1. DETERMINATION OF THE REFRIGERATING EFFECT, C.O.P OF CLOSED SYSTEM WITH THERMOSTATIC EXPANSION VALVE

AIM:

To determine the refrigerating effect, C.O.P of a closed system with thermostatic expansion valve.

SYSTEM COMPONENTS:

COMPRESSSOR CONDENSOR DRIER AND FILTER CHILLER AND FREEZER (EVAPORATOR) HIGH PRESSURE AND LOW PRESSURE CUT-OUTS AUTOMATIC EXPANSION VALVE

PROCEDURE:

1. Fill in the chiller tank (evaporator tank) with a measured quantity of clean water.2. Ensure that the thermostat is ON.3. Switch on the main supply.4. Keep the thermostatic expansion valve as the expansion device.5. Check the capillary tube is kept in fully closed condition.6. Run the system and note down the following.

T 1=¿ Temperature of water.T 2=¿ Temperature after expansion.T 3=¿ Temperature after evaporation.T 4=¿ Temperature after compression.T 5=¿ Temperature after condensation.

S. NO Description Reading/units

1. Mass of the water in the chiller (m) kg2. Initial temperature of water. (Tw1) ˚C3. Final temperature of water. ( Tw2) ˚C4. No of flickering in energy meter light. (N) 105. Time taken for 10 (N) flickering in energy meter t 2 sec6. Pressure of refrigerant at suction. (GREEN-after

evaporation)Bar

7. Pressure of refrigerant at discharge. (RED-after compression)

Bar

8. Pressure of refrigerant at inlet. (BLUE-after expansion)

Bar

9. Pressure of refrigerant at liquid line. (YELLOW-after Bar

condensation)10. Temperature of refrigerant after evaporation °C11. Temperature of refrigerant after compression °C12. Temperature of refrigerant after expansion °C

13. Temperature of refrigerant after condensation °C

Pressure gauge reading : pressure gauge Reading∈psi

14.28 bar

Tw1 = Take initial temperature of water before starting. Tw2 = Take final temperature of water after decreasing 5˚C. (Tw1- Tw2= 5˚C)t 1 = Time taken for fall in temperature.t 2 = Time taken for 10 flickering of energy meter. Definition With Formulas:

1. CARNOT COP = T 2

T2−T 1

2. ACTUAL REFRIGERATING EFFECT:

= m. C pw .(Tw1−Tw2)

t 1

KW

m = mass of water in the chiller in kg.C pw = specific heat capacity of water at constant pressure = 4.187 KJ/Kg KTw1 = Take initial temperature of water before starting. Tw2 = Take final temperature of water after decreasing 5˚C. (Tw1- Tw2= 5˚C)t 1 = Time taken for fall in temperature of water.

3. WORK DONE BY COMPRESSOR = N X 3600t2 X EMC KW

Where, N = No of flickering of energy meter light. (10 flickering). EMC= Energy meter constant (3200 lmp/Kw hr) t 2 = Time taken for 10 flickering of energy meter.

4. ACTUAL COP = ACTUAL REFRIGERATING EFFECT

WORK DONE BY COMPRESSOR

5. THEORETICAL COP = THEORETICAL REFRIGERATING EFFECT

COMPRESSORWORK

Where Theoretical refrigerating effect = (h1−h4) in KJ/Kg of refrigerant Compressor work = (h2−h1) in KJ/Kg of refrigerant h1 , h2 ,∧h4 are the enthalpies taken from the p – h chart of the corresponding refrigerant.

h1 - Specific enthalpy of vapour refrigerant at suction in KJ/Kg h2 - Specific enthalpy of vapour refrigerant at discharge in KJ/Kgh3 - Specific enthalpy of vapour refrigerant at inlet in KJ/Kgh4 - Specific enthalpy of vapour refrigerant at liquid line in KJ/Kg

Relative COP = ACTUALCOP

THEORETICALCOP

Efficiency of the cycle = ACTUALCOPCARNOT COP

RESULT:The test was conducted on the given refrigerator using thermostatic expansion valve as an

expansion device and the performance of the system as follows.1. Actual refrigerating effect = 2. Actual COP = 3. Theoretical COP = 4. Relative COP =

5. Efficiency of the cycle =

2. DETERMINATION OF THE CAPACITY OF WINDOW AIR CONDITIONER(GAS: R22)

AIM:

To determine the capacity of the given window air conditioner.TOOLS, EQUIPMENTS & GAUGES:

U – TUBE MANOMETER WINDOW TYPE AIR CONDITIONER

PROCEDURE:

7. Check the wiring connections of the given window air conditioner.8. Find out DBT (dry bulb temperature) and WBT (wet bulb temperature) of ambient

air using sling psychrometer.9. Measure the diameter of the orifice plate.10. First switch ON the main switch and keep the knob in fan. After 30 seconds switch

ON the compressor.11. Run the system for 10 minutes and then find out the DBT and WBT of conditioned

air.12. Find out the velocity of air by using u – tube manometer connected to the duct.13. Note the time taken for 10 flickering of energy meter to calculate the input energy.14. By using psychrometric chart, find out the heat and moisture removed from the

conditioned air.15. Calculate the cop of the given air conditioner.

NOTE: Ensure the wet bulb temperature wick is always immersed in water and if necessary add

water to the wick. Check all the valves in the pipe lines are in open condition. Refrigerant pressures from the pressure gauges and temperature from the thermocouples

can be used to study the vapour pressure at various points in the refrigerant cycle and prepare a enthalpy – pressure diagram

T 1=¿ Temperature after expansion.T 2=¿ Temperature after evaporation.T 3=¿ Temperature after compression.T 4=¿ Temperature after condensation.

Warning:When the compressor turns off (by the thermostat) or is switched off manually, do

not turn on the power immediately. Allow a few minutes for the pressure in the compressor inlet

to equalize. The time delay provided in the voltage stabilizer is for this purpose only. Immediate starting will cause undue load on the compressor and may even lead to burst out.

