Final Heat Power Lab Manual14

7/31/2019 Final Heat Power Lab Manual14

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THE KAVERY ENGINEERING COLLEGE

A Observation Manual

On

Heat power Laboratory

VI SEMESTER

ANNA UNIVERSITY - COIMBATORE

DEPARTMENT OF MECHANICAL ENGINEERING,THE KAVERY ENGINEERING COLLEGE,

M.Kallipatty, Mecheri 636 453


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SYLLABUS


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THERMAL CONDUCTIVITY APPARATUS

Aim

To find the thermal conductivity of the specimen by two specimens by two slabs guarded

hot plate method.

Description

The apparatus consisting of a guarded hot plate and cold plate. A specimen whose thermal

conductivity is to be measured is sandwiched between the hot and cold plate. Both hotplate and

guard heaters are heated by electrical heaters. A small trough is attached to the cold plate to hold

coolant water circulation. A similar arrangement is made on the other side of the heater as shown

in the figure. Thermocouples are attached to measure temperature in between the hot plate and

specimen plate, also cold plate and the specimen plate.

A multipoint digital temperature indicator with selector switch is provided to note the

temperatures at different locations. A regulator is provided to control the input energy to the main

heater and guard heater. An ammeter and voltmeter are provided to note and vary the input energy

to the heater. Both heaters are electrically connected in parallel; however it is wired in such a way

the voltmeter and ammeter readings indicated only the power to the main heater. The whole

assembly is kept in an enclosure with heat insulating material filled all around to minimize the heat

loss.

Formula used

The power input Q = V*A/2 Watts, since the heat flows on either side of the heater.

dt = (T1+T2+T3+T4 /4) (T5+T6/2)

Since Q = K*A dT/dx, the value of K may calculated

K= thermal conductivity of the specimen W/m2K

Procedure

Connect the power supply to the unit. Turn the regulator knob clockwise and the power to

the main heater to any desired value.

Allow water through the cold plate at a steady rate. Note the temperature at different

locations when the unit reaches steady state.


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THERMALCONDUCTIVITY APPARATUS

Sl.No

VoltmeterReadings

V

Volts

AmmeterReadings

I

Amps

Temperature Readings

T1oc T2

oc T3oc T4

oc T5oc T6

oc T7oc T8

oc


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For different power inputs in ascending order only the experiment may by repeated and

readings are tabulated as below.

The guard heater enables the heat flow in uni-direction as the temperature of this almost

equal to the main heater temperature.

Thermocouple 1, 2, 3&4 are connected to the heater.

Thermocouple 5&4 are connected to the interface of specimen and cold plat.

Thermocouple 7&8 are connected to the guard heater.


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Model calculation

The thickness of the specimen (dx) = 0.006 m

The effective heat transfer area diameter = 0.125 m

Area = 3.14/4*(0.125*0.125)

= 0.01227m2

The power input Q = V*A/2 Watts, since the heat flows on either side of the heater.

K= thermal conductivity of the specimen

dt = (T1+T2+T3+T4 /4) (T5+T6/2)

Hence Q = K*A dT/dx or K = Qdx/Adt =


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Result

Thermal conductivity of metal rod found out to be


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HEAT TRANSFER THROUGH THE LAGGED PIPE

Aim

To determine the thermal conductivity of saw dust by using lagged pipe apparatus.

Description

The insulation is defined as a material, which retards the heat flow with reasonable

effectiveness. Heat is transferred through insulation by conduction, convection, radiation or by the

combination of these three. There is no insulation that is 100% effective to prevent the flow of heat

under temperature gradient.

The experiment set up in which the heat is transferred through insulation by conduction is

under study in the given apparatus.

The apparatus consisting of a rod heater with asbestos lagging. The assembly is inside an

MS pipe. Between the asbestos lagging and MS pipe saw dust is filled.

The set up as shown in the figure. Let r1 be the radius of the heater, r2 be the radius of the

heater with asbestos lagging and r3 be the inner radius of the outer MS pipe.

Formula used

The heat flow through lagging material is given by

Q = K1 2L t / ln (r2 / r1) Or K2 2L t / ln (r3 / r2)

Where t is the temperature across the lagging.

K1 = Thermal Conductivity of in asbestos lagging W / m2 K

K2 = Thermal Conductivity of saw dust in W / m2 K

L = Length of the cylinder

r1 = Radius of heater

r2 = Radius of asbestos lagging

r3 = Radius of saw dust


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LAGGED PIPE

S

L.

