95396211 heat-transfer-lab-manual


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95396211 heat-transfer-lab-manual

  2. 2. (For Private Circulation only) 2
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  7. 7. HEAT TRANSFER LAB LIST OF EXPERIMENTS 1. Thermal conductivity by guarded hot plate method 2. Heat exchanger test – parallel and counter flow 3. Heat exchanger test – shell and tube heat exchanger 4. Emissivity measurement 5. COP of a refrigerator 6. Heat transfer from fins-natural and forced convection 7. Thermal conductivity of insulating material 8. Heat transfer through composite walls 9. Heat transfer by free and forced convection 10. Stefan – Boltzman apparatus 11. Boiler trial 7
  8. 8. Tabulation Sl. No. Inner Heater Outer Heater Cooling Plate Thermal Conductivity W/mk V Volts I Amp T1 °C T2 °C V2 Volts I2 Amp T3 °C T4 °C T5 °C T6 °C Formula Thermal Conductivity ( ) ( ) 1 / h C W x L K W mk A T T = − Where W1 = Input to the inner heater in watts = V × I L = Specimen thickness in metre A = Area = 2 4 D π in m2 D = Diameter of the heater plate in metre X = Width of gap between the heater plates in metre 1 2 3 4( ) 4 h T T T T T + + + = °C = 5 6( ) 2 C T T T + = 8
  9. 9. Ex. No: 1 Date: THERMAL CONDUCTIVITY BY TWO SLAB GUARDED HOT PLATE METHOD Aim To determine the thermal conductivity of the given specimen by using guarded hot plate method. Apparatus Required Thermal conductivity apparatus Central heater Guarded heated ring Description The heater plate is surrounded by a heating ring for stabilizing the temperature of the primary heater and prevents heat loss completely around its edges. The primary and guard heater are made up of mica sheets in which is wound closely with equal space nichrome wire and packed with upper and lower mica sheets. These heaters together form a flat which together with upper and lower copper plates and rings form the heater plate assembly. Two thermo couples are used to measure the hot face temperature at the upper and lower central heater assembly copper plates two thermocouples are used to check the balance in both the heater inputs. Specimens are held between the heater and cooling unit on each side of the apparatus. Measure the temperature of the upper cooling plate and lower cooling plate respectively. The heater plate assembly together with the with cooling plates and specimen held in position by 3 vertical studs and nuts on a base plate are shown in the assembly drawing. The cooling chamber is a composite assembly of grooced aluminum casting and aluminum cover with entry and exit adaptors for water inlet and outlet. 9
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  11. 11. Procedure The specimen is placed on either side of the heating plate assembly, uniformly touching the cooling plates. Then the outer container is filled with lose fill insulation such as glass wool (supplied in small cloth packets). The cooling circuit is opened then calculated input is given to central and guard heaters through separate single phase supply lines with a dimmer, stat in each line and it is adjusted to maintain the desired temperature. The guard heater input is adjusted in such away that there is no radial heat flow which is checked form thermocouple reading and is adjusted accordingly. The input to the central heater and the thermocouple readings are reordered in every 10 minutes till a reasonably steady state condition is reached. Result: Thus the thermal conductivity of the specimen is determined. 11
  12. 12. Tabulation Type of flow Hot Water Cold Water LMTD °C Heat transfer co- efficient kJ/hr m2 k Effectiv eness (e) Time taken for 1 liter water flow in sec. Inlet temp T3°C Outl et temp T4°C Time taken for 1 liter water flow in sec. Inlet temp T1°C Outlet temp T2°C 12
