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Transcript of Heat Fare NEW
EXPERIMENT NO: 01 DATE: 25-07-12
COMBINED RADIATION AND CONVECTION
AIM
To determine the heat transfer due to radiation and convection from cylinder.
To find emissivity at different temperatures.
To find natural convection coefficient.
APPARATUS REQUIRED
Thermometer
Heater
Stopwatch
Sand paper
THEORY
If a surface, at a temperature above that of its surroundings, is located in stationary air at the
same temperature as the surroundings then heat will be transferred from the surface to the air
and surroundings. This transfer of heat will be a combination of natural convection to the air
(air heated by contact with the surface becomes less dense and rises) and radiation to the
surroundings. A horizontal cylinder is used in this exercise to provide a simple shape from
which the heat transfer can be calculated.
Note: Heat loss due to conduction is minimised by the design of the equipment and
measurements mid way along the heated section of the cylinder can be assumed to be unaffected
by conduction at the ends of the cylinder. Heat loss by conduction would normally be included
in the analysis of a real application.
12 | P a g e
In the case of natural (free) convection the Nusselt number Nu depends on the Grashof and
Prandtl numbers and the heat transfer correlation can be expressed in the form:
Nusselt number, Nu = f (Gr, Pr)
Rayleigh number, Ra = (Gr Pr)
The average heat transfer coefficient for radiation Hrm can be calculated using the following
relationship:
H rm=ΣζF(T s¿¿ 4−T a
4)(T s−T a)
¿
The average heat transfer coefficient for natural convection Hcm can be calculated using the
following relationship:
T film=T s+T a
2
β= 1T film
N gr=gβ (T s−Ta)D3
υ2
Rad = (Grd × Pr)
Num = b(Rad)n, (c and n can be obtained from the table below)
H fm=kN um
D
Note: k, Pr, and n are physical properties of the air taken at the film temperature Tfilm.
The actual power supplied to the heated cylinder, Qin = V×I
Table for c and n values for natural convection on horizontal cylinder.
Rad c N
10-9 to 10-2 0.675 0.058
10-2 to 102 1.02 0.148
102 to 104 0.850 0.188
104 to 107 0.480 0.250
107 to 1012 0.125 0.333
13 | P a g e
Alternatively a simplified equation may be used to calculate the heat transfer coefficient for free
convection : H cm=1.32× ¿¿
The value for Hcm should be calculated using both the original and simplified equations and the
values compared.
EXPERIMENTAL PROCEDURE
The surface of the cylindrical brass specimen is polished using the sand paper.
The dimensions of the cylinder are noted down.
Now with the help of thermometer the room temperature is noted.
The cylinder is now heated till it reaches a temperature of about 200 ̊ C.
Now with the help of tongs the cylinder is taken out of the heater and thermometer is
inserted into the space in the centre of it.
Now the temperature is noted at an interval of 1 minute till it drops down to 90 ̊ C.
OBSERVATIONS
Minutes Fall in temperature
temperature
Minutes Fall in temperature
temperature 0 200 11 128
1 197 12 121
2 190 13 114
3 184 14 110
4 176 15 106
5 169 16 102
6 162 17 98
7 155 18 95
8 148 19 92
9 142 19.40 90
10 135
14 | P a g e
Weight of cylinder = 0.995 kg
Diameter of cylinder = 4 cm
Length of cylinder = 125 cm
MODEL CALCULATION
T film=T s+T a
2 =
473+3012 = 387 K
Properties of air at Tfilm, Cp = 0.0294×103 J/kg μ = 13.7 Pas k = 0.03227 W/m2K ρ = 0.03314×103 mol/m3
β= 1T film
¿1
387 = 2.5839×10-3 K-1
N gr=gβ (T s−Ta)D3
υ2 ¿ 9.8 × .0028539 ×172× 0.1253
0.172 = 0.05
N pr=μ Cp
k ¿
13.7× 29.40.03227 = 12.48×106
Nubottom = 0.27(Ngr×Npr)0.25 = 7.591
Nutop = 0.54(Ngr×Npr)0.25 = 14.177
Nulateral = 0.59(Ngr×Npr)0.25 = 15.489
htop=kNutop
D ¿
0.03227 ×7.5910.04 = 12.248 W/m2K
hbottom=kNubottom
D ¿ 0.03227 ×14.177
0.04 = 6.124 W/m2K
hlateral=kNulateral
D ¿
0.03227 ×7.5910.125 = 3.998 W/m2K
Qcon = ((htop×Atop) + (hbottom×Abottom) + (hlateral×Alateral))
= ((12.248×0.001256) + (6.124×0.001256) + (3.998×0.000157))
= 14.75 W
Qtotal=mC
p dTdt
¿0.955 ×502 × (200−90)19 ×60
= 46.25 W
15 | P a g e
Qrad = Qtotal – Qcon = 46.25 – 14.75 = 31.25 W
hrad=Qrad
A ∆T ¿ 31.5
0.018× 172 = 10.66 W\m2K
RESULT
Natural Convection coefficient = 10.66 W/m2K
16 | P a g e
EXPERIMENT NO: 02 DATE: 01-08-12
SHELL AND TUBE HEAT EXCHANGER
AIM
To study of heat transfer in shell and tube heat exchanger.
INTRODUCTION
To Heat exchanger is deviced in which heat is transferred from one fluid to another. The
necessity for doing this arises in a multitude of industrial applications. Common examples of
heat exchangers are the radiator of a car, the condenser at the back of a domestic refrigerator
and the steam boiler of a thermal power plant.
