phenolphthalein indicator till the end point of pale pink is reached.


1. Room temperature = __________ºC.

2. Molecular weight of Acetic acid = 60 g

3. Density of Benzene = 0.879 gm/cc.

TABULAR COLUMN:

Sl.no:n Volume of feed

(ml)

Volume of water

(ml)

Volume of aqueous

layer (ml)

Volume of NaOH run down

(ml)

1. 50 20

2. 50 40



STANDARDISATION OF NaOH:

TABULAR COLUMN:

Burette reading Ι ΙI

Initial burette reading

Final burette reading


(N1V1)NaOH = (N2V2) Oxalic acid

(N1)NaOH = (N2V2) Oxalic acid

(V1) NaOH

Where N2 = Normality of oxalic acid = 0.1N.

V2= Volume of oxalic acid = 10 ml.

V1= Volume of NaOH run down.

Burette: NaOH solution.

Conical flask: 10 ml oxalic acid + phenolphthalein indicator (2-3) drops.

Indicator: phenolphthalein indicator.

End point: colorless to pale pink.

CALCULATION: FOR (50 ml feed solution + 20 ml water)

Normality of Acetic acid (CH3COOH) in extract phase:

TABULAR COLUMN:

Burette reading Ι ΙI

Initial burette reading

Final burette reading


(N1V1)Extract = (N2V2) NaOH



(N1) Extract = (N2V2) NaOH

(V1) Extract

Where N2 = Normality of NaOH.

V1= Volume of aqueous layer pipetted out = 10 ml.

V2= Volume of NaOH run down.

Burette: NaOH solution.

Conical flask: 10 ml of Extract + phenolphthalein indicator (2-3) drops.

Indicator: phenolphthalein indicator.

End point: colorless to pale pink.

Amount of Acetic acid present in the Extract phase =

N1 * Molecular weight of Acetic acid * volume of Extract collected

1000

Concentration in Extract phase =

Y' = _C__

B+C

Where C = Amount of Acetic acid in Extract phase.

B + C = Amount of Extract (aqueous layer).

Feed solution = 50 ml.

Weight of feed solution = Volume * Density

= 50 * 0.879

= 43.95 gm.

A (1+XF) = 43.95 gm.

A = 43.95

1 + XF

XF = Amount of Acetic acid taken * Density of Acetic acid

Volume of benzene taken * Density of benzene

= 4 * 1.049

496 * 0.879

A = 43.95

1+

Weight of Acetic acid in feed = 43.95-A.

%Extraction = Amount of Acetic acid present in Extract phase * 100

Amount of Acetic acid in feed



Result: Percentage recovery =___________

Remarks:



1. What is solvent, solute and solution in the given experiment?

2. What is raffinate and extract?

3. Write a formula to find the percentage of extraction.

4. Write a Material balance for liquid – liquid extraction.

5. Write the application of liquid – liquid extraction in Biotechnology.




CENTRIFUGATION

Aim:

To determine the percentage recovery of various sample by differential RPM using centrifugation.


Yeast, centrifuge tubes, Distilled water, Test tubes, volumetric flask, Beakers.

Theory:

The basic principle involved in centrifugation separation of solids is the density difference between the solids and the surrounding fluid. Normally a suspension of solids in a liquid on standing settles down slowly under the influence of gravity. The process is known as sedimentation. In centrifugation, the process of settling is aided by centrifugal forces where a solid particle moves through a viscous medium its velocity is affected by two opposing forces, the gravitational resulting from the density difference between the particles and the surrounding fluid. The gravitational force, Fg, accelerate on spherical particles is quantitatively given by –

Fg = π/6 [d3 (ρs - ρ)]g - (1) Where d is the sphere’s diameter, ρs and ρ are the densities of sphere and the fluid respectively and g is gravity. Equation (1) is obtained by the buoyant force and the gravitational force acting on the particle.