  13. 13. Ex. No: 2 Date: PARALLEL AND COUNTER FLOW HEAT EXCHANGER Aim To study and compare the temperature distribution, heat transfer rate and overall heat transfer coefficient in parallel and counter flow heat exchanger. To calculate the effectiveness of parallel flow and counter flow heat exchanger. Apparatus Required Stop watch Measuring tape Thermometers (0-100) °C Description The apparatus consist of concentric tube that exchanger. The hot fluid (i.e. hot water) is obtained form an electric geyser and it flows through the inner tube. The cold fluid is cold water and 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 values provided. Temperatures of the fluids can be measured using thermometers. Flow rate can be measured using stop watch and measuring flask. The out tube is provided with adequate asbestos rope insulation to minimize the heat loss to the surroundings. Procedure Parallel flow 1. Parallel flow means the direction of cold and hot water flow is the same 2. Adjust the valves of pipes and maintain the same desired direction of flow 3. On the heater and slightly open the inlet valves of cold and hot water and measure the mass flow of cold and hot water at outlet per litre. 4. Take inlet and outlet temperature of cold and hot water 5. Increase the valve opening and measure the above readings in steps of 200 ml/min. to 1 lit/min. 6. The measured values are tabulated and the required results are obtained. 13
  14. 14. Formula: Heat transfer from hot water, 3 4( ) /h h hQ m C T T kJ hr= × × − Where, mh = Mass flow rate of hot water kg/hr Ch = Specific heat of hot water = 4.187 kJ/kgK T3 = Inlet temperature of hot water, °C T4 = Outlet temperature of hot water, °C Heat transfer from cold water 2 1( ) /C C CQ m C T T kJ hr= × × − Where, mc = Mass flow rate of cold water kg/hr = 6 1 1 3600 10 C t ρ−  × × ×    ρc = Density of cold water Cc = Specific heat of cold water = 4.187 kJ/kgK T1 = Inlet temperature of hot water, °C T2 = Outlet temperature of hot water, °C Logarithmic Mean Temperature Difference (LMTD), log ( ) i o m c i o T T T T T ∆ − ∆ ∆ = ∆ − ∆ iT∆ = Temperature difference at inlet, °C 0T∆ = Temperature difference at outlet, °C Overall Heat Transfer Coefficient, CQ U LMTD A = × Where, A = Area of the tube in m2 D = Outer diameter of the tube = 12.5mm L = Length of the tube = 1200mm Effectiveness, 2 1 2 1 3 1 3 1 ( ) ( ) ( ) ( ) C C C C m Cp T T T T E m Cp T T T T × − − = = × − − 14
  15. 15. Counter flow 1. Counter flow means the direction of cold and hot water is in opposite direction 2. Adjust the valves of pipe and maintain the flow of cold and hot water in the opposite direction to each other 3. Take the readings of the inlet and outlet temperatures of cold and hot water at various levels. 4. The measured values are tabulated and the required results are obtained. 15
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  17. 17. Result Thus the temperature distribution, heat transfer rate, overall heat transfer co efficient and effectiveness of the parallel flow and counter flow heat exchangers are calculated. 17
  18. 18. Tabulation Cold Water temp °C Steam Inlet Temp T3°C Cond ensat e Quali ty T4°C Condensate Quality Heat Transfer to water KJ/hr Cold Water flow rate LMTD °C Heat Transfer Co- efficient KJ/hr m2 °C Effecti veness (E) Inlet T1 Out let T2 ml tC Sec ml/sec Kg/h r 18
  19. 19. Ex. No: 3 Date: SHELL AND TUBE HEAT EXCHANGER Aim To determine the overall heat transfer coefficient to determine the heat exchanger effectiveness. Apparatus Required Stop watch Thermometers (0°C-100°C) 50 ml beaker Description In the present setup the unit operates as a condenser with steam condensing over the tubes and cooling water flowing through the tubes. The design of condensers for such varied applications as steam power plants, chemical processing plants and nuclear power plants and for space vehicles involved a variety of heat transfer and fluid flow problems associated with the condensation of vapours. The banks of smooth horizontal round tubes configurations and with the high vapour velocities which are normally associated with steam condenser, the