FM 301

Part A: FLOW OF POWDERS FROM BINS AND HOPPERS Introduction: Several chemical processes involve solid raw materials or products in the form of powders (e.g. cement manufacture, pharmaceuticals, etc). The powders need to be transported during their processing, hence their flow properties are of considerable importance. An application, which is very common, is the flow of powders from storage bins & hoppers, under the influence of gravity. In the case of free flowing solids, the bin may be used to give a nearly constant flow rate of solids. The flow rate obtained in such a case however will depend on the geometrical parameters of the system, as well as the nature and size of powder particles. Objectives:

i) To study the effect of the powder level above the exit on the mass flow rate. ii) To study the effect of particle diameter and orifice diameter on the mass flow

rate. Procedure:

1) Fill the bin nearly to the top with the sand keeping the exit closed. 2) Release the sand. 3) Collect about 150 cm3 of sand in a beaker noting the time interval as well as the

sand levels at the start and the end of the interval( Wait for the flow to become steady before starting the collection of sand)

4) Accurately weigh the sand collected in the beakers. 5) Repeat (1) to (4) for 2 different sand levels. 6) Repeat (1) to (4) for the different orifice supplied. 7) Repeat the above procedure for the sand with the second particle size. 8) Note orifice diameter and the mean particle size for each of the experiments.

Theory and Analysis: Inspite of an early start in the study of the flow of solids (Hagen 1852), the subject has not received as much attention as fluid flow. Particulate flow behaviour is complex, and is a blend of liquid-like and solid like characteristics. Particulate systems take the shape of the container they occupy, exert pressure on container walls and low through orifices, which is typical of liquids. Unlike liquids, the shearing stress is proportional to the normal load rather than to the rate of deformation. Like solids, they can sustain a shear stress, though the magnitude of the shearing stress at a point is generally indeterminate.

A considerable amount of work has been carried out to study the the flow patterns of discharge from bins and hoppers (Nedderman et al 1982). The experiments have shown that the discharge rate depends mainly on the geometry of orifice, the nature of the powder and is nearly independent of the height of the powder above the exit, and the vessel diameter. If the effect of particle diameter is ignored, then based on the dimensional analysis, the mass flow rate is given by

25

021

DgCW ρ= Where 0D is the orifice diameter, g the acceleration due to gravity and ρ, the bulk density of solids. C is a constant. Beverloo et al. (1981) have shown that the mass flow rate is given by

25

021

)( kdDgCW −= ρ where d is particle diameter, k and C are constants.

Plot 52

W versus D0 and check whether straight lines are obtained for the two different particle sizes. Obtain the constants C and k from the slopes and intercept of the line, if you obtain a straight line from your data.

Points for discussion:

1) Compare the behaviour of powder flow to that of fluid in a similar geometry.

2) Discuss why the mass flow rate is nearly independent of height, and diameter of the vessel.

3) Compare the constants obtained for the two different cases. 4) Can you suggest a physical justification for beverloo’s correlation?

References: R. M. Nedderman, U. Tuzun, S. B. Savage and G. T. Houlsby (1982), Chem Engg. Sci., 37, 1597-1609.

Part B: PNEUMATIC CONVEYING Introduction:

Pipeline transport of particulate solids has a number of advantages over conventional transportation systems such as conveyor belts, the former systems are compact and require much less maintenance. They are commonly used for in-plant transports of solids (e.g. in cement plants) as well as for loading and unloading materials from trucks and ships (e.g. food grains)

The flow patterns of gas and solids in pipeline are complex and depend on the particular requirement of operation (ratio of gas to solids flow rates). From the point of view of design and operation of pneumatic conveying systems, it is important to understand the basic phenomena involved in such transport as well as to estimate the energy requirements. Objectives:

i) To study qualitatively the different flow regimes for different gas and solids flow rate for horizontal pneumatic conveying.

ii) To measure pressure drops for gas-solid flow for different gas and solids flow rates.

Procedure:

1) Adjust the gas flow rate to a fixed value using the rotameter. 2) Open the solids flow valve to allow for a small flow rate of the particles. After

steady state I reached, measure the mass flow rate of solids by collecting the solids in the beaker for a fixed time interval.

3) Make qualitative observations of the flow patterns. 4) Repeat (2) – (3) for increasing solids flow rates. 5) Repeat (1) – (4) for increasing gas flow rates. 6) Note particle diameter and density.

Theory and Analysis: (a) Dilute Phase Transport:

When the gas volume flow rates are much higher that that of the solids, the transport of solids is said to be in the dilute phase. In this case there is a nearly uniform distribution of solids across the pipe cross-section of the pipe and along its axis. In this case the pressure drop for the flow is given by

fdt PPP ∆+∆=∆ Where f∆Ρ is the pressure drop due to flow of gas alone and d∆Ρ is the extra

pressure drop due to the pressure of the particles. f∆Ρ can easily be evaluated from friction factor charts. An expression for calculating d∆Ρ is

Lmd

uuPCPd

p

pf

p

fds

7.422 )(

43 −∞

−=∆

ρ

where =ρ density, u = velocity, ∞= volume fraction of gas in the pipe and the subscripts f and p refer to fluid and particle respectively, L is the length of the pipe, and

fp GGm /=

is the ratio of solid to gas mass flow rates, Cds is the drag coefficient for a single particle moving through the fluid at a velocity (uf - up). Compare your data with the predictions of the above equation.

(b) Dense Phase Transport When the gas velocity becomes low (below the ‘Salfation Velocity’ the particles settle at the bottom of the pipe. The settled particles are transported by partly sliding along the pipe. The distribution of solids in such cases where mass flow rate (m) is high generally non-uniform and this results in large pressure fluctuations. An empirical correlation for the pressure drop in dense phase transport is

2/145.0 )/(5.2 DddsLuPt pp ρ=∆ where D is the pipe diameter and dsρ is the mass per unit volume of solids in the pipe. The letter can be obtained experimentally by closing the solids valve and weighing the particles in the tube. Points for Discussion:

1) Explain qualitatively why the particular flow patterns occur. 2) What are the physical forces that contribute to pressure drop in dilute phase

transport? 3) Would the flow phenomena in vertical pneumatic transport be very different? 4) What flow phenomena would you expect for pneumatic transport in the absence

of gravity? 5) From your experimental results, can you suggest the best gas and solid flow rates

for the existing system. Give reasons for your choice.

