s 7 - heat transfer lab

HEAT TRANSFER LAB

DEPARTMENT OF

MECHANICAL ENGINEERING

HEAT TRANSFER LABLAB MANUAL

S N M INSTITUTE OF MANAGEMENT AND TECHNOLOGYMALIANKARA P O, MOOTHAKUNNAM, ERNAKULAM-683 516

Dept. of Mechanical 1 SNMIMT

HEAT TRANSFER LAB

DEPARTMENT OF

MECHANICAL ENGINEERING

HEAT TRANSFER LABLAB MANUAL

Prepared by: PRAMOD K. B. & KIRAN M.

Checked by:

Revision No.

Date

S N M INSTITUTE OF MANAGEMENT AND TECHNOLOGYMALIANKARA P O, MOOTHAKUNNAM, ERNAKULAM-683 516


HEAT TRANSFER LAB

CONTENTSExperiments Name Page No.

CYCLE – 1

1. Determination of Overall Heat Transfer Coefficient of Composite Wall……….. 5

2. Emissivity Measurement Apparatus…………………………………………….. 7

3. Heat Transfer In Forced Convection………………………………………….... 10

4. Free Or Natural Convection………………………………………………..…… 13

5. Heat Exchanger – Parallel Flow………………………………………..………. 16

CYCLE – 2

6. Heat Exchanger- Counter Flow……………………………….…………….…. 20

7. Heat Transfer Through Lagged Pipe…………………………….……………… 23

8. Performance Test On Refrigerating Unit……………………….……………… 25

9. Air Conditioning Test Rig…………………………………….……………….. 27


HEAT TRANSFER LAB

CYCLE - 1


HEAT TRANSFER LAB

Experiment No: 1

DETERMINATION OF OVERALL HEAT TRANSFER COEFFICIENT OF COMPOSITE WALL

AIM

To determine the overall heat transfer coefficient of a composite wall

THEORY

Let

Q = K1 A1(T 1−T 2)

L1 =

K2 A2(T 2−T 3)L2

= K3 A3(T3−T4)

L3

Where

K1 = conductivity of mild steel, L1-Thickness of mild steelK2 = conductivity of wood, L2-Thickness of woodK3 = conductivity of Bakelite L3-Thickness of bakelite

A=π d2

4

Where d = 300 mm

K1=Q L1

A1(T 1−T 2)=

Q L2

A2(T 2−T3)=

Q L3

A3(T3−T 4)

Overall heat transfer coefficient

U0= 1

1/ A (L 1K 1

+L 2K 2

+L 3K 3

)

SPECIFICATION;-

1) Mild steel 20 mm thick of 300 mm diameter =2 nos2) Wood 20 mm thick of 300 mm diameter = 3 nos3) Bakelite 19 mm thick of 300 mm diameter =2 nos4) Heater = 500 watts5) Thermocouples = 7 nos(iron- constantan)

APPARATUS


HEAT TRANSFER LAB

The apparatus consists of three circular slabs of different materials of same thickness, clamped in the centre using screw rod. A central heater is sandwiched between two sheets .three types of slabs are provided on both the sides of heater, which forms a composite structure. A small hand press frame is provided to ensure perfect contact between the slabs. A dimmerstat is provided for varying the input to the heater and measurements of input is carried out by a Wattmeter.

Thermocouples are embedded between interfaces of the slabs to read the temperature at the surface. The experiment can be conducted at various values of heater input and calculation can be made accordingly.