Fg = π/6 [d3 ρ]g

Fg = π/6 [d3 ρg ]g Fd = 3πdηv Where η is the viscosity of the medium and v is the velocity of the spherical particles. The equation holds good only when the sphere is small so that Reynolds number, characteristically the flow around the sphere is less than 1. This condition is almost always satisfied for biological solutes as the particle size is very small.



Vg = d2 (ρs - ρ)g 18η Sedimentation is centrifugal force is similar but the force acting on the spherical particle is centrifugal force and hence the constant g is replaced by ω2r. The terminal velocity Vc is centrifugal separation of solids is expressed in terms of angular velocity, ω is radius/sec and the radial distance, r from the centre of centrifuge into the spherical particle in cm.

Vc = d2 (ρs - ρ) ω2r 18η During the circular motion of a centrifuge, the centrifugal force Fc acting as a particle is related to the angular velocity ω and the radial distance, r (m) of the particle from the center of rotation as –

Fc = ma = m ω2r Where ω2r is acceleration due to centrifugal force. The tangential velocity of the particle V is given by the relationship –

V = ωr

Fc = m (V/r)2r = mV2/r The rotational speed of a centrifuge is generally expressed in terms of number of revolutions per min. the centrifugal force may be expressed in terms of number revolutions of centrifuge by the relationship – ω = 2πn/60

Fc = m (2πn/60)2r = mV2/r

The centrifugal force may also be expressed in terms of gravitational force Fg = mg where g is the acceleration due to gravity as

Fc/Fg = ω2r/g = V2/rg

Thus the force developed in ω2r/g times are large as the gravity force and is often expressed as equivalent to many g forces.



Labscale centrifuge: Laboratory centrifuges are used for small scale separations and clarification (removal of particles from liquids). Typically liquid volumes handled by such devices are in range of 1 – 5000 ml. The magnitude of induced gravitational force is measured in terms of G values.

G = ω2r/g = 2πrn2 g where r = distance from the axis of rotation (m) ω = Angular velocity (rad/sec)

g = Acceleration due to gravity n = rotational speed

Differential centrifugation is a common procedure in microbiology and cytology used to separate certain organelles from whole cells for further analysis of specific parts of cells. In the process, a tissue sample is first lysed to break the cell membranes and mix up the cell contents. The lysate is then subjected to repeated centrifugations, each time removing the pellet and increasing the centrifugal force. Finally, purification may be done through equilibrium sedimentation, and the desired layer is extracted for further analysis. Separation is based on size and density, with larger and denser particles pelleting at lower centrifugal forces. As an example, unbroken whole cells will pellet at low speeds and short intervals such as 1,000g for 5 minutes. Smaller cell fragments and organelles remain in the supernatant and require more force and greater times to pellet High g-force makes sedimentation of small particles much faster than Brownian diffusion, even for very small particles. When a centrifuge is used, Stokes' law must be modified to account for the variation in g-force with distance from the center of rotation.

{\displaystyle D={\sqrt {\frac {18\eta \,\ln(R_{f}/R_{i})}{(\rho _{p}-\rho _{f})\omega ^{2}t}}}}where

D is the particle diameter (cm) η is the fluid viscosity (poise) Rf is the final radius of rotation (cm) Ri is the initial radius of rotation (cm) ρp is particle density (g/ml) ρf is the fluid density (g/ml) ω is the rotational velocity (radians/s) t is the time required to sediment from Ri to Rf (s)



Procedure: 1. 1.5 g of baker’s yeast is taken and dissolved in distilled water. Emulsion was obtained 1 g of CaCO3 is obtained 1g of CaCO3 is taken and dissolved in distilled water. 2. Each sample is centrifuged at different RPM – 400, 800, 1200, 1600, 1800 and 2000 for 5 minutes at room temperature. 3. The supernatant is discarded and the pellet is dried for one day until powder of baker yeast is obtained. 4. The final recovered dry baker’s yeast powder is weighed and the percentage recovery is calculated. 5. Percentage recovery Vs. RPM is plotted and unknown percentage recovery is found.

