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HEAT TRANSFER

LAB MANUAL

NIRMA UNIVERSITY

INSTITUTE OF TECHNOLOGY

CHEMICAL ENGINEERING DEPARTMENT

Chemical Engineering Department

Institute of Technology, Nirma University

Heat Transfer Operation- Lab Manual

List of Experiments:

1. Thermal conductivity apparatus

2. Thermal conductivity of metal rod

3. Thermal conductivity of insulating powder

4. Heat Transfer in natural convection

5. Heat Transfer in forced convection

6. Extended surface equipment

7. Parallel flow heat exchanger

8. Counter flow heat exchanger

9. Shell and Tube heat exchanger

10. Finned tube heat exchanger

11. Emissivity measurement apparatus

12. Drop wise and film wise condensation apparatus




Date: Roll No:

Practical No:

THERMAL CONDUCTIVITY APPARATUS

Objective: After this experiment student will able to understand how to determine the Thermal

Conductivity of given specimen.

Apparatus: Two slab guarded hot plate thermal conductivity apparatus, Specimen, Insulation

(Glass wool) Packets.

Utility: Water, Electric Supply.

Theory:

Principle of the guarded hot plate method:

A sketch of the apparatus is shown in Fig. (1). The essential parts the Hot plate, the cold plate,

the heater assembly, thermocouples and the specimen, in position, are shown in the same figure.

For measurement of the thermal conductivity K what is required is to have one dimensional heat

flow through the flat specimen, an arrangement for maintaining its faces at the constant

temperature and metering method to measure the heat flow through a known area.

To eliminate the distortion caused by the edge losses in unidirectional heat flow from the central

plate, it is surrounded by a guard ring heater separately. Temperatures are measured by calibrated

thermocouples, attached to the plates or to the specimen at the hot and the cold faces. Two

specimens are used to ensure that all the heat comes out to the specimen only.

Knowing the heat input to the central plate heater, the temperature difference across the

specimen, its thickness and the area, one can calculate the K by the following formula.

Where,

K Thermal Conductivity of the sample, W /m C

q Heat flow rate in the specimen, W

A Area of the specimen, m2

Th Hot plate temperature, C

Tc Cold plate temperature, C

L Thickness of the specimen, m

If the specimen thickness are different and the respective hot and cold temperatures are

different than,

Where suffix 1 stands for upper specimen and 2 stands for lower specimen.

)(*2

*

ch TTA

LqK

c2h2

2

c1h1

1

T - TT - T2

LL

A

qK




Apparatus description:

The heater plate is surrounded by a guard heater for stabilising the temperature of the primary

heater and to prevent heat loss radially around the edges. The primary and guard heaters are made

of Nichrome wire packed between upper and lower mica sheets. These heaters together with

upper and lower copper plates and rings from the heater plate assembly.

Two thermocouples (1 & 2) are used to measure the hot face temperature at the upper and lower

central plate assembly copper plates. Two more thermocouples (3 & 4) are used to check balance

in both the heater inputs (see figure 1).

Specimens are held between the heater and cooling unit on each side of the apparatus.

Thermocouples (5 & 6) measure the temperature of the upper cooling plate and lower cooling

plate respectively.

The cooling chamber is a composite assembly of spiral grooved Aluminium casting and

aluminium cover with entry and exit adopters for water inlet and outlet.

Procedure:

The specimens are placed on either side of the heating plate assembly uniformly touching the

cooling plates. The outer container is filled with loosely filled insulation such as glass wool

supplied in small packets. The cooling circuit is started. Then calculated heat input is given to the

central and guard heaters through separate single phase power supply lines with dimmerstat in

each line and it is adjusted to maintain the desired temperature (For ensuring no radial heat

transfer, generally outer heater input is 2.5 to 3.0 times more than the central heating input). The

guard heater input is adjusted in such a way that there is no radial heat flow, which is checked

from thermocouples readings are recorded accordingly. The input of the central heater (current &

voltage, watts) and the thermocouple readings are recorded every 10 minutes till a reasonably

steady state condition is reached. The readings are recorded in the observation table. The final

steady state values are taken for calculations.

Precautions:

Keep the dimmerstat to 0 voltage position at start.

Increase the voltage gradually of the two heaters during initial set-up experimentation.

Start the cooling circuit before switching on the heaters and adjust the flow rates so that

practically there is no temperature rise in the circulation fluid.

Keep the heater plate undisturbed and adjust the cooling plates after keeping the samples with

the help of nuts gently.

Keep the loose filled in insulation (glass wool) packets gently and remove them slowly so that

they do not disturb the thermocouples’ terminals and heater wires.




Observation table:

Sr.

No.

Central Heater Guard Heater Cooling Plate

T1(0C) T2(0C) T3(0C) T4(0C) T5(0C) T6(0C)

The difference between the temperatures of central heater and guard heaters should not be more

than 1 0C.

V (volts)

I (amp)

q(watt)

L (m) 0.019 0.019 0.019

D (m) 0.18 0.18 0.18

A (m2)

K (Watt / m oC)

Calculation:

1. Area of Heat transfer A = ( / 4) * D2

2. Thermal Conductivity of specimen

Where,

Th, av = (T1 + T2 ) / 2

Tc, av = (T5 + T6 ) / 2

L = Thickness of slab

Result:

Conclusion:

)av Tc, - av Th,(2

*

A

LqK




Quiz:

1. Write the Fourier rate equation for heat transfer by conduction. Give physical significance of

each term.

2. Why there is a negative sign in the Fourier’s law of heat conduction?

3. What is meant by one-dimensional steady state heat conduction?

4. List some good conductors of heat; some poor conductors.

5. What is the function of guard heater in Two Slab Guarded Hot Plate Method?




Date: Roll No:

Practical No: THERMAL CONDUCTIVITY OF METAL ROD

Objective: After this experiment student will able to understand how to determine the thermal

conductivity of given metal rod.

Apparatus: Thermal conductivity apparatus, metal rod.


Theory:

The thermal energy is conducted in solids by the following modes:

(i) Lattice vibration

(ii) Transport by free electrons

In good electrical conductors a large number of free electrons move about in the lattice structure

of the material. They carry thermal energy from a high temperature region to a low temperature

region. Energy may also be transmitted as vibrational energy in the lattice structure of the

material. In general, this latter mode of transfer is not as large as the electron transport. With

increase in the temperature, however the increased lattice vibration come in the way of the

transport by free electrons and for most of the pure metals the thermal conductivity decreases

with increase in temperature.