PROCEDURE

1) First ensure that plates are symmetrically arranged on both sides of the heater plates2) Operate the hand press properly to ensure perfect contact between the plates3) Close the box by the cover sheet to achieve steady environmental conditions4) Start the supply of heater by varying the dimmerstat ,adjust the input at the desired

values5) Take readings of all the thermocouples after steady state conditions are achieved6) Note down readings in observation table.7) Repeat the procedure for different heat inputs

PRECAUTIONS

1) Keep dimmerstat to zero before start2) Increase input heat supply slowly3) Keep the assembly insulated4) Remove air gap between plates by moving hand press gently5) Operate selector switch of temperature indicator gently

OBSERVATIONS

SL NO

Wattmeter Readings(WATT)

T1

(OC)T2

(OC)T3

(OC)T4

(OC)T5

(OC)T6

(OC)T7

(OC)U0

W/m2 K

12345

RESULT

The overall heat transfer coefficient is ……………


HEAT TRANSFER LAB

INFERENCE-

Experiment No: 2

EMISSIVITY MEASUREMENT APPARATUS

AIM

To determine emissivity of test surface

THEORY

All substances emit radiation at all temperatures.Thermal radiation is an electromagnetic wave and doesn’t require any material medium for propogation. All bodies can emit radiation and have capacity to absorb all ,or part of radiation coming from the surroundings towards it.

An idealized black surface is one which absorbs all incident radiation with reflectivity and transmittivity equal to zero.T he radiant energy per unit time per unit area from the surface of the body is called as the emissive power and is denoted by ‘C’.the emissivity of the surface is the ratio of the emissive power of the surface to the emissive power of the black surface at the same temperature.

ε=ε s

εb

Absorpitivity of a black body is one and by Kirchoff’s law of emissivity of the black body becomes unity.Emissivity being a property of the surface depends on the nature of the surface and temperature It is obvious from the Stefan boltzman law that the value of emissive power of the surface require knowledge about the values of its emissivity and therefore much experimental research in radiation has been oncentrated on measuring the value of emissivity as function of surface temperature.the present experimental set up is designed and fabricated to measure the property of emissivity of the test plate surface at various temperatures.

SPECIFICATIONS

Diameter of test plate = 150 mm

Diameter of black plate = 150 mm

Enclosure size = 550 mm × 300 mm × 300 mm

APPARATUS


HEAT TRANSFER LAB

The experimental set up consists of two circular copper plates, identical in size and is provided with heating coils sandwitched.the plates are mounted on brackets (asbestos cement sheet) and are kept in an enclosure ,so as to provide undisturbed natural convection surroundings.the heat input to the heater is varied by separate dimmerstats and is measured by using an wattmeter. The temperature of plates are measured by three thermocouples placed at ends of the plate.another thermocouple is provided in the enclosure to read the ambient temperature of enclosure.

Plate one is blackened by a thick layer of lamp black to form the idealized black surface wheras plate two is the test plate whose emissivity is to be measured .the heater inputs to the two plates are dissipated from the plates by conduction,convention and radiation.the experimental set up is set up is designed in such a way that under steady state conditions the heat dissipated by conduction and convention is same for both the plates when the surface temperature are same for both the plates and the difference in the heater input readings is because of the difference in the heater input readings is because of the difference in the radiation characteristics due to this different emissivities.the schematic arrangement of the set up is as shown in the figure.

PROCEDURE

1) Gradually increase the input to the heater to black plate and test plate and adjust it to test values.

2) Check the temperature of the two plates with small time interval and adjust the input of test plate only, by the dimmerstat so that the two plates will be maintained at the same temperature.

3) Wait until steady state conditions approach and record the temperatures of both plates (T1 , T2, T3 , T4 ,T5 , T6 and T7)

4) Repeat same procedure for various surface temperature in increasing order and tabulate the result

PRECAUTIONS

1) Use stabilized A.C single phase supply.2) Always keep the dimmerstat at zero position before start.3) Gradually increase heater inputs.4) Ensure that black plate is completely covered with lamp black uniformly.5) Do not increase power beyond 100w

OBSERVATION

Dimmerstat readings

Test plate Black plateAmbient

temperature

W1

(watts)W2

(watts)T1

(oC)T2

(oC)T3

(oC)T4

(oC)T5

(oC)T6

(oC)T7(oC)


HEAT TRANSFER LAB

CALCULATIONS

1) Emissive power of test plate εg =σΔT4

σ = 5.67 × 10 -8 w/m2ok

ΔT - average surface temperature

ΔT = T1+T 2+T 3

3 + 273 – (T7+273)

