BHARATHIDASAN ENGINEERING COLLEGE

NATTRAMPALLI – 635 854

DEPARTMENT OF MECHANICAL ENGINEERING

ME6411-THERMAL ENGINEERING LAB-II

YEAR / SEMESTER : III/ V

DEPARTMENT : Mechanical

REGULATION : 2013

Name : ………………………………………

Reg No : ………………………………………

Branch : ………………………………………

Year & Semester : ………………………………………

LABORATORY MANUAL

THERMAL CONDUCTIVITY APPARATUS-GUARDED HOT PLATE METHOD Ex.No: 1 Date AIM: To find the thermal conductivity of the specimen by two slab guarded hot plate method. DESCRIPTION OF APPARTUS:

The apparatus consists of a guarded hot plate and cold plate. A specimen whose thermal conductivity is to be measured is sand witched between the hot and cold plate. Both hot plate 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 multi point digital temperature indicator with selector switch is provided to note the temperatures at different locations. An electronic 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.

The whole assembly is kept in an enclosure with heat insulating material filled all around to minimize the heat loss. SPECIFICATION: Thickness of specimen = mm Diameter of specimen (d) = mm MODEL CALCULATIONS:

FORMULA USED:

Since the guard heater enables the heat flow in uni direction

q = KA dT/dx

Where A = Surface area of the test plate considered for heat flow = m2

dx = Thickness of the specimen plate = m

dt = Average temperature gradient across the specimen = °C

q = Q/2 since the heat flow is from both sides of the heater = watts

Tavg1 = (T1 + T2 )/ 2 ; Tavg2 =( T3 +T4 )/ 2

Q = V.I. Watts

Q = K1 A. dT / dx (for lower side)

Q = K1. π d 2/4 (Tavg1 – T5)/dx

Where dx = mm = m

Diameter of specimen

d = cm = m

Q = K2 πd2/4 (Tavg2 –T6)/dx ( for upper side)

K avg = (K1 + K2 )/ 2 PROCEDURE:

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

main heater to any desired value.

2. Adjust the guard heater’s regulator so that the main heater temperature is less than or

equal to the guard heater temperature.

3. Allow water through the cold plate at steady rate. Note the temperatures at different

locations when the unit reaches steady state. The steady state is defined, as the

temperature gradient across the plate remains same at different time intervals.

4. For different power inputs is in ascending order only the experiment may by repeated

and readings are tabulated as below. RESULT:

The thermal conductivity of the specimen is found to be ------------- W/mK.

HEAT TRANSFER THROUGH COMPOSITE WALLS

Ex.No:2 Date: Aim:

To determine the rate of heat transfer through different layers of composite wall Description of Apparatus:

When heat conduction takes place through two or more solid materials of different

thermal conductivities, the temperature drop across each material depends on the resistance offered to heat conduction and the thermal conductivity of each material.

The experimental set-up consists of test specimen made of different materials aligned

together on both sides of the heater unit. The first test disc is next to a controlled heater. The

temperatures at the interface between the heater and the disc is measured by a thermocouple,

similarly temperatures at the interface between discs are measured. Similar arrangement is

made to measure temperatures on the other side of the heater. The whole set-up is kept in a

convection free environment. The temperature is measured using thermocouples (Iron-Cons) with multi point digital temperature indicator. A channel frame with a screw rod arrangement

is provided for proper alignment of the plates.

The apparatus uses a known insulating material, of large area of heat transfer to enable

unidirectional heat flow. The apparatus is used mainly to study the resistance offered by

different slab materials and to establish the heat flow is similar to that of current flow in an

electrical circuit. The steady state heat flow Q = t/R Where t = is the overall temperature drop and

R is the overall resistance to heat conduction.

Since the resistance are in series

R = R1 + R2 Where R1, R2 are resistance of each of the discs.

TABULATION:

Sl.No. Voltmeter Ammeter T1 T2 T3 T4 T5 T6 T7 T8

reading reading

COMPOSITE WALLS

WOOD T8

ASBESTOS T7

T6

MS T5

HEATER

T4

MS

ASBESTOS T3

T2

WOOD

T1

SPECIFICATION:

1. Thermal conductivity

Of sheet asbestos = 0.116 W/MK

Thickness = 6mm

2. Thermal conductivity of wood = 0.052W/MK

Thickness = 10mm

3. Dia. Of plates = 300mm

4. The temperatures are measured from bottom to top plate T1,T2,………….T8.

PROCEDURE

1. Turn the screw rod handle clockwise to tighten the plates.

2. Switch on the unit and turn the regulator clockwise to provide any desired heat input.

3. Note the ammeter and voltmeter readings.

4. Wait till steady state temperature is reached.

5. (The steady state condition is defined as the temperature gradient across the plates

does not change with time.)