OBSERVATION:d = dia of orifice. = (50 mm)cd = 0.62a = area of orifice = 1.963 X 10−3 m2

DBT of ambient air T DB1 = ℃ WBT of ambient air T wB1 = ℃

DBT of conditioned air T DB 2 = ℃ WBT of conditioned air T wB2 = ℃

CALCULATIONS: 1) AIR FLOW RATE :

Manometer level difference = (h¿¿1−h2)¿ m of water.2) Volume flow rate of air Q = cd X a X √2 gH m3/ sec Where,

a = area of orificed = dia of orifice. (50 mm)cd = 0.62g = 9.81 m /sec 2

H = ( h1−h2

1000 )Xρw

ρa

ρw=1000 kg/m3, ρa=Density of air = 1.16 kg /m3

3) Mass flow rate of air ma = 1.16 X Q kg ⁄ sec

4) From the psychometric chartEnthalpy of ambient air using¿) h1= KJ/Kg of dry airEnthalpy of conditioned air using¿) h2= KJ/Kg of dry air

Specific humidity of ambient air¿) w1= Kg/Kg of dry air Specific humidity of conditioned air¿) w2= Kg/Kg of dry air

5) Capacity of window air conditioner = ma X (h1−h2¿ KW

6) Input energy to the air conditioner = N X 3600t X EMC

kw

N = no of light flickering. (10 flickering)

Energy meter constant (EMC) = 3200 imp / kwhTime taken for no of flickering, t = seconds

COP of air conditioner:

COP of air conditioner = Refrigeration effect

Input energy

RESULT:The test was conducted on the given air conditioner and the following were determined.1. COP of air conditioner = 2. Capacity of window air conditioner = Kw

3. THERMAL CONDUCTIVITY OF SOLIDS

AIM: To determine the thermal conductivity of a poor-conducting material, say asbestos sheet by guarded hot plate method.

RELEVANT THEORY: Thermal conductivity is a specific property of any conducting material which is defined below for a homogeneous solid as the quantity of heat conducted across a unit area normal to the direction on unit time and for unit temperature gradient along the flow.

q dLK = ----------- A dT

Where

q = heat conducted in watts

dL = thickness of plate (m)

A = area of conduction heat transfer, m2

dT = temperature difference across the length dL (deg. C)

BASIS OF MEASUREMENT: Experimental measurement of thermal conductivity of solid can be accomplished by a variety of methods, all based on the observation of the temperature gradient across a given area of the material conducting heat at a known rate. Each of these methods has certain unique limitations, and the choice of one over another is governed by the general temperature level at which ‘K’ is measured by physical structure of the material in question, and by whether the material is a good or poor conductor.

In measuring the thermal conductivity of poor conductors, the specimens are takenin the form of sheets in order that the heat flow path be short and the conducting area large (low dL, higher A).

Guarded-hot plate method is generally used to conduct such an experiment. Figure 1 illustrates this method. In this method, electrically heated thermal guards are placed adjacent to the exposed surface of the source H1, specimen S, and sink Ho, These thermal guard plate are independently maintained at the same temperature as the

Adjacent surface, to ensure ideally no heat leakage occurring from source, specimen or the sink boundaries. The enclose drawings given actual dimensions of various components of the apparatus.

DESCRIPTION OF APPARATUS

Enclosed drawing and various specifications associated with all the components. The test specifications associated with all the components. The test section assemblies consisting of the asbestos specimen, heaters as well as thermocouples are shown separately 9 thermocouples are available to measure temperatures of heaters, the two specimens and the water inlet and outlet. It should be noted that (T3-T4) gives the temperature gradient across the top asbestos sheet and (T5-T6) gives the corresponding Quantity for the bottom specimen. Provision is also made to measure the cooling water flow rate. The whole assembly is enclosed in an insulating layer of mineral wool to prevent radiation and connective losses to the maximum extent possible.

EXPERIMENTAL PROCEDURESupply a small quantity of energy to the source ‘H’ (the main heater MH). Now adjust the input to the guard heaters such that the temperatures are same as that of the main heater. Allow water through the cooling circuit slowly. Allow 20-30 minutes for the temperatures to stabilize, note down all the temperatures, V, I, the water flow rate (cm3/min). Repeat the experiment at different at temperature values by adjusting appropriately the input conditions.

SPECIFICATIONS:

Material; Asbestos sheet (commercial grade)

Specimen dia, d = 150 mm or 0.15m

Specimen thickness, ∆L =12mm (2 pieces, one at the top and one at the bottom) ПArea of specimen = -------- (0.15)2 m2 =0.018 m2

4

Heat input Hi =0.86 VI watts

Heat input is from two contributions: Main heater and guard heater.

Both these contributions are to be summed up to obtain the total heat that is dissipated by the heaters. However only half of this heat input will pass (ideally through each of the two specimens) It may also be noted that since the heaters are sandwiched between 2 layers of commercial grade asbestos sheet, heat will be dissipated through both the layers. For the purpose of calculations, average temperature gradient [(T3-T4)+ (T5-T6) ] ÷ 2 is used.

OPERATIONAL PROCEDURE:

The panels consist of voltmeter, ammeter, and temperature indicator (all digital). Dimmer controls, volt and ammeter selector (common) switch, thermocouples selector switch and schematic diagram.

a) Connect the three pin top to 230v, 50 Hz, 5amps power supply socket, dimmers in OFF position.

b) Keep the voltmeter and ammeter selector switch in 1 position. Turn the dimmer one in clockwise and adjust the power input to main heater to any desired value by looking at the voltmeter and ammeter

c) Allow the unit to stabilize (anywhere from 20-30 minutes.)

d) Note down the temperature indicated by the digital temperature indicator by turning the thermocouples selector switch clockwise step by step (1,2,3,4,5,6,7,8,9)

e) Repeat the experiment for different power inputs to the heater.

f) Tabulate all the readings and calculate for different conditions.

g) After the experiment is over turn all the dimmer knobs anti clockwise direction, bring back the voltmeter, ammeter and thermocouple selectorSwitches to their original position, disconnect the three pin plug top from top the mains.