N

O

Vol

ts

Cur

rent

Heater

temperatur

e oc

T1 T2 T3

Avg.

temperat

ure oc

Asbestos

temperature

oc

T4 T5 T6

Avg.

temperatu

re oc

Saw dust

temperat

ure oc

T7 T8

Avg.

tempera

ture oc

Knowing the thermal conductivity of one lagging material the thermal conductivity of the other

insulating material can be found


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Procedure

1. Switch on the unit and check if all channels of temperature indicator showing proper

temperature.

2. Switch on the heater using the regulator and keep the power input at some particular value.

3. Allow the unit to stabilize for about 20 to 30 minutes.

4. Now note down the ammeter, voltmeter reading which gives the heat input. Temperatures

1, 2 and 3 are the temperature of heater rod 4, 5 and 6 are the temperature on the asbestos

layer. 7 and 8 are temperatures on the saw dust lagging. The average temperature of each

cylinder is taken for calculation.

5. The temperatures are measured by thermocouple (Fe/Ko) with multipoint digital

temperature indicator.

6. The experiment may be repeated for different heat inputs.

Observation


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L = Length of the cylinder = 20mm

r1 = Radius of heater = 10mm

r2 = Radius of asbestos lagging = 0.5 m

r3 = Radius of saw dust = 40mm

Average temperature of heater = T1 + T2 + T3/3

Average temperature of Asbestos lagging = T4 + T5 + T6/3

Average temperature of sawdust lagging = T7 + T8 + /2

The heat flow from the heater to outer surface of Asbestos lagging q = (K1*2**L (t) / ln(r2/r1))

Watts

Where

K1 = Thermal Conductivity of in asbestos lagging W / m2 K from data book @ 548oc = 0.1105 W /

m2 K

Substituting the values, q = Watts

Substituting the values, q to find thermal conductivity of sawdust, q = (K2*2**L (t) / ln(r3/r2))

Watts

K2 =


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Result

Thus the thermal conductivity of saw dust can be determined in the lagged pipe apparatus.


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HEAT TRANSFER IN NATURAL CONVECTION


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Aim

To determine the heat transfer coefficient for a given apparatus.

Apparatus required

1. Heated vertical rod

2. Vertical duct

Introduction

Convection is a mode of heat transfer where by a moving fluid transfer heat from a surface.

When the fluid movement is caused by density difference in the fluid due to temperature

variations, it is called FREE orNATURAL CONVECTION.

This apparatus provides students with a sound introduction to the features of freeconvection heat transfer from a heated vertical rod. A vertical duct is fitted with a heated vertically

placed cylinder. Around this cylinder air gets heated and becomes less dense, causing it to rise.

This in turn gives rise to continuous flow of air upward in the duct. The instrumentation provided

gives the heat input and the temperature at different points on the heated cylinder.

Formula used

The power input to heater = V*I = h*A*t

Bulk mean temperature

Where A = Area of heat transfer = *d*l m2

d = diameter of heater rod = 40mm

l = length of the heater rod = 500mm

Average temp. of heater = T2+T3+T4+T5


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NATURAL CONVECTION APPARATUS

Sl.No Voltmeter

Readings

V

Volts

Ammeter

Readings

I

Amps

Temperature Along The Heated Cylinder

T1oc T2

oc T3oc T4

oc T5oc T6

oc


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Average temp. of air = T1+T6

t = Average temp. of heater rod - Average temp. of air = oc

h*A*t = Watts

h = overall heat transfer coefficient = power / (A*t)

Procedure

1. Switch On The Unit And Adjust The Regulator To Provide Suitable Power Input.

2. Allow some time for the unit to reach steady state condition.

3. Note the temperature of inlet air, outlet air and temperature along the heater rod.

4. Note ammeter and voltmeter readings.

5. For different power inputs the experiments may be repeated.


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Result

Heat coefficient for a vertical tube losing heat by natural convection is found out to be


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HEAT TRANSFER IN FORCED CONVECTION

Aim

To determine the heat transfer coefficient in forced convection of air in a tube.


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Introduction

The important relationship between Reynolds number, prandle number and nusselt number

in heat exchanger design may be investigated in this self contained unit.

The experiment

The experimental setup consists of a tube through which air is sent in by a blower. The test

section consists of a long electrical surface heater on the tube, which serves as a constant heat flux

source on the flowing medium. The inlet and outlet temperature of the flowing air are measured by

thermocouple and also the temperatures at several locations along the surface heater from which an

average temperature can be obtained. An orifice meter in the tube is used to measure the airflow

rate with a U tube water manometer.

An ammeter and a voltmeter are provided to measure the power input to the heater.

A power regulator is provided to vary the power input to heater.

A multi point digital temperature indicator is provided to measure the above thermocouples input.

