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THERMAL CONDUCTIVITY APPARATUS

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ME 1355 THERMAL LAB II

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TABULATION:

Sl. No.

Voltmeter reading (V)

Ammeter T1 reading (A)
0

T2
0

TAvg 1

T3
0

T4
0

TAvg 2

T5
0

T6
0

T7
0

T8
0

K W/m.k

C

C

C

C

C

C

C

C

KAvg =

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Ex.No:

Date

THERMAL CONDUCTIVITY APPARATUS-GAURDED HOT PLATE METHOD 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 = 2.5mm Diameter of specimen (d) = 20cm

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THERMAL CONDUCTIVITY APPARATUS

T T T

T

T T

T

T

GUARD HEATER

MAIN HEATER

SPECIMEN PLATES

MODEL CALCULATIONS:

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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. пd2/4 (Tavg1 – T5)/dx Where dx = 2.5mm = 0.0025m Diameter of specimen d = 20cm = 0.2m Q = K2 пd2/4 . (Tavg2 –T6)/dx ( for upper side) KAvg = (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.

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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.

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Ex.No: Date:

HEAT TRANSFER THROUGH COMPOSITE WALLS
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 = Δt/R Where Δt = is the overall temperature drop and R is the overall resistance to heat conduction. Since the resistance are in series R = R1 + R 2 Where R1 , R2 are resistance of each of the discs.

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

COMPOSITE WALLS

WOOD

T8

ASBESTOS

T7

T6 MS T5 HEATER T4 MS

ASBESTOS

T3

T2 WOOD

T1

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SPECIFICATION: 1. Thermal conductivity Of sheet asbestos Thickness 2. Thermal conductivity of wood Thickness 3. Dia. Of plates = 0.116 W/MK = 6mm = 0.052W/MK = 10mm = 300mm

4. The temperatures are measured from bottom to top plate T 1,T2,………….T 8. 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

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Q = (T 4 – T1)/R =(T2 –T1)/R1=(T3 – T2)/R2=(T4 – T3)/R3

COMPOSITE WALLS V A T1 76 T2 75 T3 72 T4 71 T5 66 T6 67 T7 50 T8 51 Time for 1 Rev. E.M

182 0.5

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 /A1 K1 = 0.005/0.07 X 69 X 10 -3 =1.03 Resistance of Wood (R2) = L2/A2 K2 = 0.008/0.07 X 52 X 10 -3 = 2.19 Heat flow Q1 = Temp. across Asbestos / R1 = 5/1.03 =4.85 Watts Q 2 = Temp. across Wood / R2 = 16/2.19= 7.3 Watts As per electrical anology Q 1 = 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

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TABULATION: Sl.No. Voltmeter Ammeter T1 reading (V) reading (A) T2 T3 T4 T5 T6

NATURAL CONVECTION
T6

T5 T4

T3 T2

T1

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Ex.No: Date: AIM:

HEAT TRANSFER BY FREE CONVECTION

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 = 50 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: -

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FORMULA USED: The power input to heater = V X A = hAΔt Where A = Area of heat transfer = χdl D = Dia. Of heater rod = 40mm L = Length of heater rod = 500mm Δt= 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 Δt / υ2 ß = 1 / Mean temp. of air + 273 K The properties of air at mean temperature = (T 1+T2+T3+…+T8 )/ 8 Hence h can be evaluated.

NATURAL CONVECTION: V 85 A 0.38 T10c 30 T20c 55 T30c 60 T40c 65 T50c 63 T60c 38

ß = 1/51.8 + 273 = 3 X 10 -3 Gr = ßgl3 Δt / υ2 Δt = [(T2 + T3 + T4 + T5) / 4 ]– [(T1+F6)/2]

= 3 X 10-3 X 9.81 X (0.5)3 X 26.75 / (17.96 X 10 -6)2 = 3.05 X 108

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where l = length of heater υ = Kinematic viscosity of air at mean temp. Pr = from data book for air mean temp. = 0.698 Hence GrPr = 2.13 X 108 Hence using free convection correlations Nu = hl / K = 0.13 (GrPr)1/3 where K is the Thermal conductivity of air at mean temp. = 72.82 Overall heat transfer co-efficient h = 72.82 X 28.26 X 10 -3 / 0.5 = 4.11 w/m2 -0 c

RESULT: The heat transfer coefficient is found to be -------------- W/m2 K

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Ex.No: Date:

FORCED CONVECTION
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.