BATCH SETTLING OF SOLID SLURRIES

Objectives:

1. Vertical Cylinders: Obtain the batch settling data for the given calcium carbonate

slurry (i.e. the settling rate versus concentration of slurry), and demarcate the

different settling regimes. (‘free settling” and ‘hindered settling’)

2. Tilted Cylinders: Obtain the batch settling data in the free settling regime for

different angles. Observe the flow patterns during the settling.

Apparatus:

Vertical settling cylinder, inclined settling cylinder, glass rod, stop watch

Reagents:

Water, Calcium Carbonate

Theory:

Introduction:

Settling is the process by which particulates settle to the bottom of a liquid and form a

sediment. Particles that experience a force, either due to gravity or due to centrifugal

motion will tend to move in a uniform manner in the direction exerted by that force. For

gravity settling, this means that the particles will tend to fall to the bottom of the vessel,

forming a slurry at the vessel base.

Figure.1 Forces acting on a particle during settling

For settling particles that are considered individually, i.e. dilute particle solutions, there are

two main forces enacting upon any particle. The primary force is an applied force, such as

gravity, and a drag force that is due to the motion of the particle through the fluid. The

applied force is usually not affected by the particle's velocity, whereas the drag force is a

function of the particle velocity.

For a particle at rest no drag force will exhibited, which causes the particle to accelerate

due to the applied force. When the particle accelerates, the drag force acts in the direction

opposite to the particle's motion, retarding further acceleration, in the absence of other

forces drag directly opposes the applied force. As the particle increases in velocity

eventually the drag force and the applied force will approximately equate, causing no

further change in the particle's velocity. This velocity is known as the terminal velocity,

settling velocity or fall velocity of the particle. This is readily measurable by examining the

rate of fall of individual particles.

The terminal velocity of the particle is affected by many parameters, i.e. anything that will

alter the particle's drag. Hence the terminal velocity is most notably dependent upon grain

size, the shape (roundness and sphericity) and density of the grains, as well as to the

viscosity and density of the fluid.

The behaviour of settling particles in slurry can be conveniently studied in small batch

experiments. The data is then useful for designing large scale settling tanks which have a

number of applications (e.g. clarification of waste water). The main information required

for design is the settling rate of the particles as a function of the system parameters such as

particle size and shape, concentration, geometry of the system, etc.

The effect of concentration on the settling slurry continuously increases with time. The

effect of geometry of the system on the settling rate can be significant. When the cylinder

is tilted, Boycott (1920) found that the settling rate increases due to shorter sedimentation

path. This phenomenon is known as the Boycott effect, and is used to enhance the rate of

settling in some applications.

(i) Vertical Cylinders:

Depending on the concentration of the slurry, two regimes of settling are possible, free

settling and hindered settling. As the name implies, in free settling, each particle is

unaffected by the motion of the neighboring ones and its terminal velocity is given by

2

1/ 2

( )

18

18

( )

24

Re

1

p pt

tp

p

D

p

g DU

UD

g

C

C

ρ ρµ

µρ ρ

ερ

−=

= −

=

= −

*All above equations are valid only for Reynolds no. Re < 1.

Where

pρ and ρ are the densities of the particle and the suspending medium respectively,

pD is the diameter of the particle,

CD is the drag coefficient

And Reynolds no.= /p t pD U ρ µ

The model assumed for describing Free Settling, has some limitations in practical

application. Such as the interaction of particles in the fluid, or the interaction of the

particles with the container walls can modify the settling behavior. Settling that has these

forces in appreciable magnitude is known as hindered settling.

In the hindered settling regime due to particle-particle interactions and up draft of liquid,

the velocity of individual particles is considerably smaller. The settling velocity ( sU ) may

be estimated by an empirical equation of the form.

ns tU U ε=

where ε is the volume fraction of the fluid and n is a constant.

From your data, plot the settling velocity which is given by ( )sU dH dt= − versus

concentration and hence estimate the concentration at which the settling crosses over from

H(t)H0

b

the free settling to the hindered regime. Assuming that the concentration is nearly uniform

over the cylinder, the concentration at any time is given by

0 0C HC

H=

where C0 is the initial concentration,

H0 is the initial height of the suspension-clear liquid interface (constant height before

stirring),

H is the height at time t.

If acceleration of the particles during the start of their fall is neglected, the rate of change of

height is simply given by

t

dHU

dt= −

Find the equivalent diameter of the particles using the free settling data (Ref. P. 139, Unit

Operations of Chemical Engineering, McCabe, Smith and Harriot for C0. Using the

equivalent diameter found by this method, calculate the exponent ‘n’ in the empirical

equation for hindered settling. Compare with values in the literature.

ii) Tilted Cylinders:

The theory for an increased rate of the settling for inclined cylinders was proposed by

Ponder-Nakamura and Kuroda (PNK) based on the increased projected area available for

settling. The Volumetric rate of increase if the clear fluid (S) according to the PNK theory

is given by

S= W Ut (b secθ + H tanθ)

where :

S is Volumetric rate of settling

W is width of the cylinder.

b is the breadth of the cylinder (b=W for square column)

H is the height of the liquid column.

θ is the inclination of cylinder with the surface.

If δ (differential height) is small, the rate of change of height with time is given by

1 sint

dH HU

dt bθ = − +

Procedure:

Vertical Cylinder:

1. Note the initial concentration of slurry.

2. Record the initial height of the slurry bed below the clear liquid before mixing.

3. The system is then thoroughly mixed with the help of glass rod.

4. After mixing wait for a minute and then observe the height below the clear liquid.

5. Repeat the step (4) after every one minute interval till a constant height is reached.

6. Simultaneously carry out similar runs for second vertical cylinder.

Tilted Cylinder:

1. Fix the angle of the cylinder to a particular value. note the initial concentrating of

slurry.