A regulator is provided to vary the speed of the blower to regulate the flow rate of air.

Formula used

The heat input Q = h*A*LMTD = m*Cp* (Temp.of tube Temp.of air)

Cp = specific heat of air= 1.005 KJ / oK Kg

ma = mass flow rate of air in Kg / sec

h = Heat Transfer Coefficient in W / m2k

LMTD = ((Avg.temp.of tube outlet air temp) (Avg.temp.of tube inlet air temp) / ((1n

(Avg.temp.of tube outlet air temp) / (Avg.temp.of tube inlet air temp))

A = area of heat transfer = *d*L


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Where

d = Inner diameter of the tube = 40mm

L = Length of the tube = 500mm

From the above, the heat transfer coefficient h can be calculated

Calculate the velocity of air in the tube using orifice meter/water manometer

The volume of air flowing through the tube Q = (cd*a1*a2 2*g*h0 / (a12 a22) m3/sec

h0 = head of air causing the flow = (h1-h2)*(w / a)

Where

w = 1000 kg/ m3

a = 1.16 kg/ m3

h1 and h2 are manometer reading in meters

a1 = area of the tube

a2 = area of the orifice

Heat transfer rate and flow rates are expressed in dimension less form of Nusselt Number and

Reynoldss Number

Re = Reynoldss Number

V Di

Re = ------------

u

Where

u = Kinematic Viscosity at bulk mean

The heat transfer coefficient can also calculated from Dittus - Boelter correlation

Nu = Nusselt Number

Nu = 0.023 x Re0.8 x Pr0.3

Where pr is the prandtl number for which air can be taken as from data book 0.696

0.688

0.684

0.681


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Average temperature of heater = (T2 + T3 +T4 + T5)/4 =

Average temperature of Asbestos lagging = (T1 + T6)/2 =

d1 = diameter of the pipe = 40mm

d2 = diameter of the orifice = 20mm

The volume of air Q = (Cd*a1*a2 2*g*h0) / ( a1*a2) m3/sec

Where

Cd =

a1 = (/4)*d12 = m2

a2 = (/4)*d22 = m2

h = (h1-h2)*(w / a) meters

= meters

Q = m3/sec

Velocity of air flow = Q/a1 = m/sec

Reynoldss Number

V Di

Re = ------------

u


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Using Forced Convection Correlation

Nu = 0.023 x Re0.8 x Pr0.3 kinematic viscosity from data book = 1.9E-05

2.3E-05

3.7E-05

4.2E-05

H = Nu*K/D m/sec

K = thermal conductivity of air at mean temp from data book0.02876

0.0321

0.03430.0374

Therefore, Heat Transfer Coefficienth = m/sec

Procedure

1. Switch on the mains.

2. Switch on the blower.

3. Adjust the regulator to any desired power input to heater.

4. Adjust the blower speed regulator to any desired flow rate of air.

5. Wait till steady state temperature is reached.

6. Note manometer reading h1 and h2.

7. Note temperature along the tube. Note air inlet and outlet temperature.

8. Note voltmeter and ammeter reading.

9. Vary the flow rate of air and repeat the experiment.

10. For various air flow rates and for various power inputs the readings may be taken to repeat

the experiments.


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Result

The heat transfer coefficient in forced convection, of air in a tube found to be h =


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HEAT TRANSFER FROM A PIN-FIN APPARATUS


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Aim

To determine the temperature distribution of the PIN-FIN for forced convection and to find

the FIN efficiency.

Theory

Consider a PIN-FIN having the shape of rod whose is attached to a wall at a surface

temperature Ts, the fin is cooled along the axis by a fluid at temperature Tamb. The fin has a uniform

cross sectional area Ao is made of material having a uniform thermal conductivity K and the

average heat transfer co-efficient between the surfaces to the fluid. We shall assume that transverse

temperature gradients are so small so that the temperature at any cross section of the fin is uniform.

Description

The apparatus consists of a PIN-FIN placed inside an open duct, (one side open) the outer

end of the duct is connected to the suction side of a blower; the delivery side of a blower is taken

up through an orifice meter to the atmosphere. The airflow rate can be varied by the blower speed

regulator and can be measured on the U tube manometer connected to the orifice meter. A heater is

connected to one end of the PIN-FIN and seven thermocouples are connected by equal distance all

along the length of the pin and the eighth thermocouple is left in the duct. The panel of the

apparatus consists of voltmeter, ammeter and digital temperature indicator. Regulator is to control

the power input to the heater. U tube manometer with connecting hoses.