εg =________

2) Emissive power of Black plate εb =σΔT4

ΔT = T 4+T5+T 6

3+ 273 – (T7+273)

εb= _______

3) Emissivity = ε g

εb

RESULT

Emissivity of non black test plate (E) is = ______

INFERENCE


HEAT TRANSFER LAB

Experiment No: 3

HEAT TRANSFER IN FORCED CONVECTION

AIM

To determine the forced convection heat transfer coefficient for flow through the given horizontal tube or cylinder ,and compare the theoritical and actual values.

THEORY

Convection is the process of energy transfer by the combined action of heat conduction, energy storage and mixing motion. When the mixing motion is induced by some external agency such as pump or a blower the process is called forced convection. the intensity of mixing motion is generally high in forced convection and consequently the heat transfer coefficients are higher than free convection.

The heat transfer coefficient h can be obtained from Newton-Rockmens law of convection as

Q= h A ΔTBy using the method of dimensional analysis on the experimental setup, results

obtained in heat transfer can be correlated by equation of forces

Nu = Φ (Re 4 (Pr))Where Φ and 4d are functions of Reynolds number and Prandtl number respectively.

SPECIFICATIONS

Outer pipe diameter (do) = 32 mmInner pipe diameter (di) = 28 mmLength of test section (L) = 400 mmOrifice diameter (d) = 25 mmCoefficient of discharge (Cd) = 0.62Dimmerstat 0-2 ampere, 230V ACBlowerTemperature indicatorWattmeterHeater

APPARATUSThe apparatus consists of a blower unit fitted with a test pipe .The test section is

surrounded by a nichrome band heater. Four thermocouples are embedded in test section and two thermocouples in air stream at the entrance and exit of the test section to measure air inlet and outlet temperatures. Test pipe is connected to the delivery side of the blower along with the orifice to measure flow of air through the pipe. Input to heater is given through


HEAT TRANSFER LAB

dimmerstat and measured by wattmeter. Air flow is measured with help of an orifice meter and manometer fitted on board.

PRECAUTIONS

1) Keep dimmerstat to zero before start2) Increase voltage slowly3) Keep the assembly insulated4) Use proper range of wattmeter5) Operate selector switch of temperature indicator gently6) Ensure there is no block in flow path of blower7) Maximum heater input -100W

PROCEDURE

1. Switch on the supply and select the range2. Switch on the blower unit and adjust the air flow rate using gate valve of blower to a

desired level difference in the manometer.3. Wait until steady conditions are attained4. Note the wattmeter readings and thermocouples readings5. Calculate the heat transfer coefficient using newtons formula6. Repeat the experiment for different heat input values.

Nu= c (Re)m (Pr)n

Prandtl number, Pr = M CP

K Heat transfer coefficient, hth= Nu K

hc

OBSERVATIONS

Sl No

Manometer reading(m)

H *10^-3 (m)

Dimmerstat reading(watts)

Surface temperature(oC)ha

(w/m2k)

(h1) (h2) T1 T2 T3 T4 T5 T6

1

2

CALCULATIONS

Specific heat of air from chart (Cp) at Ta = ______

Ta = T1 +T6

2Density of air at Ta in oC from chart (þ) = ________


HEAT TRANSFER LAB

Density of water (þw) = ________

Coefficient of discharge (Cd) = ________

Acceleration due to gravity (g) = ________

Air discharge Q = CdA × square root (2gwþw) ÞVelocity of air at outlet V= Q/A = ______

Mass flow rate of air Ma = þa ×Q = ______

Heat carried away by air, Qa = Ma Cp (T6-T1)

Mean temperature Ts = T2 +T3 + T4 + T5

4The value of surface heat transfer coefficient neglecting end losses using equation

h = Q ____ A (Ts-Ta)

Heat transfer area = πDL =

Thermal conductivity at Ta from chart =

Actual nusslet number = h × Di K

Kinematic viscosity (ν) =

Reynolds number (Re) = V Dν

Prandtl number =

Theoretical Nusselt number = 0.023 ×Re 0.8×Pr 0.4

RESULT

Theoretical nusselet number = Actual nusselet number = Heat transfer coefficient =

INFERENCE


HEAT TRANSFER LAB

Experiment No: 4

FREE OR NATURAL CONVECTION

AIM

To determine the natural convection heat transfer coefficient for vertical tube exposed to atmospheric air.