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TABULATION: Sl No Inlet temp. of air Outlet temp. of air Temperatures along the duct

FORCED CONVECTION
ORIFICE DIA = 20 mm

PIPE DIA. = 40 HEATER BLOWER

T6

T5 T4

T3

T2

T1

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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. 6. Note manometer readings h1 and h2. 7. Note temperatures along the tube. Note air inlet and outlet temperatures 8. Note volt meter 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. 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) = (cd212 2√2gh0 ) / (√a12 – a22 ) m3 / sec. ho = heat of air causing the flow. = (h1 – h2 )℮w/ ℮a These

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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 / a 1 m/sec 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 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 Where Pr is the Prandtl number for which air can be taken as 0.7. The Prandtl number represents the fluid properties. The results may be represented as a plot of Nu exp/ Nu corr. Vs Re which should be a horizontal line. Re = Dv/ υ

FORCED CONVECTION V 50 A 1 T1 35 T2 42 T3 45 T4 46 T5 47 T6 38 h1 cm h2 cm 9 19

Avg. Temp. Of heater = (42 + 45 + 46 + 47) / 4 = 45 oC Avg. Temp. of Air = (35 + 38) / 2 = 36.5 oC Vol. Of air flow Q = (Cda1a2√2gh) / (√a12 – a22 ) Cd = 0.6 A1 = π/4 (0.04)2 = 1.256 X 10-3 A2 = π/4 (0.02)2 = 3.14 X 10-4 H = ℮ water/℮air (h1 – h 2) mtrs = 1000/1.16 (0.1) = 86.2 mtrs.

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Q = [0.6 X 1.256 X 10 -3 X 3.14 X 10-4 √(2 X 9.81 X 4)] / √(1.256X10-3)2 – (3.14X10-4)2 = 9.73 X 10-6 / 1.256 X 10-3 = 8.002 x 10-3 Velocity of air flow = Q / a 1 = 6.37 m/sec Re = D/ r = 15023 R = kinematic viscosity at mean temp. Using forced convection correlation Nu = hD /k = 0.023 Re 0.8 Pr 0.4 Pr at mean temp = 0.699 = 0.023(15023)0.8 (0.699)0.4 hD/k = 43.7 k= Thermal conductivity of air at mean temp h = 43.7 X 28.56 X 10 -3 / (0.04) = 31.2 w/mc.

RESULT: The heat transfer coefficient is found to be ---------------- W/m2 K

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Ex.No: Date:

STEFAN – BOLTZMAN APPARATUS
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, there by 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 indicator and

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 2. Dia. of the disc 3. Material of the disc 4. Cp = 0.005 kg. = 0.020 m. = copper = 381 J/KgK

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TABULATION: Avg.temp. of Sl.No. T1 T2 T3 hemisphere Th T4 Time Steady temp. of the disc. Td

HEATER

WATER

T2 T1 T3

T4

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PROCEDURE: 1. Allow water to flow through the hemisphere. Remove the disc from the bottom of the hemisphere. temperature. 2. Note down the temperatures T1, T2 and T3. The average of these temperatures is the hemisphere temperature (T h). 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). Switch on the heater and allow the hemisphere to reach a steady

CALCULATIONS : Q = ∑σ (T h4 – Td4) A.

σ= Q / ∑(Th4 – Td4) A and ∑=1.
The readings may be tabulated as follows:

T1 40.4

T2 40.1

T3 40.5

T4 33.8 33.9 34 34.1 34.2 . 34.7

Time 15 30 45 60 75

225

Final Temp of the disc σ = Q / Σb (T h4 – Td4) A.

34.7

Q = mass of the disc X Cp of disc X d/c Cp = 381J/Kgo k

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Q = 6.35 X 10-3 Avg. Temp. of hemisphere = 40.4 + 40.1 + 40.5 / 3 = 40.33 o C + 273 = 3 Td = 34.8o C = 307.8o k -4

A = Area of the disc = П / r (0.02)2 = 3.14 X 10-4 dT/dt = 0.1 / 30 J = 6.35 x 10-3 / [(315.3)4 – (307.8)4] x 3.14x10-4 = 2.228 x 10-8

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

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Ex.No: Date:

HEAT EXCHANGER TEST – PARALLEL FLOW AND COUNTERFLOW
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.