Formula used

Volume of air flowing through the duct

Vo = (cd*a1*a2 2*g*h0 / (a12 a22) m3/sec

Where cd = coefficient of orifice = 0.6

g = gravitational constant = 9.81 m/sec2

h0 = heat of air (w/a)h

a1 = area of the pipe

a2 = area of the orifice = 12


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PIN-FIN APPARATUS

S

Sl.

No

Votmeter

Readings

VVo

lts

Ammeter

Readings

IAm

ps

PIN-FIN surface Temperature

T1o

c T2 T3o

c T4o

c T5o

c T6o

c T7o

c

Amb.t

emp

T8

oc

Man

ometer

readingsh

1

c

m

h

2

c

m

Model calculation

Volume of air flowing through the duct

Vo = (cd*a1*a2 2*g*h0 / (a12 a22) m3/sec

Where cd = coefficient of orifice = 0.6

g = gravitational constant = 9.81 m/sec2

h0 = heat of air (w/a)h = 68.96552 m


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h = manometer differential head

Velocity of air in the duct = Vo/ (W*B)

Where W = width of the duct

B = breadth of the duct

REYNOLDS NUMBER OF AIR FLOW

Reynolds number re = (L*Va*a)/ a

Where Va = velocity of air in the duct

a = density of air in the duct

a = viscosity of air at Toc

PRANDTL NUMBER OF AIR FLOW

prandtl number = (cpa* a)/ Ka

where

cpa = specific heat of air

a = viscocity of air

Ka = thermal conductivity of air

Heat transfer coefficient h =Nmu*(Ka/L)

Ka = thermal conductivity of air

L = length of fin

Efficiency of the PIN-FIN

Efficiency of the PIN-FIN = tan h (ML) / (ML)

h = heat of the pin transfer coefficient

L = length of the pin

Temperature distribution = Tx = (cos M (L-X) / cos (ML))* (To-Ta) + Ta

Fin length L = 14.5

Fin diameter Df


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a1 = area of the pipe = (/4)*d12 = 0.001257 m2

a2 = area of the orifice = (/4)*d12 = 0.000314 m2

h = manometer differential head = 68.96552 m

w = 1000 kg/m3

a = 1.16 kg/m3

Diameter of pipe d1 = 40mm

Diameter of orifice d2 = 20mm

Mean temp = T1+T2+T3+T4+T5+ T6 + T7 + T8 /8

x = distance between thermocouple and heater

distance between thermocouple = 20mm


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EVALUATION OF THE HEAT TRANSFER COEFFICIENT (H)

Nnav = average Nusselt number

= (hD)/K

D = diameter of fin

K = thermal conductivity of air

Procedure

Connect the three pin plug to a 230V, 50Hz, 15A power and switch on the unit.

Keep the thermocouple selector switch in first position.

Turn the thermocouple knob to clockwise and set the power to the heater to any desired

value by looking at the voltmeter and ammeter.

Allow the unit to stabilize

Switch on the blower, and regulate the speed for any desired flow rate of air.

Set the airflow rate to any desired value looking at the difference in U tube manometer

limb levels.

Note down the temperature indicated by temperature indicator

Repeat the experiment by varying the airflow rate and keeping the power input to the heater

constant.

Varying the power input to the heater and keeping the airflow rate constant.

Tabulate the readings and calculate for difference conditions.

After all the experiment is over, put off the blower switch, turn the energy.

Regulator knobs anti clockwise put off the main switches and disconnects the power

supply.

Specification

Duct width b = 150mm

Duct height w = 100mm

Orifice diameter d0 = 20mm

Orifice coefficient cd = 0.6


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Result

The fin efficiency of the PIN-FIN apparatus by forced convection is


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STEFAN BOLTZMANN APPARATUS


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Aim

To find the Stefan Boltzmann constant in the Stefan Boltzmann apparatus.

Apparatus required

1. Hemispherical radiation

2. Hot water source

3. Control valve

4. Small disk

5. Multipoint temperature indicator

6. Thermocouple

Theory

Stefan - Boltzmann law which establishes the dependence of integral hemispherical

radiation on temperature may be verified in this unit. The experimental set up consisting of

concentric hemispheres with provision for the hot water to pass through the annulus. A hot water

source is provided. The water flow may be varied using the control valve provided, thereby to

control the hot water temperature. A small disk is placed at the bottom of the hemisphere which

receives the heat radiation and can be removed (or) refitted while conducting the experiment. A

multipoint digital temperature indicator and thermocouples (Fe/Ko) are provided to measure

temperature at various points on the radiating surface of the hemisphere and on the discharge.