6. When steady state is reached note temperatures and find the temperature gradient

across each slab.

7. Since heat flow is from the bottom to top of the heater the heat input is taken as Q/2

and the average temperature gradient between top and bottom slabs from the heater to

be taken for calculations. Different readings are tabulated as follows.

CALCULATION: Now the resistance ( R ) offered by individual plates for heat flow. R1 = L1/AK1 R2 = L2 / AK2 R3 = L3/AK3 Where A = Area of the plate

K = Thermal Conductivity

L = Thickness of the plate. Knowing the thermal conductivities Q = (T4 – T1)/R =(T2 –T1)/R1=(T3 – T2)/R2=(T4 – T3)/R3

COMPOSITE WALLS

V A T1 T2 T3 T4 T5 T6 T7 T8 Time for 1 Rev.

182 0.5 76 75 72 71 66 67 50 51 E.M

heater ms 71.5 ashess 66.5 wood 50.5

Area of the plate п / 4 (0.3)2 = 0.07m2 Resistance of Asbestos (R1) = L1 /A1K1 = 0.005/0.07 X 69 X 10-3

=1.03 Resistance of Wood (R2) = L2/A2K2 = 0.008/0.07 X 52 X 10-3 =

2.19 Heat flow Q1 = Temp. across Asbestos / R1 = 5/1.03 =4.85 Watts

Q2 = Temp. across Wood / R2 = 16/2.19= 7.3 Watts

As per electrical anology Q1 = Q2 = Q3 Total Resistance R3 = 1.03 + 2.19 = 3.22

Q3 =(Temp. across Asbestos + Wood) / R3 = 21/3.22 = 6.521

As we have find the inside heat transfer co-efficient for heat flow from heater to MS plate, we consider only the second and third layer. RESULT: The rate of heat transfer through different materials are found to be

a. MS section = ------------- W b. Wood section = ------------- W c. Asbestos section = --------------W

Ex.No: 3 HEAT TRANSFER BY FREE CONVECTION Date: AIM:

To find the heat transfer coefficient under natural convection environment. DESCRIPTION OF APPARATUS:

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

surface. When the fluid movement is caused by density differences in the fluid due to

temperature variations, it is called FREE or NATURAL CONVECTION.

This apparatus provides students with a sound introduction to the features of free

convection heat transfer from a heated vertical rod. A vertical duct is fitted with a heated

vertical placed cylinder. Around this cylinder air gets heated and becomes less dense, causing

it to rise. This in turn gives rise to a continuous flow of air upwards in the duct. The

instrumentation provided gives the heat input and the temperature at different points on the

heated cylinder. SPECIFICATION:

Length of cylinder = cm 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 temperatures along the heater rod.

4. Note ammeter and voltmeter readings.

5. For different power inputs the experiments may be repeated. The readings are tabulated as below: - FORMULA USED: The power input to heater = V x A = hA t

Where A = Area of heat transfer = πdl

D = Dia. Of heater rod = mm

L = Length of heater rod = mm

t= Avg. temp. Of heater rod – Avg. temp. of

air. H = Overall heat transfer co-efficient.

THEORETICAL METHOD

Using free convection correlations for vertical cylinders. Nu = hl / K = 0.53(GrPr)1/4 for GrPr < 105 Nu = hl / K = 0.56(GrPr)1/4 for 105 < GrPr < 108 Nu = hl / K = 0.13(GrPr)1/3 for 108 < GrPr < 1012 Characteristic length is the height of the cylinder (l)

K = Thermal conductivity of air P = Prandtl number of air Gr = ßgl3 t / υ2 ß = 1 / Mean temp. of air + 273 K The properties of air at mean temperature = (T1+T2+T3+…+T8)/ 8 Hence h can be evaluated.

NATURAL CONVECTION:

V A T10 c T2

0 c T30 c T4

0 c T50 c T6

0c

ß = 1/51.8 + 273 = 3 X 10-3

Gr = ßgl3 t / υ2 t = [(T2 + T3 + T4 + T5) / 4 ]– [(T1+F6)/2]

=

=

Where l = length of heater

υ = Kinematic viscosity of air at mean temp. Pr = from data book for air mean temp.

= Hence GrPr = Hence using free convection correlations Nu = hl / K = 0.13 (GrPr)1/3 where K is the Thermal conductivity of air at mean temp.