Apparatus Description:

The experimental set up consists of metal bar, one end of which is heated by an electric heater

while the other end of the bar projects inside the cooling water jacket. The middle portion of the

bar is surrounded by cylindrical shell filled with asbestos magnesia insulating powder. The

temperature of the bar is measured at different sections (figure1) from 1 to 7 while the radial

temperature distribution is measured by separate thermocouples at two different sections in the

insulating shell.

The heater is provided with a dimmerstat for controlling the heat input. Water under constant

heat condition is circulated through the jacket and its flow rate and temperature rise is noted.

Procedure:

Start the electric supply. Adjust the room temperature in the temperature indicator by means of

rotating knob, for compensation of the temperature equal to the room temperature (normally this

is readjusted). Give input to the heater by slowly rotating the dimmerstat and adjust it to voltage

equal to 80 to 120 volts. Start the cooling water supply through the jacket and adjust it about 350

cc/ minute. Go on checking the temperatures at some specified time intervals say 5 minutes and

continue this till a satisfactory steady state condition is reached. Note the mass flow rate of water

in kg/ min. and temperature rise in it and also note the temperature readings from 1 to 11 using

temperature indicators.




Observations:

1. Length of the metal bar (total) 450 mm

2. Size of the metal bar (diameter) 38 mm

3. Test length of the bar 230 mm

4. No. of thermocouples mounted on the bar 7

5. No. of thermocouple in the insulation shell 4

6. Heater coil (band Type) Nichrome

7. Cooling Jacket Diameter 90 mm

8. Radial distance of thermocouple in insulating shell ro = 55 mm=r2

ri = 35 mm =r1

9. Mass flow rate of water ‘m’ = ___________litre/ minute

___________kg/ minute

10. Inlet temperature of water Ti = ___________0C

11. Outlet temperature of water To= ___________0C

Observation Table:

Sr. No: V ( volt ) I ( amp ) W ( watt )

T/C

No.

1 2 3 4 5 6 7 8 9 10 11

Temp

(0C)

Plot the graph of the temperature distribution (at steady state) along the length of the metal rod

using observed values 1 o 7 (as show in fig 2), for determining the slope at BB and CC sections,

and also at AA section.

Slope is dt / dx at various desired points on the plot of T vs. Distance. Nature of the graph will be

probably as shown in figure 3.

Heat transmitted in Radial direction by the cylinder can be calculated by following formula:

Calculations:

1. Heat transferred to cooling water = Heat passed from AA section

qw=qAA = m * Cp * (To - Ti) = -KAA * (dT / dx)AA * A

KAA = _____________ w/ m K

kL

rr

TTQ oi

2

)/ln( 12




2. Heat passed through = Heat passed in linear + Heat passed in redial

BB section direction from AA section direction

KBB = _____________w/ m K

3. Heat passed through = Heat passed in linear + Heat passed in redial

CC section direction from BB section direction

KCC = _____________w/ m K

Results:

Conclusion:

Quiz:

1. Define thermal conductivity. What is the approximate range of thermal conductivity of solids,

liquids and gases?

2. State the effect of impurities on the thermal conductivity of a metal.

3. Point out and explain various factors which affect the thermal conductivity of material.

4. Explain the mechanism of thermal conduction in gases, liquids and solids. Discuss the effect

of temperature on thermal conductivity.

AdxdTkrr

TTkLqq BBBB

io

WBB *)/(*/ln

)(21110

AdxdTkrr

TTkLqq CCCC

io

BBCC *)/(*/ln

)(2 98




Date: Roll No:

Practical No:

THERMAL CONDUCTIVITY OF INSULATING POWDER

Objective: After this experiment student will able to understand how to determine the thermal

conductivity of given insulating powder.

Apparatus: Conductivity Instrument, Insulating Powder

Utility: Electric Supply


The apparatus consists of two thin walled concentric copper spheres. The inner sphere houses the

heating coil. The insulating powder (Asbestos powder-lagging material) is packed between the

two shells. The power supplied to the heating coil is varied by using a dimmerstat and is

measured by a wattmeter or voltmeter and ammeter. Iron-Constantan thermocouple is used to

measure the temperature. Thermocouple couple 1 to 4 are embedded on inner sphere and 5 to 10

are embedded on the outer shell (Figure 1). Position 1 to 10 is as shown on change over switch of

temperature indicator. Under steady state condition the temperature 1 to 10 are noted and also

particular power input reading is recorded. These readings in turn enable to find out the thermal

conductivity of insulating powder packed between two shells. We assume the insulating powder

as an isotropic material and the value of the thermal conductivity to be constant. The apparatus

assume one-dimensional radial heat conduction across the powder layer and thermal conductivity

can be determined as above under steady state condition.

Observations:

1. Radius of the inner copper sphere (ri) = 50 mm

2. Radius of the outer copper sphere (ro) = 100 mm

3. Thermocouples 1 to 4 embedded on inner sphere to measure Ti

4. Thermocouples 5 to 10 embedded on outer sphere to measure To

5. Insulating powder Asbestos Magnesia packed between the two spheres

Procedure:

1. Insert male socket of control panel and connect thermocouple band on temperature indicator

and test set-up in proper position and start the main switch of control panel.

2. Increase slowly the input to heater by the dimmerstat starting from 0 volts position.

3. Adjust input equal to 40 watts maximum by voltmeter and ammeter or watt meter.

4. See that this input remains constant throughout the experiment.

5. Wait till a satisfactory steady state condition is reached. This can be checked by reading

temperatures of thermocouples 1 to 10 & note changes in their readings with time.

6. Note down the readings in the observation table as given below.




Observation Table:

Voltmeter Reading: V

Ammeter Reading: amp.

Heater Input: W

Inner Sphere:

Thermocouple

No.

T1 T2 T3 T4 Mean Temp.(Ti)

(T1 + T2 + T3 + T4)) /

4

Temp (C)

Outer Sphere:

Thermo-

couple No.

T5 T6 T7 T8 T9 T10 Mean Temp. (To)

(T5 + T6 + T7 + T8 + T9 + T10)

6

Temp (C)

Calculations:

Heat transfer through hollow sphere can calculate by following formula

Result:

Conclusion:

rr*K* * 4

)(

)T - (T

0i

oi

io rrQ

cmWrrQ

K oio /)T - (T r*r* * 4

)(*

oiio




Quiz:

1. Which aspect makes the thermal conductivity of insulating materials lower than that of

metals?

2. State the assumptions made with reference to insulating material.

3. Write commonly used insulations for heating and cooling applications.

4. Explain why quilt is better insulator than a woollen blanket of the same thickness.




Date: Roll No:

Practical No:

HEAT TRANSFER IN NATURAL CONVECTION

Objective: After this experiment student will able to understand how to determine the surface

heat transfer coefficient for a vertical tube losing heat by natural convection.