THEORY

In contrast to forced convection, natural convection phenomenon is due to the temperature difference between the surfaces and the fluid, and is not created by any external agency.The present experiment setup is designed and fabricated to study the natural convection phenomenon from a vertical cylinder in terms of variation of the local heat transfer coefficient and its comparison with the value obtained by using an approximate correction.

The value of surface heat transfer coefficient h= QA (T¿¿ s−T a)¿

Q = rate of heating in wattsTs = average surface temperature

Theoretical value of heat transfer coefficient can be obtained with the correlation equations below

hL = 0.56 (Gr.Pr)0.25 for 104 <Gr Pr<108

K

hL = 0.13 (Gr.Pr)1/3 for 104 <Gr Pr<108

K

SPECIFICATION:-

1) Diameter of tube (d) = 38 mm2) Length of tube (L) = 500mm3) Duct size = 250 mm × 250 mm × 900 mm4) Number of thermocouples = 6 nos.

APPARATUS

The apparatus consists of brass tube fitted in a rectangular duct in a vertical position.The duct is open at the top and bottom and forms an enclosure and serves the


HEAT TRANSFER LAB

purpose of undisturbed surroundings.An electric heating element is kept in vertical tube which in turn heats the tube surface.the heat is lost from the tube to the surroundings air by natural convection.the temperature of the vertical tube is measured by seven thermocouples .the heat input to the heater is measured by a wattmeter and is varied by a dimmerstat.the tube surface is polished to minimize radiation losses.

PRECAUTIONS

1) Keep dimmerstat to zero before start2) Increase voltage slowly3) Keep the assembly insulated4) Use proper range of wattmeter5) Operate selector switch of temperature indicator gently

PROCEDURE

1) Switch on supply and adjust the dimmerstat to obtain the required heat input.2) Wait until steady state conditions are attained3) Note down the readings of thermocouples4) Note down ambient temperature5) The experiment is repeated for different heat input values.

OBSERVATIONS

Sl noWattmeterReadings

(watt)

T1

(OC)T2

(OC)T3

(OC)T4

(OC)T5

(OC)T6

(OC)T7

(OC)

Ambient temperature

(OC)

ha

w/m2K

1

2

3

4

5

CALCULATIONS

1) The value of surface heat transfer coefficient neglecting end losses using equation

h= QA (T¿¿ s−T a)¿

Q = rate of heating in wattsTs = average surface temperature

Ts = T1+T 2+T 3+T 4+T 5+T6+T 7

7 K


HEAT TRANSFER LAB

A = ΠdlD = diameter of cylinder in m.L = Length of the tube in m.

ha= _______

2) Compare the experimentally obtained value with the predictions of the correlation equations below

hL = 0.56 (Gr.Pr)0.25 for 104 <Gr Pr<108

KhL = 0.13 (Gr.Pr)1/3 for 104 <Gr Pr<108

K

Tf = Ts + Ta =______ 2β = 1 = ______ Tf

Grashof number (Gr) = gβL 3 ΔT ν2

Prandtl number (Pr) = µCp

K Gr Pr = ________

hL = 0.56 (Gr.Pr)0.25 K

hth =

RESULT

Heat transfer coefficient by natural convection is =

INFERENCE


HEAT TRANSFER LAB

Experiment No: 5

HEAT EXCHANGER

AIM

To determine the log mean temperature difference and overall heat transfer coefficient of heat exchanger

THEORY

Heat exchanger is a device in which heat is transfefered from one fluid to another, the temperature of fluid change in the direction of flow causing the flow of heat, the hot fluid is hot water and cold fluid is cold water.