2. Record the initial height of the slurry bed below the clear liquid before mixing.

3. The system is then thoroughly mixed with the help of glass rod.

4. After mixing wait for a minute and then observe the height below the clear liquid..

5. Repeat the step (4) after every one minute interval till a constant height is reached

6. Repeat for 100 and 150.

FM 306: SIZE REDUCTION AND SIEVING Introduction: Reduction of particle size is an important operation in many chemical and other industries. The important reasons for size reduction are:

Easy handling Increase in surface area per unit volume Separation of entrapped components

The operation is highly energy intensive; hence a variety of specialized equipment is available for specific applications. The equipment may utilize one or more of the following physical mechanisms for size reduction: (i) Compression, (ii) Impact, (iii) Attrition, (iv) Cutting. Estimation of energy for the operation is important and is usually done by empirical equations. Enormous quantities of energy are consumed in size reduction operations. Size reduction is the most inefficient unit operations in terms of energy, as 99% of the energy supplied goes to operating the equipment and producing undesirable heat and noise, while less than 1% goes in creating new interfacial area. Reduction to very fine sizes is much more costly in terms of energy as compared to relatively coarse products. Sieving refers to the separation of a mixture of particles of different sizes using sieves each with a uniform sized opening. Standard sieves of specified opening sizes are used. Sieves are stacked with the sieve with the largest opening on the top and the material is separated into fractions by shaking. The material between two sieves is smaller than the upper sieve opening but larger than the smaller sieve opening. Objectives:

1. To grind the given limestone material to a smaller size using a ball mill and to obtain the size distribution of the initial and final mixture by sieving.

2. To estimate the energy required for the grinding operation. 3. To analyze the results using available theories.

Procedure:

1. Weigh the given limestone sample and obtain the initial size distribution by sieving.

2. Grind the sample in the ball mill for 30 minutes noting the energy consumed during grinding.

3. Measure the size distribution by sieving. 4. Repeat steps 2 and 3.

Theory and Analysis: The minimum energy required for crushing is the energy required for creating fresh surface. In addition, energy is absorbed by the particulate material due to deformation, friction, etc., which results in an increase of the material temperature. Defining the crushing efficiency as

n

wawbsc W

AAe

materialbyabsorbedEnergycreatedenergySurface )( −

==η (1)

Where is the surface energy per unit area and is the energy absorbed. We can experimentally find

se nW

cη . The range of cη is between 0.06 – 1.00%. If mη is the mechanical efficiency, the energy input is

mc

wawbs

AAeW

ηη)( −

= (Since WW mn η= ) (2)

Finally, the grinding energy used per unit mass is

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

saasbbpmc

s

DDe

mW

φφρηη116

(3)

where m is mass of material being ground. In the above equationφ is the sphericity, sD is the surface volume diameter and the subscripts a and b refer to the initial and final states, respectively. Experiments show that the first term in Eq. (3) is not independent of sD , and as a result the above equation is difficult to use for analysis. Instead a number of empirical laws have been proposed for calculation the energy requirements for crushing. The laws can be unified in a differential form as follows:

ns

s

DDd

kmWd −=⎟

⎠⎞

⎜⎝⎛ (4)

The different laws for the different values of the exponent are

n = 1 : ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sb

saK D

DK

mW ln (Kick’s law) (5)

n = 2 : ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

sasbR DD

KmW 11 (Rittinger’s Law) (6)

n = 3/2 : ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−=

ab

BDD

KmW

8080

11 (Bond’s Law) (7)

Note that the definition of particle size in Bonds law is different: 80D = Particle size such that 80% by weight of the sample is smaller than it. Bonds law is often written in terms of the work index (Wi) as,

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−=

ab

iDD

WmW

8080

1110 (8)

Where the work index is defined as the energy required per unit mass in kWh/ton to reduce an infinitely large particles to 80D = 100 µm. In the above equation, unit of 80D is µm, of W is kWh and of m is ton. Values of the work index: obtained from experiments for different materials are given in the table below.

Material Wi (kWh/ton) Basalt 20.41 Coke 73.8 Limestone 11.6 Mica 134.5 Glass 3.08 Calcined clay 1.43

Dry Grinding work index is 1.34 times wet grinding index. Open circuit work index is 1.34 times closed circuit work index. Questions:

1. Plot the initial distribution and distributions obtained after sieving. 2. Calculate the surface volume diameter in each case. 3. Obtain the diameter 80D for all three distributions. 4. Obtain the coefficients KK, KR and the work index, Wi for all the runs. Are there

any variations in coefficients / working indices in the runs? 5. Assuming reasonable values of cη and mη estimate es. 6. Do you have any suggestions to improve the experiments?

FM-308 manual

Will be uploaded soon

HEAT TRANSFER IN AGITATED VESSEL

AIM (a) To determine coil side overall heat transfer coefficient (h) for different agitation speeds in

given agitated vessel.

(b) Determination of coil side heat transfer coefficient while transferring heat from an agitated

liquid in a vessel to cold water flowing through the coil, submerged in the vessel under

steady state conditions.

APPARATUS 1. An agitated vessel fitted with an electrical heater, a cooling coil and a variable speed

agitator with a suitable blade for agitating liquid in the vessel.

2. Two digital thermometers with 0.1 oC accuracy for measuring the inlet and outlet

temperatures of the cooling water circulating through cooling coil.

3. Liquid in glass thermometer to measure the bath temperature.

4. Variable speed-pump to circulate cooling water through cooling coil at a constant flow

rate.

5. Stopwatch and a bucket to measure the flow rate of cooling water.

6. Tachometer to measure the speed of agitator

PROCEDURE 1. Fill the given agitated vessel with the given test liquid to about 85-90 % of its capacity.

2. Start the agitator motor and set its speed at the desired r.p.m. by manipulating its speed

regulator.

3. Connect the inlet of the cooling water circulation pump to cooling water supply line, and

start the pump. Adjust the flow rate of the cooling water at the desired level by adjusting

its speed regulator.

4. Start the heaters in the agitated vessel and set the desired temperature on the thermostat,

so as to keep temperature in the agitated vessel at a constant level. Throughout the given

set of readings keep this temperature at this level.

5. Allow sufficient time for the steady state to be attained. After steady state is attained note

down inlet and outlet temperatures of the cooling water. Also measure the flow rate of the

cooling water.

6. Repeat step (5) for different flow rates of the cooling water keeping the agitation speed

constant. [Alternatively the experiment can be carried out keeping flow rate of cooling

water constant and changing the speed (r.p.m.) of the agitator motor.]