Formula used

Stefan Boltzmann constant () = Q/b * (Th4-Td4)*A

Where,

Q = Mass of the disk

Th = Avg. Temp of Hemisphere

Td = Steady. Temp of disk

A = Area of disk = (/4)*d2


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STEFAN BOLTZMANN APPARATUS

Sl.no Hemisphere temp.

T1oc T2

oc T3oc

Avg.temp of

the

hemisphere

Thoc

T4oc

Time in sec Steady state

of the disc

Tdoc

Mass of the disc = 0.

Procedure


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Allow water to flow through the hemisphere. Remove the disc from the bottom of the

hemisphere.

Switch on the heater and allow the hemisphere to reach a steady temperature.

Note down the temperatures T1, T2 and T3. The average of these temperatures is the

hemisphere temperatures (Th).

Refit the disc at the bottom of the hemisphere and start the stop clock. The raise in

temperature T4 with respect to time is noted.

Also note down the disc temperature at T4 when steady state is reached (Td). A is the area

of the disc receiving the heat radiation.

Model calculation


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Q = mass of the disc*cp of the disc* dT/dt Diameter of the disc = 0.02kg

Q = material of the disc = copper m

cp = 0.091

cp = 381 Kcal/kgoc

Th = (T1+T2+T3)/3

=

Td =

=

Area of the disc = (/4)*d2 = 0.000314159 m2

A =

= Q/b * (Th4-Td4)*A

hence,

Stefan Boltzmann constant =

=


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Result

Thus the Stefan Boltzmann constant was determined.

The Stefan Boltzmann constant =


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TEST ON EMISSIVITY APPARATUS


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Aim

To measure the emissivity of the test plate surface.

Theory

An ideally black surface is one, which absorbs the radiation falling in it. Its reflectivity and

transivity is zero. The radiation is emitted per unit time per unit area from the surface of the body

is called emissive power.

The emissive power of a body to the emissive power of black body at the same temperature

is known as emmissivity of the body. For a block body absorptivity is 1 and by kirchoffs law its

emmissivity is also 1.emissivity depends on the surface temperature and the nature of the surface.

Description

The experimental set up consists of two circular aluminium plates identical in size and is

provided with heating coils at the bottom.

The plates are mounted on thick asbestos sheet and kept in an enclosure so as to provide

undisturbed natural convection surroundings. The heat input to the heater is varied by a regulator

and is measured by an ammeter and voltmeter. Since the heaters are electrically connected I

parallel the same power input is given to both heaters any time. The temperatures of the plates are

measured by Ir/Con thermocouples. Each plate is having three thermocouples.Hence an average temperature may be taken. One thermocouple kept in the enclosure to

read the chamber temperature. One plate is blackened by layer of enamel black paint to form the

idealized black surface where as the other plate is the test plate. The heat dissipation by condition

is same in both cases.

Apparatus Required

1. Two Circular Aluminium Plates

2. Thick Asbestos sheet3. Ir/Con thermocouples.


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EMMISSIVITY APPARATUS

Voltag

e

(V)

Volts

Curr

ent

(A)

Amp

s

Black body

temperature oc

T5 T6 T7

Average

tempera

ture oc

Tb

Polished

body

temperature

oc

T1 T2 T3

Average

temperatu

re oc

Tp

Chamber

temperatu

re oc

T4

Emissivit

y

p

Model Calculations

Temperature of black body in absolute unit Tba =Tb + 273

Temperature of polished body in absolute unit Tpa = Tp + 273

Formula Used


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Power input Q = pA [Tp4-Ta4] = pA [Tb4-Ta4]

Since the power input is same for both heaters and area of radiating surface (A) is also same,knowing b =1

Emissivity (p) = b* [(T4ba-T4ca) / (T4pa-T4ca)]

Where b emissivity of black body

Procedure

1) Connect the three pin plug to the 230V, 50Hz, 15Amps main supply and switch on the unit.

2) Keep the thermocouple selector switch in first position.

3) Adjust the regulator to provide the required power input to both heaters.4) Allow the unit to stabilize.

5) Turn the thermocouple selector switch clockwise step by step and note down temperatures

indicated by temperature indicator from channel 1 to 7.

6) Tabulate the readings and calculate.

7) For the various power inputs repeat the experiment.


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Result

Emissivity of nonblack test plat surface is found to be =


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PARALLEL FLOW / COUNTER FLOW HEAT EXCHANGER


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Aim

To find the overall heat transfer co-efficient in parallel flow and counter flow.

Apparatus required

1. Concentric heat exchanger2. Digital temperature indicator

3. Stop clock

4. Asbestos rope insulation

Specifications

Length of the heat = 1800 mm

Inner copper tube ID = 12mm

OD = 15mm

Outer GI tube ID = 40mm

Theory

Heat exchangers are devices in which heat is transferred from one fluid to another. Common

examples of the heat exchangers are the radiator of a car, condenser at the back of domestic

refrigerator etc. Heat exchangers are classified mainly into three categories: 1. Transfer type 2.