Tabular column: Sl. No. RPM YEAST CaCO3 Percentage recovery Final Initial Final Initial

Results:

The Percentage recovery of yeast and calcium carbonate ___________ as RPM _____________.

Applications: In general, one can enrich for the following cell components, in the separating order in actual application:

Whole cells and nuclei; Mitochondria, chloroplasts, lysosomes, and peroxisomes; Microsomes (vesicles of disrupted endoplasmic reticulum); and Ribosomes and cytosol.



Remarks:



1. What are the methods for production of ethanol? Name any microorganism responsible for

it.

2. What are the methods of estimation of ethanol? Explain any one.

3. Potassium dichromate is a powerful oxidizing agent. How?

4. What is the unit of specific gravity?

5. Briefly explain the principle of specific gravity method for the estimation of ethanol.




CELL DISRUPTION TECHNIQUE: YEAST CELL DISRUPTION USING

HOMOGENIZER

Aim:

To disrupt yeast cells by mechanical method using homogenize and estimation of total protein

content by Biuret method.

Apparatus/Components required: Centrifuge tubes, Commercial Baker's yeast, 0.1 M phosphate buffer, pH 6.9, Homogenizer,

Cooling centrifuge, 20%TCA, Acetone , Biuret reagent, Standard protein solution (5mg/ml), 0.1 N

NaOH, Test tubes, Measuring cylinder Theory:

Various mechanical and non-mechanical methods are available for '.the disruption of microbial

cells for product recovery of intracellular contents like enzymes, metabolic products and

recombinant DNA products. Homogenizers have been found to be particularly effective to provide

effective release of cellular constituents while maintaining a high degree of biological activity in the

product.

A homogenizer is a piece of laboratory equipment used for the homogenization of various types of material, such as tissue, plant, food, soil, and many others. Many different models have been developed using various physical technologies for disruption. The mortar and pestle, already used for thousands of years, is a standard tool even in modem laboratories. More modem solutions are based on blender type of instruments (also known in the kitchen), bead mills, ultra sonic treatment, high pressure, and many other physical forces. Whereas older technologies just focused on the disruption of the material, newer technologies also address quality or environmental aspects, such as cross-contamination, aerosols, risk of infection, or noise.

Homogenization is a very common sample preparation step prior to the analysis of nucleic acids,

proteins, cells, metabolism, pathogens, and many other targets. The degree of cell disruption and

the release of the products depends on microorganism used, location of the product and the type of

homogenizes.

Under constant conditions the rates of product release for bacteria and yeast can be described by

first order kinetics. The concentration of the released product (Cr) will be proportional to the



concentration of disrupted cells. The rate of release dCr /dt will be proportional to the concentration

of unreleased product, Cr max - Cr, so

dCr = Kh (Cr max - Cr) - (1)

dt

Where Kh is the first order release constant for homogenizer, which depends on operating pressure and t being the homogenization time.

Integration of (l) gives,

ln (Crmax ) = - Kh t - (2)

(Cr max-Cr)

Cr = (1- exp (- Kh t) - (3)

Crmax

Cr = Cr max (1 – exp (- Kh t) - (4)

The release of intracellular enzymes is more complex compared to the release of proteins because

the enzymes may be located at different regions within the cell. The rate of release of an enzyme to

the total protein release is independent of disruption pressure, temperature and cell concentration.

Enzymes located outside the cell membrane are released at a rate faster than the total protein, and

those soluble in the cytoplasm like glucose -6- phosphate dehydrogenate are released approximately

at the same rate as total protein. Membrane bound enzymes such as fumarase are released more

slowly than total protein and enzymes such as inverts and a- glycosidase, located in the periplasmic

space are released faster than cytoplasm proteins. Thus the rate of release of the enzyme can be an

indication of its location within the cell.

Procedure:

PREPARATION OF REAGENTS:

1. Biuret reagent::

Dissolve 3g. of copper sulphate and 9g of sodium potassium tartarate in 500ml of 0.2M sodium

hydroxide; add 5g of potassium iodide and make up to 1 litre with 02M sodium hydroxide.