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TABULATION : FOR PARALLEL FLOW

Time for 1 Lit. of Time for 1 Lit. of Sl.No. Hot Water (sec) cold water (sec) T1 T2 T3 T4

FOR COUNTER FLOW Time for 1 Lit. of Time for 1 Lit. of Sl.No. Hot Water (sec) cold water (sec) T1 T2 T3 T4

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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 ID OD Outer GI tube ID = = = 12mm 15mm 40mm

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 lit./min. and cold water flow approximately 5 lit./min. 4. Switch ON the geyser. Allow the temperature to reach steady state. 5. Note temperatures T 1 and T2 (hot water inlet and outlet temperature respectively).

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COLD Tci

PARALLEL FLOW HEAT EXCHANGER

Thi Tho

HOT

Tco

PARALLEL FLOW
Tho

Thi

Length of the Exchanger

Tci

Ti

To Tco Tm = Ti Log e To Ti To

27

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6. Under parallel flow condition T 3 is the cold-water inlet temperature and T 4 is the cold water outlet temperature.Note the temperatures T 3 and T4.Under counter flow condition T4 is the cold-water inlet temperature T 3 is the cold-water outlet temperature. 7. Note the time for 1 liter flow of the hot and cold water. Calculate mass flow rate Kg/sec. 8. Change the water flow rates and repeat the experiment.

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COUNTER FLOW HEAT EXCHANGER
Tco

Thi Tho

HOT

Tci COLD

COUNTER FLOW
Tho

Thi

Ti To Tm = Tci Ti Log e To Ti To

Length of the Exchanger

Tco

29

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CALCULATIONS: Refer drawing and find LMTD (Δtm) = Δt1 – Δto / ln (Δt1 / Δto ) Please not Δt1 and Δto 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 (T hi – T ho) cp = specific heat of water (j/kgc) A = Outer area of hot water pipe. Mh = mass of hot water (kg/sec) 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) Neglecting the thickness of inner tube and film resistance where ho and h1 are the co-efficient of heat transfer of hot and cold side respectively. 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.

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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.

Parallel flow / Counter flow Heat exchanger. Parallel Flow Hot Water Inlet T1 Thi 40 Inlet T1 Thi 37 Cold Water Inlet T3 Tci 28 Outlet T4 Tco 31 24 Sec. 21 Sec. Time Time for Hot Cold water flow 1 lit. for Water

flow 1 lit.

LMTD = (T hi – Tci) – (Tho – Tco) / ln (Thi – Tci / Tho – Tco) = (40 – 28) – (37 – 31) / ln 40 – 28 / 37 – 31 Heat input qb = A.U LMTD Hence the overall heat transfer co-efficient, U = qb / A L M T D

qb = mb cb (T hi – T ho) = 1/24 X 4.178 X 3 = 0.522 kw U = 0.522 / K

Theoretical Method: 1/U = 1/hi + 1/ho

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hi = Volume of hot water flow = (1/24) / 1000 m3 / sec. = 4.166 X 10-5 m3 / sec. Velocity of hot water flow = 4.166 X 10 -5 / Π/4(0.012) 2 m/sec = 0.368 m/sec Re = Dυ / υ = 0.012 X 0.368 / 0.75 X 10 -6 = 5888. Using the heat transfer correlation Nu = 0.023 (Re)0.8 (Pr)0.3 = 0.023 (5888)0.8 (5.5)0.3 = 39.79 hi = 39.79 X 605 X 10 -3 / 0.012 = 2006 = hiD/k

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

ho = Volume flow rate of Cold water (1/21) / 1000 m3 / sec. Qc = 4.76 X 10-5 m3 / sec. Velocity of Cold water flow Vc = Q c / Ac Ac = Annulus area i.e. Π/4(0.04)2 - Π/4 (0.015)2 = 1.08 X 10-5 Vc = 4.76 X 10-5 / 1.08 X 10-3 = 0.044 m/sec Re = Dυ / υ = (0.04 – 0.015) X 0.044 / 0.75 X 10 -6 = 1466

Since the flow is not turbulent we can using the following equation.

Nu = 0.37(Re)0.6 (Pr)0.33 hoDc / k = 51.5 h o = 1247. 1/U = 1/hi + 1/ho = 1/2006 + 1/1247 Dc = Annulus dia. (0.04 – 0.015) = 0.025

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U = 769 W/m2 o c. This procedure is repeated for counter flow heat exchanger, however care to be taken while calculating LMTD.

RESULT : (i) Parallel flow Overall heat transfer coeffient by theoretical method ----------- W/ m2 K Overall heat transfer coeffient by prctical method (i) Counter flow Overall heat transfer coeffient by theoretical method ----------- W/ m2 K Overall heat transfer coeffient by prctical method ----------- W/ m2 K ----------- W/ m2 K

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Ex.No: Date:

THERMAL CONDUCTIVITY OF INSULATING MATERIAL - LAGGED PIPE
AIM : To find the thermal conductivity of different insulating material.