Heat exchangers are classified in three categories:
Transfer Type
Storage Type
Direct Contact Type
THEORY
A transfer type of heat exchanger is one on which both fluids pass simultaneously trough the
device and heat is transferred through separating walls. In practice, most of the heat exchangers
used are transfer type ones.
The transfer type heat exchangers are further classified according to flow arrangements as –
Single pass
17 | P a g e
Multiple pass
A sample example of transfer type heat exchanger can be in the form of a tube type arrangement
in which one of the fluids is flowing through the inner tube and the other through the annulus
surrounding it. The heat transfer takesplace across the walls of the inner tube.
The heat lost by the hot fluid can be calculated
qh = Heat transfer rate to the hot water.
qh = mh Cph (Thi –Tho)
Heat taken by the cold fluid can also be calculated
qc = Heat transfer rate to the cold water
qc = mc Cpc (Tco – Tci )
Qavg = qc + qh
2
Uo = Qavg
Ao ∆ Tm
18 | P a g e
DESCRIPTION
The apparatus consists of 1- 2 pass Shell and Tube heat exchanger. The hot fluid is hot water,
which is attained from an insulating water bath using a magnetic drive pump and it flow
through the inner tube while the cold water flowing through the annuals. For flow measurement
rotameters are provided at inlet of cold water and outlet of hot water line. The hot water bath is
of recycled type with digital temperature controller.
UTILITIES REQUIRED
Water supply 20 lit/min (approx.)
Drain.
Electricity supply: 1 phase, 220 V AC, and 4 Kw.
Floor area of 1.5 m× 0.75 m
EXPERIMENTAL PROCEDURE
19 | P a g e
STARTING PROCEDURE
Clean the apparatus and make water bath free from dust. Close all the drain valves provided.
Fill water bath ¾ with clean water and ensure that no foreign particles are there.
Connect cold water supply to the inlet of cold water rotameter line.
Connect outlet of cold water from Shell to Drain.
Ensure that all On/Off switches given on the panel are at OFF position.
Now switch on the main power supply.
Switch on heater by operating rotary switch given on the panel.
Set temperature of the water bath with the help of digital temperature controller.
Open flow control valve and by-pass valve for hot water supply.
Switch on magnetic pump for hot water supply.
Adjust hot water flow rate with the help of flow control valve and rotameter.
Record the temperatures of hot and cold water inlet & outlet when steady state is achieved.
CLOSING PROCEDURE
When experiment is over, switch off heater first.
Switch of magnetic pump for hot water supply.
Switch off power supply to panel.
Stop cold water supply with the help of flow control valve.
Drain cold and hot water from the shell with the help of given drain valves.
Drain water bath with the help of drain valve.
SPECIFICATIONS
Water Shell.
Material = S.S.
Dia. = 220 mm
Length = 500 mm
25% cut baffles at 100 mm distance 4 Nos.
Tube
20 | P a g e
material =S.S.
OD = 16 mm
ID = 13 mm
Length of tubes = 500 mm
Nos. of tubes = 24
Temperature Controller = Digital 0- 199.90 C
Temperature Sensors = RTD PT- 100 type (5 nos.)
Temperature Indicator = Digital 0 to 199.90 C with multi-channel switch.
Electric Heater = 2 kW (2 Nos.)
Flow measurement = Rotameter (2 Nos.)
Water bath = Material: SS insulated with ceramic wool and powder coated MS outer shell
fitted with heating element.
Pump = FHP magnetic drive pump (max. operating temp. 850 C).
FORMULAE
Rate of heat transfer from hot water
Qh = MhCph(Thi – Tho ), Watt
Mh = Fh∗10−3∗ρh
3600 , Watt
Rate of heat transfer from hot water
Qc = McCpc(Tco – Tci ), Watt
Mc = Fc∗10−3∗ρc
3600 , Watt
Average heat transfer
Q =Qh+Qc
2 , Watt
LMTD
ΔTm = ΔTi−ΔTo
ln ( ΔTiΔTo
)
Where, ΔTi = Thi - Tc
And, ΔTo = Tho - Tci
21 | P a g e
Note that in a special case of counter flow exchanger exists when the heat capacity rates Cc
and Ch are equal, then Thi – Tco = Tho– Tci thereby making ΔTi = ΔTo. In this case LMTD is
of the form 0/0 and so undefined. But it is obvious that since ΔT is constant throughout the
exchanger, hence
ΔTm = ΔTi = ΔTo
Overall Heat Transfer coefficient
Ui =Q
AiΔTm , W/m2°C
Uo =Q
AoΔTm , W/m2°C
OBSERVATIONS
Ai = 3.187×10-3 m2
Ao = 4.827×10-3 m2
Sl NO Hot Water Side Cold Water side
Fh (LPH) Thi °C Tho °C Fc (LPH) Tci °C Tco °C
1 155 50.2 40 115 28 35
2 155 50.8 39 165 28 34
3 115 50.5 37 165 28 33
CALCULATION
Mh = 125× 10−3 ×988.5423600
= 0.034 kg/sec.
Qh = (0.034 × 4180.5 ×(322.5−312)) = 1493.44 W
Mc = 155× 10−3 ×995.673600
= 0.042 kg/sec.