2. 0.1N NaOH:

Dissolve 0.4 grams of NaOH in 100ml of distilled water.

3. Standard protein: solution (5mg/ml):

Dissolve 0.5 grams of BSA in 20ml of distilled water. Transfer this solution to 100ml volumetric

flask and! make up to the mark with distilled water.



4. 0.1 M phosphate buffer:

Solution 1: Prepare 200ml of 0.1M NaH2P04. Solution

2: Prepare 200ml of 0.1M Na2HP04.

Mix 68.5ml of solution 1 with 31.5ml of solution 2 and check the pH of the solution using pH

meter.

5. 20% TCA:

Dissolve 20grams of TCA in 50mi of distilled water. Transfer this solution to 100ml volumetric

flask and make up to the mark with distilled water.

Procedure:

1. Preparation of homogenate:

1. Weighed two grams of commercial baker's yeast into 5 .different centrifuge tubes labeled

1min, 2min, 3min, 4min and 5 min.

2. Added 5ml of phosphate buffer to each of the centrifuge tube prior to homogeni2ati0n.

3. The yeast solution in the centrifuge rube was homogenized for different times as labeled

using a homogenizer keeping the rpm constant. 4. The homogenized suspension was centrifuged at 12,000 rpm for 10 min at 4°C using a

cooling centrifuge.

5. Supernatant was measured and was made-up to 5 ml using phosphate buffer.

2. Extraction of protein from the homogenate.

1. Added 5ml of 20% TCA to 2mi of each homogenate and ^incubated for 30 minutes on ice to

precipitate out proteins.

2. Centrifuged at 12,000 rpm for 10 min at 4°C.

3. Discarded the supernatant and the protein pellet was washed .with 5ml of acetone.

4. Centrifuged at 12,000 rpm for 10 min at 4°C.

5. Discard the acetone and dissolved the pellet in lml.

6. The protein solutions obtained at different time interval were stored in refrigerator for

estimating the total protein.

3. Preparation of Standard BSA carve.

1. Pipette out standard BSA solution in the range of 0.0 to 1.0ml to dT ssnt test tubes.

2. Made up the contents to 4.0ml using distilled water.

3. Added 6.0ml of Biuret reagent to all the test tubes followed by keeping al est tubes at room

temperature for ten minutes.

4. O.D was read at 520 nm against the blank solution.

5. Standard graph was plotted using O.D v\s amount of protein in mg.



4. Estimation of total protein by Biuret method.

1. Pipette out 1ml of protein solutions obtained at different time interval to different test tubes

and for the blank test tube 1ml of 0. 1N NaOH.

2. Added 3ml of distilled water to all test tubes followed by the addition of 6ml of Biuret

reagent.

3. Incubated all the test tubes at room temperature for ten minutes.

4. O.D was read at 520nm against the blank.

5. Estimated the total protein present in 1ml each of extract (=2ml of each homogenate) in mg

using standard graph.

6. Calculated the total protein present in 5ml each of homogenate.

7. Graph was obtained by plotting amount of total protein in mg Vs. homogenization time.

Results: The yeast cells were disrupted using a homogenizer; proteins were precipitated out and estimated. It

was found that amount of total protein increased with increase in the homogenization time.

Applications: Sonication can be used for the production of nanoparticles, such as nanoemulsions, nanocrystals, liposomes and wax emulsions, as well as for wastewater purification, degassing, extraction of plant oil, extraction of anthocyanins and antioxidants, production of biofuels, crude oil desulphurization, cell disruption, polymer and epoxy processing, adhesive thinning, and many other processes. In biological applications, sonication may be sufficient to disrupt or deactivate a biological

material. For example, sonication is often used to disrupt cell membranes and release cellular

contents. This process is called sonoporation. Small unilamellar vesicles (SUVs) can be made by

sonication of a dispersion of large multilamellar vesicles (LMVs). Sonication is also used to

fragment molecules of DNA, in which the DNA subjected to brief periods of sonication is sheared

into smaller fragments.