DESCRIPTION OF APPARATTUS : The insulation defined as 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. There is no insulation, which is 100% effective to prevent the flow of heat under temperature gradient. The experimental set up in which the heat is transferred through insulation by conduction is under study in the given apparatus. The apparatus consisting of a rod heater with asbestos lagging. The assembly is inside as 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 2 ΠL(Δt) / (In(r2)/r1) or = K2 2 ΠL(Δt) / (In(r3)/r2) Where Δt is the temperature 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.

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TABULATION : Asbestos Heat temperature S.No 1 2 3 avg 4 5 6 avg 7 8 avg temperature Sawdust temperature Volts Amps

LAGGED PIPE

SAW DUST

ASBESTOS

T1

HEATER

T3

ASBESTOS T4 SAW DUST T7 T4 T1 T7 = 20 mm = 80 mm T5 T8 T8 T6 T3 T5 T6

d1 - HEATER DIA DUST DIA

d2 - HEATER WITH ASBESTOS DIA = 40 mm d3 - ASBESTOS & SAW LENGTH = 500mm

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SPECIFICATION: Diameter of heater rod Diameter of heater rod with asbestos lagging Diameter of heater with asbestos lagging and saw dust The effective length of the above set up of cylinders = 20mm = 40mm = 80mm = 500mm

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 reading which given the 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 sawdust 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

90

0.4

108

117

89

51

59

53

41

41

Avg. Temp. of heater

= T1 +T2 +T3 / 3 = 104.6 o c

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Avg. Temp. of Asbestos lagging Avg. Temp. of sawdust lagging

= T4 + T5 + T6 / 3 = 54.3 o c = T7 + T8 / 2 = 41 o c

The heat flow from heater to outer surface of asbestos lagging = q k1 = k1 2 Πl (Δt) / ln (r2 / r1) = Thermal conductivity of asbestos lagging from data look at = 110.5 X 10-3 w/mo k. = 54o c r2 r1 l = Radius of the asbestos lagging = 20 = Radius of the heater = 10 mm = Length of the heater = 0.5 mtrs.

Substituting these values

q = 110.5 X 10-3 X 2 X Π X 0.5 X

Substituting this value of q to find the thermal conductivity of sawdust.

25.19 = k2 X 2 X Π X l X 13.3 / ln (r3/r2)

k2 = 25.19 X ln (40/20) / 2XΠX13.3X8. = 0.417

RESULT : Thermal conductivity of (i) (ii) Asbestas---------------W/mK Sawdust----------------W/mK

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Ex.No: Date:

HEAT TRANSFER FROM FINS

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 T AMB. 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 eigth 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.

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SPECIFICATIONS: Duct width Duct height Orifice dia. Orifice co-efficient Fin length Fin diameter (Characteristic length) b w do cd L df = = = = = = 150 mm 100 mm 20 mm 0.6 14.5cm 12mm

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

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TABULATION:

Manometer Sl.No. reading h1 h2

Fin surface temp. Amb. temp. T1 T2 T3 T4 T5 T6

PIN - FIN APPARATUS
T7 T6 T5 T4 T3 T2 T1

HEATER

BRASS PIN - FIN T8 ORIFICE DIA PIPE DIA = 20 mm = 40 mm LENGTH = 145 mm DIA = 12 mm

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CALCULATIONS :

Volume of air flowing through the duct

Vo

= cd a1a2 √2gha / √a1 2 a2 2

Where cd g ha a1 a2 h

= co-efficient of orifice = 0.6 = gravitational constant = 9.81 m/sec2 = heat of air = (lw /la)h = area of the pipe. = area of the orifice. = manometer differential head.

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

Where W D

= width of the duct. = breadth of the duct.

REYNOLD’S NUMBER OF AIRFLOW:

Reynold’s number Re = (L X Va X Pa) / μa

Where Va Pa μa L

= Velocity of air in the duct. = density of air in the duct. = Viscosity of air at to C. = length of fin.

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PRANDTL NUMBER OF AIRFLOW Prandtl number = (cpa X μa ) / ka

Where cpa μa ka

= specific heat of air. = viscosity of air = thermal conductivity of air.