(* Note: Students must ensure that a ‘steady state’ (in terms of temperatures as well as flow

rate) is attained before noting the observation readings)

THEORY Tube coils afford one of the cheapest means of obtaining heat transfer surface. They are

usually made by rolling lengths of copper, steel or alloy tubing into helixes or double helix

coils in which inlet and outlets are conveniently located side by side. Helical coils of either

type are frequently installed in vertical cylindrical vessels with or without an agitator,

although free space is provided between the coil and the vessel wall for circulation. When

such coils are used with mechanical agitation, the vertical axis of the agitator usually

corresponds to the vertical axis of the cylinder. However very limited data are available for

predicting heat transfer coefficient from submerged coil to the surrounding fluid in natural

convection although the coefficients are undoubtedly lower. A mechanical agitation can

improve the heat transfer coefficient between fluid in the agitated vessel and the coil. Chilton,

Drew and Jebens have published an excellent correlation on both jacketed vessels and coils

under batch and steady-state conditions and employing ‘j’ factor with a Reynolds number

modified for mechanical agitation. Although much of the work was carried out on a vessel 1.0

feet in diameter, checks were also obtained on vessels five times those of experimental setup.

The deviations on runs with water were highest for the fluids tested, which included lube oils

and glycerol, and were in some instances off by 17.5 %. Their correlation for heat transfer to

fluids in vessel with mechanical agitation heated or cooled by submerged coils is as follows,

0.142 / 3 1/ 32

0.87 pc

J w

Ch L ND K K

µρ µµ µ

⎡ ⎤⎡ ⎤ ⎡ ⎤= ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦ ⎣ ⎦ [1]

where Dj = inside diameter of the agitated vessel [m]

hC = coil side heat transfer coefficient [Kcal/hr m2 OC] L = agitator diameter [m]

N = agitator speed [rev/sec]

ρ = density of fluid in the vessel [kg/m3]

K = thermal conductivity of fluid in the vessel [Kcal/hr m OC]

µ = viscosity of fluid in the vessel [kg/m hr]

µw= viscosity of fluid in the vessel at coil wall temp. [kg/m hr]

It can be noticed from equation [1] that for the given vessel with the given fluid and coil the

heat transfer coefficient will be proportional to N2/3.

As far as the inside coefficient for the coil is concerned because of the increased turbulence

due to circulatory path the heat transfer coefficient will be greater than those calculated for

straight pipes. For ordinary use McAdams suggests that straight tube equations such as Dittus-

Boelter equation or Sider-Tate equation can be used, when the value of ‘h’ so obtained is

multiplied by 1 + 3.5[D/DC] where D is the inside diameter of the tube and DC is the diameter

of the coil helix.

GRAPHS Plot the graph of [1/U] versus [1/V0.8] on the linear scale. Intercept of this graph will give the

value of [1/ho] in case of constant agitator speed experiment. The intercept of this graph will

give the value of 1/ho. For the experiment with constant cooling water flow rate and variable

agitator speed plot the graph of [1/U] versus [1/N] on a log-log graph. Slope of this graph

should be around 2/3.

RESULTS Comment on the nature of the graph obtained. CONCLUSION

OBSERVATIONS 1) Temperature of liquid in the agitated vessel (T) _________ OC 2) Length of the coil immersed in the agitated vessel (L) ____2.59__ m 3) Inside diameter of the coil tube (di) ____0.009_ m 4) Outside diameter of the coil tube (do) ____0.012_ m 5) Area of coil available for heat transfer = π x do x L = _________ m2 6) Coil helix diameter __________ m 7) Specific heat of water (Cp) 1 Kcal/kg OC OBSERVATION TABLE Obs. No.

Inlet temperature cooling of water

t1 [oC]

Outlet temperature Of cooling water

t2 [oC]

Weight of water collected

W [Kgs]

Time of collection

t [Sec] 1

2

3

4

5

6

7

8

9

CALCULATIONS Specimen Calculation for Reading No. : ________ . 1. Mass flow-rate of water (m) = (W/t) x 3600 = _________ kg/hr.

2. ∆t = t2 – t1 = ___________ oC

3. Q = m x Cp x ∆t = ___________ Kcal/hr

4. ∆T1 = T – t1 = ___________ oC

5. ∆T2 = T – t2 = ___________ oC

6. LMTD (∆Tlm) = (∆T1- ∆T2)/[ln(∆T1/∆T2)] = ___________ OC

7. U = Q/(A x ∆Tlm) = ___________ Kcal/hr. m2. OC RESULT TABLE

Obs. No.

Amount of heat

transferred Q [Kcal/hr]

Log mean temperature difference

[oC]

Overall heat transfer

coefficient [Kcal/hr m2 oC]

1/U

1/V0.8

1 2 3 4 5 6 7 8 9

Expt. HT 307

Finned Tube Heat Exchanger

Objective

To determine the efficiency of given longitudinal/pin fin and compare it with the theoretical value for thegiven fin.

Apparatus

1. Longitudinal fin heat exchanger.

2. Pin fin heat exchanger.

3. Bare pipe without fins.

4. Steam generator to generate steam at constant pressure. The steam generator is also provided withtemperature indicator and a dead weight safety valve.

Procedure

1. IMPORTANT INSTRUCTIONS: Follow instructions 1 and 2 without fail, otherwise electrical heaterwill burn out.

2. Open the drain valve provided at the bottom of steam generator and drain out the water from steamgenerator completely.

3. Close the drain valve and charge 4 lit. of water through charging valve provided at the top of the steamgenerator and close it. Ensure that the dead weight safety valve is free.

4. Start the electrical heater of steam generator. Initially supply full voltage to the electrical heater.Steam will start forming within about 15-20 min. of switching on the heater. During this period,keep open the valve to one of the test sections (either longitudinal fin heat exchanger or pin fin heatexchanger). Also keep the needle valve at the end of test section open. Once the steam generationstarts, the finned tube heat exchanger will start getting heated up and condensate will start coming outof the needle valve provided at the bottom of condensate collector. When the test section (finned tubeheat exchanger) is fully heated up, steam will start coming out of the needle valve. Now regulate theneedle valve in such a way that only condensate comes out of it. At this point of time also regulatethe voltage supplied to the electrical heater so as to keep the pressure in the steam generator constant.The pressure can be regulated between 0-1 atm. gauge as per the requirement.