Apparatus: Vertical brass tube with housing.

Utility: Electric Supply.

Theory:

When a hot body is kept in a still atmosphere, heat is transferred to the surrounding fluid by

natural convection.

The fluid layer in contact with the hot body gets heated, rises up due to the decrease in its density

and the cold fluid rushes in to take place. The process continues and the heat transfer takes place

due to the relative motion of hot and cold fluid particles.

The heat transfer coefficient is given by:

h = q / [ AS * ( Ts - Ta ) ]

Where,

h = Average surface heat transfer coefficient (kcal / hr.m2.0C),

AS = Area of heat transfer surface , m2

Ts = Average surface temperature , 0C

Ta = Ambient temperature of the duct . 0C

The surface heat transfer coefficient of a system transferring heat by natural convection depends

on the shape, dimensions and orientation of the fluid and the temperature difference between heat

transferring surface and the fluid. The dependence of h on all the above mentioned parameters is

generally expressed in terms of non-dimensional groups as follows:

Where, hL / k = Nusselt number,

g L3 T / 2 = Grasshof number,

Cp / k = Prandtl number .

A and n are constants depending on the shape and orientation of the heat

transferring surface.

L = Characteristic dimension of the surface, m.

k = Thermal conductivity of the fluid , w/(m 0C)

µ = Dynamic viscosity of fluid, kg/m s .

= Kinematics Viscosity, m2/s

CP = Specific heat of fluid, J/kg 0C

= Coefficient of volumetric expansion of fluid, K-1(0C-1)

g = Acceleration due to gravity , m/s2

n

**2

3

k

C

v

TngLA

k

hL p




T = Ts - Ta, 0C

For gases , = _____1_____ where Tf = Ts + Ta_

Tf + 273 2

For a vertical cylinder losing heat by natural convection, the constants A and n have

been determined and the following empirical correlation obtained ,

McAdams equation:

h L = 0.59 ( Gr * Pr ) 0.25 for 104 < Gr * Pr < 109

k

h L = 0.13 ( Gr * Pr )1/3 for 109 < Gr * Pr < 1012

k

where L = Length of the cylinder.

All the properties of fluid are determined at the mean film temperature Tf .


The apparatus consist of a brass tube fitted in a rectangular duct in a vertical fashion. The duct is

open at the top and bottom and forms an enclosure and serves the purpose of undisturbed

surroundings. One side of the duct is made up of perplex for visualisation. An electric heating

element is kept in the vertical tube which in turn heats the tube surface. The heat is lost from the

tube to the surroundings to the air by natural convection. The temperature of vertical tube is

measured by 7 thermocouple. The heat input to the tube is measured by an ammeter and

voltmeter and is varied by dimmerstat.

The vertical cylinder with the thermocouples, position are shown in Figure 1; while the possible

flow patterns and also expected variation of local heat transfer coefficient shown in figure 2. The

tube surface is polished to minimise the radiation losses.

1. Diameter of the tube (D) 38 mm

2. Length of tube (L) 500 mm

3. Duct size 20 cm * 20 cm * 0.75 m length

4. No. of thermocouples 7 and are shown as 1 to 7

Thermocouple 8 reads the ambient temperature and is kept in the duct.

Procedure:

1. Switch on the supply and adjust the dimmerstat to obtain the required heat input.

2. Wait till the steady state is reached which is confirmed from temperature readings T1 to T7.

3. Measure surface temperatures at the various points at T1 to T7.

4. Note the ambient temperature T8.

5. Repeat the experiment at different heat inputs.

Precautions:

1. Adjust the temperature indicator to ambient level by using compensation screw before

starting the experiment.

2. Keep dimmerstat to zero volt position and increase it slowly.




3. Use the proper range of ammeter and voltmeter.

4. Operate the change over switch of temperature indicator gently from one position to other,

i.e. from 1 to 8 positions.

5. Never exceed 80 watts.

Observation Table: Input to heater

Sr. No: V ( volt ) I ( amp ) W ( watt )

Sr.

No.

T1

( 0C )

T2

( 0C )

T3

( 0C )

T4

( 0C )

T5

( 0C )

T6

( 0C )

T7

( 0C )

T8 = Ta

( 0C )

Calculations:

Experimental Calculations:

The average surface heat transfer coefficient, neglecting end losses using the equation,

h = q / (As [Ts - Ta])

q = V I cos

=______ Watt

As = D L

= m2

Ts = ( T1 + T2 + T3 + T4 + T5 + T6 + T7 ) / 7

= _________ 0C

Theoretical Calculations :

Compare the experimentally obtained value with the prediction of the correlation equation,

McAdams Equation:

Laminar Flow over vertical plates and cylinder:

h* L / K = 0.59 (Gr * Pr) 0.25 for 104 < Gr * Pr < 108

Turbulent Flow over vertical plates and cylinder:

h * L / K = 0.13 (Gr * Pr) 0.33 for 109 < Gr * Pr < 1012

Calculate and plot the variation of local heat transfer coefficient h along the length of the

tube L.




Result:

Sr. No h ( Experimental )

( W/m2.0C )

h ( Theoretical )

( W/m2.0C )

1

2

3

Conclusion:

Quiz:

1. Differentiate between mechanisms of heat transfer by free and forced convection. Mention

some of the areas where these mechanisms are predominant.

2. Give a general equation for the rate of heat transfer by convection, and hence define the

coefficient of heat transfer. List the various factors on which the value of these coefficient

depends.

3. What is film temperature? How does it differ from bulk temperature?

4. Write the correlation that has been suggested for natural convection over a vertical plate or

cylinder in turbulent flow region.




Date: Roll No:

Practical No:

HEAT TRANSFER IN FORCED CONVECTION

Objective: After this experiment student will able to understand how to determine average

surface heat transfer coefficient for pipe losing heat by forced convection.

Apparatus: Forced convection apparatus equipped with blower, manometer.

Utility: Electric Supply.


The apparatus consists of a blower unit fitted with the test pipe. The test section is surrounded by

Nichrome bend heater. 4 Thermocouples are embedded on the test section and 2 thermocouples

are placed in the air stream at the entrance and the exit of the test section to measure air

temperatures. Test pipe is connected to the delivery side of the blower along with the orifice to

measure the flow of air through the pipe. Input to the heater is given through dimmerstat and is

measured by meter. It is also noted that only a part of the total heat supplied is utilised in heating

the air. A temperature indicator with cold junction compensation is provided to measure the

temperature in pipe wall in test section. Air flow is measured with the help of an orifice meter

and the water manometer fitted on the board. Schematic diagram of the setup is shown in the

figure 1 and details of the test pipe are given in figure 2.