HEAT TRANSFER CO-EFFICIENT CALCULATIONS

NUSSELT NUMBER (Nu)

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 X (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)

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= Tan Hyperbolic ML/ML

Where, h L M P

= = = =

heat transfer co-efficient length of the fin √hp/(kb X A) perimeter of the fin (π X dia of the fin)

A kb

= =

cross sectional area of the fin. thermal conductivity of brass rod.

Temperature distribution = Tx = [cosh M (L-X) /cosh ML (T o - 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 < 10 4 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 K = = Dia. of fin thermal conductivity of air.

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Gr = Grashof number = gβ ΔT

D 3 / r2

β = 1/ (Tav + 273) ΔT= (Tav – Tamb)

Pr = Prandtl Number = (μ Cp) / K

PIN-FIN

V 135

A 0.6

T1 67

T2 61

T3 59

T4 56

T5 49

T6 47

T7 46

T8 29

h1cm h2cm 75 15.5

Mean Temp = 51.75 o C

Vol. of airflow thro’ duct = Q = Cd a 1 a2 √2gh / √a12 - a22

a1 = Π/4 (0.04)2 = 1.256 X 10-3 a2 = Π/4 (0.02)2 = 3.14 X 10-4 h =℮w / ℮a ( 0.155 – 0.075) = 68.96m

Q = 8.704 X 10-6 / 1.216 X 10-3 = 7.158 X 10-3 m3/sec

Velocity of air flow thro’ duct = Q/A

A = Length X Breadth of the duct = 0.15 X 0.1 = 0.015 m2 Velocity = 0.477 m/sec

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Re = Dυ / υ = D = Length of the Fin = 0.145 = 0.145 X0.477/ 17.95 X 10 -6 = 3853

Using the correlation For 40 > Re <4000 Nu = 0.683 (Re)0.466 (Pr)0.33 = 0.683(3853)0.466 (0.698)0.33 = 28.436

Pr = hl / K K= Thermal conductivity of airflow at mean time

Heat transfer coefficient, h

= 28.436 X 28.26 X10 -3 / 0.145 = 5.54 w/m2 o c

M

= √hp / kb A = √ 5.54 X 0.0376 / 110.7 X Π/4(0.012)2 = 4.078 kg/m

Fin efficiency Temp. distribution T2

= Tan G ML/ML =0.89 = 89 o = [cosh M (L-X) /cosh ML (T o - Ta)] + Ta = [cosh M (0.022) /1.17997 (67 - 29)] – 0.85088(38) = T3 T4 T5 T6 T7 = = = = = 61.33o c 0.8588(32) + 29 = 56.22 o c 0.8588(30) + 29 = 54.5 o c 0.8588(27) + 29 = 51.9 o c 0.8588(20) + 29 = 46 o c 0.8588(18) + 29 = 44 o c

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RESULT : The efficiency of the fin is found to be ---------------------Temperature at x = 20mm, T 20 = ------------Temperature at x = 40mm, T 40 = ------------Temperature at x = 60mm, T 60 = ------------Temperature at x = 80mm, T 80 = -------------

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Ex.No: Date:

TEST ON EMISSIVITY APPARATUS

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 Kirchaff’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 = 150mm

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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) f) g) Tabulate the readings and calculate. After the experiment is over turn both the energy regulators 1 & 2. For various power inputs repeat the experiment.

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TABULATION :

Black Sl.No.

body Average Temp. Tb

Polished body temperature T1 T2 T3 Average Chamber Temp. Tp Temp. T4

Emmissivity ε

temperature T5 T6 T7

EMISSIVITY APPARATUS

CHAMBER T5

T1

T4

T2

T3

T6

T7

TEST PLATE DIA. - 150 mm

BLACK PLATE DIA. - 150 mm

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CALCULATIONS: Temperature of the black body in absolute unit T ba Temperature of the polished body in absolute unit T pa Temperature of the chamber in absolute unit T ca Emissivity εp = εb X T4 ba - T4 ca / T4 pa - T4 ca = T b + 273 = T p + 273 = T 7 + 273

Where εb, emissivity of black body which is equal to 1. Emmissivity apparatus : V 100 A 0.4 T1 89 T2 92 T3 90 T4 40 T5 79 T6 80 T7 81

Avg. temp. of polished plate = 363.3 o k = (89 + 92 + 90 / 3) + 273 Avg. temp. of Black plate = (79 + 80 + 81 / 3) + 273 = 353 o k. Chamber temp. = 40 + 273 = 313 o k

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) = 3534 - 3134 / 3634 - 3134 = 5.9 X 109 / 7.76 X 109 = 0.76

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


				
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