Temperature effectiveness includes the performance of heat exchanger with respect to temperature alone.it is defined as the ratio of temperature difference that one fluid undergoes maximum temperature difference prevailing across the heat exchanger

It is denoted by p1

P1=T co−T ci

T hi−Tho

Effectiveness of heat exchanger, E =Q

Qmax

SPECIFICATION

Inside diameter of the outer tube = 35 mm

Length of the pipe = 1500 mm

Inside diameter of inner pipe = 12.5 mm

APPARATUS

The apparatus consists of two concentric horizontal tubes.Hot water flows in the inner tube and cold water flows around the inner tube in the annular space.Electric Geyser is used to heat the water.Flow of water is measured by stop watch and jar provided.Thermocouples are provided and insulations are provided to minimize heat loss.

THEORY

a) Parallel flow :-

Heat loss by hot water, Qh = mhCph(Thi-Tho)

Heat gained by cold water, Qc =mc Cpc(Tco-Tci)


HEAT TRANSFER LAB

LMTD=ΔT max−ΔT min

log(ΔT max / ΔTmin)

ΔT max = Thi-Tci

ΔT min = Tho-Tco

Q=Qh+Qw

2

Area of pipe, A= ПDL

Where-D=diameter of inner tube

L= length of the equipment

Overall Heat transfer coefficient, U = (Q/A) (LMTD)

Effectiveness of heat exchanger, E= Q/Qmax

Q=mh c p(T hi−Tho)

¿¿

PROCEDURE:-

1) Open the valve to start the flow of cold water in the outer pipe. Select the type of flow ,either parallel or counter by adjusting the valve.

2) Put on the switch of electric geyser for hot water circulation3) Put on the selector switch of the temperature indicator to the required temperature4) Wait until steadt y state conditions is reached5) Note down the temperature readings of water flow6) Measure the flow rate of hot water and cold water

OBSERVATIONS

Sl. No

(Mh) kg/s.

(Mc)kg/s.

hot water(oC)

Thi Tho

cold water(oC) Tci Tco

LMTDU

w/m2kE

123

CALCULATIONS


HEAT TRANSFER LAB

Heat lost by hot water -Qh

Mh = ______ Kg/sec.

Qh=MhCph (Thi-Tho)

Heat gained by cold water-Qc

Mc = …………Kg/sec

Qc = McCpc(Tco-Tci)



ΔT max = Thi-Tci = _________

ΔT min = Tho-Tco =__________

Area of pipe with diameter 12.5 mm = ПdL =

Heat Transferred in The Equipment, Q =_Qh+Qc

2Overall heat transfer coefficient in paralell flow- Up

Up = _Q___ A*(LMTD)

Effectiveness, E = _Q_ Qmax

Qmax = (mCp) min × ΔT max

RESULT

Determined LMTD of the heat exchanger in paralell flow

Calculated the actual value of Overall heat Transfer Coefficient

Determined the Effectiveness of the equipment in paralell flow

INFERENCE


HEAT TRANSFER LAB

CYCLE - 2

Experiment No: 6


HEAT TRANSFER LAB

HEAT EXCHANGER- COUNTER FLOW

AIM

To determine the log mean temperature difference and overall heat transfer coefficient of heat exchanger

THEORY

Heat exchanger is a device in which heat is transfefered from one fluid to another, the temperature of fluid change in the direction of flow causing the flow of heat, the hot fluid is hot water and cold fluid is cold water.

Temperature effectiveness includes the performance of heat exchanger with respect to temperature alone. It is defined as the ratio of temperature difference that one fluid undergoes maximum temperature difference prevailing across the heat exchanger

It is denoted by p1

P1=T co−T ci

T hi−Tho

Effectiveness of heat exchanger, E =Q

Qmax

SPECIFICATION

Inside diameter of the outer tube = 35 mm

Length of the pipe = 1500 mm

Inside diameter of inner pipe = 12.5 mm

APPARATUS

The apparatus consists of two concentric horizontal tubes.Hot water flows in the inner tube and cold water flows around the inner tube in the annular space.Electric Geyser is used to heat the water. Flow of water is measured by stop watch and jar provided.Thermocouples are provided and insulations are provided to minimize heat loss.