Qc = (0.042×4180.5×(310-301)) = 1580.23 W
22 | P a g e
Q = 1493.44+1580.23
2 = 1536.8345 W
ΔTi = Thi – Tco = 12.5
ΔTo = Tho- Tci = 11
ΔTm =ΔTi−ΔTo
ln ( ΔTiΔTo
) = 17.73K
Ai = 0.49m2
Ao = 0.603 m2
Ui = Q
Ai× ΔT m W/m2°C
= 1536.83
.49× 17.73
= 176.89 W/ m2°C
Uo = Q
Ao × ΔT m W/m2°C
= 1536.83
.603× 17.73
= 143 W/m2°C
NOMENCLATURE
Ai =inside area of heat transfer,m2
Ao=outside area of heat transfer,m2
Cpc=specific heat of cold fluid at mean temp. ,J/kgoc
Cph=specific heat of hot fluid at mean temp.,J/kgoc
Do=outer dia.of S.S tube,m
FC=flow rate of cold water,LPH
Fh=flow rate of hot water,LPH
L=length of the tube,m
MC=cold water flow rate,kg/s
23 | P a g e
Mh=hot water flow rate,kg/s
Ph=thermal conductivity of hot fluid,W/moC
Q=average heat transfer,W
QC=heat gained by cold water,W
Qh=heat lost by hot water,W
Rh=density of hot fluid, kg/m3
Tc=mean temp. of cold water oC
Tci=cold water inlet temp. oC
Tco=cold water outlet temp. oC
Th=mean temp. of hot water. . oC
Thi=hot water inlet temp. oC
Tho=hot water outlet temp. oC
Ui=inside overall heat transfer coefficient, W/m2
Uo=outside overall heat transfer coefficient, W/m2 oC
ρc=density of cold water at mean temp.kg/m3
ρh=density of hot water at mean temp.kg/m3
∆Tm=logarithm mean temp. difference, oC
PRECAUTIONS
Never switch on main power supply before ensuring that all the on/off switches given on
the panel are at off positions.
Never switch on heaters before filling water bath ¾ with clean water. It may damage heaters
Never run the pump at voltage less than 180 & above 230 volts
Never fully close the delivery and by-pass line valves simultaneously.
Always keep apparatus free from dust.
To prevent cloggig of moving parts, run pump at least onces in a fortnight.
RESULT
24 | P a g e
The total heat transfer coefficient for fc: Lph is,
Ui = 176.89 W/m2K Uo = 143.00 W/m2K
EXPERIMENT NO: 03 DATE: 08-08-12
PARALLEL AND COUNTERFLOW HEAT EXCHANGER
OBJECTIVE
To study the heat transfer phenomena in parallel / counter flow arrangements.
AIM
25 | P a g e
To To calculate rate of heat transfer, LMTD and overall heat transfer coefficient for both
type of heat exchanger.
To compare the performance of parallel and counter current flow heat exchanger.
INTRODUCTION
To Heat exchanger is deviced in which heat is transferred from one fluid to another. The
necessity for doing this arises in a multitude of industrial applications. Common examples of
heat exchangers are the radiator of a car, the condenser at the back of a domestic refrigerator
and the steam boiler of a thermal power plant.
Heat exchangers are classified in three categories:
Transfer Type
Storage Type
Direct Contact Type
THEORY
A transfer type of heat exchanger is on which both fluids pass simultaneously through the
device and heat is transferred through separating walls. In practice most of the heat exchangers
used are transfer type ones.
The transfer type exchangers are further classified according to flow arrangement as-
Parallel flow in which fluids flow in the same direction.
Counter flow in which they flow in opposite direction and
Cross flow in which they flow at right angles to each other.
A simple example of transfer type of heat exchanger in the form of a tube type arrangement in
which one of the fluids is flowing through the inner tube and the other through the annulus
surroundings it. The heat transfer takes place across the walls of the inner tube.
DESCRIPTION
The apparatus consists of a tube in tube type concentric tube heat exchanger. The hot fluid is hot
water which is obtained from an insulated water bath using a magnetic drive pump and it flow
26 | P a g e
through the inner tube while the cold fluid is cold water flowing through the annuals. The hot
water flows always in one direction and the flow rate of which is controlled by means of a
valve. The coid water can be admitted at one of the end enabling the heat exchanger to run as a
parallel flow apparatus or a counter flow apparatus. This is done by valve operations. For flow
measurement rotameters are provided at inlet of cold water and outlet of hot water line. A
magnetic drive pump is used to calculate the hot water from a recycled type water tank, which is
fitted with heaters and digital temperature controller.
UTILITIES REQUIRED
Electricity supply: single phase, 220 VAC, 50Hz, 5-15Amp socket with earth connection.
Water supply: 10 lit/min (approx.)
Drain
Bench area required: 2m x 0.6 m
EXPERIMENTAL PROCEDURE
Put water in bath and switch on the heaters.
Adjust the required the temp. of hot water using DTC.
Adjust the valve. Allow hot water to recycle in bath through by-pass by switching on the
magnetic pump.
Start the flow through annulus and run the exchanger either as parallel flow or counter flow
unit.
Adjust the flow rate on cold water side by rotameter.
Adjust the flow rate on hot water side by rotameter.
Keeping the flow rates same, wait till the steady state conditions are reached.
Record the temperatures on hot water and cold water side and also the flow rates accurately.
Repeat the experiment with a counter flow under identical flow conditions.