CAUTION: The equipment should be operated between 60 and 100 V.

S.No V I WATTS VI

T 3 T 4 T 5 T 6 CONDUCTIVITY K

MODEL CALCUATIONS:

Insulating material : Asbestos sheet (commercial grade)

Specimen dia : 150mm =0.15 m

ПArea of specimen : ---------- (0.15)2 = 0.018m2

4

Specimen thickness : ∆L=12mm = 0.012m

A) SET OF READINGS

V = 80VoltsI = 0.75 Amps

q = 0.86 * 80 * 0.75 = 51.6 Watt

∆T = { (T3-T4) + (T5-T6)} ÷ 2 = {(193-56) + (178-53)} ÷ 2 = (137 +125) ÷ 2 = 131 Deg. C.

∆L = 0.012 m A = 0.0176 m2

RESULT:THERMAL CONDUCTIVITY OF SPECIMEN:

q ∆L 51.6 X 0.012 K = --------- = ---------------------- A ∆ T 0.0176X131

= 0.268 W/m2 K

4. PARALLEL AND COUNTER FLOW HEAT EXCHANGER

AIM:

To determine LMTD, the effectiveness and the overall heat transfer coefficient for parallel and counter flow heat exchanger.

APPARATUS:

Heat exchanger, supply of hot and cold water

THEROY:

Heat exchanger is a device in which heat is transferred from one fluid to another, common examples of heat exchanger are:

i) Condensers and boilers in steam plant.ii) Intercoolers and pre-heatersiii) Automobile radiatorsiv) Regenerators.

CLASSIFICATION OF HEAT EXCHANGERS:

I. BASED ON THE NATURE OF HEAT EXCHANGE PROCESS:i) Direct contact type: Here the heat transfer takes place by direct mixing of hot

and cold fluids.ii) Indirect contact heat exchangers: Here the two fluids are separated through a

metallic wall. Example.Regenerators, Recuperators etc.

II. BASED ON THE RELATIVE DIRECTION OF FLUID FLOW:i) Parallel flow heat exchanger: Her both hot and cold fluids flow in the same

direction.

ii) Counter flow, heat exchanger: Here the hot and cold fluids flow in opposite direction.

iii) Cross – flow heat exchangers: Here the two fluids cross one another.

LOGARTHEMIC MEAN TEMPERATURE DIFFERENCE (LMTD):

This is defined as that temperature difference which, if constant, would give the same rate of heat transfer as usually occurs under variable condition of temperature difference.

For parallel flow:

LMTD = (T ho−Tco )−(Th i−T ci)¿(T h o−T co)/(T h i−T ci)

Where

T h i = Inlet temperature of hot fluid

T h o = Outlet temperature of hot fluid

T ci= Inlet temperature of cold fluid

T co = Outlet temperature of cold fluid

For counter flow:

LMTD = (T h i−T ci )−(T h o−T ci)¿(T h i−T co)/(T h o−T ci)

OVERALL HEAT TRANSFER COEFFICIENT:

The rate of heat transfer between hot and cold fluid is given by

Q =U o Ao

LMTDWhere,

U o= overall heat transfer coefficient based on outer surface area of tubes, W/m²-K

Ao= The total outer surface area of tubes, m²

EFFECTIVENESS:

Effectiveness of a heat exchanger is defined as the ratio of actual heat transfer rate to the theoretical maximum possible heat transfer rate.

Effectiveness: ε = Q

Qmax

It can be shown that

ε = Th i−T ho

Th i−T ci if mh c p h<mc cpc

And ε = Tco−T ci

T hi−T ci if mc cpc<mh cp h

Where,

mh And mh is the mass flow rate of hot and cold fluids respectively in kg/s

c p h And c pcis the specific heat of hot and cold fluids respectively in J/kg–K.

DESCRIPTION OF THE APPARATUS:

The apparatus consists of a concentric tube heat exchanger. The hot fluid namely hot water is obtained from an electric geyser and it flows through the inner tube. The cold fluid is, cold water can be admitted at any one of the ends enabling the heat exchanger to run as a parallel flow or a counter flow exchanger. This can be done by operating the different valves provided. Temperature of the fluid can be measured using thermocouples with digital display. Flow rates can be measured using stop clock and measuring flask. The outer tube is provided with insulation to minimize the heat loss to the surroundings.

SPECIFICATIONS:

1) Inner tube material – copperInner diameter d i = 9.5 mm

Outer diameter do = 12.5 mm

2) Outer tube material G. IInner diameter = 25 mm

Outer diameter = 33 mm

3) Length of the heat exchanger = 1 m

PROCEDURE:

1) Start the flow of the hot water side.2) Start the flow on cold water through the annulus and run the exchanger as flow unit.3) Put on the geyser. Adjust the flow rate of hot water side say 500 ml/minute4) Keep the flow rate same till the steady state conditions are reached.5) Note down the temperature on hot and cold water sides. Also measure the flow rate.6) Repeat the experiment for different flow rates as well as for parallel flow.

OBSERVATION TABLES:

PARELLEL FLOW

Sl. No

Hot water flow rate mh

Cold water flow rate mc

Temp. of Cold water in oC

Temp. of Hot water in oC

Inlet Outlet Inlet Outlet

Kg/s Kg/s T ci T co T hi T ho

COUNTER FLOW

Sl. No

Hot water flow rate mh

Kg/s

Cold water flow rate mc

Kg/s

Temp. of Cold water in oC

Temp. of Hot water in oC

InletT ci

OutletT co

InletT hi

OutletT ho

EQUATIONS USED:

1) Heat transfer from hot waterQh = mh c p h(Th i−T h o) watts

mh = mass flow rate of hot water kg/sec

c p h = Specific heat of hot water = 4186.8 J kg-K

2) Heat gain by the cold fluid

Qc = mc cpc (T co−T ci) watts

mc = Mass flow of cold fluid, kg/s c pc= Specific heat of cold fluid = 4186.8 J/kg -K

3) Q = Qh+Qc

2 Watts

4) LMTD = θ2−θ1

¿(θ2/θ1) o C

θ1 = T hi−T ci and θ2 = T ho−T co for parallel flow heat exchanger.