overall heat transfer co-efficient is primarily a function of cooling water velocity for clean, bright, new horizontal tubes with no contamination wither on the steam side or on the water side. If the tubes are of a material other than admiralty metal or have a wall thickness other than 18BWG correction factors indicated in table one should be employed. The condenser is a horizontal shell and tube heat exchanger with steam condensing over the tubes. Cooling water flows through tow tube passes. Steam pressure and temperature at inlet to the exchanger are monitored. The condensate leaves at the bottom through a valve. The condensate temperature is also monitored. Water inlet and outlet temperatures are measure by dial thermometers. A manometer is provided for evaluating the pressure drop in the cooling water circuit. Water flow rate is measured by either measuring the quantity of water collected in bucket in a known time or by means of rot meter quantity of steam condensed is measured by collecting the condensate in a measuring jar. 19
  20. 20. Formula mw = Mass flow rate of water in kg/hr = w w w T q ρ××× −6 103600}{ Heat transfer to water, )( 12 TTCmQ Pccw −×= in kJ/hr T1 = Inlet temperature of cold water in °C T2 = Outlet temperature of cold water in °C CPc = Specific heat of cold water = 4.187 kJ/kgK ρw = Density of water = 1000 kg/m3 Heat given out by steam = c c c s L T Q Q ρ××××= − 360010 6 Where, L = Latent heat of steam Qc = Quantity of condensate collected in ml. tc = Time for collecting condensate in seconds ρc = Density of steam at temperature T3 °C Logarithmic Mean Temperature Difference (LMTD) = 43 13 12 )(log TT TT TT T e m − − ∆−∆ =∆ Heat transfer = NLd t ××× 0π Where, N = Number of tubes d0 = Outside diameter of tube in metre L = Effective length in metre Overall heat transfer coefficient = LMTD A Qw × kJ/hrm2 K Where, A = Area of tube in m2 D = Outer diameter of tube 12.5mm L = Length of tube 1200mm Effectiveness, )( 13 min TT C qc −×=ε Where, T3 = Temperature of steam at inlet, °C T1 = Temperature of water at inlet, °C 20
  21. 21. Procedure 1. Switch on the boiler heaters 2. In the pump and adjust the mass flow of cold water to 25ml/sec and wait until the boiler pressure reduces to 0.5kgfcm2. 3. When the boiler pressure is 0.5kgf/cm2 4. Then note the steam inlet temperature and coldwater inlet and outlet temp, and time taken for 50ml condensate and steam outlet temp. 5. Then varying the mass flow of the cold water to 45, 65,…105 ml/sec. corresponding readings are tabulated. 21
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  23. 23. Result Thus the overall heat transfer co-efficient and effectiveness are determined for the given shell and tube heat exchanger. 23
  24. 24. Tabulation Sl. No. Black Plate Test Plate Ambient Temperature °C Emissivity of test plate V1 volts I1 Amp T1 °C V2 volts I2 Amp T2 °C 24
  25. 25. Ex. No: 4 Date: EMISSIVITY MEASUREMENT Aim To determine the emissivity of the given specimen. Apparatus Required Emissivity apparatus Description The experimental set up consists of two circular aluminum plates identical in size and is provided with heating coils sandwiched. The plates are mounted on brackets and are kept in an enclosure so as to provide undisturbed natural convection surroundings. The heat input to the heater is varied by separate dimmer stats and is measured by using an ammeter and a voltmeter with the help of double through switches. The temperature of the plates is measured by thermocouples. Separate wires are connected to diametrically opposite points to get the average surface temperature of plates. Another thermocouple is kept in the enclosure to read the ambient temperature of enclosure. Plate 1 is blackened by a thick layer of lamp black to form the idealized black surface where as the plate whose emissivity is to be determined. The heater inputs to the two plates are dissipated from the plates by conduction, convection and radiation. The experimental set up is designed in such a way that under steady state conditions the heat dissipation by conduction and convection is same for both plates, when the surface temperatures are same and the difference in the heater input readings is because of the difference in radiation characteristics due to their different emissivity’s. 25