Procedure:

1. Start and adjust the flow by means of a valve to some desired difference in the manometer

levels.

2. Start the heating of the test section with the help of dimmerstat and adjust the desired heat

input.

3. Take the readings of all thermocouples at an interval of 10 minutes until the steady state is

reached.

4. Wait for steady state and take readings of all six thermocouples at steady state.

5. Note down the heater input with the help of ammeter and voltmeter.

Precautions:

1. Adjust the temperature indicator to ambient level by using compensation screw before

starting the experiment.

2. Keep dimmerstat to zero volt position and increase it slowly.

3. Use the proper range of ammeter and voltmeter.

4. Do not stop the blower between the testing period.

5. Do not disturb the thermocouples while testing.

6. Operate the changeover switch of temperature indicator gently from one position to other, i.e.

from 1 to 6 positions.

7. Never exceed 200 watts.




Data:

1. Outer Diameter of the tube (D) 34 mm

2. Inner Diameter of the tube (D) 28 mm

3. Length of the test section (L) 500 mm

4. Blower motor 1 H.P.

5. Orifice Diameter 14 mm connected to water manometer

6. Coefficient of discharge 0.64

Observations:

Sr.

No.

T1A T6A T6A -

T1A

= T

Ta AV = (T1A + T6A)/2

T2S T3S T4S T5S TS,,AV

=(T2S+ T3S+

T4S+ T5S) /4

1.

2.

3.

4.

5.

Calculations:

Pressure differences, velocities, and mass flow rate of air,

Va = Velocity of air =

Where,

Cd Coefficient of discharge = 0.64

g 9.81 m / sec2

h Differential pressure expressed in meters of air = (H) * (w,T / air,T)

H Manometer reading, m

w,T Density of water at temperature at temperature T, kg/m3

air,T Density of ait at temperature at temperature T, kg/m3

QA Volume of air flow, m3/ sec

Sr.

No.

H w,T air,T h Va QA = A* Va

(m3/ sec)

m = QA *

air,T

(kg / sec)

1.

2.

3.

4.

5.

sec / m **2 hgCV d




Rate of air heating and heat transfer Coefficient hT at temperature T,

Air heating rate = qA = m * Cp * T , kcal / sec

Where Cp = Sp. heat of air at temperature (T1A + T6A ) / 2

Heat transfer coefficient of wall = h = qA / (Awall [Ts,AV - Ta, AV])

Sr.

No.

m

(kg / sec)

Cp

(kcal / kg C)

T

(C)

qA

kcal / sec

H(practical)

(w/m2 0C)

1.

2.

3.

4.

5.

Theoretical Calculation:

Reynolds Number (Re)

Re = Va * Di / air, TAV , (air, TAV Kinematic Viscosity of air at TAV)

Nusselt Number Nu = (ha, T * Di) / KTAV

Where ha, T =Average heat transfer coefficient of film over the length of the pipe

KTAV = Thermal conductivity of air at TAV = ( T1A + T6A) / 2

Sr.

No.

Di air,

TAV

Re ha, T KTAV

Nu (Re)0.8 (Pr)0.4 Nu,tur

1.

2.

3.

McAdams Equation:

Turbulent flow inside tube:

Nu, TUR = 0.023 * (Re)0.8 * (Pr) 0.4

Result:

Conclusion:




Quiz:

1. Why the heat transfer coefficient for natural convection is much less than that for forced

convection?

2. Why are heat transfer rates higher in turbulent flow inside a tube? Why?

3. Write the correlation that has been suggested for forced convection over a vertical plate or

cylinder in laminar or turbulent flow region.




Date: Roll No:

Practical No:

EXTENDED SURFACE EQUIPMENT

Objective: After this experiment student will able to understand how to study the temperature

profile along with length in case of an extended surface.

Apparatus: Hot Plate with attached rod, dimmerstat, thermocouples.

Utility: Electric supply.

Theory:

When one of the two fluid streams has a much lower heat transfer coefficient than the other,

extended surface heat exchangers are used. In this exchangers , the outside area of the tube is

extended by fins, pegs, disks and other appendages and the outside area in contact with the fluid

thereby made much larger than the inside area .

Ui = ______1_______

1 + _ Ai_

hi Ao ho

This equation shows that if ho is small and hi is large , the value of Ui will be small , but

if A0 is made much larger than Ai , then Ui increases .

The temperature profile of extended surface is given by the following equation:

T - Ta = ( T0 - Ta) cos h [ m ( L : x ) ]-

cos h [ m L ]

_________

m = (h P / KA)

Where,

T0 = Temperature of the base plate.= Ts

Ta = Temperature of the ambient air = T∞

T = Temperature at distance x from base plate ( theoretical )

h = film coefficient of air

K = thermal conductivity of extended surface

L = Height of the extended surface

P = Perimeter of the extended surface

A = Area of cross section of the extended surface


The apparatus consists of a hot plate with a M.S rod attached to the centre of the plate. A

groove is made at the centre of the plate and the M.S rod is joined to it. The temperature of the

hot plate can be varied by the dimmerstat. The temperature at various points of the rod is

measured by thermocouples. One thermocouple measures the ambient temperature of air.




Apparatus Specifications:

Diameter of the rod , D = 25 mm.

Total length of the rod, L = 60 cm.

Distance between the thermocouple = 10 cm.

Procedure:

1. Switch on the power supply and adjust the dimmerstat to obtain the required heat input.

2. Keep the hot plate temperature at 250 0C maximum.

3. Wait until the steady state is reached, which is confirmed from the consistent temperature

readings T1 to T8 .

4. Measure the surface temperature at various points of the extended surfaces, base of the hot

plate and ambient temperature of air by rotating the selector switch.

5. Repeat the experiment at different heat inputs.

Precautions:

1. Keep dimmerstat to zero volt position before switching on the supply and then increase it

slowly.

2. Use proper range of ammeter and voltmeter.

3. Operate the selector switch of the temperature gently from 1 to 8.

4. Take the readings only when steady state is reached.

Data:

Thermal conductivity of M.S rod , K = 40 W / m.0K

Heat transfer coefficient of air , h = 0.4 W/m2. 0K

Observations:

Distance from base x

(cm)

0 10 20 30 40 50 60 Ambient

air

Temperature of rod

( 0C)

T7 =

T0

( base

plate )

T1 T2 T3 T4 T5 T6 T8 =

Ta

Obs.1 ( 150V)

Obs.2 ( 200V )

Obs.3 (250V )




Calculations:

Perimeter of the rod , P = 3.14 * D

= m.