Sonication is commonly used in nanotechnology for evenly dispersing nanoparticles in liquids.

Additionally, it is used to break up aggregates of micron-sized colloidal particles.

Remarks:


Probable viva questions: 1. What is the principle behind the homogenizer? 2. What are the methods of cell disruption? 3. Briefly explain the enzymatic cell disruption. 4. What is the importance of cell disruption? 5. Briefly explain the extraction process of protein.



Experiment No: _____12._____ Date: ____________

ESTIMATION OF ETHANOL FROM FERMENTED BROTH

Aim:

To estimate the amount of ethanol present in an unknown (given) sample calorimetrically.


Standard Alcohol solution (0.2 to 1.0 %), Potassium dichromate, Cone H2S04, Distilled water, Test tubes, volumetric flask, Beakers.

Theory:

Potassium dichromate is a powerful oxidizing agent. K2Cr2O7 is reduced by organic substance

forming green solution containing chromium salt with alcohol in pressure of cone. H2SO4 it gives

green colored chromic sulphate and acetaldehyde. The green color is calorimetrically read at

600nm.

K2Cr207+4 H2SO4—*K2 SO4+ Cr2 (S04)3 + 4 H20+3[0] Chromic salt C2H5OH+ [O] ------- ►CH3CHO + H20 Acetaldehyde Most of the chemical oxidation methods are based on the complete oxidation of ethanol by

dichromate in the presence of sulfuric acid with the formation of acetic acid. This reaction is

popular because potassium dichromate is easily available in high purity and the solution is

indefinitely stable in air. The theoretical reaction stoichiometry is shown below:

2Cr2O7-- + 3C2H5OH + 16H

+ -----> 4Cr

+++ + 3CH3COOH + 11H2O

Dichromate (Cr2O7--, Cr(VI)) is yellowish in color and the reduced chromic product (Cr

+++, Cr(III))

is intensely green. Because the absorption spectra of dichromate and chromic ions overlap

significantly, Beer's law is not obeyed. Instead, the spectra of the solution of interest must be

analyzed at multiple wavelengths to calculate the individual concentrations of dichromate and

chromic ions in a mixture subject to the material balance that the total number of chromium atoms

must be conserved.

Other methods of determination primarily based on the above reaction are commonly used. In these

methods, any one of the reactants or products participating in Reaction (1) can be analyzed through



another separate reaction. For example, the excess dichromate remaining in the solution can be

further reduced by titration with other oxidizing reagents such as ferrous ammonium sulfate:

Cr2O7-- + 6Fe

++ + 14H

+ -----> 2Cr

+++ + 6Fe

+++ + 7H2O

To enhance the visualization of titration endpoint, organic indicators such as sodium diphenylamine

sulfonate and 1,10-o-phenanthroline are added. Another similar method is based on iodometric

titration.

Proper concentration of sulfuric acid in the surrounding solution will direct the oxidation of ethanol

toward acetic acid instead of acetaldehyde or the excess dichromate can be analyzed.

Procedure: 1. Blank Preparation: 10ml of distilled water and 10ml of K2Cr2O7 served as the reagent blank

(taken in test tube).

2. 10ml of std. alcohol solution of cone. 0.2, 0.4 ............................ 1.0% were taken in the

test tubes, numbered as 2 to 6. To each of the test tube 10ml of potassium dichromate reagent

was added.

3. To the given sample (10 ml), 10ml potassium dichromate reagent was added.

The test tubes were air sealed with foil. All the tubes were incubated at 60°C

for 20min.The absorbance of the content in each of the tube was read

calorimetrically at 600nm against reagent blank.

Reagents Preparation:

1. Potassium dichromate solution:

Solution A. 34 gm of potassium dichromate was dissolved in 400 ml of distilled water.