= Overall heat transfer co-efficient h = = W/m2-0c RESULT:

The heat transfer coefficient is found to be -------------- W/m2K

FORCED CONVECTION

Ex.No: 4 Date : AIM:

To find the heat transfer coefficient under forced convection environment. DESCRIPTION OF APPARATUS:

The important relationship between Reynolds number, Prandtl number and Nusselt

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

The experimental set up (see sketch) 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 temperatures of the

flowing medium are measured by thermocouples 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 is 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 valve is provided to regulate the flow rate of air.

OBSERVATION TABLE Length of the test section L = 300 mm Co- efficient of discharge of the orifice plate, Cd = .62 Diameter of the orifice plate, d = 20 mm Inside dia. of the pipe, D = 25 mm Specific gravity of water ρw = 1000 Specific gravity of water ρa = 1.28 Orifice pr. In mm of H (h2 – h2 ) =

FORCED CONVECTION:

V A T10 c T2

0 c T30 c T4

0 c T50 c T6

0c T70c

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 position of the valve to any desired flow rate of air.

5. Wait till steady state temperature is reached, for 5min

6. Note manometer readings h1 and h2.

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

8. Note voltmeter and ammeter reading.

9. Adjust the position of the valve and vary the flow rate of air and repeat the experiment.

10. For various valve openings and for various power inputs and readings may be taken to

repeat the experiments. The readings are tabulated

The heat input Q = h A L M T D = m Cp (Temp. of tube – Temp. of air) M = mass of air.

Cp = specific heat of air. LMTD = (Avg temp of tube – outlet air temp) – (Avg. temp of tube – inlet air temp.)

1n x (Avg. temp of tube – outlet temp. of air)

(Avg. temp of tube – inlet temp. of air)

H= Heat transfer co-efficient. A = Area of heat transfer = T1d1 From the above, the heat transfer co-efficient ‘h’ can be calculated. These experimentally

determined values may be compared with theoretical values.

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

The volume of air flowing through the tube (Q) = (cd a1 a2√ 2gh0 ) / (√ a12 – a2

2 ) m3 /

sec. ho = heat of air causing the flow.

= (h1 – h2) ρw/ ρa h1 and h2 are manometer reading in

meters. a 1= area of the tube. a2 = area of the orifice.

Hence the velocity of the air in the tube, V = Q / a1 m/sec.The heat transfer rate and

flow rates are expressed in dimension less form of Nusselt number and Reynolds number

which are defined as

Nu = h D/K Re = Dv/ υ

D = Dia. of the pipe

V = Velocity of air

K = Thermal conductivity of air.

The heat transfer co-efficient can also be calculated from Dittus-Boelter

correlation. Nu = 0.023 Re 0.8 Pr 0.4

RESULT: The heat transfer coefficient is found to be -------------- W/m2K

STEFAN – BOLTZMAN APPARATUS

Ex.No:5 Date: AIM:

To find Stefan-Boltzman constant.

DESCRIPTION OF APPARATUS:

Stefan – Boltzman law which establishes the dependence of integral

hemispherical radiation on temperature. We can verify this phenomenon 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 multi point digital temperature

indicatorand thermocouples (Fe/Ko) are provided to measure temperature at

various points on the radiating surface of the hemisphere and on the disc.

SPECIFICATIONS:

1. Mass of the disc = kg.

2. Dia. of the disc = m.

3. Material of the disc = copper

4. Cp = 381 J/KgK

TABULATION:

Avg.temp. of Steady

temp. of

Sl.No. T1 T2 T3 hemisphere T4 Time

the disc.

Th

Td

PROCEDURE:

1. 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.

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

hemisphere temperature (Th).

3. Refit the disc at the bottom of the hemisphere and start the stop clock.

4. 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).

CALCULATIONS :

Q = ∑ σ (Th4 – Td 4) A.

σ = Q / ∑ (Th4 – Td 4) A and ∑ =1.

The readings may be tabulated as follows:

T1 T2 T3 T4 Time

Final Temp of the disc σ

= Q / Σb (Th4 – Td

4) A.

Q = Mass of the disc X Cp of disc X d/c

Cp = 381J/Kgo K

Q = 6.35 x 10-3

Avg. Temp. of hemisphere = = o C + 273 = Td = RESULT:

Stefan Boltzman constant is found to be------------W/m2 K4

HEAT EXCHANGER TEST – PARALLEL FLOW AND COUNTERFLOW

Ex.No:6 Date: Aim:

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

DESCRIPTION OF APPARATUS:

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 type of heat exchangers are most widely used. A transfer type of heat

exchanger is one in which both fluids pass simultaneously through the device and head is

transferred through separating walls. Transfer type of exchangers are further classifies as

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

2. Counter flow type in fluids 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 fluid flowing through the inner tube and the other through the annulus surrounding it.