Area of cross-section of rod , A = (3.14 * D2 ) / 4

= m 2.

_________

m = (h P / KA)

= m-1

T - Ta = ( T0 - Ta) cos h [ m ( L - x ) ]

cos h [ m L ]

Where

T0 = Temperature of the base plate.

Ta = Temperature of the ambient air

T = Temperature at distance x from base plate ( theoretical )

Distance

from base

plate

x cms.

Temperature

( experimental ) 0C

Temperature

( Theoretical ) 0C

% Error

(TExp – TTheo)/(TExp)

Obs.1 Obs.2 Obs.3 Obs.1 Obs.2 Obs.3 Obs.1 Obs.2 Obs.3

10

20

30

40

50

60

Graph:

(1) Temperature vs Distance (experimental)

(2) Temperature vs Distance (theoretical)

Result:

Conclusion:




Quiz:

1. Give a few practical and specific examples of use of fin in heat transfer.

2. How does a fins enhance heat transfer at a surface?

3. Mention the most common types of fins and sketch them.

4. Enumerate the various assumptions made in the formation of energy equation for one-

dimensional heat dissipation from an extended surface.

5. Fins are generally made of aluminium? Why?




Date: Roll No:

Practical No:

PARALLEL FLOW HEAT EXCHANGER

Objective: After this experiment student will able to understand how to (1) determine overall

heat transfer coefficient for parallel flow pattern in double pipe heat exchanger (2) temperature

distribution in parallel flow heat exchanger.

Apparatus: Double pipe heat exchanger assembly, Geyser, Thermometers.



The apparatus consists of a tube in tube type (concentric) heat exchanger. The hot fluid (hot

water) obtained from an electric geyser flows through inner tube while the cold fluid (cold water)

flowing through the annulus.

The direction of hot water flow can not be altered while the direction of cold water can be altered

using valves so as to make the heat exchanger to function as parallel or counter flow heat

exchanger s shown in Figure 1.

The experiment is conducted by keeping the identical flow rates while running the unit as a

parallel flow and/or counter flow exchanger.

The temperatures are measured by mercury in glass thermometers and flow rates by graduated

measuring flasks and stop clock. The readings are recorded when the steady state is reached. The

outer tube is provided with adequate thermocol insulation to minimise the heat losses.

Procedure:

1. Place thermometers in the positions and note down their readings when they are at room

temperature and no water is flowing at either side this is required to correct the temperature.

2. Start the flow on hot water side.

3. Start the flow through annulus and run the exchanger as parallel flow unit. Put on the geyser.

4. Adjust the flow rate on hot water side, between the rate of 1.5-4.0 litre/ min.

5. Adjust the flow on cold water side between range of 3.0-8.0 litres/ min.

6. Keeping the flowrate same, wait till the steady state conditions are reached.

7. Record the temperatures on hot water and cold water side and also the flow rates accurately.

8. Repeat the experiment with a counter flow under identical flow conditions.

Observations:

1. Inner Tube Material: Copper I.D. (di) 10 mm O.D. (do ) 12 mm

2. Outer Tube Material: G. Iron I.D. (Di) 27.5 mm O.D. (Do ) 33.5 mm

3. Length of the heat exchanger (L) 2 m




4. Thermal conductivity of copper (k) 30 kcal/h m 0C

5. Thermometers: 0- 50 0C 2 Nos. (for cold water side)

0-100 0C 2 Nos. (for hot water side)

Observation Table:

PARALLEL FLOW RUN:

Hot Water Side Cold Water Side

Sr. No: Flow Rate

mh (kg/ h)

Thi

( 0C)

Tho

( 0C)

Flow Rate

mc (kg/ h)

Tci

( 0C )

Tco

( 0C)

1

2

3

Calculations:

Heat Transfer Rate, is calculated as,

qh = Heat transfer rate from hot water = mh Cph (Thi - Tho)

qc = Heat transfer rate to the cold water = mc Cpc (Tco - Tci)

q = (qh + qc)/2 W

(Assume Cph = Cpc= 1 kcal/kg C)

LMTD - Logarithmic Mean Temperature Difference. The temperature distribution in the two

runs is plotted. LMTD can be calculated as

Overall Heat transfer coefficient can be calculated using,

q= U*A*Tm

U = q/ (A * Tm)

Calculated Ur i based on Ai = * di * L

Uro based on Ao = * do * L

Compare the values of Tm and q in the parallel and counter flow runs.

Note that if experiment is conducted very carefully then the superiority of counter flow

arrangement in terms of higher value of Tm and excess value of q for same flow rates conditions

can be revealed. The value of overall heat transfer coefficient U is more or less same for both the

runs.

oi

oim

TT

TTTLMTD

/ln




The overall heat transfer coefficient value can be predicted by using the force convention heat

transfer correlations for flow through the tube and annulus and can be compared with the

experimentally determined values.

1 / UO = 1 / ho + ( ro ln (ro / rI )/ k ) + ( (ro / ri) (1 / hi) )

Where,

hi heat transfer coefficient for inner tube

ho heat transfer coefficient for outer tube

k thermal conductivity of material (for copper pipe) 30 kcal / hr m C

hi is calculated by using the correlation NuDi = 0.023 (ReDi)0.8 (Pr)0.3

Evaluate properties of water at average bulk mean temperature (Thi + Tho) / 2

ho is calculated by using the correlation Nu (Di - do) = 0.023 (Re (Di - do))0.8 (Pr)0.4

Evaluate Di - do properties at average bulk mean temperatures (Tci + Tco) / 2 for cold water.

Result:

Sr.

No.

Parallel Flow run

Pract. U (W/m2 0C) Theo. U (W/m2 0C)

1

2

3

4

Conclusion:

Quiz:

1. Discuss the importance of heat exchangers for industrial use.

2. What do you mean by overall heat transfer coefficient? What is the significance of overall

heat transfer coefficient?

3. State the condition when the logarithmic temperature difference approximately equals the

arithmetic mean temperature difference.

4. What are the requirement of good heat exchangers?

.




Date: Roll No:

Practical No:

COUNTER FLOW HEAT EXCHANGER

Objective: After this experiment student will able to understand how to (1) determine overall

heat transfer coefficient for counter flow pattern in double pipe heat exchanger

(2) temperature distribution in counter flow heat exchanger.

Apparatus: Double pipe heat exchanger assembly, Geyser, Thermometers.



The apparatus consists of a tube in tube type (concentric) heat exchanger. The hot fluid (hot

water) obtained from an electric geyser flows through inner tube while the cold fluid (cold

water) flowing through the annulus.