NOMENCLATURE
Ai = inside heat transfer area, m2
27 | P a g e
Ao = outside heat transfer area, m2
Cph= specific heat of hot fluid at mean temp., kJ /kg 0C
Cpc = specific heat of cold fluid at mean temp., kJ/kg 0C
Do = outer diameter of tube , m
Di = inner diameter of tube, m
Fh = flow rate of hot water, LPH
Fc = flow rate of cold water, LPH
L = length of tube , m
Mh = mass flow rate of the hot water, kg/s
Mc = mass flow rate of the cold water, kg/s
Q = average heat transfer from the system, W
Qc = heat gained by the cold water,W
Qh = heat loss by the hot water, W
Th = mean temp. of hot water, 0C
Tc = mean temp. of cold water, 0C
Tho = outlet temp. of the hot water, 0C
Thi = inlet temp. of the hot water, 0C
Tco = outlet temp. of the cold water, 0C
Tci = inlet temp. of the cold water, 0C
ΔTm = log mean temp. difference, 0C
Ui = inside overall heat transfer coefficient,W/m2 0C
Uo = outside overall heat transfer coefficient, W/m2 0C
ρc = density of cold water at mean temp., kg/m3
ρh = density of hot water at mean temp., kg/m3
OBSERVATIONS
Sl. No: Flow Fh LPH Thi 0C Tho 0C Flow Fc LPH Tci 0C Tco 0C
1Counter
Parallel
115
115
53.8
53.6
48.2
47.9
Counter
Parallel
135
135
30.1
29.2
34.8
33.8
28 | P a g e
2Counter
Parallel
165
165
53.3
53.2
48.5
48.5
Counter
Parallel
135
135
29.5
29.2
34.7
34.1
3Counter
Parallel
165
165
53.3
52.3
47.6
48.1
Counter
Parallel
165
165
29.5
29.3
34.2
33.7
CALCULATIONS
Find the properties of water at
Th = T ho+T hi
2
= 52.4+46.1
2
= 49.2 ◦C
Tc = ( T co+T ci
2 )
= 33+27.1
2
= 30 ◦C
Parallel flow,
Cpc = 4.184 KJ/kg◦ C
ρc = 995 kg/m3
ρh = 988.6 kg/m3
Mh = F h∗ρ h
3600× 1000
= 95∗988.6
3600× 1000
= 0.026 kg/s
Qh = Mh Cph (Thi-Tho)
= 0.026×4.180*103×(52.4-46.1)
= 684.68 W
29 | P a g e
Mc = F c∗ρ c
3600× 1000
= 100∗995
3600× 1000
= .0276
Qc = Mc Cpc (Tci-Tco)
= 0.0276×4.184×103×(33 - 27.1)
= 681.3
Q =684.68+681 ,.3
2
= 683 W
ΔTm =
ΔT 1 – ΔT 2
ln ΔT 1ΔT 2
= 19.4 – 19
ln 19.419
= 19.2 ◦C
ΔT1 = Thi-Tci
= 46.1- 27.1
= 19 ◦C
ΔT2 = Tho-Tci
= 52.4 – 33
=19.4 ◦C
Ai = 0.048 m2
Ao = 0.064 m2
Ui = Q
A i ΔT m
= 683
.048× 19.2
= 741.1 W/m2 oC
30 | P a g e
Uo = Q
A o ΔT m
= 683
.0642×19.2
= 554.1 W/m2 oC
Similarly for the counter current flow,
Ui = Q
A i ΔT m = 775 W/m2 oC
Uo = Q
A o ΔT m = 579 W/m2 oC
RESULT
For Parallel flow,
Ui = 741.1 W/m2K
Uo = 554.1 W/m2K
For Counter flow,
Ui = 775.0 W/m2K
Uo = 579.0 W/m2K
31 | P a g e
EXPERIMENT NO: 04 DATE: 05-09-12
HEAT TRANSFER IN FORCED CONVECTION
AIM
To study the heat transfer in forced convection
INTRODUCTION
Whenever a fluid is being forced, over the heated surface, forced convection heat transfer
occurs. The Dynamic apparatus consists of a circular pipe, through which cold fluid ie air is
forced. Pipe is heated by a band heater outside the pipe. Temperature of pipe is measured with
thermocouples attached to pipe surface. Heater input is measured by a voltmeter and ammeter.
Thus, heat transfer rate and heat transfer coefficient can be calculated.
SPECIFICATION
Test pipe-33mm ,1.0 ,500mm long
Band heater for pipe
Multichannel digital temperature indicator 0-3000C using chromel /alumel Thermocouples.
Dimmerstat 2amps,240volts for heater input control.
Volt meter 0-200V
Ammeter 0-2A
Blower to force the air through test pipe.
Orifice meter with water manometer.
EXPERIMENTAL PROCEDURE
Put on mains supply.
Adjust the heater input with the help of dimmerstat.
32 | P a g e
Start the blower and adjust the air flow with valve.
Wait till steady state is reached and note down the reading in observation table.
NOTE: The calculated values and actual values may differ appreciably because of heat losses.
The heat loss through natural convection, conduction and heat losses through insulation over the
heater is not considered, but they are present.
OBSERVATION
Sl.NO. Volt(V) I(A)Temperature 0c Manometer
difference,hw(cm of water)
Temperature °C
T2 T2 T2 T2 T2 T2 T2
1. 90 0.40 32 42 45 48 39 34 35 6.5
2. 98 0.44 33 43 46 49 39 36 36 6.5
3. 118 0.53 34 50 54 58 43 38 39 6.5
CALCULATION
Air inlet temperature, T 1 = 33 0C
Air outlet temperature, T7 = 36 0C
Density of air, ρa = 1.299× 273
273+T 1
= 1.149 Kg/m3
Diameter of orifice = 33 mm
Manometer difference = Water head = 0.041 m of water
Air head , ha = hw(ρ wρ a ) =
.041×10001.149
= 35.68 m of air
Air volume flow rate, Q = Cd × a0 × √2gh
= 0.64 × п/4 × (0.033)2 × √(2×9.81×35.68) = 1.4 × 10-2 m3/s
33 | P a g e
Mass flow rate of air , ma = Q× ρa
= 1.4 × 10-2 × 1.149
= 1.66 × 10-2 kg/m3
Velocity of air, V = Q
a p
= 1.66× 10−23.14 × .0332
= 16.37 m/sec
Heat gained by air , q = ma ×Cp × a(T7 – T1 )
= 1.66 × 10-2 ×1.047 × 103 × (36-33)
= 52.14 Watt
TS = T 2+T 3+T 4+T 5+T 6
5
= 43.80C
Tm = T 1+T 7
2
= 34.50C
Heat loss by radiation = q1
= 0.4× A × (TS4- Tm
4) σ , A = 0.518m2 , σ = 5.67x10-8
= 0.4 × 0.518 × 5.67 × 10-8 (316.84 – 307.54)
= 13.29 Watt
Actual heat loss = q –q1 = 16.408 Watt
hexp = q-q /A X (TS – Tm) = 38.85 W/m2 k
Reynolds no: , NRe = ρ× V × D
µ
= 16.37 × .03347.85 ×10−6
= 11289.65
For turbulent flow , Nu = 0.23(Re)0.8 (Pr)n = 0.23× 11289.65 0.8 × 0.7010.4
= 34.42, n = 0.4 for heating fluid
hth = 1.865 Watt/m2k
34 | P a g e
RESULT
Heat transfer in forced convection has been studied. Experimental and theoretical values of heat transfer coefficient have been calculated.