Solution B. 325 ml of cone sulphuric acid was added slowly with constant stirring to 200 ml

of distilled water and cooled. Solution A and B were mixed and the volume was made up to

1000 ml with distilled water.

2. Std. alcohol solutions: Std. alcohol solutions of concentration 0.2 to 1.0 % were prepared using

absolute alcohol.

Results:

The concentration of ethanol in the given unknown sample was found to be _____________.

Applications:



Estimation of alcohol in beverages and other fermentation samples Estimation of food spoilage based on alcohol content produced Clinical analysis of ethanol in biological fluids.

Remarks:



1. What are the methods for production of ethanol? Name any microorganism responsible for

it.

2. What are the methods of estimation of ethanol? Explain any one.

3. Potassium dichromate is a powerful oxidizing agent. How?

4. What is the unit of specific gravity?

5. Briefly explain the principle of specific gravity method for the estimation of ethanol.



Reference: 1. Unit operations in Chemical Engineering by McCabe W.L. and Smith J.C, McGraw-Hill, (ISBN 0-

07-112738-0), Year of publication: 1993.

2. Transport Process Principles and Unit Operations by Christie Geankoplis, Prentice Hall of India,

(ISBN-13: 978-0131013674), Year of publication: 1993 .

3. Introduction to chemical Engineering by Badger and Banchero, T M H Publication, ISBN – (ISBN-

10: 0070850275), Year of publication: 2002.

4. Fluid Mechanics by K L Kumar, S Chand & Company Ltd. Mechanics of fluids by B.S. Massey,

Chapman & Hall Publishers, (ISBN-10: 9380027656), Year of publication: 1995

5. Biochemical Engineering Fundamentals by Bailey J.E. and Oillis K, McGraw Hill, (ISBN-10:

0070032122), Year of publication: 1986.

6. Principles of Unit Operations by Alan S Foust, L.A. Wenzel, C.W. Clump, L. Maus, and L.B.

Anderson , John Wiley & Sons, (ISBN 10: 0471268976), Year of publication: 1960.

7. Chemical Engineering by Coulson and Richardson. Vols I & II. Elsevier Science, (ISBN - 13: 978-

0750625586), Year of publication: 1977 .

8. Chemical Engineers Hand Book by Perry, McGraw Hill Publications, (ISBN-10: 0070498415), Year

of publication: 1934.

9. Process Heat Transfer by Kern, McGraw Hill, (ISBN-10: 0074632175), Year of publication: 1950.

10. Heat Transfer by J P Holman, McGraw Hill International Ed.,( ISBN 10: 0071069674), Year of

publication: 1977

11. Mass Transfer Operations by Robert E. Treybal. McGraw-Hill Education, (ISBN-13: 978-

0070651760), Year of publication: 1955.

12. Protein Purification by Scopes R.K., IRL Press.

13. Rate controlled separations by Wankat P.C., Elsevier.

14. Bioseparations by Belter P.A. and Cussier E., Wiley.

15. Product Recovery in Bioprocess Technology - BIOTOL Series, VCH.

16. Separation processes in Biotechnology by Asenjo J. and Dekker M.

17. BIOSEPARATION S: SCIENCE & ENGINEERING by Roger G Harrison, Paul Todd, Scott R

Rudge, Demetri P Petrides, Oxford University Press.



Dayananda Sagar College of Engineering Department of Biotechnology

Bengaluru-560078

BIOPROCESS CONTROL & AUTOMATION LABORATORY (10BTL78)

PROBABLE/SUGGESTED QUESTION BANK

1. What are the modes of heat transfer?

2. When you say heat is conducted?

3. What is the difference between conduction and convection?

4. Give practical example of natural convection?

5. What do you mean by forced convection?

6. What is the importance of Heat transfer in chemical engineering?

7. What is Fourier law of heat conduction?

8. What is the significance of thermal conductivity?

9. Define insulation?

10. What is the significance of individual heat transfer coefficient?

11. What is overall heat transfer coefficient?

12. When extended surface type of heat exchangers is used?

13. Why LMTD is required?

14. Justify the countercurrent flow over co-current flow?

15. What are condensers?

16. What are the advantages and disadvantages of DPHE?

17. Tell some industrial applications of heat exchangers?

18. Define radiation?

19. Define the heat law governing the radiation mode of heat transfer?

20. Why tubular heat exchangers (STHE) are preferred over DPHE?

21. What is the advantage of plate type of heat exchangers?

22. Explain the heat transfer operation when water is heated in a container up to its

boiling point?