Storage type 3. Direct contact type. Transfer types of heat exchangers are most widely used. A

transfer types of heat exchanger is one in which both fluids pass simultaneously through the

devices and heat is transferred through separating walls. Transfer types of exchangers are further

classified as.

1. Parallel flow type in fluids flow in the same direction.

2. Counter flow types in fluid flow in the opposite direction.

3. Cross flow type in which fluids flow at any angle to each other.

A simple heat exchanger of transfer type can be in the form of a tube arrangement. One fluidflowing through the inner tube and the other through the annulus surrounding it. The heat transfer

takes place across the walls of the inner tube.


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Description

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

from an electric geyser and flows through the inner tube. The cold fluid i.e. cold water can be

admitted at one of the ends, enabling the heat exchanger to run as a parallel flow apparatus or a

counter flow apparatus. This can be done by operating the different valves provided. Flow rate can

be measured using stop clock and measuring flask. The outer tube is provided with adequate

asbestos rope insulation to minimize the heat loss to the surroundings.

Formula used

LMTD (tm) = (td-to) / (ln (ti /to))

= ((Thi-Tci) - (Tho-Tco)) / Ln ((Thi-Tci) / (Tho-Tco))=

Heat input qh = A*U*LMTD

Hence,

The overall heat transfer coefficient

U = (qh/ (A*U*LMTD)

qh = mh*Ch*(Thi Tho)

=

Where,

mh is mass of hot water = 0.041667

Density of water = 1000 kg/m3

Cp = 4.178 KJ/Kg

Area A = dl = 0.084823 m2

d is outer diameter of inner pipe = 15mm

l is the length of heat exchanger = 1800mm

Inner ID diameter = 12mm

Outer GI tube ID d2 = 40mm

U = w/m2 oc


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Procedure

Connect water supply at the back of the unit. The inlet water flows through geyser and

inner pipe of the heat exchanger and flows out. Also the inlet water flows through the

annulus gap of the heat exchanger and flows out.

For parallel flow open valve V2, V4 and V5.

For counter flow open valve V3, V1 and V5.

Control the hot water flow approximately 2lit. /min and cold water flow approximately 5lit.

/min.

Switch ON the geyser. Allow the temperature to reach steady state.

Note temperature T1 and T2 (hot water inlet and outlet temperature respectively)

Under parallel floe condition T3 is the cold water inlet temperature and T4 is the cold water

outlet temperature.

Note the temperature T3 and T4.

Under counter flow condition T4 is the cold water inlet temperature T3 is the cold water

outlet temperature.

Note the temperature T3 and T4.

Note the time for 1 liter flow of hot and cold water. Calculated mass flow rate Kg/sec.

Change the water flow rates and repeat the experiment.

Precautions

1. Do not put on heater unless water flow is continuous.

2. Once the flow is fixed, do not change it until note down the readings for that flow.

3. The thermocouples should keep in pockets

4. There should make the oil well in pockets of thermocouple.

5. Equipment should be earthed prop

6. Once the experiment is completed drain out the water remains in both the tubes.

Constants

1. Cpc = Specific heat of cold water = 4.174 KJ / KG k

2. Cph = Specific heat of hot water = 4.174 KJ / KG k


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PARALLEL FLOW / COUNTER FLOW HEAT EXCHANGER

Sl.No Hot Water

Inlet

Temp. Thi

oC

Hot Water

Outlet

Temp. Tho

oC

Cold

Water

Inlet

Temp. Tci

oC

Cold

Water

Outlet

Temp. Tco

oC

Time for

hot water

flow 1 liter

sec

Time for

cold water

flow 1 liter

sec


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Result

Thus the overall heat transfer co-efficient is determined by using in parallel flow and

counter flow method.


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AIR CONDITION TEST RIG

Aim: to determine the co-efficient of performance of an air conditioning system.

Introduction

UTE air conditioning for human comfort or industrial process required certain processes to becarried out on air to vary the physchometric properties of air to requirements. These processesmay involve the mixing of air streams, heating of air, cooling o the air, humidifying air, anddehumidifying air and combination of the process. All such processes are studied with the givenair-condition test rig.

Description

The vapour compression air condition test rig is designed for exclusive study of refrigerantproperties while conditioning an environment of correct temperature, humidity and air movement.The test air rig consist of a closed type compressor, energy meter to measure electrical input to the

compressor and heater, pressure gauges, and thermocouple sensor fitted at respective locations,digital temperature indicator, air cooled condenser, an expansion valves, etc. air cooled typeevaporator. The air is passed by a blower unit through the duct. The expand refrigerant passesthrough the coil fixed in the duct. The passing air comes in contact with the coil.