35 | P a g e
Sl.NO. Voltage (V) Hexp w/m2k Hthe. W/m2k
1. 90 3.911 2.827
2. 98 3.858 2.8201
3. 118 4.17 3.15
EXPERIMENT NO: 05 DATE: 12-09-12
COMPOSITE WALL APPARATUS
AIM
To determine the total thermal resistance of composite wall and to plot the temperature gradient
along composite wall surface
INTRODUCTION
The apparatus consists of plates of different materials sandwiched between two aluminum
plates. Three types of slabs are provided on both sides of heater which forms a composite
structure. A small hand press frame is provided to ensure the perfect contact between the slabs.
A dimmer stat is provided for varying the input to the heater and measurement of input is
carried out by a voltmeter and ammeter. Thermocouples are embedded between interfaces of
input slabs, to read the temperature at the surface. The experiment can be conducted at various
values of input and calculation can be made accordingly.
EXPERIMENTAL PROCEDURE
Arrange the plates properly (symmetrical) on both side of heater plate. See that plates are
symmetrically arranged on both sides of heater plate (arranged normally).
Operate the hand press properly to ensure perfect contact between the plates.
Close the box by cover sheet to achieve steady environmental conditions.
Start the supply of heater. By varying the dimmerstat, adjust the input (range 30-70 watts)
and water supply.
Take readings of all the thermocouples at an interval of 10 minutes until steady state is
reached.
Note down the steady state readings in the observation table.
36 | P a g e
OBSERVATIONS
WALL THICKNESS CONDUCTIVITY
M.S. = 2.5 cm 0.46 w/m °K
Bakelite = 1.0 cm 0.12 w/ m °K
Brass = 1.0 cm 110 w/m °K
Sl. no.
Heat supplied (W) Temperature (°C)
Voltmeter (V) Ammeter (A) T1 T2 T3 T4 T5 T6 T7 T8
1 120 0.43 44 44 43 43 32 32 31 31
2 129 0.46 49 49 48 48 33 33 32 32
3 140 0.50 54 54 53 53 32 32 31 31
CALCULATION
For Sl. No. 2:
Mean Readings
T a=T1+T 2
2 =
49+492 = 54 °C
T b=T3+T 4
2 =
48+482 = 53 °C
T c=T 5+T 6
2 =
33+332 = 32 °C
T d=T 7+T8
2 =
32+322 = 31 °C
Rate of heat supplied, Q = VI = 140×0.5 = 70 W
37 | P a g e
Heat flux, q ¿QA ¿
700.0122 = 5737.70 N/m2
Total thermal resistance of composite slab, R = Ta+T d
q=¿
54−315737.70 = 4×10-3 m2K/W
Thermal conductivity of composite slab, K composite = qh
Ta−T d =
5737.70× 0.04554−31 = 11.22 W/mK
RESULT
Thermal Conductivity of composite slab = 11.22 W/mK
Total thermal resistance of composite slab = 4×10-3 m2K/W
38 | P a g e
EXPERIMENT NO: 06 DATE: 21-09-12
STEFAN BOLTZMAN APPARATUS
AIM
To determine and verify the value of the Stefan – Boltzmann constant.
THEORY
All substances emit thermal radiation. When heat radiation is incident over a body, part of
radiation is absorbed, transmitted through and reflected by the body. A surface which absorbs
all thermal radiation incident over it, is called black surface. For black surface, transmissivity
and reflectivity are zero and absorptivity is unity. Stefan Boltzmann Law states that emissivity
of a surface is proportional to fourth power of absolute surface temperature ie;
e∝T 4 T4
e = σεT4
Where, e = emissive power of surface.
T = absolute temperature.
Σ = Stefan Boltzmann constant
ε = Emissivity of the surface
σ = 5.667 × 10-8 W/m2K4
For a black body,
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ε = 1, hence above equation reduces to
e = σT4
APPARATUS
The dynamic apparatus consists of a water heated jacket of hemispherical shape. A copper test
disc is fitted at the center of jacket. The hot water is obtained from a hot water tank, fitted to the
panel, in which water is heated by an electric immersion heater. The hot water is taken around
the hemisphere, so that hemisphere temperature rises. The test disc is then inserted at the
center. Thermocouples are fitted inside hemisphere to average out hemisphere temperature.
Another thermocouple fitted at the center of test disc measures the temperature of the test disc.
A timer with a small buzzer is provided to note down the disc temperatures at the time intervals
of 5 seconds.