The direction of hot water flow cannot be altered while the direction of cold water can be

altered using valves so as to make the heat exchanger to function as parallel or counter flow

heat exchanger s shown in Figure 1.

The experiment is conducted by keeping the identical flow rates while running the unit as a

parallel flow exchanger and/or counter flow exchanger.

The temperatures are measured by mercury in glass thermometers and flow rates by graduated

measuring flasks and stop clock. The readings are recorded when the steady state is reached.

The outer tube is provided with adequate thermocol insulation to minimise the heat losses.

Procedure:

9. Place thermometers in the positions and note down their readings when they are at room

temperature and no water is flowing at either side this is required to correct the

temperature.

10. Start the flow on hot water side.

11. Start the flow through annulus and run the exchanger as parallel flow unit. Put on the

geyser.

12. Adjust the flow rate on hot water side, between the rate of 1.5-4.0 litre/ min.

13. Adjust the flow on cold water side between range of 3.0-8.0 litres/ min.

14. Keeping the flowrate same, wait till the steady state conditions are reached.

15. Record the temperatures on hot water and cold water side and also the flow rates

accurately.

16. Repeat the experiment with a counter flow under identical flow conditions.

Observations:

4. Inner Tube Material: Copper I.D. (di) 10 mm O.D. (do ) 12 mm

5. Outer Tube Material: G. Iron I.D. (Di) 27.5 mm O.D. (Do ) 33.5 mm

6. Length of the heat exchanger (L) 2 m




4. Thermal conductivity of copper (k) 30 kcal/h m 0C

5. Thermometers: 0- 50 0C 2 Nos. (for cold water side)

0-100 0C 2 Nos. (for hot water side)

Observation Table:

PARALLEL FLOW RUN:


Sr. No: Flow Rate

mh (kg/ h)

Thi

( 0C)

Tho

( 0C)

Flow Rate

mc (kg/ h)

Tci

( 0C )

Tco

( 0C)

1

2

3

COUNTER FLOW RUN:


Sr. No: Flow Rate

mh (kg/ h)

Thi

(0C)

Tho

(0C)

Flow Rate

mc (kg/ h)

Tci

(0C)

Tco

(0C)

1

2

3

Calculations:

Heat Transfer Rate, is calculated as,

qh = Heat transfer rate from hot water = mh Cph (Thi - Tho)

qc = Heat transfer rate to the cold water = mc Cpc (Tco - Tci)

q = (qh + qc)/2 W

(Assume Cph = Cpc= 1 kcal/kg C)

LMTD - Logarithmic Mean Temperature Difference. The temperature distribution in the two

runs is plotted. LMTD can be calculated as

Overall Heat transfer coefficient can be calculated using,

q= U*A*Tm

oi

oim

TT

TTTLMTD

/ln




U = q/ (A * Tm)

Calculated Ur i based on Ai = * di * L

Uro based on Ao = * do * L

Compare the values of Tm and q in the parallel and counter flow runs.

Note that if experiment is conducted very carefully then the superiority of counter flow

arrangement in terms of higher value of Tm and excess value of q for same flow rates

conditions can be revealed. The value of overall heat transfer coefficient U is more or less

same for both the runs.

The overall heat transfer coefficient value can be predicted by using the force convention heat

transfer correlations for flow through the tube and annulus and can be compared with the

experimentally determined values.

1 / UO = 1 / ho + ( ro ln (ro / rI )/ k ) + ( (ro / ri) (1 / hi) )

Where,

hi heat transfer coefficient for inner tube

ho heat transfer coefficient for outer tube

k thermal conductivity of material (for copper pipe) 30 kcal / hr m C

hi is calculated by using the correlation NuDi = 0.023 (ReDi)0.8 (Pr)0.3

Evaluate properties of water at average bulk mean temperature (Thi + Tho) / 2

ho is calculated by using the correlation Nu (Di - do) = 0.023 (Re (Di - do))0.8 (Pr)0.4

Evaluate Di - do properties at average bulk mean temperatures (Tci + Tco) / 2 for cold water

Result:

Sr.

No.

Counter Flow run

Pract. U (W/m2 0C) Theo. U (W/m2 0C)

1

2

3

4

Conclusion:

Quiz:

1. Compare parallel flow and counter flow heat exchanger.

2. What is meant by fouling factor? How does it affect the performance of a heat exchanger?

3. What are the selection criteria of heat exchangers?




Date: Roll No:

Practical No: SHELL AND TUBE EXCHANGER

Objective: After this experiment student will able to understand how to determine Overall

Heat Transfer Coefficient for the shell and tube heat exchanger.

Apparatus: 1-2 pass Shell and Tube heat exchanger, Thermometer



Shell

Material M.S.

Inner Diameter. Di 208 mm

Thickness 6 mm

Length 500 mm

25% cut baffle plates at 100 mm apart – 4 numbers

Tubes

Material Copper

Inner Diameter. Di 13 mm

Outer Diameter. D0 16 mm

Length L 500 mm

No. of Tubes N 32

Fluid Water

Hot water from the geysers enters the heat exchanger at the bottom into the header. The

header has a partition at the centre. The hot water flows through the bottom bank of tubes and

comes to the other end, changes direction and passes through the top bank of tubes and leaves

the chamber at top. The cooled water enters at the bottom of shell and after passing through

the five compartments made by four baffle plates leaves the shell at top.

Procedure:

Adjust the flow rate such that there is applicable temperature difference between the outlet

and inlet temps. of two fluid. Put on the geyser and set up pre determined hot water temp.

After steady state read four thermometer readings and measure the hot and cold flow rates.

Observation:

1. Heat transfer area of tubes and baffle plates = 1m2

2. Correction factor for heat exchanger (F) = ____

3. Inlet temperature of hot water (thi) = ____ C

4. Cold water flow rate for 1 Lit. = ____ Kg/s

5. Inlet temperature for cold water = ____C




6. Outlet temperature for cold water(tco) = ____C

7. Hot water flow rate for 1 Lit. = ____ Kg/s

8. Outlet temperature for hot water (tho) = ____C

Observation Table:


Sr.

No:

Flow Rate

mh (kg/ hr)

Thi

( 0C)

Tho

(0C)

Flow Rate

mc (kg/ hr)

tci

( 0C)

tco

( 0C)

1

2

3

Sample Calculations:

For Shell and Tube heat exchanger,

(LMTD) 1-2 pass shell & tube = F *(LMTD) Counter Flow

where, LMTD = Log mean temperature difference

F correction factor (From LMTD correction factor chart)

*(LMTD) Counter Flow = [ ( Thi – tco ) – ( Tho – t ci ) ] / ln [ ( Thi – tco ) / ( Tho – t ci ) ]

= _______ 0 C

Now q is given by:

q = mc Cp (tco - tci ) for cold water.

where Cp =Specific heat of water =4.1868 kJ/kg-k

Now for Overall Heat Transfer coefficient ‘U’

q = U*A*F (LMTD) Counter Flow.