θ1 = T ho−T ci and θ2 = T hi−T co for counter flow heat exchanger.

5) Overall heat transfer coefficient based on outside surface area of inner tube

U o =Q

Ao LMTD W/m² K

Where AO = π dOl m2

dO = Outer diameter of the tube = 0.0125 m

l = Length of the tube = 1.0 m

6) Effectiveness:

ε = Th i−T ho

Th i−T ci if mh c p h<mc cpc

And ε = Tco−T ci

T hi−T ci if mc cpc<mh cp h

This is applicable both for Parallel and counter flow heat exchanger.

RESULT:

EFFECTIVENESS USING NUMBER OF TRANSFER UNITS (NTU) METHOD.

NTU = U o Ao

cmin

Note: If c p h < c pc then c p h <cmin , c pc < cmax

And ifc pc< c p h then c pc<cmin , c p h < cmax

5. INSTRUCTION MANUAL FOR MEASUREMENT OF EMISSIVITY APPARATUS

AIM: To determine the Emmissivity of a gray surface.THEROY: When heat is supplied at constant rate to any body heat loss takes place by condition, convection and radiation, if two bodies made of same geometry are heated under identical condition, the heat loss by conduction and convection can be assumed same for both the bodies. The heat loss by radiation depends on

a) Characteristic of the material b) Geometry of the surfacec) Temperature of the surface.

The heat loss by radiation when one body is completely enclosed by the other body is given by A1(T1

4 - T24)Q = -------------------------------

1/1 + A1/A2(1/2 -1) If the body is loosing heat surrounding atmosphere then A2 >> A1. Thus if any body loosing heat by radiation to the surrounding atmosphere equation (1) takes the form.

Q = A11 (T14 - T24)

Where = Stefan Boltzman constant = 5.67 x 10-8W/m 2K4

A1 = Surface area m2

1 = Emmissivity T1 = Surface temperature of the body K T2 = Surrounding atmospheric temperature K

When two bodies one is the standard body (black body) and the other is the test body (Gray body) are heated under identical condition. The condition and convection heat loss remains the same.Qs and Qt is the heat supplied to standard and test body respectivelyQs = Conduction heat loss + Convection loss + radiation lossQt = Conduction heat loss + Convection loss + radiation lossIf heat input to both the bodies are sameQs = Qt

Heat loss by radiation by the standard body = Heat loss by radiation by the test bodyAss (Ts

4 - Ta4) = Att (Tt

4 - Ta4)

A = As = At and s = 1

❑t = T s

4−T a4

T t4−T a

4

DESCRIPTION OF THE APPARATUS: The experimental set up consists of two circular plates of identical dimensions. One of the plates is made black by applying a thick layer of lamp black while the other plate whose Emmissivity is to be measured is a gray body. Heating coils are provided at the bottom of the plates the plates are mounted on asbestos cement sheet and kept in a enclosure to provide undisturbed natural convection conditions. Three thermocouples are mounted on each plate to measure the temperature. One thermocouples is in the chamber to measure the ambient temperature or chamber air temperature.

SPECIFICATIONS: Diameter of plates = 150 mm.PROCEDURE:

1) Connect the two heaters to the electric mains.2) Operate the dimmer stat and give the power input to both the plates.3) Adjust the same heat input to both the plates say 50V using toggle switch for heater 1&24) When the steady state is reached note down the temperature T1 to T6 by rotating the

temperature selection switch.5) Note down the air temperature in the chamber T7 = Ta

6) Repeat the experiment for different heat input.

OBSERVATION TABLE:

Heater input Temperature of gray

Surface oC

Temperature of black

surface oC

Ambient Temperature

T7 = Ta oC

VoltmeterReading

VVolts

AmmeterReading

IAmps

Watts

VI T1 T2 T3 T4 T5 T6

SPECIMEN CALCULATIONS:

1) Temperature of the test surface (gray body)

T t= T1+T 2+T 3

3 + 273 K

2) Temperature of the standard surface (black body)

T s = T 4+T5+T 6

3 + 273 K

3) Chamber air temperature

T a= T 7+ 273 K4Heat input to the coils = V x I Watts

5) Emissivity of test plate (gray body)

❑t = T s

4−T a4

T t4−T a

4

RESULT:Emissivity of test plate (gray body) =

6. HEAT TRANSFER THROUGH LAGGED PIPE

AIM:To plot the radial temperature distribution in the composite cylinder and to

determine the thermal conductivity of the pipe insulation.

DESCRIPTION:The apparatus consists of three concentric pipe mounted on suitable stands. The

inside pipe consists of the heater. Between the first two cylinders the insulating material with which lagging is to be done is asbestos and in between second and third pipe is wooden dust. The thermocouples are attached to the surface of cylinders to measure the temperature. The input to the heater is varied through a dimmer stat and measure on a voltmeter, Ammeter. The Experiment can be conducted at various values of input and calculations can be made accordingly.

SPECIFICATIONS:Location of thermocouples 1, 2, 3 at a radius =25 mm (INNER) Location of thermocouples 4, 5, 6 at a radius = 50mm Location of thermocouples 7, 8, 9 at a radius = 75mm (OUTER)

Length of the pipe = 500 mm Insulating material= Glass wool

Thermal conductivity (glass wool) K2 = 0.04 W/m K.

LIMITS AND PRECAUTIONS:1) Keep dimmer stat to zero position before start.