EXPERIMENTAL PROCEDURE
See that water inlet cock of water jacket is closed and fill up sufficient water in the heater tank.
Put ON the heater.
Blacken the test disc with the help of lamp black & let it cool
Put the thermometer and check water temperature.
Boil the water and switch OFF the heater.
See that drain cock of water jacket is closed and open water inlet cock
See that there is sufficient water above the top of hemisphere( A piezometer tube is fitted to
indicate water level )
Note down the hemisphere temperatures (i.e up to channel 1 to 4)
Note down the test disc temperature (i.e channel No.5)
Start the timer, Buzzer will start ringing. At the start of timer cycle, insert test disc into the
hole at the bottom of hemisphere.
Note down the temperature of disc, every time the buzzer rings. Take atleast 4-5 readings.
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Hemisphere
temperature(0C)Time interval (s) Test disc temperature (0C)
T1 = 67.9 0 29.0T2 = 67.8 5 29.2T3 = 67.7 10 29.4T4 = 68.2 15 29.5
20 29.725 29.930 30.135 30.340 30.545 30.650 30.760 30.865 30.970 31.075 31.180 31.385 31.490 31.595 31.7100 31.9105 32.1110 32.2115 32.3120 32.4125 32.5
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130 32.6135 32.7140 32.8145 32.9150 33.1
OBSERVATIONS
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CALCULATION
Weight of test disc = 5.2×10-3 kg
Hemisphere temperature, T H=T 1+T 2+T 3+T 4
4 =
67.9+67.8+67.7+68.24 = 67.9 °C =
341.05 K
Initial test disc temperature, TD = T5 = 31.725 °C = 304.875 K
As area of hemisphere is very large as compared to that of test disc, we can put, qexp = Aσε(Th
4 – Tb4) = 5.55×10-8 Js-1
qcal = me dTdt
= 5.2×10-3×381× 30.3−30.15
= 0.07294 Js-1
σ=me( dT
dr)
A ¿¿ =
0.072943.14 ×0.0001 × (341.14−304.8754 ) = 5.14×10-8 W/m2K
Theoretical value of σ is 5.667 × 10-8 W/m2 K
In the experiment, this value may deviate due to reasons like convection, temperature drop of
hemisphere, heat losses, etc.
From graph, (dT/dt) = 0.4 Ks-1
PRECAUTIONS
Never turn ‘ON’ the heater before putting water in the tank.
Turn ‘OFF’ the heater before draining the water from heater tank.
Drain the water after completion of experiment.
Operate all the switches and controls gently.
RESULT
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The value of the Stefan – Boltzmann constant is determined = 5.14×10-8 W/m2K
EXPERIMENT NO: 07 DATE: 19-08-12
HEAT TRANSFER NATURAL CONVECTION
AIM
To determine the surface H.T coefficient for a vertical tube losing heat by natural convection.
INTRODUCTION
In contrast to the forced convection, natural convection phenomenon is due to the temp.
between the surface and the fluid and is not created by any external agency. The present
experiment set up is designed and fabricated to study the natural convection phenomenon from
a vertical cylinder in terms of the variation of local heat transfer coefficient along the length and
also the average heat transfer coefficient and its comparison with the value obtained by using an
appropriate correlation.
APPARATUS
The apparatus consists of a brass tube fitted in a rectangular vertical duct. The duct is open at
the top and bottom and forms an enclosure and serves the purpose of undisturbed surrounding.
One side of the duct is made up of Perspex for visualization. An heating element is kept in the
vertical tube which in turn heats the tube surface. The heat is lost from the tube to the
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surrounding air by natural convection. The temp. of the vertical tube is measured by seven
thermocouples. the heat input to the heater is measured by an ammeter and voltmeter and is
varied by a dimmestat. The vertical cylinder with the thermocouple is shown in fig 2. while the
possible flow pattern and also the expected variation of local heat transfer coefficient is shown
in fig 3. the tube surface is polished to minimize the radiation losses.
SPECIFICATIONS
Diameter of the tube (d) = 38 mm.
length of tube (L) = 500mm.
dust size 200 mm. × 200 mm.×800mm.length
multichannel digital temp. indicator 0-300oc using chromel/ alumel thermocouple.
ammeter 0-2 Amp. And voltmeter 0-200 volts.
dimmersat 2 Amp. 240 volts.
THEORY
When a hot body is kept in still atmosphere , heat is transferred to the surrounding fluid by
natural convection. The fluid layer in contact with the hot body get heated , rises up due to the
decrease in its density and the cold fluid rushes in to take place. The process is continuous and
the heat transfer takes place due to the relative motion of hot and cold fluid particles.
The heat transfer coefficient is given by :
h= (q – q1) / As * (Ts – Ta)
where,
h= average surface heat transfer coefficient (W/ m2 o C)
q= heat transfer rate (watts)
As= area of the heat transferring surface = π .d. l
TS= average surface temp = (T1+T2+T3+T4+T5+T6+T7) / 7 OC
Ta= ambient temp in the duct = T8 o C
q1 = Heat loss by radiation = σ. A.e. (TS4 – TS
4)
Where,
Σ= Stephen boltzman constant= 5.667 × 10-8 W/ m2. k4
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A= surface area of pipe. = 0.0597 m 2
e= emissivity of pipe material
Ts & Ta = surface and ambient temperatures in o K respectively.