The heat transfer takes place across the walls of the inner tube.

TABULATION :

FOR PARALLEL FLOW

Sl.No. Time for 1 Lit. of Time for 1 Lit. of

T1 T2 T3 T4

Hot Water (sec) cold water (sec)

FOR COUNTER FLOW

Sl.No. Time for 1 Lit. of Time for 1 Lit. of

T1 T2 T3 T4

Hot Water (sec) cold water (sec)

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 any 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. Temperatures of the fluids can be measured using thermometers. 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.

SPECIFICATIONS:

Length of the heat exchanger

Inner copper tube inner diameter = mm

Outer diameter = mm

Outer GI tube ID = mm PROCEDURE:

1. 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.

2. For parallel flow open valve V2, V4 and V5.

For counter flow open valve V3, V1 and V5.

3. Control the hot water flow approximately 2 l/min. and cold water flow approximately

5 l/min.

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

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

6. Under parallel flow condition T3 is the cold-water inlet temperature and T4 is the cold

water outlet temperature. Note the temperatures T3 and T4.Under counter

flowcondition T4 is the cold-water inlet temperature T3 is the cold-water outlet

temperature.

7. Note the time for 1 liter flow of the hot and cold water. Calculate mass flow rate in kg/s.

8. Change the water flow rates and repeat the experiment.

CALCULATIONS: Refer drawing and find LMTD ( tm) = t1 – to / ln (Δt1 / Δto) Please note t1 and to to be calculated as per drawing for Parallel flow and Counter flow. Qh = A U (L M T D) Hence the overall Heat transfer co-efficient U = Qh / A L M T D Where Qh = mh Cp (Thi – Tho) Cp = Specific heat of water (J/kg0 C) A = Outer area of hot water pipe. Mh = Mass of hot water (kg/s) Effectiveness of Heat exchanger = Actual heat transfer/ Max. possible heat transfer = (tco – tci) / (thi – tci) THEORETICAL METHOD: The overall Heat transfer co-

efficient 1/U = (1/ho) + (1/h1) Neglect the thickness of inner tube and film resistance. h1 = Inside heat transfer co-efficient (from hot to inner surface of the inner tube) ho = Out side heat transfer co-efficient (from outer wall of the inner tube to the cold

fluid). Re = hot water flow = Dυ / υ υ = Velocity of hot water. Knowing the mass flow rates (υ) may be calculated for hot and cold water.

Nu = 0.023 (Re)0.8 (Pr)0.3 = (hiD) /K

K = Thermal conductivity of water.

In a similar manner ho can also be calculated. However for finding ho the

characteristic dia. is taken as the annulus which is given by the (ID of the outer pipe – OD of

outer pipe). Hence, ‘U’ the overall Heat transfer co-efficient is evaluated for Parallel flow / Counter flow

Heat exchanger.

Parallel Flow

Hot Water Temperature Cold Water Temperature Time taken

Time taken for for 1 litre

Inlet outlet Inlet Outlet

1litre Hot water Cold Water

T1 T1 T3 T4 flow. flow.

Thi Thi Tci Tco

LMTD = (Thi – Tci) – (Tho – Tco) / ln (Thi – Tci / Tho – Tco)

=

Heat input Qb = A.U LMTD

Hence the overall heat transfer co-efficient, U = Qb / A L M T D

Qb = mb Cb (Thi – Tho)

= Theoretical Method:

1/U = 1/hi + 1/ho

hi = Volume of hot water flow = m3 / sec.

= m3 / sec.

Velocity of flow of hot water = m/sec

= m/sec

Re = Dυ / υ =

Using the heat transfer correlation

Nu = 0.023 (Re)0.8 (Pr)0.3 = hiD/k

k = Thermal conductivity of water Pr = Values from data book RESULT:

THERMAL CONDUCTIVITY OF INSULATING

MATERIAL - LAGGED PIPE

Ex.No:7 Date: AIM :

To find the thermal conductivity of different insulating materials. DESCRIPTION OF APPARATUS :

The insulator is a material, which retards the heat flow with reasonable effectiveness.

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

combination of these three.

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

conduction is under study.

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.