2) Increase voltage gradually.3) Keep the assembly undisturbed while testing.4) Operate selector switch of temperature indicator gently.

EXPERIMENTAL PROCEDURE : 1) Starts the supply of heater and by varying dimmer stat adjusts the input for desired values

by using voltmeter and ammeter.2) Take reading of all the 6 thermocouples when the steady state is reached.

OBSERVATION TABLE:S.No Voltmeter

ReadingV

Volts

AmmeterReading

IAmps

HeatInput

ViWatts

Thermocouples reading oC

T1 T2 T3 T4 T5 T6 T7 T8 T9

1) Mean temperature

Ti = T1+T 2+T 3

3 oC`

To =T7+T 8+T 9

3 oC

2) Heat supplied = V x I Watts

3) Heat conducted through composite cylinder

Q = 2l(T i – T o)

¿ r 2r 1

/ K Watts

Where,T i , T o- inner and outer wall temperature. r1, r2-inner and outer radii of the pipe.

4) Effective thermal conductivity

Q =

2l K eff (T i –T o)

¿r2

r1

Keff = Q∈

r2

r1

2l(T i –T o) W/mK

GRAPH: Using the above equation find the temperature for various values of ׳r׳ with the composite cylinder and tabulate theoretical and measured temperature. Then plot the temperature profile.

RESULT: Thermal conductivity of the pipe insulation =

7. STEFAN BOLTZMAN APPARATUS

AIM: To determine the value of Stefan Boltzman constant for radiation heat transfer.

THEORY:Stefan Boltzmann law states that the total emissive power of a perfect black body is

proportional to fourth power of the absolute temperature of black body surface Eb = σT4

Where, σ = Stefan Boltzman constant = 5.6697 x 10-8 W/ (m² K4)

DESCRIPTION:The apparatus consists of a flanged copper hemisphere fixed on a flat non-conducting

plate. A test disc made of copper is fixed to the plate. Thus the test disc is completely enclosed by the hemisphere. The outer surface of the hemisphere is enclosed in a vertical water jacket used to heat the hemisphere to a suitable constant temperature. Three Cr-Al thermocouples are attached at three strategic places on the surface of the hemisphere to obtain the temperatures. The disc is mounted on an ebonite rod which is fitted in a hole drilled at the center of the base plate. Another Cr-Al thermocouple is fixed to the disc to record its temperature. Fill the water in the SS water container with immersion heater kept on top of the panel.

SPECIFICATIONS:Specimen material : Copper

Size of the disc : 20mm x 0.5mm thicknessBase Plate : 250mm x 12mm thickness (hylam)Heater : 2 kW capacity, immersion type

Copper Bowl : 200mm Digital temperature indicator : 0 -199.9° CThermocouples used : 3 nos. on hemisphere Stop Watch : Digital typeOverhead Tank : SS, approx. 12 liter capacityWater Jacket : 230 mm, SSMass of specimen, ‘m’ : 5 gm

PROCEDURE:1. Remove the test disc before starting the experiment.2. Heat the water in the SS container to its boiling point.3. Allow the boiling water into the container kept at the bottom containing copper

hemisphere until it is full. Allow sufficient time to attain thermal equilibrium which is indicated by the four thermocouples provided on the hemisphere.

4. Insert the test disc fixed on the ebonite rod sleeve completely inside and lock it. Start the stop clock simultaneously.

5. Note down the temperature of the test disc at an interval of about 15 sec for about 15 to 20 minutes.

OBSERVATION TABLE:

Let Td = Temperature of the disc before inserting into the plate in K

ThermocoupleTemperature of the

Copper hemisphere ° CT1

T2

T3

Temperature – time response of test disc:Time t sec Temperature T5 ° C

CALCULATIONS:1. Plot the graph of temperature of the disc v/s time to obtain the slope (dT/dt) of the line,

which passes through/nearer to all points.

2. Average temperature of the hemisphere

Tavg = (T1 + T2 + T3) + 273.15 K 33. Td = Temperature of the disc before inserting to test chamber º K (ambient)

4. Rate of change of heat capacity of the disc = m Cp dT

dtNet energy radiated on the disc = σ Ad (T4

avg – T4d)

Where, Ad = area of the disc = πd² m2

d = 20 mm 4Cp = specific heat of copper = 0.38 kJ/kg–K

Rate of change of heat capacity of the disc = Net energy radiated on the disc

m Cp dT = σ Ad (T4avg – T4

d) dt

Thus ‘σ’ can be evaluated as shown m Cp dT

dt σ =

Ad (T4avg – T4

d)

RESULT:

The value of Stefan Boltzman constant for radiation heat transfer =

8. HEAT TRANSFER BY NATURAL CONVECTION

AIM:

To find out heat transfer coefficient and heat transfer rate from vertical cylinder in natural convection.

THEORY:

Natural convection heat transfer takes place by movement of fluid particles within to solid surface caused by density difference between the fluid particles on account of difference in temperature. Hence there is no external agency facing fluid over the surface. It has been observed that the fluid adjacent to the surface gets heated, resulting in thermal expansion of the fluid and reduction in its density. Subsequently a buoyancy force acts on the fluid causing it to flow up the surface. Here the flow velocity is developed due to difference in temperature between fluid particles.

The following empirical correlations may be used to find out the heat transfer coefficient for vertical cylinder in natural convection.

¼ 5

Nu = 0.53 (Gr. Pr) for Gr.Pr<10 ¼ 5 8

Nu = 0.56 (Gr.Pr) for 10< Gr. Pr <10 1/3 8 12

Nu = 0.13 (Gr.Pr) for 10 < Gr.Pr<10

Where,Nu = Nusselt number = hL k

Gr = Grashof number = L3 β g (Ts –Ta)ν²

Pr = Prandtl number = µcp

kβ = Volumetric coefficient of thermal expansionFor ideal gases β = 1

Tf

Where ‘Tf’ is the absolute film temperature at which the properties are taken.