HT 307-1

5. Once the test section (finned tube heat exchanger along with bare pipe without fins) is fully heated,drain out completely the condensate it any. Close the needle valve on condensate drain line simul-taneously starting the stop-watch. Collect the condensate accumulated at an interval of 15 min. forfinned tube heat exchanger and 30 min. for bare pipe. If the quantity of condensate collected is samefor 2 to 3 consecutive readings (within experimental accuracy), note down the volume of condensatecollected and time interval.

6. Repeat the procedure given in 4 for pin fin heat exchanger as well as bare pipe.

Theory

In a heat exchanger, the two fluids namely; hot and cold, are separated by a metal wall. Under this conditionthe rate of heat transfer will depend on the overall resistance to heat transfer given by the equation:

1UiAi

=1

hiAi+

xKAlm

+1

hoAo(1)

where,

Ui = Overall heat transfer coefficient based on inner area [Kcal/hr m2 ◦C ]

Uo = Overall heat transfer coefficient based on outer area [Kcal/hr m2 ◦C ]

hi ,ho = Inside and outside film heat transfer coefficients [Kcal/hr m2 ◦C ]

Ai ,Ao =Inside and outside surface area [m2]

When viscous liquids are heated in a double pipe heat exchanger or any standard tubular heat exchanger bycondensing steam or hot fluid of low viscosity, the film heat transfer coefficient of the viscous liquid willbe much smaller than that on the hot fluid side and will therefore, become controlling resistance for heattransfer. This condition is also present in case of air or gas heaters where the gas side film heat transfercoefficient will be very low (typically of the order of 0.01 to 0.005 times) compared to that for the liquidor condensing vapour on the other side. Since, the heat transfer coefficient of viscous fluid or gas cannotbe improved much, the only alternative is to increase the area available for heat transfer on that side so thatits resistance to heat transfer can be reduced. To conserve space and to reduce the cost of equipment inthese cases, certain type of heat exchange surfaces, called extended surfaces, have been developed in whichoutside area of tube is increased many fold by fins and other appendages.

Two types of fins, are in common use viz; longitudinal fins and transverse fins. Longitudinal fins are usedwhen the direction of flow of the fluid is parallel to the axis of tube and transverse fins are used when thedirection of the flow of the fluid is across the tube. Spikes, pins, studs or spines are also used for eitherdirection of flow.

The outside are of a finned tube consists of two parts: the area of fins and the area of bare tube not coveredby the bases of fins. A unit area of fin surface is not as efficient as a unit area of bare tube surface becauseof the added resistance to the heat flow by conduction through the fin at its base. The expression for finefficiencies can be derived by solving the general differential equation of heat conduction with suitableboundary conditions. Generally three boundary conditions are used;

1. Fin of infinite length so that there is no heat dissipation from its tip, or in other words temperature atthe tip of fin is same as that of the surrounding fluid.

HT 307-2

2. Insulated tip. This condition even though cannot be realized in practice, but considering that the tiparea is negligible as compared to the total fin area, heat dissipated from tip can be neglected and hence,dt/dx is assumed to be zero at the tip.

3. Finite heat dissipation from the tip. Even though the assumption of insulated tip is invalid, most ofthe fins are treated under this category, and longitudinal fin efficiency for this case is given by theexpression:

η f in =tanh(mL)

mL(2)

where

m =√

(hC/KA)

h = film heat transfer coefficient from the fin surface [Kcal/hrm2◦C ]

C = circumference of the fin [m]

K = thermal conductivity of fin material [Kcal/hr m◦C ]

A = cross-sectional area of fin [m2]

From the above equation, it can be seen that the fin efficiency is a function ofmL, and as the value ofmLincreases, the fin efficiency decreases. A reasonable value of fin efficiency will be around 50 to 75% forwhich mL should have a value between 1 and 2. If the fin heightL should be sufficient (of the order of 5 to8 cm), then it can be seen that the value ofh should be around 10 to 20 which can be given by air in naturalconvection. The value of film heat transfer coefficient for any other liquid in natural convection, or any gasin forced convection will be much higher than 20. Thus, the given set-up is used for heat transfer to air innatural convection region.

Observations

1. Finned Tube:

1. Height of fin (L) : cm.

2. Width of fin (W) : cm.

3. Thickness of fin (b) : cm.

4. Number of fins (N) : 4

5. O.D. of fin tube : cm.

6. Thermal conductivity of fin material (K) : Kcal/hr m◦C

2. Bare Tube:

1. Length of tube (l) : cm.

2. O.D. of tube : cm.

3. Tambient : ◦C

HT 307-3

Calculations

1. Circumference of fin (C):C = 2(w + b) = m. (3)

2. Cross-sectional area of fin (A):A = wxb= m2 (4)

3. Fin area available for heat transfer:

AF = CxLxN= m2 (5)

4. Tube area available for heat transfer in finned tube heat exchanger:

AB = (ΠD − Nb)xw = m2 (6)

5. Total area of finned tube heat exchanger:

At = AF + AB = m2 (7)

6. Heat given out by steam through finned tube heat exchanger (Q1):

Q1 = (m1x)xλ = Kcal/hr (8)

7. Heat given out by steam through bare tube (Q2):

Q2 = (m2x)xλ = Kcal/hr (9)

whereλ = latent heat of vaporization of water at steam pressure (Kcal/Kg)

8. Film heat transfer coefficient from bare tube (h):

h = Q2/(Ax∆T) = Kcal/hrm2◦C;

A = ΠDL = m2;

∆T = (Tsteam− Tambient) =◦C

9. m=√

(hC/KA) =

10. mL=

11. η f in (Theoretical)= tanhmL/mL

12. Amount of heat actually dissipated by fin:

Qf in = Q1 − (AB × h× ∆T) = Kcal/hr (10)

13. Amount of heat that can be dissipated by ideal fin:

Qideal = AF × h× ∆T = Kcal/hr (11)

14. Observed value of fin efficiency:

η(Observed) =Qf in

Qideal= (12)

Conclusion

HT 307-4

Expt. HT 310

Plate Heat Exchanger

Aim

1. To determine the overall heat transfer coefficient in a plate heat exchanger,

2. To study its scaling dependence on the hot fluid flow rate,

3. To determine the laminar turbulent transition, and

4. To suggest a correlation function for this dependence in various regimes (laminar, transition, andturbulent).

Apparatus

The setup employed for this experiment is as follows.