THEORY

Counter flow:-Heat loss by hot water, Qh = mhCph(Thi-Tho)

Heat gained by cold water, Qc =mc Cpc(Tco-Tci)


HEAT TRANSFER LAB



ΔT max = Thi-Tco

ΔT min = Tho-Tci

Q=Qh+Qw

2

Area of pipe, A= ПDL

Where-D=diameter of inner tube

L= length of the equipment

Overall Heat transfer coefficient, U = (Q/A) (LMTD)

Effectiveness of heat exchanger, E= Q/Qmax

Q=mh c p(T hi−Tho)

¿¿

PROCEDURE:-

1) Open the valve to start the flow of cold water in the outer pipe. Select the type of flow, either parallel or counter by adjusting the valve.

2) Put on the switch of electric geyser for hot water circulation.3) Put on the selector switch of the temperature indicator to the required temperature.4) Wait until steadt y state conditions is reached.5) Note down the temperature readings of water flow.6) Measure the flow rate of hot water and cold water.

OBSERVATIONS

Sl. No

(Mh) kg/s.

(Mc) kg/s.

hot water(oC)Thi Tho

cold water(oC) Tci Tco

LMTDU

w/m2kE

1

2

3

CALCULATIONS

Heat lost by hot water -Qh

Mh = ______ Kg/sec.

Qh=MhCph (Thi-Tho)


HEAT TRANSFER LAB

Heat gained by cold water-Qc

Mc = …………Kg/sec

Qc = McCpc(Tco-Tci)



ΔT max = Thi-Tco =

ΔT min = Tho-Tci =

Area of pipe with diameter 12.5 mm = ПdL =

Heat Transferred in the Equipment, Q =_Qh+Qc

2Overall heat transfer coefficient in counter flow

Up = Q___ A*(LMTD)

Effectiveness, E = _Q_ Qmax

Qmax = (mCp) min × ΔT max

RESULT

LMTD and effectiveness of the heat exchanger in counter flow is calculated.

LMTD of heat exchanger in counter flow = ……………………………

Effectiveness of the equipment in counter flow = ………………………….

INFERENCE

Experiment No: 7


HEAT TRANSFER LAB

HEAT TRASFER THROUGH LAGGED PIPE

AIM

To determine the thermal conductivity of the insulating material in the lagged pipe and plot the variation of temperature with radius.

THEORY

Fourier’s equation for one dimensional steady state heat condution is

Q ∝ AdTdx

= - k A dTdx

Q=heat transfer rate∈watts

A=area∈m2

dTdx

= temperature gradient in °C / m

k = thermal conductivity of material in w/ m°C

SPECIFICATIONLength of pipe, l = 900 mm

Radius of pipe, R1 = 41 mm

R2 = 75 mm

R3 = 125 mm

PRECAUTIONS

1) Always keep the dimmerstat at zero position before starting.2) Gradually increase heater inputs.3) Do not increase power beyond 100w4) Allow steay state conditions to be attained before taking readings.

PROCEDURE:-

1. Before switching on the mains see that the dimmerstat is in zero position.

2. Switch on the apparatus and turn the dimmerstat to desired value.

3. Allow the apparatus to attain steady state condition.


HEAT TRANSFER LAB

4. With the help of selector switch take the temperature readings.

5. Change the heater input and repeat the experiment.

6. Tabulate the readings and calculate the thermal conductivity.

7. Plot the graph of temperature verses radius.

OBSERVATIONS

Sl no.