A multiple air heater is provided in the duct for dehumidification. A steam boiler is provided toallow the steam to mix with air passing through the duct for humidification. Wet and dry bulbtemperatures are measured at the inlet and outlet section of the duct. The flow of air is measuredusing an anemometer.The whole unit with panel and air circulation system mounted on a trolley with caster wheels. Thepipelines are suitably colour coded to indicate different states of the gas.

Procedure

Switch on the mains. Switch on the condenser fan and blower,

Keep the manually operated valves in proper position.

Switch on the compressor and allow the unit to stabilize after adjusting the airflow through theduct.

Note down the following

Pressure of supper heated vapour at the exit on the compressor (delivery) p1Pressure at entry to throttle valve p2Pressure after throttling p3Pressure of super heated vapour at suction to the compressor p4Note the corresponding temperatures T1,T2,T3 and T4 at respective state points

S. T T T3 T P P P P Inlet DBT WBT DBT Time for Anemom


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No 1 2 4 1 2 3 4 WET 10 rev ofenergymeter ofcompressor

eterreading

Model calculation

Energy meter constant = 1200rev/kwhDiameter of orifice fixed to the duct =0.08mArea of the orifice fixed to the duct =0.005m2Density of air =1.16The mass flow rate of air = velocity air flow m/sec * area of the orifice *density of air

= 0.031 kg/sec

From psychometric chart at inlet DBT & WBT enthalpy = kj/kg

From psychometric chart at outlet DBT & WBT enthalpy = kj/kg

Hence the drop in enthalpy due to the refrigerating effect = kj/kg

For the mass flow rate of air measured the drop in enthalpy of the refrigerating effect =

Power input = 3600 * 10 KW/E* t

Where E is the energy meter constant.

Performance of the unit = refrigerating effect/power input to the compressor


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To find the COP using p-h chart.

Convert the pressure gauge readings to absolute pressures in bar.Note 14.5psi = 1 kgAbsolute pressure = gauge pressure in kg+ 1.03 bar.

Plot different state points on the p-h chart (R22) using the absolute pressure and temperaturereadings. Note the enthalpy at each point. P1, T1 corresponds to point 1, and p2, T2, p3, T3 andp4, T4, corresponds to points 2,3 and 4 respectively.

State 4-1 represents the adiabatic compression.State 3-4 represents the isobaric evaporation.

Theoretically sub cooling and super heating of suction vapour states is neglected.

From p h chart

H1 = KJ/KgH2 = KJ/Kg

H3 = KJ/Kg

H4 = KJ/Kg

Saturation cycle COP = H4 H3/H1 H4

To find the COP by experimental method.

To find the refrigerating effect / cooling capacity of the system

For calculating the efficiency of the cycle experimentally, note the dry bulb & wet bulbtemperatureof air a t inlet and outlet section of the duct.

The mass of air is calculated as follows.Using anemometer note velocity airflow m/sec. knowing the area of the orifice through which airis sucked in ot the duct, and the density of air, calculate mass flow rate of air of air Kg/sec.

Using a psychrometric chart note the enthalpy drop of air from the inlet to the outlet section of theduct, KJ/Kg, multiplying this value with the mass flow rate of air, the refrigerating effect can befound KJ/sec.

Note the time taken (t) for 10 Rev. of the energy meter disc for the compressor and calculate thepower as follows.

Power input = 3600 * 10 KW/E* t


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Where E is the energy meter constant.

Now the performance of the unit refrigerating effect / power to compressor.

Note. The compressor should be restarted only after the pressure is equalized.


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Result

Thus the COP of the system is found


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REFRIGERATION TEST RIG

Aim: to determine the co-efficient of performance of a given refrigerating system.

Introduction

The vapour compression refrigeration test rig is introduced for the exclusive demonstration ofmechanical refrigeration system and its electrical controls.

The system exposes functions of compressor, condenser, evaporator, driver, expansion valve,capillary, thermostat for effecting pressure and temperature changes.

The test rig mounted on caster wheels is of study MS tubular frame with a laminated panel board,compressor, condenser with fan, and evaporator in the form of SS vessel with coil wound over theouter surface with proper insulation mounted on the base of the unit. The panel houses allelectrical/refrigeration controls with throttling devices with copper pipes properly color-coded to

indicate different state of the refrigeration gas. Provision is made to measure the temperature andpressure at each of these state points.