1. A stainless steel plate heat exchanger with a facility to measure inlet and outlet temperature of hotand cold fluid with an accuracy of 0.1 ◦C. The plates are planar (not corrugated), There are a total of10 plates making 11 chambers for the fluid transport–six for the cold fluid and five for the hot fluid.The total heat transfer area available is equal to that of the number of plates (10). (TASK: Measurethe dimensions of the plate heat exchanger)

2. The cold fluid used here is water and the hot fluid is ethylene glycol. (TASK: Determine the depen-dence of both the fluids’ properties–viscosity, thermal conductivity, and specific heat–with tempera-ture in the range 40–80 ◦C).

3. A stainless steel insulated tank with a heater to act as a reservoir for the hot fluid.

4. Hot fluid circulation pump with a speed control potentiometer.

5. Cold fluid inlet from the water supply tap.

6. Four temperature sensors at the inlet and outlet points for each of the two fluids. The hot-fluid inletthermometer is also a thermostat control, which controls the heater connected to the reservoir by asimple relay mechanism.

7. Rotameters for fluid flow measurements.

HT 310-1

Procedure

General setup

1. The zero correction of the thermometers are determined by measuring steady the fluid inlet and outlettemperature under the following conditions (without switching on the heater).

• Stationary (Assuming the equipment is at equilibrium, before the start of the experiment, all thethermometers should indicate the same temperature. Any deviation indicates the error of thethermometer/sensor combination)

• Allow minimal flow of the hot fluid (≈ 25 lph) and measure any temperature difference (whichis more than the above error). If the outlet temperature is greater, it indicates viscous dissipation.

• Set the pump to maximum capacity flow rate (≈ 550 lph), and measure the temperature differencebetween the outlet and inlet of the hot fluid. (TASK: Calculate the Brinkman number Br =µ u2/K∆T , where µ is the fluid viscosity at the mean temperature, u is the velocity of the hotfluid one chamber, and K is its thermal conductivity, and ∆T is the temperature increase. Br is ameasure of the heating due to viscous dissipation.)

2. Set the temperature of the inlet hot fluid in the dual temperature indicator cum controller. The setpoint should be set around 65 to 75 ◦C.

3. Provide cooling water supply to the plate heat exchanger so that the flowrate is between 13–15 lpm.This will ensure that the temperature rise is restricted to about 2–3 ◦C. Keep this flow rate constantthroughout the experiment.

4. Connect the 15 A and 5 A plug pins to a stable 230 V A.C. electric supply. Care should be taken toconnect these two pins in different phases of the power supply. Switch on the heater power supply.

5. Adjust the flow rate of hot fluid through the heat exchanger by adjusting the speed of hot fluid circu-lation pump. Note down the flow rate of hot fluid as indicated by the rotameter. If during the courseof any experiment, the flow rate changes (due to power fluctuations, or due to temperature changes),make minor adjustments to the potentiometer (which controls the pump speed) to manually reset theflow rate to the desired set value. This kind of adjustments should be done for all the experiments tofollow to ensure that the flow rate is maintained at a constant value.

Determination of characteristic settling time

Adjust the set point temperature to a temperature around T = 60◦C. Set the flow rate to an intermediatevalue, V̇ = 300lph. Through out the measurement make sure the flow rate is at this value. Measure the inletand outlet temperatures for about 15 minutes at 30 second intervals. Use a graph sheet to plot the variationin temperature. Use this plot to obtain an estimate of the time it takes for the inlet and outlet temperaturesto settle down to a constant value or to a constant periodic oscillation. Note down if there is any time lag inthe behaviour of the outlet temperature variation with respect to that of the inlet.

Both these readings, settling time and time lag, should be used in the main experiment: The readingsshould be taken down only after the settling time (usually one or two time periods of oscillation) and theoutlet temperature after the time lag.

HT 310-2

Determination of the overall heat transfer coefficient

1. Set the inlet temperature to a high value (≈ 70◦C).

2. Set the flow rate to the highest possible value. Note down the value and maintain it constant (seeabove on how to do this).

3. Wait for the predetermined characteristic settling time, and then note down the steady inlet and outlettemperatures of both the fluids.

4. Repeat steps from 2 (at constant temperature) for at least 10 different flow rates ranging from a maxi-mum of about 550 lph to a minimum of about 25 lph. It is useful to place the interval in the flow ratein a geometric progression (GP), which will give equally spaced data in a logarithmic scale. Note thatit is not possible to set the value of the flow rate to arbitrary precision as calculated by the GP. Use avalue that is closest to the resolution provided by the flow meter. For example, if the resolution of theflow meter is 40, then approximate 28.73 to 40. For higher flow rates, you can also play around withthe value of the spacing in GP to a smaller value (say half of that in the low flow rates) so that thereare values for at least 10 flow rates.

5. In order to ensure reproducibility, for a given set of inlet temperature and flow rate, take at least threesets of readings of the stream temperatures. The three sets can be obtained by starting (i) at the highestflow rate and descending to the lowest, (ii) ascending to the highest, and (iii) down again to the lowest.

Theory

The plate heat exchanger normally consists of corrugated plates assembled into a frame. The hot fluidflows in one direction in alternating chambers while the cold fluid flows in true counter-current flow inthe other alternating chambers. A schematic diagram of the flow is shown in Figure 1. The fluids aredirected into their proper chambers either by a suitable gasket or a weld depending on the type of exchangerchosen. Traditionally, plate and frame exchangers have been used almost exclusively for liquid to liquidheat transfer. The best example is in the dairy industry. Today, many variations of the plate technology haveproven useful in applications where a phase change occurs as well. This includes condensing duties as wellas vaporization duties. Plate heat exchangers are best known for having overall heat transfer coefficients(U-values) in excess of 3–5 times the U-value in a shell and tube designed for the same service.

Plate heat exchanger is an attractive option when more expensive materials of construction can be employed.The significantly higher U-value results in far less area for a given application. The higher U-values areobtained by inducing turbulence between the plate surfaces. Owing to this they are also known to minimizethe fouling.

Heat transfer correlation

In general, the heat transfer correlation for a fluid flow past a solid surface is expressed in a dimensionlessform

Nu = Nu(Re,Pr). (1)

where Nu is the non-dimensional heat transfer coefficient Nu = h D/k. For a heat transfer in a laminar fluidflow past a solid surface, with constant fluid properties, the steady state temperature profile is a function onlyof Re, and Pr. The heat transfer coefficient is a function of the temperature profile. Therefore, the above

HT 310-3

Hot In

Hot Out

Cold Out

Cold In

External Wall

Plates

Figure 1: Schematic diagram of a one pass counter current heat exchanger showing the flow pattern.

relationship. This expression is often used in situations where the properties vary with temperature, and forturbulent flows.