Q(Watts)

T1

(°C)T2

(°C)T3

(°C)T4

(°C)T5

(°C)T6

(°C)K1

(w/mk)K2

(w/mk)

1

2

3

CALCULATIONS

Q=2 π k1 L (T1−T2 )

ln( R2

R1)

=2 π k 2 L (T2−T 3 )

ln(R3

R2)

where T1 = T1+T 2

2 , T2 =

T3+T 4

2, T3 =

T5+T 6

2

RESULT

Average thermal conductivity of insulating material k1 = …………………w/mC

Average thermal conductivity of insulating material k2 = …………………w/mC

INFERENCE

Experiment No: 8


HEAT TRANSFER LAB

PERFORMANCE TEST ON REFRIGERATING UNIT

AIM

To determine the coefficient of performance of refrigerating unit.

THEORY

The vapour compression refrigerating system has four major components namely evaporator, compressor, condenser and a expansion valve, the working fluid known as refrigerant undergoes thermodynamic cycle while transferring heat from low temperature reservoir to hot temperature reservoir with the help of external work.

Wet vapour refrigerant enters the compressor (point 1) and is compressed isentropically to saturated vapour (point 2). This is then passed through condenser where it is condensed to saturated liquid (point 3) at constant pressure by giving out latent heat to the surroundings. This liquid refrigerant is throttled through a expansion device (point 4). After expansion it is passed through evaporator where it absorbs the latent heat of vaporization and produces cooling or refrigerating effect.

Performance of refrigerating unit is expressed in terms of coefficient of performance (C. O. P.).

C. O. P. = HEAT ABSORBED∈THE EVAPORATOR

WORK DONE BY COMPRESSOR

SPECIFICATION

Refrigerant – R 134 a

Chiller capacity – 10 litres

PROCEDURE:-1. Fill evaporator container with pure water.2. Switch on the main and wait for sometime so that red light on stabilizer glows.3. Note initial energy meter reading and initial temperature of water.4. Switch on the compressor of refrigerating system.5. Run machine for 30 – 40 minutes.6. Finally note down the temperature of water and also energy meter reading.7. Change the water and repeat the experiment.

OBSERVATIONS

Duration of test, T = ………………….

Initial energy meter reading, E1 = …………………

Initial temperature of water in chiller, t1 = ………………..

Final energy meter reading, E2 = …………………


HEAT TRANSFER LAB

Final temperature of water in chiller, t2 = ………………..

Temperature at inlet of compressor, T1 = …………………

Temperature at outlet of compressor, T2 = …………………

Temperature at outlet of condenser, T3 = …………………

Temperature at outlet of expansion valve, T4 = ………………

Suction pressure, P1 = ………………bar

Delivery pressure, P2 = ………………bar

CALCULATIONS

Heat extracted, HE = mcp ΔT

ΔT = t1 – t2

Work done by compressor, WC = (E2 – E1) x 3600

Actual C. O. P. = H E

W C

Theoretical C. O. P. = h1−h4

h2−h1

where, h1 = specific enthalpy corresponding to P1 T1

h2 = specific enthalpy corresponding to P2 T2



h1, h2, h3, and h4 are found out from p – h diagram of R – 134 a.

RESULT

Coefficient of performance of refrigerating system is calculated.

Actual C. O. P. = ……………………….

Theoretical C. O. P. = ………………….

INFERENCE


HEAT TRANSFER LAB

Experiment No: 9

AIR CONDITIONING TEST RIG

AIM

To determine the properties of moist air at inlet and outlet of duct in the air conditioning test rig and determine the Coefficient of Performance.

THEORY

Air conditioning is that branch of engineering science which deals with the study of conditioning of air i. e., supplying and maintaining desirable internal atmospheric conditions (like temperature, humidity, purity and distribution) for human comfort.

Psychrometric chart is used for finding out the properties of air in the field of air conditioning to eliminate lot of calculations. Psychrometric chart is graphical representation of the various thermodynamic properties of moist air.

In Psychrometric chart, dry bulb temperature (DBT) is taken as abscissa and specific humidity as ordinate. The saturation curve represents 100% relative humidity at various DBT. It also represents the wet bulb temperature (WBT) and dew point temperature.