  26. 26. Formula )()( 4 3 4 21 TTEEAWW sb −×−××=− σ Where, W1 = Heater input to black plate = V1×I1 Watts W2 = Heater input to test plate = V2×I2 Watts A = Area of two plates = 2 4 2 d×× π m2 Diameter of the black and test plate (d) = 160mm Ts = Surface temperature of discs = T1 + 273K (or) T2 + 273K T3 = Temperature of enclosure = T3 + 273K Eb = Emissivity of black plate = 1 σ = Stefan Boltzmann constant = 5.67 × 10-8 W/m2 K4 By using Stefan Boltzmann Law: 86.0 )()( 4 3 4 21 TTEEA WW sb −×−×× =− σ 26
  27. 27. Procedure 1. Give power supply to T.P. (230 v signal phase) and adjust the reading in it equal to room temperature by roating the compensation knob (normally this is pre-adjusted). 2. Select the proper range of voltage on voltmeter 3. Gradually increase the input to the heater to black plate and adjust it to some value viz., 0,50,75 watts. And adjust the heater input to test plate slightly less than the black plate 27, 35, 55 watts etc., 4. Check the temperatures of the two plates with small time intervals and adjust the input of test plate only, by the dimmer stat so that the two plates will be maintained at the same temperature. 5. This will require some trial and error and one has to wait sufficiently (more than one hour or so) to obtain the study state condition. 6. After attaining the steady state condition record the temperatures, and voltmeter and ammeter readings for both the plates. 7. The same procedure is repeated for various surface temperatures in increasing order. 27
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  29. 29. Graph: Graph is drawn between Surface temperature (X axis) Vs Emissivity (Y axis) Result Thus the emissivity of the given specimen is calculated and corresponding graph has been drawn. 29 Emissivity Surface Temperature in Kelvin
  30. 30. Tabulation Sl. No. Compressor work time taken for 30 revolution in second t1 Fan work time taken for 10 revolution in seconds t2 Time taken for one lit water flow in seconds t3 Water Temperature °C Co-efficient of performance (COP) Inlet T1 Outlet T2 T1-T2 Formula Heat extracted from water Q = mw × CPw × (T1 - T2) kJ/hr Where, w w w w T q m ρ×××= −6 103600}{ kg/hr T1 = Inlet temperature of water in °C T2 = Outlet temperature of water in °C CPw = Specific heat of water = 4.186kJ/kgK ρ = Density of water = 1000kg/m3 Coefficient of Performance, W Q COP = Where, W = W1+W2 W1 = Compressor work, W2 = Fan work 30
  31. 31. Ex. No: 5 Date: PERFORMANCE TEST ON A REFRIGERATOR Aim To determine the co-efficient of performance of the refrigerator Apparatus Required Refrigerator test rig Thermometer Measuring jar Stop watch Description The refrigeration is the process of cooling a space less than the surrounding temperature. The working substance used in the refrigerator is known as refrigerant. The refrigerant passes through the compressor to increase the pressure. Then it flows to condenser where it is condensed at constant pressure. The pressurized liquid refrigerant expands in the expansion valve. Thus the low pressure and low temperature liquid evaporates in the evaporator by absorbing the latnt heat of evaporation form the space to be cooled at constant pressure. Now the vapour again passes through the compressor and the cycle is repeated again. Procedure Before starting the unit, ensure that valves are closed. The water is allowed to flow at constant rate into the container by opening inlet supply tap. The refrigeration can work with either solenoid valve or thermostat. First the fan is switched on. Then the following procedure is followed. The solenoid valve with thermostat expansion is started, the valve s2, s4 and s5 are closed and solenoid valve switch is put on. The valves s2, s4 and s5 are closed if the capillary is put off. then the thermostat is rotated in clockwise direction. The thermostat is inserted in their respective places. The refrigerator is allowed to run for stabilization. The no of revolutions of the disc in the energy meter of the compressor and fan are noted down for a known time. 31