For fully developed laminar flows (internal flows), we expect the Nusselt number Nu to be constant, howeverfor a developing flow its is expressed as:

Nu = C1 Reα Prβ (2)

The value of β ≈ 0.4. The value of α is found to be around 0.3 for developing laminar flow and around0.64 for turbulent flow. The transition from laminar to turbulent region occurs between 10 / Re / 100 forcorrugated plates. It can be expected to be higher for plain plates.

The heat transfer coefficient appearing in the Nusselt number can be calculated from the overall heat transfercoefficient U, which is given by

1U=

1hh+∆xKp+

1hc

(3)

where, hh is the hot fluid heat transfer coefficient and hc is the cold fluid heat transfer coefficient, Kp is thethermal conductivity of the metal plate and ∆x is its thickness. Once the heat exchanger material and itsgeometry are fixed, then the metal wall resistance (∆x/Kp) becomes constant. Similarly, if the flow rate ofcold fluid is fixed and its mean temperature does not differ much for different flow rates of hot fluid, thenthe resistance of the cold fluid will remain almost constant. Thus, the overall heat transfer coefficient willdepend upon the value of the hot fluid heat transfer coefficient alone. If the bulk mean temperature does notdiffer much for different flow rates, then all the physical properties will remain nearly the same and Eq. (3)can be re-written in combination with Eq. (2) as

1U=

1hh+C =

muα+C (4)

where m and C are constants. hh can therefore be evaluated from the intercept of the plot of 1/U vs 1/uα.Since the value of α is not known, it has to be estimated first. A plot of log d(1/U)/du vs log u will eliminatethe constant C and the slope will give (−α−1). The constant m can also be evaluated with this intercept. Thena plot of 1/U vs 1/uα will provide the intercept value C, which is then used to calculate the heat transfercoefficient from Eq. (4). The Nusselt number correlation can then be found. For the sake of simplicity, it isoften assumed that α = 1/3. This can be verified if the plot of 1/U vs 1/u1/3 is a straight line for a largerange in the small u limit.

HT 310-4

Plots

The following data need to be plotted. A sample calculation to obtain the values of the variables is shownbelow.

1. Plot of the temperature transient for the hot fluid (at the outlet and the inlet) when the heater is turnedon till the temperature attains a steady state value

2. Plot of 1/U vs 1/u1/3: In this plot, for each value of flow rate, U values corresponding to the threeindependent readings should be shown, apart from the U value computed from the average of thetemperature measurement. The average should be used for fitting and computing the intercept. Sincethis involves several calculations, it is suggested that programmable spreadsheets be made use of(OpenOffice.org or MS Exel)

3. Plot of Nu/Pr0.4 vs Re

References

1. G H Hewitt, G L Shires, and T R Bott, “Process Heat Transfer”, CRC Press, NY, 1994

Observations

Height of Plate H = 10cm

Width of Plate W = 5 cm

Gap between two plates b = 1 mm

Number to plates N = 10

Number of hot fluid chambers Nh = 4

Number of cold fluid chambers Nc = 5

Zero error of hot fluid digital thermometers δTz =◦C

HT 310-5

Observation table

Obs. No. Flow rate V̇ (lph) Hot Fluid Temperature (◦C) Cold Fluid Temperature (◦C)

Inlet (T1) Outlet (T2) Inlet (t1) Outlet (t2)

1

2

3

4

5

6

7

8

9

10

Parameters estimation

Total heat transfer area of heatexchanger

A = N H W =

Cup mean temperature (use anytypical value)

Tm =12 (T1 + T2) =

Density of Ethylene glycol atTm

ρ =

Specific heat of Ethylene glycolat Tm

Cp =

Viscosity of Ethylene glycol atTm

µ =

Thermal conductivity of Ethy-lene glycol at Tm

K =

Prandtl number for hot fluid Pr =Cp µ

K=

Equivalent diameter De =2 W bW + b

=

HT 310-6

Sample calculation

Flow rate V̇ =

Velocity of hot fluid in a cham-ber

u =V̇

W b Nh=

Total heat transferred Q = ρCp V̇ (T1 − T2) =

Brinkman number Br =µ u2

K(T1 − T2)=

Log mean Temperature differ-ence (LMTD)

∆TLM =(T1 − t2) − (T2 − t1)

ln [(T1 − t2)/(T2 − t1)]=

Overall heat transfer coefficient U =Q

A∆TLM=

Reynolds number Re =De u ρµ

=

Intercept of 1/U vs 1/u1/3 plot C =

Hot fluid heat transfer coeffi-cient

1hh=

1U−C =

Nusselt number Nu =hh De

K=

Student information

Date of Experiment =

Batch Number =

Roll Numbers

HT 310-7

Expt. MT 302

Differential Distillation

Aim

To verify the Rayleigh’s equation for a differential distillation in a binary system

Theory

In the case of a differential distillation, the vapour at any time is in equilibrium with the liquid from whichit rises but changes continuously in the composition. Thus, the mathematical approach used must be differ-ential. Assume thatL mol of liquid in the still of compositionx mol fractionA and that an amountdD molof distillate is vaporized, of mol fractiony∗ in equilibrium with the liquid.

Material Balance

The rate of depletion of liquid is equal to the rate of distillate output. The instantaneous rate of depletion ofa component in the liquid is therefore, In - out= accumulation

0− dD = dL (1)

Taking balance on more volatile component,

0− y∗dD = d(Lx) (2)

0− y∗dD = xdL+ Ldx (3)

y∗dL = xdL+ Ldx (4)

Therefore rearrangement gives, ∫ F

W

dLL=

∫ xF

xW

dxy∗ − x

(5)

This equation can be integrated to get the following form which is called the Rayleigh’s Equation,

lnFW=

∫ xF

xW

dxy∗ − x

(6)

MT 302-1

where,F = moles of feed of compositionxF , W = moles of residual liquid of compositionxW, W andxwcan be obtained by material balance,

F = D +W (7)

FxF = DxD +WxW (8)

The integral in eqn. 5 can be solved analytically (provided the relationship betweeny∗ andx is available) orgraphically (calculating the area under the curve for the plot of 1/(y∗ − x) vs x.