Coefficient of Performance = COOLING EFFECT

WORK DONE

SPECIFICATION

Area of duct = 12cm 12cm

Refrigerant used = R-22

APPARATUS

The compressed refrigerant from the compressor is sent to an air cooled condenser and then the condensate is passed through an expansion valve or a capillary tube for throttling. Due to throttling, temperature of refrigerant falls and the cold refrigerant passes through a heat exchanger to absorb heat from the air to be chilled or conditioned. The heat exchanger is known as evaporator. The warm refrigerant is then returned to compressor.

A thermostat is provided to cutoff compressor when the air temperature reaches the required value. Wet bulb and Dry bulb thermometers are provided to measure the temperature of ambient air and the conditioned air. An energy meter is provided to measure the energy input to compressor and fan. A voltmeter and ammeter is provided to monitor the power supply conditions.


HEAT TRANSFER LAB

PROCEDURE

1. Switch on the air conditioning test rig.2. Make sure the water level in the Wet bulb thermometer.

3. After steady state condition reaches, note down T1, T2, T3, &T4 and P1, & P2 and rotameter readings

4. Also note down the WBT and DBT of air at inlet and outlet.5. Using inlet DBT and WBT determine the properties of moist air at inlet of duct from

psychrometric chart.6. Using outlet DBT and WBT determine the properties of moist air at outlet of duct

from psychrometric chart.7. Measure the air velocity at exit of duct using anemometer.

OBSERVATIONS

Temperature at inlet of compressor, T1 = …………………

Temperature at outlet of compressor, T2 = …………………

Temperature at outlet of condenser, T3 = …………………

Temperature at outlet of expansion valve, T4 = ………………

Suction pressure, P1 = ………………bar

Delivery pressure, P2 = ………………bar

Time for ‘n’ revolution of energy meter disc, t = ……..

Velocity of air, v =

Inlet

WBT = ………………C

DBT = ………………C

Outlet

WBT = ………………C

DBT = ………………C

CALCULATIONS

Coefficient of Performance, C.O.P. = Q

W c

Cooling effect, Q = mcpΔT = m (h2 – h1)


HEAT TRANSFER LAB

h2 = specific enthalpy at exit kJ/ Kg

h1 = specific enthalpy at inlet kJ/ Kg

m = ρ A v

ρ = density of air, Kg/ m3

A = Cross sectional area at exit, m2

v = velocity at exit, m/ s

Work done, WC = n ×3600

t × N

N = energy meter constant

Actual Coefficient of Performance of the unit = Q

W c = ……………….

Properties of moist air at inlet and outlet of duct in the air conditioning test rig is determined using psychrometric chart.

At inlet

Relative humidity = ……………….. Kg/ Kg of air

Dew point temperature = …………. C


HEAT TRANSFER LAB

Enthalpy = …………….kJ/ Kg

Humidity ratio = …………..

Partial pressure of vapour = ……………..bar

Saturation pressure corresponding to DBT = …………….bar

Saturation pressure corresponding to WBT = …………….bar

Volume of dry air = ………………….

At outlet

Relative humidity = ……………….. Kg/ Kg of air

Dew point temperature = ………….C

Enthalpy = ……………. kJ/ Kg

Humidity ratio = …………..

Partial pressure of vapour = …………….. bar

Saturation pressure corresponding to DBT = ……………. bar

Saturation pressure corresponding to WBT = ……………. bar

Volume of dry air = ………………….

The theoreitical (cycle) C.O.P. is detremined using the formula

C. O. P. = h1−h4

h2−h1

where, h1 = specific enthalpy of R-22 at P1 T1

h2 = specific enthalpy of R-22 at P2 T2



h1, h2, h3, and h4 are found out from p – h diagram of R – 22.

RESULT Actual C.O.P. of the unit =

Cycle C.O.P. of the unit =

INFERENCE


s 7 - heat transfer lab

Documents