The surface heat transfer coefficient, of a system transferring heat by natural convection
depends on the shape, dimensions and orientation of the fluid and the temp. difference between
heat transferring surface and the fluid. The dependence of h on all the above mentioned
parameters is generally expressed in terms of non-dimensional groups as follow :
h × L / k = A × [g.L3.β.ΔT / v2 × Cp.μ / k]
where,
h × L / K ----------is called the Nusselt no:
g . L3. β.ΔT /v3 ------is called to Grashof no:
c.p.μ / k -------is the prandtl no:
A and n are constant depending on the shape and orientation of the heat transferring surface.
Where,
L = a characteristic dimension of the surface.
K= thermal conductivity of fluid.
V = kinematic viscocity of fluid.
μ= dynamic viscosity of fluid.
Cp= specific heat pf fluid.
β= coefficient of volumetric expansion for the fluid.
g = acceleration due to gravity.
ΔT= [Ts – Ta]
For gases ,
β = 1 / ( Tf + 273) /O K
Tf= (Ts +Ta) / 2
For a vertical cylinder losing heat by natural convection, the constants A and n of equation (2)
have determined and the following empirical correlations obtained.
h × L / K =0.59 (Gr.Pr.) 0.25 for 104 < Gr.Pr. <108 -------(3)
h ×L / K =0.13 (Gr.Pr) <1017 ---------(4)
46 | P a g e
L = length of the cylinder.
All the properties of the fluid are determined at the mean film temp.(Tf)
PROCEDURE
put ON the supply and adjust the dimmerstat to obtain the required heat input (say 40W,
60W, 70W, etc)
wait till the steady state is reached , which is confirmed from temp. readings.- (T1-T2)
measure surface temp. at the various points i.e T1 – T7
note the ambient temp. i.e T8
repeat the experiment at different heat inputs (do not exceed 80W)
OBSERVATION
O.D. cylinder = 38 mm.
length of cylinder = 500 mm
input to heater = V × I watts
Sl
No
Volt
(V)
Amp
(A)
TEMPERATURE,ºC
T1 T2 T3 T4 T5 T6 T7 T8
1 106 0.9 155 159 164 167 172 164 165 39
1. 91 0.7 127 130 134 136 136 132 130 38
2. 100 0.8 140 144 148 151 150 146 142 39
PROPERTIES OF AIR
T(ºC) Ρ (kgm/m3)Cp
(kJ/kgm.K)
μ×10-3 (N-
sec /m2)k (w/ m.K) Pr
V×10-6
(m2/ sec)
0 1.293 1.005 17.2 0.0244 0.707 13.2810 1.247 1.005 17.7 0.0251 0.705 14.16
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20 1.205 1.005 18.1 0.0259 0.703 15.0630 1.165 1.005 18.6 0.0267 0.701 16.0040 1.128 1.005 19.1 0.0276 0.699 16.9650 1.093 1.005 19.6 0.0283 0.698 17.9560 1.060 1.005 20.1 0.0290 0.696 18.9770 1.029 1.009 20.6 0.0297 0.694 20.0280 1.000 1.009 21.1 0.0305 0.692 21.0990 0.972 1.009 21.5 0.0313 0.690 22.10100 0.946 1.009 21.9 0.0321 0.688 23.13120 0.898 1.009 22.9 0.0334 0.686 25.45140 0.854 1.013 23.7 0.0349 0.684 27.80
CALCULATION
Model calculation for Sl No. 1:
q = V × I = 106 × .9
= 95.4 W
As = π ×d× l = 3.14 × 38× 500×10-6 = 5.969 ×10-2 m2
Ts = T 1+T 2+T 3+T 4+T 5+T 6+T 7
7
= 163.71 0C
Ta = T8 = 39ºC = 312.15 K
q1 = σ × A × e × (Ts4 – Ta
4)
= 5.667 × 10-8 ×5.969 ×10-2 ×0.4× (436.434 – 312.154) = 36.2W
hexp = (q – q1)
A s×(T s – T a)
= 95.4−36.2
.0596×(436.86−312.15)
= 7.964 W/m2 K
Tf = T s+T a
2
= 436.86+312.15
2
= 374.5 0 K
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Β = 1
T f +273¿¿
= 0.00267 K-1
ΔT = Ts – Ta = 441.86 – 314.15 = 127.71 K
According to correlation,
¿¿= A × [(g× L3 × β . × ΔT
ν2 ) × (Cp×μ / k)]n
Gr = g × L3 × β .× ΔT
ν2
= 2.67 ×10−3 ×5003 ×10−9× 9.81× 124.7122.8 ×10−62
= 785.456 × 106
Pr = 0.690
Gr.Pr = 0.690 × 785.456 ×106 = 541.96×106
hcorr = [k ×0.13 ×(Gr × Pr )1/3
L]
= 6.32 W/m2.K
Local heat transfer coefficient
H = Q
A s× ¿¿
h1 = (q– q1)A s׿¿
= 8.97
Similarly,
h2 = 8.78 W/m2.K
h3 = 8.32 W/m2.K
h4 = 8.02 W/m2.K
h5 = 7.98 W/m2.K
h6 = 8.46 W/m2.K
h7 = 8.41 W/m2.K
PRECAUTIONS
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Proper earthing is necessary for the equipment.
keep dimmerstat to zero volt position before putting on main switch and increase it slowly.
keep at least 200 mm . space behind the equipment.
operate the change over switch of temp. indicator gently from one position to another , i.e
from 1 to 8 positions.
never exceed input above 80 watts.
RESULT
The heat transfer coefficient is having a maximum value at the beginning as expected because
of the just starting of the building of the boundary layer and it decreases as expected in the
upward direction due to thickening of layer and which is laminar one. This trend is maintained
up to half of the lengths (approx.) and beyond that there is little variation in the value of local
heat transfer coefficient because of the transition and turbulent boundary layers. The last point
shows somewhat increase in the value of heat transfer coefficient which is attributed to end loss
causing a temperature drop.