MT 302-2

Apparatus

Figure MT 302.0.1: Schematic of the setup

Procedure

1. Prepare a calibration plot of mole fraction (x) vs. refractive index (h) of pure componentsA andB.

2. Weigh 8 nos. of tagged stoppered conical flasks.

3. Start the flow of water through the condenser.

4. Fill 3/4th (approx. 300 ml) volume of the distillation flask with a mixture ofA and B of knowncomposition (xF). The mixture is weighed (w) before charging in the distillation flask.

5. Start heating at a slow rate. When the mixture starts boiling, collect the distillate in a weighed 50 mlflask. After approximately 30 ml of the distillate has been collected, remove the flask and collect next8 to 10 drops of the distillate in tagged test-tube and then put another flask for the collection of thedistillate. This procedure should be repeated for collecting 8 distillate samples.

6. Measure the refractive indices (RI) of the samples collected in the test-tubes (η1t, η2t,...,η12t). Weighthe samples collected in the conical flasks (w1, w2... w12). Measure the RI of the bulk from each ofthe flasks (eta1b, eta2b, ...,η12b).

Observations and Calculations

Data from the literature

1. Moelcular weights ofA andB.

2. Refractive indices ofA andB.

3. Densities ofA andB.

4. Vapour liquid equilibrium data forA andB at atmospheric pressure.

MT 302-3

Calibration data for mole fraction vs. RI

S.N. Mole fraction ofA RI

1 0

2 0.1

3 0.2

4 0.3

5 0.4

6 0.5

7 0.6

8 0.7

9 0.8

10 0.9

11 1.0

Calculations

1. CalculateD (amount of distillate) (from weighed 30 ml sample) andxD (distillate composition) (fromrefractive index of 30 ml sample) for each sample.

2. CalculateW (amount of residue still left in the flask) andxW (composition of residue) using Eqn. 7and 8 for each fraction.

3. Calculatey∗ (vapor phase composition) for each sample (from Refractive index of 8 drops collected.)

4. Calculatex (liquid phase composition in equilibrium withy∗) using Raoult’s law.

5. Complete the following Table.

MT 302-4

S.N. F xF D xD W xW y∗ x ln(F/W)i 1/(y∗ − x)

1

2

3

4

5

6

7

8

6. Calculate ln(F/W) = ln(F/W)1 + ln(F/W)2+ ... + ln(F/W)8

7. Plot 1/(y∗ − x) vs x and measure the area under the curve.

8. Now verify Eqn. 6.

9. Calculate

% Error =

∣∣∣∣∣∣∣∣ln(F/W) −

∫ xF

xwdx/(y∗ − x)

ln(F/W)

∣∣∣∣∣∣∣∣ × 100

MT 302-5

1

CL-333 Manual

MT 303: Batch Distillation

Objectives:

To determine the height equivalent to number of theoretical stages and the number of

transfer units for a packed column.

Theory:

Distillation is the most widely used separation technique in the chemical and petroleum industry

and is performed in either tray or packed columns. In a packed column, HETP (Height

Equivalent to a Theoretical Pate) or HTU (Height of a Transfer Unit) are used to relate the

column height with the number of theoretical stages obtained by standard design methods such

as McCabe-Thiele or Ponchon-Savaritt. Distillation operated at total reflux yields maximum

possible separation. The number of theoretical stages provided by the column is the minimum

number of theoretical stages for the top and bottom composition. Minimum number of stages

may be calculated using graphical methods or by Fenske’s equation given by (1).

1log( )

11

log

WD

D WM

av

xx

x xN … (1)

av top bottomX … (2)

NM = number of theoretical stages at total reflux, and αav = average relative volatility

Experimental procedure:

1. Prepare a calibration chart of RI (Refractive Index) vs. mole fractions of components A

and B.

2. Fill approximately 2/3rd

volume of the reboiler kettle with a mixture of A and B.

3. Start water circulation through the condenser.

4. Close the bottom sample “draw out” valve to prevent the overflow of liquid through the

sample line during the experiment. Switch on the heating mantle and allow the flask

temperature to rise.

5. When the vapors rise to the top of the column, increase the heating rate if necessary.

After about 30 min., note the temperatures along the length of the column. If some milky

liquid appears in the reflux line, drain it in a small glass vessel and discard. This could be

2

due to the formation of emulsion of the organic liquid with moisture already present in

the system.

6. When the temperature readings along the column length become steady (based on four

successive readings taken at 10 min time interval) it indicates that the system has attained

steady state and the samples may be withdrawn.

7. 2 to 4 ml of top sample is collected in a sample bottle after purging off some quantity.

The bottle should be immediately closed to avoid losses. Determine the RI of the sample.

8. Step 7 should be repeated now for the bottom sample.

9. Count the number of actual plates in the column.

HETP may be calculated by (3) once the number of theoretical stages (NM) provided by the

column is known.

Height of the column

M

HETPN

… (3)

HTU can be calculated from (4) once NTU (Number of Transfer Units) for the column is known.

NTU for a packed column is determined using (5).

Height of the columnHTU

NTU … (4)

*

D

B

y

y

dyNTU

y y … (5)

# Please read standard text books (e.g. Treybal) on mass transfer operations to understand the

difference between HTU and HETP.

Observations

Data from the literature

1. Molecular weights of A and B.

2. Refractive indices of A and B at ambient temperature.

3. Densities of A and B at ambient temperature.

Calibration data for mole fraction vs. RI

S. No. Mole fraction of A RI

3

Figure 1: Set-up for Batch Distillation

4

Observations

1. Number of actual stages in the column.

2. Top temperature.

3. Bottom temperature.

4. RI of top and bottom samples.

5. Mole fraction of top sample (xD).

6. Mole fraction of residue or bottom sample (xB).

Calculations, Results and Discussion

1. Calculate the number of theoretical stages by McCabe Thiele method and calculate

the HETP using equation (3).

2. Calculate the number of stages by Enthalpy-Composition (Ponchon-Savaritt) method

and calculate the HETP using (3).

3. Find the number of stages by Fenske’s equation (1) and compare the result with that

obtained from Step 1 above.

4. Determine the NTU using (5).

5. Comment on the precaution and sources of error.