SPECIFICATIONS:

Specimen : Stainless Steel tube,

Size of the Specimen : O.D 50mm, 500mm length

Heater : Nichrome wire type heater along its length

Thermocouples used : 8nos.

Ammeter : Digital type, 0-2amps, AC

Voltmeter : Digital type, 0-300volts, AC

Dimmerstat for heating coil : 0-230 V, 2 amps, AC power

Enclosure with acrylic door : For visual display of test section (fixed)

APPARATUS:

The apparatus consists of a stainless steel 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 purpose of undisturbed surroundings. One side of the duct is made of acrylic sheet for visualization. A heating element is kept in the vertical tube, which heats the tube surface. The heat is lost from the tube to the surrounding air by natural convection. Digital temperature indicator measures the temperature at different points with the help of seven temperature sensors, including one for measuring surrounding temperature. The heat input to the heater is measured by Digital Ammeter and Digital Voltmeter and can be varied by a dimmerstat.

PROCEDURE:

1. Ensure that all ON/OFF switches given on the panel are at OFF position.2. Ensure that variac knob is at zero position, provided on the panel.3. Now switch on the main power supply (220 V AC, 50 Hz).4. Switch on the panel with the help of mains ON/OFF switch given on the panel.5. Fix the power input to the heater with the help of variac, voltmeter and ammeter provided.6. Take thermocouple, voltmeter & ammeter readings when steady state is reached.7. When experiment is over, switch off heater first.8. Adjust variac to zero position.9. Switch off the panel with the help of Mains On/Off switch given on the panel.10. Switch off power supply to panel.

PRECAUTIONS:

1. Never switch on main power supply before ensuring that all on/off switches given on the panel are at off position

2. Never run the apparatus if power supply is less than 180 or above 200 volts.

TABULAR COLUMN:

Sl. No.

V Volts

I Amps

VI watts

Thermocouple readings °C

T1 T2 T3 T4 T5 T6 T7 T8

chamber

CALCULATIONS:

1. Ts = T1 + T2 + T3 + T4 + T5 + T6 + T7 + 273 K

7

Ta = Surrounding ambient temperature = T8 = + 273 K

2. Obtain the properties of air at a mean temperature of Tm = (Ts + Ta) K

2

3. Volumetric coefficient of thermal expansionβ = 1

Tm

4. Grashof Number, Gr = L3 β g (Ts – Ta) ν²

Where,ν =kinematic viscosity

L= length of the specimen(500 mm)

6. Rayleigh Number Ra = Gr.Pr

7. Nusselt Number Nu = hL

k

The following correlations are used to find Nusselt Number

Nu = 0.53 (Ra)1/4 for Ra < 105

Nu = 0.56 (Ra) 1/4 for 105 < Ra < 108

Nu = 0.13 (Ra) 1/3 for 108 < Ra < 1012

8. Free convective heat transfer coefficient

h = Nu k W/m²–K

L9. Heat transfer rate by convection

Qc = h A (Ts – Ta)

Qc = h π d L (Ts –Ta) watt

10. Heat Input to the coil

Qi = V x I watt

RESULT:

Heat transfer coefficient =

Heat transfer rate =

9. HEAT TRANSFER BY FORCED CONVECTION

AIM: To determine the convective heat transfer coefficient and the rate of heat transfer by forced convection for flow of air inside a horizontal pipe.

THEORY:

Convective heat transfer between a fluid and a solid surface takes place by the movement of fluid particles relative to the surface. If the movement of fluid particles is caused by means of external agency such as pump or blower that forces fluid over the surface, then the process of heat transfer is called forced convection.

In convectional heat transfer, there are two flow regions namely laminar & turbulent. The non-dimensional number called Reynolds number is used as the criterion to determine change from laminar to turbulent flow. For smaller value of Reynolds number viscous forces are dominant and the flow is laminar and for larger value of Reynolds numbers the inertia forces become dominant and the flow is turbulent. Dittus–Boelter correlation for fully developed turbulent flow in circular pipes is,

Nu = 0.023 (Re) 0.8 (Pr) n

Where, n = 0.4 for heating of fluid

n = 0.3 for cooling of fluid

Nu = Nusselt number = hd

K

Re = Reynolds Number = Vd

υ

Pr = Prandtl Number = μ cp

k

DESCRIPTION OF THE APPARATUS:

The apparatus consists of a blower to supply air. The air from the blower passes through a flow passage, heater and then to the test section. Air flow is measured by an orifice meter placed near the test section. A heater placed around the tube heats the air, heat input is controlled by a dimmerstat. Temperature of the air at inlet and at outlet is measured using thermocouples. The surface temperature of the tube wall is measured at different sections using thermocouples embedded in the walls. Test section is enclosed in a asbestos rope where the circulation of rope is avoid the heat loss to out side.

PROCEDURE:

1. Start the blower after keeping the valve open, at desired rate.2. Put on the heater and adjust the voltage to a desired value and maintain it as constant3. Allow the system to stabilize and reach a steady state.4. Note down all the temperatures T1 to T7, voltmeter and ammeter readings, and manometer

readings.5. Repeat the experiment for different heat input and flow rates.

SPECIFICATIONS :Specimen : Copper Tube

Size of the Specimen : I.D. 25mm x 400mm long

Heater : Externally heated, Nichrome wire Band Heater

Ammeter : Digital type,0-20amps, AC

Voltmeter : Digital type, 0-300volts, AC

Dimmerstat for heating Coil : 0-230v, 2amps

Thermocouple Used : 7 nos.