Specifications

Compressor:1/3 HP sealed reciprocating compressor

Condenser:forcd air-cooled condenser.

Evaporator: SS vessel with copper coil wound and soldered around. Properly insulated. Providedwith a thermocouple sensor to measure the water temperature.The diameter of this vessel =295mmEnergy meter constant =rev/kwh

Location of temperature sensor and pressure gauges.

P1,T1 corresponds to state point 1, and p2 T2, T3,and p4, T4, corresponds to points 2,3and 4respectively.

Thermostat to cut off the compressor when the water temperature .

Indication lamp glows when safety device operates.

Expansion devices: capillary tube automatic expansion valve.

Valves: diaphragm type hand shut off valves and an electrically operated solenoid valve.


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Vapour compression refrigeration test rig

Intial

temp. ofwater

Intial

temp. ofwater

Durat

ionofexp.

Delivery

pressure

Delivery

temp.

Condenser

outletpressure

Condenser

outlettemp.

Pressure

afterthrottling

Temp. after

throttling

Suction

pressure

Suction

temp.

Timefor 10

rev ofcompressorenergymeter

Height

ofwaterinthevessel

Calculations.

a. To find the COP by experimental method.

To find the refrigerating effect.Mass of water in the evaporator vessel (m) w (/4* d2) h. kg, where h is the height of water level,w is the density of water, d is the diameter of the vessel.

Rate of fall in temperature of water =

T5 initial T5 final / duration of experiment in seconds. dt/dT oC/sec.

Cp of water = 4.182 KJ/kg oC

Hence the refrigerating effect can be calculated as m* cp* dt/dT. Kw.The work done by the compressor may be calculated using the time taken (seconds) for 10revolution of energy meter disc. E is the energy meter constant. (Rev/KW.hr.)

The work done by the compressor =3600(10)/t(E) KW, for time taken for 10 rev. of energy meter.

Hence the COP of the refrigerating system = refrigerating effect/work done

b. To find the COP using p-h chart

Convert the pressure gauge readings to absolute pressures in bar.Note. 14.5psi = 1 kg.

Absolute pressure = gauge pressure in kg+1.03b

Procedure.


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Check for any loose connection in the electrical circuit.

For expansion through capillary, open hand shut off valve before and after capillary, closeall other valves. Solenoid in off position.

Put on the main switch, condenser fan, and temperature indicator.

Fill water in the evaporator vessel to 3/4th level. Note the level of water using a measuringscale (h).

Note the initial temperature of water T5

Note the starting time, and switch on the compressor.

Allow the unit to stabilize.

Note the time for 10 revolution of compressor energy meter reading.

When the temperature T3 and T4 are almost equal note the following.

Note the closing time, and temperature of water T5

Note pressure gauge reading p1,p2,p3 and p4 at different state points.

Note temperatures at different state points T1, T2, T3, and T4 using the selector switchprovided on the temperature indicator.

The experiment may be repeated for a different water level/temperature. Please note forrepeating the experiment the water needs to be changed.

For throttling through the automatic expansion valve, close all hand valves. Put on thesolenoid switch, however the main hand valve in the line must be open.

Repeat the above procedure and note down the readings.


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Plot different state points on the p-h chart using the absolute pressure and temperature readings.Note the enthalpy at each point. P1, T1, corresponds ot point 1, and p2, T2, p3, T3, and p4, T4,corresponds to points 2,3, and 4 respectively.

State 4-1 represents the adiabatic compressionState 3-4 represents the isobaric evaporation

Theoretically sub cooling and super heating of suction vapour states is neglected.

From p-h chart.

H1 = KJ/Kg

H2 = KJ/Kg

H3 = KJ/Kg

H4 = KJ/Kg

Saturaqtion cycle COP = H4 H3/H1 H4

Condenser to evaporator heat transfer ratio cycle.

Theoretical = H1 H2/H4 H3

Model calculation

Evaporator vessel dia D = 295 mmEnergy meter constant E= 1200rev/KWH

By actual experimental method

The COP of the refrigeration unit may be calculated as below.

To find the mass of water in the evaporator vessel

m = v* kg/sec.

where v is volume of water in evaporator= (/4* D2) h.

D = diameter of evaporator vessel


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h= water level in the vessel

is the density of water which is equal to=1000kg/m3

heat absorbed from evaporator water = m*cp*(T5ini-T5fin)/dT KW

cp = 4.18 kj/kgoC

heat absorbed from evaporator water (refrigeration effect) +

work done by the compressor =3600 /E*10/t KW

hence COP = refrigeration effect/ work done

Result

Thus the COP of the system is found =


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Documents

Final Heat Power Lab Manual14