U = q / A* F *(LMTD)Counter flow

= ______ KW/m2 C

Result:

Conclusion:

Quiz:

1. Sketch a shell and tube type heat exchanger.

2. Why baffles are used in shell and tube type heat exchanger

3. Sketch a two shell pass, four tube pass, reversed current heat exchanger. Label the

different part.




Date: Roll No:

Practical No:

FINNED TUBE HEAT EXCHANGER

Objective: After this experiment student will able to understand how to study and compare

temperature distribution, heat transfer rate, overall heat transfer coefficient of the finned tube

heat exchanger.

Apparatus: Stop clock, measuring flask, thermometers.

Utility: Water, electricity supply.

Theory:

Finned tube heat exchangers are used when there is a huge difference between the values of

the heat transfer coefficients of the fluids involved e.g. heat transfer between air and water. So

to increase the total transmission of heat, extended surfaces like fins are added. The fins are

of many types like longitudinal, transverse, pegs or studs, spines etc.

Apparatus description:

The apparatus consists of a concentric tube heat exchanger. The hot fluid i.e. Hot water is

obtained from an electric geyser and it flows through the inner finned tube, in both directions.

The direction can be changed by operating different valves. The cold air is admitted through

one end. Temperatures of the fluids can be measured using stop and measuring flask. The

outer tube is provided with adequate asbestos rope insulation to minimise the heat loss to the

surrounding.

Apparatus Specification:

Length of the heat exchanger = 1m

Inner copper finned tube I.D = 0.016 m, O.D = 0.02 m with longitudinal fin 0.022m height

and 1m. long.

Total number of fins = 8

Outer tube, I.D = 0.064 m, O.D = 0.075 m

Fin thickness = 0.00166 m

Annulus flow area = 1.288 m2

Perimeter of the fin = 0.4016 m

Procedure:

Keep the thermometers in position.

Start the hot water flow. Start the flow of cold air.

Put on the electric geyser.

Adjust the flow rate on hot water side (approx.: 500 ml/min.) .Keep the flow rate same till

the steady state condition is reached.

Note the temperatures and measure the flow rate.

Repeat the experiment for different flow rates.




Observation:

Sr. No Type

of flow

Hot Water Air LMTD 0C Mass

Flow

(kg/s)

Mh

Inlet

temp 0C

Thi

Outlet

temp 0C

Tho

Mass

Flow

(kg/s)

mc

Inlet

temp 0C

Tci

Outlet

temp 0C

Tco

Calculation:

Heat transfer from hot water :

Qh = mh * Ch * ( Thi - Tho )

Heat transfer from cold air :

Qc = mc * Cc* ( Tco – Tci )

Where,

Ch = specific heat of hot water , kJ/kg

Cc = specific heat of cold air , kJ/kg

mh = mass flow rate hot water , kg/hr

mc = mass flow rate cold air , kg/hr

LMTD = 0C

Area of heat transfer (based on outer diameter)

A = ( 8 0.022 1 ) + ( 0.02 1) + 0.025

Overall heat transfer coefficient U = Q / ( A * LMTD )

= W/m2.0C

Effectiveness = mc * Cc* ( Tco – Tci _) =

MH * CH * ( THI - THO )

Result:

Sr. No Heat Flow ( Watts ) Heat transfer

coefficient

( W/ m2.0C )

Effectiveness




Quiz:

1. What are the advantages of finned tube heat exchanger?

2. Give applications of finned tube heat exchanger.

3. Define effectiveness and number of transfer units in the context of heat exchanges.




Date: Roll No:

Practical No: EMISSIVITY MEASUREMENT APPARATUS

Objective: After this experiment student will able to understand how to determine emissivity

of test plate.

Apparatus: Black plate, Test plate (aluminium)

Theory:

All the bodies emit and absorb the thermal radiation to and from surroundings. The rate of

thermal radiation depends upon the temperature of body. Thermal radiation is electromagnetic

wave and does not require any material medium for propagation.

When thermal radiation strikes a body, part of it is reflected, part of it is absorbed and part of

it is transmitted through body.

The fraction of incident energy, reflected by the surface is called reflectivity ().

The fraction of incident energy, absorbed by the surface is called absorptivity ().

The fraction of incident energy transmitted through body is called transmissivity ( ).

The surface which absorbs all the incident radiation is called a black surface.

FOR A BLACK SURFACE (=1)

The radiant flux, emitted from the surface is called emissive power (e).

The emissivity of a surface is ratio of emissive power of a surface to that of black surface at

the same temperature.

Construction of apparatus:

The apparatus uses comparator method for determining the emissivity of test plate. It consists

of two aluminium plates, of equal physical dimensions. Mica heaters are provided inside the

plates. The plates are mounted in an enclosure to provide undisturbed surroundings.

One of the plates is blackened outside for use as a compurgator because black surface has

=1. Another plate is having natural surface finish. Input to heaters can be controlled by

separate dimmerstats. Heater input is measured on common ammeter and voltmeter. One

thermocouple is fitted on surface of each plate to measure the surface temperature with digital

temperature indicator. By adjusting input to the heaters, both the plates are brought to the

same temperature, so that conduction and convection losses from both the plates are equal

and difference in input is due to different emissivity.

Holes are provided at backside bottom and at the top of enclosure for natural circulation of air

over the plates.

The plate enclosure is provided with Perspex acrylic sheet at the front. Thus, =e/eb




Procedure:

1. Blacken one of the plates with the help of lamp black (normally this is blackened at the

works, but if blacking is wiped out, then blackening is necessary)

2. Keep both the dimmer Knobs at ZERO position.

3. Insert the supply pin-top in the socket (which is properly earthen) and switch ON the main

supply.

4. Switch ON the main switch on the panel.

5. Switch the meter selector switch (toggle switch) in downward position.

6. Adjust dimmer of black plate, so that around 110-120 volts are supplied to black plate.

7. Now, switch the meter selector switch in upward position.

8. Adjust test plate voltage slightly less than that of black plate (says 100- 110 volts)

9. Check the temperature (after, say 10 minutes) and adjust the dimmers so that temperature

of both the plates are equal and steady. Normally, very minor adjustments are required for

this.