Centrifugal Blower : Single Phase 230v, 50 hz, 13000rpm

Manometer : U-tube with water as working fluid

Orifice diameter, ‘do’ : 25mm

G. I pipe diameter, ‘dp’ : 45 mm

OBSERVATION TABLE:

Room Temperature TR = ………. + 273.15 K

Sl. No Heater input

Diff. in Manometer reading hm

mm

Air temp. °C

Tube surface Temperature °C

Voltmeter reading V

volts

Ammeter reading I

amps

VI watts

Inlet T1

Outlet

T7

T2 T3 T4 T5 T6

SPECIMEN CALCULATIONS:

1. Mass density of air ρa = P kg/m³ RTR

Where, P = Atmospheric Pressure = 101325 N/m²

R = Gas constant for air = 287 J/kg K

TR = Room temperature in K

2. Pressure drop across orifice meter in ‘m of air ρm hm

ha = ρa

where, ρm = Mass density of water = 1000 kg /m3

hm = Differential manometer reading of water

3) Velocity of air at the orifice

Vo = Cd 2gha m/s

1- (do/dp)4

4) Velocity of air in the tube

Vo (d²o/4) Vo d²o m/s

Va = =

(d²s/4) d²s

(Note: Change in density of air with temperature is neglected i.e., ρa = constant)

5. Average surface temperature of the tube

Ts = T2 + T3 +T4 +T5 +T6 + 273.15 0K

5

6. Mean temperature of air

T∞ = T1 + T7 + 273.15 0K

2

Properties of air are taken at Tm = Ts + T∞ ………0K

2

At temperature Tm, kinematic viscosity ‘ν’, Prandtl number ‘Pr’ and thermal conductivity ‘k’ are taken from properties of air table

6. Reynolds Number Re = Va x ds

ν

8. Nusselt number Nu = 0.023 Re0.8 Pr

0.3

9. Nu = h x ds

k

Forced convective heat transfer h = Nu k W/m²-K

ds

10. Rate of heat transfer

Q = h A (T∞ – Ts)

Q = h π ds ls (T∞ – Ts)…… watt

RESULT:

The convective heat transfer coefficient =

The rate of heat transfer =

10. HEAT TRANSFER FROM PIN-FIN APPARATUS

AIM: To determine the value of heat transfer co-efficient under forced condition and to finda) Theoretical values of temperatures along the length of fin

b) Effectiveness and efficiency of the Pin-Fin for insulated and boundary condition

THEORY:

The heat transfer from a heated surface to the ambient surrounding is given by the relation, q = h A T. In this relation hc is the convective heat transfer coefficient, T is the temperature difference & A is the area of heat transfer. To increase q, h may be increased or surface area may by increased. In some cases it is not possible to increase the value of heat transfer coefficient & the temperature difference T & thus the only alternative is to increase the surface area of heat transfer. The surface area is increased by attaching extra material in the form of rod (circular or rectangular) on the surface where we have to increase the heat transfer rate. "This extra material attached is called the extended surface or fin."

The fins may be attached on a plane surface, and then they are called plane surface fins. If the fins are attached on the cylindrical surface, they are called circumferential fins. The cross section of the fin may be circular, rectangular, triangular or parabolic.

Temperature distribution along the length of the fin is

T-T cosh [m(L-x)]

= = 0 To-T cosh (mL)

Where, T = Temperature at any distance x on the fin

T0 = Temperature at x = 0

T = Ambient temperatureL = Length of the fin

where, h = convective heat transfer coefficient

hc P P = Perimeter of the fin

m = A = area of the fin

K A K = Thermal conductivity of the fin

Heat flow, ‘q’ = 0 hc PKA tanh mL

Effectiveness of a fin is defined as the ratio of the heat transfer with fin to the heat transfer from the surface without fins.

For end insulated condition

h PKA tan h mL

= 0 --------------------------

h A 0

PK

= ------- tanh (mL)

hA

The efficiency of a fin is defined as the ratio of the actual heat transferred by the fin to the maximum heat transferred by the fin if the entire fin area were at base temperature.

h PKA tan h mL

f = 0

h PL0

tan h mL

f =

mL

SPECIFICATIONS:

Length of the fin, ‘L’ = 150mm

Diameter of the fin, ‘df’ = 12mm

Thermal conductivity of fin material (brass) = 110.7 W/m2–K

Diameter of the orifice, ‘do’ = 25 mm

Width of the duct, ‘W’ = 150 mm

Breadth of the duct, ‘B’ = 100 mmCoefficient of discharge of the orifice, ‘Cd’ = 0.62

Density of manometer fluid (water) = 1000 kg/m3

PROCEDURE:1. Connect the equipment to electric power supply.2. Keep the thermocouple selector switch to zero position.3. Turn the Variac (dimmerstat) clockwise and adjust the power input to the heater to the

desired value and switch on the blower.4. Set the air–flow rate to any desired value by adjusting the difference in mercury levels in the

manometer and allow the unit to stabilize.5. Note down the temperatures, T1 to T6 from the thermocouple selector switch.6. Note down the difference in level of the manometer and repeat the experiment for different

power inputs to the heater.

OBSERVATION TABLE:

CALCULATIONS:

d0

= d0 = Diameter of the Orifice

dp dp = Diameter of the pipe

Velocity of Orifice

2gh (m –a) 1

Vo = Cd x a (1-)

m = density of manometric fluid = 13.6 x 10³ kg/m³

a = density of air = 1.17 kg/m³

Velocity at orifice x cross sectional area of orifice

Va = Velocity of air in the duct =

Cross sectional area of duct

V0 x (d0²)/4

Va =

W x B

where, dp = diameter of pipe

d0 = diameter of orifice

W = Width of the duct

B = Breath of the duct

Sl. No.