23. Define resistance to heat transfer by conduction and by?

24. Define emissivity?

25. What is black body radiation?

26. What do you mean by monochromatic radiation?

27. What is absolute pressure?

28. Explain the mode of heat transfer in a boiler?

29. What is evaporation? Is it heat transfer operation or mass transfer operation?

30. Define critical thickness of insulation?

31. Explain laminar and turbulent flow?

32. How turbulence of shell side fluid is increased in a STHE?



33. What is triangular and square pitch and which is advantageous, why?

34. What are the standard dimensions of tubes, shell, baffles in a STHE as specified by

TEMA?

35. What do you mean by fixed sheet, floating head STHE?

36. What do you mean by 1-1, 1-2, 2-4 heat exchangers? Draw a schematic representation of each

one of them?

37. What is the advantage of having multiple effect evaporators?

38. What do you mean by boiling point elevation and freezing point depression?

39. What are forward feed, backward feed and mixed feed evaporators?

40. Why baffles are provided in STHE construction?

41. What is dirt factor and how it affects rate of heat transfer?

42. What is fouling in case of heat exchangers?

43. How to encounter fouling problems in case of heat exchangers?

44. How the allocation of fluids done in case of DPHE and STHE?

45. What is LMTD correction factor? Why it is employed?

46. Mention the heat transfer resistances for the case of boiling of liquid?

47. Explain the heat transfer operation from SUN to Earth?

48. What is baffle cut and what is its standard dimension?

49. What is the major mode of heat transfer in case of furnaces?

50. Why heat exchangers are required?

51. What are cooling towers?

52. Explain the process of sun drying of cloth?

53. Explain the process of condensation?

54. What do you mean by film type and drop wise condensation? Which one is preferred

industrially?

55. Give few examples for film type and drop wise condensation?

56. What do you say about the type of condensation in your condenser experiment?

57. What is equivalent diameter? How do you find the same mathematically?

58. When we go for equivalent diameter?

59. What is hydraulic radius?

60. Define a heat exchanger?



DAYANANDA SAGAR COLLEGE OF ENGINEERING DEPARTMENT OF BIOTECHNOLOGY, BENGALURU-560078

CONTINUAL EVALUATION FORMAT

HEAT AND MASS TRANSFER LABORATORY (17BT5DLHMT)

(Academic Year - 2017)

Semester /Section : Batch :

S

No

.

USN student Name Expt. No: 1 Expt. No:2 Expt. No:3 Expt. No:4

Date: Date: Date: Date: Viva

(05)

Record

(10)

Total

(15)

viva

(05)

Record

(10)

Total

(15)

Viva

(05)

Record

(10)

Total

(15)

Viva

(05)

Record

(10)

Total

(15)

Faculty Signature With Date

Name of the Faculty Incharge (1) (2) (3)

Note:

(1) Viva questions to be asked w.r.t the current experiment of the particular week.

(2) The above same page format is used for next set of experiments i.e. 5, 6,….expts.

(3) Separate sheets must be used for different batches.



DAYANANDA SAGAR COLLEGE OF ENGINEERING DEPARTMENT OF BIOTECHNOLOGY, BENGALURU-560078

FINAL IA MARKS FORMAT

HEAT AND MASS TRANSFER LABORATORY (17BT5DLHMT)

Year: 2017

Semester /Section : V Batch :

SN USN Name Of The Student

Continual Evaluation

Marks (15)

IA Test Marks (10)

Final Marks (25)

Signature of Student

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