Now the heat flow through the lagging materials is given by

Q = K1 2π L (Δt) / (ln (r2)/r1) or

= K2 2π L(Δt) / (ln(r3)/r2) Where t is the temperature difference across the lagging. K1 is the thermal conductivity of asbestos lagging material

and K2 is the thermal conductivity of saw dust. L is the length of the cylinder. Knowing the thermal conductivity of one lagging material the thermal conductivity of the

other insulating material can be found.

TABULATION :

Heater temperatures

Asbestos Sawdust Applied Current

temperatures temperatures Voltage

S.No

volts Amps

T1 T2 T3 avg T4 T5 T6 avg T7 T8 avg

LAGGED PIPE

SAW DUST

ASBESTOS

HEATER T1 T3

ASBESTOS T4 T5 T6

SAW DUST T7 T8

T4 T1 T7 T5 T8 T6 T3

d1 - HEATER DIA = 20 mm d2 - HEATER WITH ASBESTOS DIA = 40 mm d3 - ASBESTOS & SAW DUST DIA = 80 mm LENGTH = 500mm

SPECIFICATION: Diameter of heater rod, d1 = Diameter of heater rod with asbestos lagging, d2 = Diameter of heater with asbestos lagging and saw dust, d3 = The effective length of the cylinder = 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. Now note down the ammeter,

voltmeter readings the product of which give heat input.

4. Temperatures 1, 2 and 3 are the temperature of heater rod, 4, 5 and 6 are the

temperatures on the asbestos layer, 7 and 8 are temperatures on the saw dust lagging.

5. The average temperature of each cylinder is taken for calculation. The temperatures

are measured by thermocouple (Fe/Ko) with multi point digital temperature indicator.

6. The experiment may be repeated for different heat inputs. The readings are tabulated as below: CALCULATIONS : Lagged Pipe:

V A T1 T2 T3 T4 T5 T6 T7 T8

Avg. Temp. of heater = T1 +T2 +T3 / 3 = o C

Avg. Temp. of Asbestos lagging = T4 + T5 + T6 / 3 = o C

Avg. Temp. of sawdust lagging = T7 + T8 / 2 = o C

The heat flow from heater to outer surface of asbestos lagging =

Q = k1 2 πl (Δt) / ln (r2 / r1)

k1 = Thermal conductivity of asbestos lagging, from data look at------------------ o C

(average temp of asbestos lagging)

= W/m K.

r2 = Radius of the asbestos lagging =

r1 = Radius of the heater = mm

l = Length of the heater = m

Substituting these values

Q = ( ) 2 π x l x (Δt) / (r2 / r1)

Substituting this value of q to find the thermal conductivity of saw dust, K2 Q= K2 x 2 π x l x (Δt) / ln (r3/r2) K2 = x ln ( )/ 2 π x x .

= RESULT : Thermal conductivity of

(i) Asbestos---------------W/mK

(ii) Sawdust----------------W/mK

HEAT TRANSFER FROM FINS

Ex.No:8 Date: AIM:

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

FIN efficiency. DESCRIPTION OF APPARATUS:

Consider a PIN-FIN having the shape of rod whose base 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 surface 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.

The apparatus consists of a Pin-fin placed inside an open duct, (one side open) the

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

is taken up through a gate valve and an orifice meter to the atmosphere. The airflow rate can

be varied by the gate valve 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.

SPECIFICATIONS:

Duct width b = mm

Duct height w = mm

Orifice dia. do = mm

Orifice co-efficient cd =

Fin length L = cm

Fin diameter df = mm

(Characteristic length)

PROCEDURE:

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

2. Keep the thermocouple selector switch in first position.

3. Turn the regulator knob to clockwise and set the power to the heater to any desired

value by looking at the voltmeter and ammeter.

4. Allow the unit to stabilize for 10min

5. Switch ON the blower.

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

manometer limb levels.

7. Note down the temperatures indicated by temperature indicator.

8. Repeat the experiment by

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

b. Varying the power input to the heater and keeping the air flow rate

constant.

9. Tabulate the readings and calculate for different conditions.

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

knob anti clockwise, put off the main switch and disconnect the power supply.

For 40 < NRe < 4000 Nnu = 0.683 (NRe) 0.466 (NPr) 0.333 For 1 < NRe < 4 Nnu = 0.989 (NRe)0.33 (NPr)0.333 For 4 < NRe < 40 Nnu = 0.911 (NRe)0.385 (NPr)0.333 For 4000 < NRe < 40000 Nnu = 0.193 (NRe)0.618 (NPr)0.333 For NRe > 40000 Nnu = 0.0266 (NRe)0.805 (NPr)0.333 Heat transfer co-efficient h = Nnu (Ka / L) Ka = thermal conductivity of air L = length of fin. Efficiency of the pin-fin = actual heat transferred by the fin

(heat which would have been transferred if entire fin where

at the base temperature)

= Tan Hyperbolic ML/ML Where, h = Heat transfer co-efficient

L = Length of the fin

M = √ hp/ (Kb X A)

P = perimeter of the fin

(π D)

D = dia of the fin

A = cross sectional area of the fin.