10. Note down the readings after the plate temperatures reach steady state.

Precautions:

1. Black plate should be perfectly blackened

2. Never put your hand or papers over the holes provided at the top of enclosure.

3. Keep at least 200 mm distance between the back side of unit and wall

Operate all the switches sand

Observations:

Enclosure temp. - T3 =__________ C

Calculation:

1. Enclosure temp.

TD = TE= ----------- C

=( T3 + 273.15 ) K

2. Flate surface temp.

T1 = T2 = C

Ts=T3 = (T1 + 273.15) K

3. Heat input to black plate

WB = V * I watts

4. Heat input to test plate

WT = V * I watts

Test plate Black plate

Input v*I

Surface

temperature

T1 T2




5. Surface area of plates

A = 2* /4{D2 + (.D.t)}

=0.0447 m2

Where,

D= dia. of plates=0.16 m

And t= thickness of plates = 0.009 m

6. For black plate

WB = WCVB+WCDB+WRB -------------------(I)

Where,

WCVB = Convection losses

WCDB= Conduction losses

WRB = Radiation losses

Similarly, for test plate,

WT=WCVT+WCDB+WRT ----------------(ii)

As both plates are of same physical dimensions, same material and at same temperature.

WCVB = WCVT and WCDB = WCDT

Subtracting equation (ii) from (i), we get

WB _- WT = WRB - WRT

= [ A B(TS4 - TD

4)]- [ A T(TS4 - TD

4)]

= A (TS4 - TD

4)( B-T)

As emissivity of black plate is 1.

WB-WT = A (TS4 - TD

4) (1-T)

Where,

T =Emissivity of test plate

= Stefan Boltzmann constant

= 5.667*10-8 W/m2 K4

Result:

The emissivity of test plate was found to be ___________at the temperature of __________

K Conclusion:

Quiz:

1. Explain the mechanism of radiation heat transfer

2. Differentiate between the radiation heat transfer and conduction / convection heat transfer

combined.

3. Define a black body. Give examples of some surfaces which do not appear black but have

high value of absorptivities.

4. Two pieces of wood are placed in sunlight; one piece is painted white and the other black.

Which piece will absorb more heat?




Date: Roll No:

Practical No: DROP WISE AND FILM WISE CONDENSATION APPARATUS

Objective: After this experiment student will able to understand how to study the dropwise

and filmwise condensation phenomena.

Apparatus: Condensers, main unit, heating elements, temperature indicator, rota meter,

pressure gauge


Condensers:

One chromium plated for dropwise condensation and one natural finish for filmwise

condensation otherwise identical in construction.

Dimensions: 15 cm dia.*12 cm length

Surface area: cm2

Fabricated from copper with reverse flow in concentric tubes Fitted with thermocouple for

surface temp. Measurements

Main unit:

M.S. fabricated construction comprising test section and steam generation section. Test

section provided with glass converse on front and rear sides of test section for visualisation of

the process.

Heating elements:

1.5KW water heater operated through 30 amps D.P.switch

Procedure:

1. Fill up the water in the main unit through bottom valve upto the sight glass fitted on the

unit.

2. After filling the water close the valve. Start the water flow rate through one of the

condensers which is to be tested and note water flow rate in the rotameter. Ensure that

during measurement water is flowing only through the condenser under test and second

valve is closed.

3. Connect supply socket to the mains and switch on the heater.

4. Slowly steam generation will start in the bottom portion of the unit and as the steam rises

to test section and get condensed on the tube and fells back in the trough.

5. Depending upon the type of condenser under test dropwise or filmwise condensation can

be visualised.

6. If water flow rate is low than steam pressure in chamber will rise and pressure gauge will

read the pressure.

7. If the water flow rate is matched than condensation will occur at more or less at

atmospheric pressure.

8. Process of dropwise and filmwise condensation can be easily viewed through the front

glass window of main unit.




9. Observation like temperatures, water flow rates, pressures are noted down in the

observation table at the end of the each set.

Observation Table:

Filmwise Condensation:

Sr.

No

Steam

Pressure

Water

Rate

Temperature

Plated

Conden.

Plain

Conden.

Steam Water

in

plated

Water

out

plated

Water

in

plain

Water

out

plain

Dropwise Condensation:

Sr.

No

Steam

Pressure

Water

Rate

Temperature

Plated Plain Steam Water

in

plated

Water

out

plated

Water

in

plain

Water

out

plain

Calculations:

Normally steam will not be pressurised, but the pressure gauge reads some pressure than

properties of steam should be taken at that pressure or otherwise atmospheric pressure will be

taken. First calculate the heat transfer coefficient inside the condenser under test. For this

properties of water are taken at bulk mean temperature of water i.e. (Twi + Two ) / 2.

Following properties are required:

Density of water 1 kg / m3

Kinematics Viscosity 1 m2/ sec

Thermal Conductivity K1 kcal / hr m C

Prandtl Number Pr

Reynolds Number ReD = 4 * mw / D1

where D1 Inner Diameter of Condenser = 2.5 cm

If this value of ReD > 2100 then flow is turbulent, below this value flow is laminar.

Normally flow will be turbulent in the tube.

Nusselt Number NuD = 0.023 (ReD)0.4 (Pr)0.4




hi = NuD * K / D1 kcal / hr m2 C Inside heat transfer coefficient

Calculate heat transfer coefficient on outer surface of the condenser ho.

For this properties of water are taken at bulk mean temperature of condensate (Ts + Tw ) / 2.

Where, Ts Temperature of steam, Tw Temperature of condenser wall

Density of water 2 kg / m3

Kinemtic Viscosity 2 m2/ sec

Thermal Conductivity K2 kcal / hr m C

Heat of evaporation ho kcal / kg

ho = 0.725 * 2 * g * K23 / ( Ts - Tw) * Do , Do Outside diameter of condenser = 2.75 cm

From these values overall heat transfer coefficient (U) can be calculated,

1/ U = 1 / hi + (Di / Do) (1 / ho) kcal / hr m2 C

same procedure can be repeated for next condenser.

Except for some exceptional cases overall heat transfer coefficient for dropwise condensation

will be higher than that of filmwise condensation. Results may vary from theory to some

degree due to unavoidable heat losses from front and rear glass windows and through walls.

Precautions:

Do not start heater supply unless water is filled in the test unit.

Operate gently the selector switch of temperature indicator to read various temperatures.

Result:

Conclusion:

Quiz:

1. What is condensation and when does it occur?

2. The rate of heat transfer in drop-wise condensation is many times larger than that in film

condensation. Why?

3. How does film-wise condensation differ from drop wise condensation? Which type has a

higher heat transfer coefficient and point out the reason thereof?

4. In the design of condensers, which of the two types of condensation is usually selected and

why?

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