The comparison of average heat transfer coefficient is also made with predicted values are
somewhat less than experimental values due to the heat loss by radiation.
Case 1:
hexp= 7.097W/m2.K
hcorr= 5.86 W/m2.K
Case 2:
hexp= 8.017 W/m2.K
hcorr= 6.067 W/m2.K
Case 3:
hexp= 8.94 W/m2.K
hcorr= 6.94 W/m2.K
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EXPERIMENT NO: 08 DATE: 10-10-12
HEAT PIPE DEMONSTRATOR
AIM
To plot the graph of heat sink water temperature raise upto 30 mins. To plot longitudinal
temperature distribution for pipes
INTRODUCTION
Heat pipe is an interesting device, which is used to transfer heat from one location to another. It
works with the help evaporation and condensation of liquid, which is filled inside the heat pipe
as a working medium.
Heat pipe basically consist of a stainless steel pipe, sealed at both the ends. It is evacuated and
filled partially with distilled water. Stainless mesh is provided at inside periphery of the pipe.
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When heat is applied at lower end of heat pipe, water inside it evaporates and vapour passes to
upper end of pipe. The heat is taken by the medium surrounding upper portion of heat pipe. The
vapour condenses giving its latent heat of evaporation to surrounding medium. The condensed
vapour returns to bottom through the mesh packing thus because circulation of vapor, hat pipe
operates at near to isothermal operation and conducts much heat than conventional conductors.
The dynamic apparatus consist of three pipes, viz, a heat pipe, copper pipe, and a stainless steel
pipe. All the pipes have same physical dimensions.
Copper and stainless steel pipes serve the purpose of comparison of heat pipe performance with
copper pipe as good conductor of heat and with stainless steel pipe as same material of
construction. All pipes are mounted vertically with a band heater at lower end and water filled
heat sink at upper end. When heaters starts heating, the pipes , begin to transfer the heat to heat
sink. Rapid rise of temp .of water in the heat pipe heat sink demonstrates high (apparent)
thermal conductivity of heat pipe.
SPECIFICATIONS
Heat pipe – stainless pipe, 25 mm O.D., 400 mm long at both ends , evacuated and filled
partially with distilled water – 1 no.
copper and stainless steel pipes of same size as that of heat pipe- 1 each.
equal capacity heater at bottom end of each pipe
water filled heat sinks at upper end of each pipe
measurements and controls
dimmerstat to control heat input to all the heaters.-1no.
a volt meter and an ammeter to measure input to heaters one each.
Multichannel digital temperature indicator to measure temp. along length of pipes five
thermo couples are provided on each pipe.
Thermometers to note down water temperatures in heat sinks- 3 nos.
EXPERIMENTAL PROCEDURE
Fill up sufficient water in each sinks.
Ensure proper earthing to the unit put the thermometers at the top of heat sinks.
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Keep dimmarstat zero position and start the electric supply to the unit.
Slowly increase the dimmer so that power is supplied to heaters. As same dimmerstsat
supplies power to all heaters and heaters are of same capacity, power input to all the heaters
remains same. This makes the comparison simpler.
Go on noting down the temperatures of water in heat sinks every 5 minutes(stir the water
before noting down the temp.
After around 30 minutes note down the longitudinal temp. of the pipes, from the
temperatures indicator. repeat the procedure at different heat inputs, but each time it is
necessary to replace the water.
Replace the water when pipes become cool lower than 45oC , otherwise removing water at
high temp. of pipe may burn the seals at the bottom of heat sinks.
If experiments is conducted for more time, it is merely to raise the water temp. & ultimately
evaporation of water. Hence it is recommended to conduct the experiment for more times
than 30 minutes.
OBSERVATION
Heat Sink Water Temperature
Time (min)Stainless steel sink
(°C)Cu heat sink Heat pipe sink
5
10
15
20
25
30
28
28
29
29
30
30
28
30
33
34
36
37
28
29
30
31
31
32
Longitudinal Temperature Distribution
Stainless steel (°C) Cu pipe (°C) Heat pipe (°C)
T6 = 22 T11 = 123 T1 = 81
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T7 = 21
T8 = 21
T9 = 20
T10 = 20
T12 = 115
T13 = 93
T14 = 76
T15 = 65
T2 = 82
T3 = 82
T4 = 82
T5 = 83
CALCULATION
Voltage, V = 130 V
Current, I = 1.25 A
Length of pipe, x = 0.4 m
Area of cross-section, A = пr2 = 3.14×(1.25×10-3)2 4.91×10-4m2
Heat, q ¿−kA dTdx
For Stainless steel, thermal conductivity, k1 = qx
A ∆ T = 162.5× 0.4
0.000491×(22−20) = 66191.4 W/m2K
For Copper, thermal conductivity, k2 = qx
A ∆ T = 162.5× 0.4
0.000491×(123−65) = 2282.4 W/m2K
For Heat pipe, thermal conductivity, k1 = qx
A ∆ T = 162.5× 0.4
0.000491×(83−81) = 66191.4 W/m2K
PRECAUTIONS
Proper earthing is necessary.
Stir the water before noting the water temp. In heat sink.
Do not remove water from heat sinks till the pipes become cool.
Operate only one meter selector switch at a time in upward position. Other two switches
must be in down ward position.
RESULT
The graph between temperature of heat sink and time was plotted. The longitudinal temperature
distribution for pipes is plotted.
The thermal conductivity values calculated for pipes are:
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For Stainless Steel, k = 66191.4 W/m2K
For Copper, k = 2282.4 W/m2K
For Heat pipe, k = 66191.4 W/m2K
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