Kb = thermal conductivity of brass rod. Temperature distribution = Tx = [cosh M (L-X) /cosh ML (To - Ta)] + Ta

X = distance between thermocouple and heater.

EVALUATION OF THE HEAT TRANSFER CO-EFFICIENT (h) Natural convection (blower off)

Nuav = (hd)/k = 1.1 (Gr Pr)1/6 for 1/10 < Gr Pr < 104 Nuav = 0.53 (Gr Pr)1/4 for 104 < Gr Pr < 109 Nuav = 0.13 (Gr Pr)1/3 for 109 < Gr Pr < 1012

Where Nuav = average Nusselt number =

(hD) / K

D = Dia. of fin

K= thermal conductivity of air.

Gr = Grashof number = gβ TD3 / r2

β = 1/ (Tav + 273)

T= (Tav – Tamb)

Pr = Prandtl Number = (μ Cp) / K

PIN-FIN

V A T1 T2 T3 T4 T5 T6 T7 T8 h1cm h2cm

Re = Dυ / υ = D = Length of the Fin =

= Using the correlation For 40 > Re <4000 Nu = 0.683 (Re)0.466 (Pr)0.33

=

= Heat transfer coefficient, h =

=

M = √ hp / Kb A = π/4

Pr = hl / K

K= Thermal conductivity of airflow at mean time

= Fin efficiency = Tan G ML/ML = =

Temp. distribution = [cosh M (L-X) /cosh ML (To - Ta)] + Ta

T2 = [cosh M

=

T3 =

T4 =

T5 =

T6 =

T7 =

RESULT : The efficiency of the fin is found to be ---------------------- Temperature at x = 20mm, T20 = ------------- Temperature at x = 40mm, T40 = ------------- Temperature at x = 60mm, T60 = -------------

Temperature at x = 80mm, T80 = -------------

TEST ON EMISSIVITY APPARATUS Ex.No:9 Date: AIM:

To measure the emissivity of the test plate surface. DESCRIPTION OF APPARATUS :

An ideal black surface is one, which absorbs the radiation falling on it. Its reflectivity

and transivity is zero. The radiation 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 emissivity of that body. For a black body absorptivity is 1 and by

Kirchhoff’s law its emissivity is also 1. Emissivity depends on the surface temperature and the

nature of the surface.

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

are provided with heating coils at the bottom. The plates or mounted on thick asbestos sheet

and kept in an enclosure so as to provide undisturbed natural convection surroundings. The

heat input to the heaters is varied by two regulators and is measured by an ammeter and

voltmeter. 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 is kept in the enclosure to read the chamber temperature. One plate is blackened

by a layer of enamel black paint to form the idealized black surface whereas the other plate is

the test plate. The heat dissipation by conduction is same in both cases. SPECIFICATION: Diameter of test plate and black surface = mm

PROCEDURE:

a) Connect the three pin plug to the 230V, 50Hz, 15 amps main supply and switch on

the unit.

b) Keep the thermocouple selector switch in first position. Keep the toggle switch in

position 1. By operating the energy regulator 1 power will be fed to black plate.

Now keep the toggle switch in position 2 and operate regulator 2 and feed power

to the test surface.

c) Allow the unit to stabilize. Ascertain the power inputs to the black and test

surfaces are at set values. i.e. equal.

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

temperatures indicated by the temperature indicator from channel 1 to 7.

e) Tabulate the readings and calculate.

f) After the experiment is over turn off both the energy regulators 1 & 2.

g) For various power inputs repeat the experiment.

TABULATION :

Black body Polished body Emmissivity

Sl.No. temperature Average

temperature Average Chamber

ε

Temp. Tb

Temp. Tp

Temp. T4

T5 T6 T7 T1 T2 T3

EMISSIVITY APPARATUS

CHAMBER

T1 T4 T5

T2 T3 T6 T7

TEST PLATE BLACK PLATE DIA. - 150 mm DIA. - 150 mm

CALCULATION S:

Temperature of the black body in absolute unit T=

Temperature of the polished body in absolute unit T=

Temperature of the chamber in absolute unit T= Emissivity εp = εb X T4 ba - T4 ca / T4 pa - T4 ca

Where εb, emissivity of black body which is equal to 1.