Heat Input Pressure drop, ‘h’

mm of water,

Temperatures, 0C

V A T1 T2 T3 T4 T5 T6

Average surface temperature of fin is given by

T1 + T2 + T3 + T4 + T5

Ts = + 273.15 K 5

T = T6 = Ambient temperature = + 273.15 K

Tm = Mean temperature = Ts + T

2

Properties of air at _____0C

ν = , Pr = , K =

Va df

Re = ----------- = Re = Reynolds number ν Pr = Prandtl number

Nu = Nusselt number

The relationship for Nu is

n 1/3

Nu = C Re Pr

For Re = 0.4 to 4.0 C = 0.989 and n = 0.33

Re = 4 to 40 C = 0.911 and n = 0.385

Re = 40 to 4000 C = 0.683 and n = 0.466

Re = 4000 to 40,000 C = 0.293 and n = 0.618

Re = 40,000 to 400,000 C = 0.27 and n = 0.805

Nu k

h =

df

Thermal conductivity of fin material, ‘K’ = 110.7 W/m–K

h P

m = --------

K A

Temperature distribution is given by

T- T Cosh m (L-x)

------ = ------------------

To-T Cosh mL

Therefore, T = T + (To- T) Cosh m (L-x)

Cosh mL

x1 = 0.045 T1 =

x2 = 0.075 T2 =

x3 = 0.105 T3 =

x4 = 0.135 T4 =

Effectiveness of fin = PK x tanh mL

h A

fficiency of fin = __tanh mL__

mL

RESULT:

11. THERMAL CONDUCTIVITY OF COMPOSITE WALL

Distance

x, m

Temperature from

Experiment °C

Temperature °C from calculation

AIM:

To determine the thermal resistance &thermal conductivity of composite slab.

APPARATUS:

Standard equipment for the determination of thermal conductivity ofComposite slab.

The apparatus consists of three disks of equal diameters but variablethickness arranged to form a slab of same diameter. A second set of the samearrangement is taken & an electric heater coil is sandwiched between these twoslabs in such a way that the nearest to farthest sequence from the heater is the same.

Two pressure plates are provided at the top & bottom of this entire assembly to ensure close contact between the adjacent plates in which case, contact resistance can be safely neglected.

A set of 8 thermocouples is incorporated to measure temperatures at everyinterface. A voltmeter & ammeter are provided to measure voltage and current tothe heater coil. A dimmer-stat is used to vary the input power to the heater.

THEORY:

A composite slab consists of slab of three different materials which are castiron, hylum & pressed wood. Slabs are circular in cross section & heating elementtoo. The heating element is placed between the three sets of slabs which are ofdifferent materials.

PROCEDURE:

1. Keep the dimmer-stat at minimum voltage position. Switch ON theelectric supply.2. Adjust the dimmer-stat to supply a particular voltage. Maintain thisconstant throughout the experiment.3. Wait for steady state to be attained.4. Note down the readings in the observation table as given below.

OBSERVATIONS:

Diameter of disks= 150 mm.Thickness of Asbestos= 12 mm.Thickness of Wood disk= 12 mm.Thickness of Mild steel= 12 mm.

OBSERVATION TABLE:

S.no V I Q=VI T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8

Mean readings; T A= T1+T 5

2

T B= T2+T 6

2

T C= T3+T 7

2

T D= T 4+T8

2

CALCULATIONS:

Read the Heat supplied Q = V x I Watts (In S. I. Units) For calculating the thermalconductivity of composite walls, it is assumed that due to large diameter of the plates,heat flowing through central portion is unidirectional i. e. axial flow. Thus forcalculation, central half diameter area where unidirectional flow is assumed isconsidered. Accordingly, thermocouples are fixed at close to center of the plates.

Now Q= V X I

2 W

A= πd2

4 (where d= diameter of disk)

1. Total thermal resistance of composite slab

R= T A+T D

Q

2. Thermal conductivity of composite slab

K=Qxt

A X (T ¿¿ A−T D)¿

t= Total thickness of composite slab

RESULT:

The total thermal resistance of the composite slab is = ºC/w.

The thermal conductivity of the composite slab is = W/mK.

12.THERMAL CONDUCTIVITY OF INSULATING POWDER

AIM:

To determine the thermal conductivity of insulating powder.

APPARATUS:

Standard equipment for the determination of thermal conductivity ofinsulating powder.

The apparatus consists of two thin walled concentric copper spheres. The inner sphere houses the heating coil. The insulating powder (in our case- Chalk Powder) is packed between the two shells. The power supply to the heating coil is adjustable. Fe-Constantan thermocouples are used to measures the temperatures. Thermocouples 1 to 4 are embedded on the inner surface of the inner sphere and thermocouples 5 to 8 are embedded on the outer shell.

THEORY:

It is advised that students should write the theory on the subject in their own words under the guidance of the teacher.

INSULATION:

Covering the surface with another surface with another material of low thermal conductivity in order to prevent excess heat transfer to the surrounding is termed as Insulation. In order to insulate material, it is poor conductor of heat andhence to cover the surface of heat. It is used where excess heat transfer is prevented.Electrical conductors are almost always good conductor of heat viz. Copper,Aluminum and Silver. & electrical conductors are good heat insulators.Commonly known heat insulators are Glass, Wood, Window glass, Saw dust,Chalk, Loosely packed or boards of sheet of asbestos.

PROCEDURE:

1. Keep the dimmer-stat at minimum voltage position. Switch ON the electric supply.

2. Adjust the dimmer-stat to supply a maximum 40W to the heating coil. Maintain this constant throughout the experiment.

3. Wait for steady state to be attained.

4. Note down the reading in the observation table as given below.

OBSERVATIONS:Radius of the inner sphere= ri =50 mm.

Radius of the outer sphere= ro =100 mm.

OBSERVATION TABLE:

S.no V I Q=VI T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8

CALCULATIONS:

Consider the transfer of heat by conduction through the wall of a hollow sphere formed by the insulating powdered layer packed between two thin copper spheres.Let, ri= Radius of inner sphere in meters.

ro= Radius of outer sphere in meters. T i= Average Temperature of the inner sphere in ºC T o= Average Temperature of the outer sphere in ◦C

Where, T i= T1+T 2+T 3+T 4

4

T o= T5+T 6+T7+T 8

4Thermal conductivity of insulating powder

K= Q(ro−r i)

4 πr i X ro(T i−T o) w/mK.

RESULT:

The thermal conductivity of insulating powder is = W/mK.