EMMISSIVITY APPARATUS :

V A T1 T2 T3 T4 T5 T6 T7

Avg. temp. of polished plate

= Avg. temp. of Black plate

= Chamber temp. =

Power Input Q = ΣpσA (Tp4 - Ta4) = ΣbσA (Tb4 - Ta4)

Since the power input is same for both heaters and area of radiating surface (A) is also

same, knowing the Σb =1. The emmissivity of polished surface

Σp = Σb (Tb4 - Ta4) / (Tp4 - Ta4)

=

=

=

RESULT :

Emissivity of the specimen is found to be ---------------

Experiments on Air Conditioning System

Ex.No:10 Date:

Introduction:

Air Conditioning for human comfort or industrial process requires certain processes to be carried out on air to vary the psychometric properties of air to requirements. These processes may involve the mixing of air streams, heating of air, cooling of the air, humidifying air, and dehumidifying air and combination of the process. All such processes are studied with the given air-condition test rig.

c] Procedure for doing the experiment:

Step No. Details of the Step 1. Switch on the mains. 2. Switch on the condenser, fan and blower. 3. Switch on the compressor and allow the unit to stabilize. 4. Note down the following.

a] Pressure P1, P2, P3 and P4 from the respective pressure gauge. b] Note the corresponding Temperatures T1, T2, T3 and T4 at the

respective state points. c] Monometer readings. d] Note DBT and WBT at the inlet of the duct [before evaporation]. e] Note DBT and WBT at the outlet of the duct [after evaporation].

FORMULA:

DBT = Dry Bulb Temperature [Td] WBT = Wet Bulb Temperature [Tw]

[1] ha = ρw hw / ρa

ρw = Density of water [1000 kg/m3]. hw = Manometer reading. ρa = Density of air [1.123 kg/m3].

[2] Va = Velocity of air

Va = √2 x g x ha

g = acceleration due to gravity 9.81m/s2.

[2] Velocity of air [Va]

Va = √2 x g x ha

.

[3] Mass of air [ma] = ρa x A x Va

[4] Refrigeration effect = ma [h2 – h1]. [5] Capacity = Refrigeration effect / 3.5

[1 tonne of refrigeration = 210 KJ/min. = 3.5 KW] [6] Carnot COP = TL / [TH - TL]

TL = Lower temperature to be maintained in the evaporator. P1 = 55 PSI = 55 x 0.07 + 1.013 = 4.863 bar. P4 = 67.5 PSI = 67.5 x 0.07 + 1.013 = 5.738 bar.

Pmin = [P1 + P4] / 2 = [4.863 + 5.738] / 2 = 5.3 bar.

From Table R – 22 TL = 20C = 275 K TH = Higher temperature to be maintained in the condenser.

P2 = 285 PSI = 285 x 0.07 + 1.013 = 20.963 bar. P3 = 270 PSI = 270 x 0.07 + 1.013 = 19.913 bar.

Pmax = [P2 + P3] / 2 = [20.963 + 19.913] / 2 = 20.438 bar. From Table Freon – 22, TH = 520C = 325 K.

Carnot COP = TL / [TH - TL]

Carnot COP =

[7] Theoretical COP

Theoretical COP = [h1–h3] / [h2–h1] [Where h1, h2, h3 are enthalpies of refrigerant taken from p-h chart.] P1 = 4.863 bar; T1 = 1.1120C; h1 = 260 KJ/kg. P2 = 5.738 bar; T2 = 48.880C; h2 = 300 KJ/kg. P3 = 19.913 bar; T3 = 48.880C; h3 = 100 KJ/kg. Theoretical COP = [260 – 100] / [300 – 260] Theoretical COP = 4.

[3] Mass of air ma = ρa x A x Va

ρa = Density of air [kg/m3] Va = Velocity of air [m/s] A = H x L

[4] Refrigeration effect = ma [h2 – h1].

h2 = Enthalpy of ambient air [KJ/kg.] h1 = Enthalpy of condition air [KJ/kg.]

[5] Capacity = Refrigeration effect /3.5 [6] Carnot COP = TL / [TH - TL]

TL = Lower temperature to be maintained in the evaporator in absolute unit [0K].

TH = Higher temperature to be maintained in the condenser in absolute unit [0K].

[7] Theoretical COP = [h1–h3] / [h2–h1] h1 corresponding to P1 and T1.

h2 corresponding to P2 and T2. h3 corresponding to P3 and T3.

[Enthalpy is to be found out from the P-h diagram of R-22]

Result: Thus the experiment on the air condition system was conducted and result were

tabulated.

Carnot COP Theoretical COP Capacity TR

Performance test on Two stage reciprocating Air Compressor Ex.No:11 Date:

OBJECTIVE [AIM] OF THE EXPERIMENT To conduct a performance test on a two stage air compressor and determine its

volumetric efficiency. FACILITIES REQUIRED AND PROCEDURE a] Facilities required to do the experiment:

Sl. No. Facilities required Quantity 1. Two stage reciprocating air compressor. 1

b] Description The air compressor is a two stage reciprocating type. The air is sucked from

atmosphere and compressed in the first cylinder. The compressed air then passes through an inter cooler into the second stage cylinder, where it is further compressed. The compressed air then goes to a reservoir through a safety valve. This valve operates an electrical switch that shuts off the motor when the pressure exceeds the set limit.

The test unit consists of an air chamber containing an orifice plate and a U – tube manometer; the compressor and an induction motor. Compressor Specification:

Diameter of low pressure piston = 70 mm. Diameter of high pressure piston = 50 mm. Stroke = 90 mm.

KC Compressor Details: Model : DPS S. No. : 317 RPM : 900

Induction Motor Details: S. No. : 1970 KW : 2.2 RPM : 1440

Precautions: 1. Check oil level in the compressor crank case. 2. The orifice should never be closed, test the manometer liquid [water] will be

sucked into the tank. 3. At the end of the experiment the outlet valve at the air reservoir should be opened

as the compressor is to be started again at low pressure to prevent undue strain on the piston.

c] Procedure for doing the experiment:

Step No. Details of the Step 1. Close the outlet valve. 2. Fill up the manometer with water up to the half level. 3. Start the compressor and observe the pressure developing slowly. 4. At the particular test pressure, the outlet valve is opened slowly and

adjusted so that the pressure in the tank is maintained constant.

5. Observe the following readings. [i] Speed of the compressor – Nc R.P.M. [ii] Manometer reading h1 and h2 cm of water. [iii] Pressure gauge reading P Kg/cm2.

FORMULA:

Volumetric Efficiency ηv = Vact / Vtheo x 100

Vact = Actual volume of air compressed. Vact = Cd x A x √2gH m3 / sec. Cd = Co-efficient of discharge of Orifice = 0.62. A = Orifice Area [Dia. = 20 mm] g = 9.81 m/sec2. h = Water head causing flow.

Theoretical Volume of air Vtheo = [3.14 x D2 x L x Nc] / 4 x 60.

D = Dia. Of piston = 0.07m.

L = Stroke length = 0.09 m. Nc = RPM of the compressor.

ρw = Density of water Kg/m3. ρair = Density of air Kg/m3.

[1] ACTUAL VOLUME OF AIR: Vact = Cd x A x √2gH m3 / sec.

Cd = 0.62. = Coefficient of discharge. d = 20 mm. A = π/4 [20/1000]2 = 0.000314 m2. g = 9.81 m/sec2

h = h1 – h2 = 10.3 – 8.3 = 2 Cm. H = [h x ρw] / 100 x ρair ρair = Density of air kg/m3.

ρw = Density of water kg/m3. = [2 x 1000] / 100 x 1.162] ρw = 1000 kg/m3.

ρair = 1.162 kg/m3. H = 17.21.

S.

No.

Receiver Pressure

Kgf/cm2

Speed of the compressor

[RPM]

Manometer Reading

h - h

cm

V m3/s

V

m3/s

Volumetric Efficiency

Vact / Vtheo x 100% h cm

h cm

Vact = Cd x A x √2gH m3 / sec.

= 0.62 x 0.000314 x √2 x 9.81 x 17.21 Vact = 3.58 x 10-3 m3 / sec

[2] THEORTICAL VOLUME OF AIR: Vtheo = 3.14 x D2 x L x Nc / 4 x 60

= 3.14 x [0.07]2 x 0.09 x 880 / 4 x 60.

D = 0.07 m.

Vtheo = 5.08 x 10-3 m3/sec L = 0.09 m. Nc = 880 rpm.

[3] VOLUMETRIC EFFICIENCY: Volumetric Efficiency ηv = Vact / Vtheo x 100

= [3.58 x 10-3] / [5.08 x 10-3] x 100

d] Result: Thus the volumetric efficiency of the reciprocating air compressor are determined.

S. No.

Pressure [Kgf/cm2]

Volumetric Efficiency [τvol]

1. 2. 3. 4. 5.