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  33. 33. Result: Thus the coefficient of performance of the given refrigerator is calculated 33
  34. 34. Tabulation Experiment Power supply Fin. Temperature °C Monometer Reading Ambient Temperatu re °C Rate of Heat Transfer Q (J/sec) Effecti veness Amp Volt T1 T2 T3 T4 T5 h1 h2 Natural Convection Forced Convection 34
  35. 35. Ex. No: 6 Date: HEAT TRANSFER THROUGH PINFIN Aim To study the temperature distribution along the length of a pinfin in natural and forced convection and to determine the heat transfer rate from the fin and the fin effectiveness. Description A brass fin of circular cross section is fitted across a long rectangular duct. The other end of the duct is connected to the suction side of a bowler and the air flows past the fin perpendicular to its axis. One end of the fin projects outside the duct and is heated by a heater. Temperature at five points of the fin measured by chromel alumel thermocouples embedded on the fine. The air flow rat is measured by an orifice meter fitted on the delivery side of the bowler. Procedure Natural convection 1. Start heating the fin by switching on the meter element and adjust the voltage on dimmer stat, to 50volts by increasing slowly from 0 onwards 2. Note down the thermocouple reading I to 5 3. When steady state condition is reached record the final reading I to 5 and also record the ambient temperature reading T6 4. Repeat the same experiment with 100 volts and 120 volts. Forced Convention 1. Start the heating of fin by switching on the heater and adjust dimmer stat voltage equal to 100 volts. 2. Start the blowe rand adjust the difference of level in the monmieter H=….cm with the help of value. 3. Note down the thermocouple reading I to 5 at the time interval of 5 minutes 4. When the steady state steady is reached, note the ambined temp reading 6. 5. Repeat the same experiment with different value of 4 volts. 35
  36. 36. Formula Natural Convection 1. T average = 5 54321 TTTTT ++++ °C 2. Tmf = mean film temperature = 2 6TTavg + 3. Grash Number Gr = where g = acceleration due to gravity = 9.81 m/sec2 β = , ΔT = Taverage- T6 T6 = Temperature of direct fluid γ = kinematic viscosity m2 /sec (From Data book in table for Tmf o C) d = diameter of the fin 1.27 x 10-2 m 4. Heat transfer coefficient h = where Nu = Nusselt Number = 1.10 x (Gr x Pr)1/6 for 10-1 < Gr x Pr < 104 36
  37. 37. = 0.53 x (Gr x Pr)1/4 for 104 < Gr x Pr < 109 = 0.13 x (Gr x Pr)1/3 for 109 < Gr x Pr < 1012 Ka = Thermal conductivity of air W/mK ( From Data book in table for Tmf o C) 37
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  39. 39. 5. m = m-1 Where h = Heat transfer coefficient W/m2 K P = Circumference of the fin = π × d in metre K = Thermal conductivity of the material (given specimen), W/mK kbrass = 110.5 W/mK, ksteel = 46.5 W/mK kAl = 232.6 W/mK A = Cross sectional area of the fin, A= m2 6. The rate of heat transfer from the fin (Q) = Where, K = thermal conductivity of air (at mean film temperature) 7. Efficiency of the fin = Where, L = Length of fin = 1.5 cm 8. Predicted temperature at each point At starting point T1, T1 (Pr)= T2 (Pr) = + T6 T3 (Pr) = + T6 T4 (Pr) = + T6 T5 (Pr) = + T6 39
  40. 40. Where, X2 = 2.5 cm, X3 = 5 cm, X4 = 7.5 cm, X5 = 10 cm, L = 15cm 40 Length of the fin cm Temperature in o C Predicted Observed Length of the fin cm Temperatureino C Predicted Observed
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  42. 42. Forced Convection 1. T average = o C 2. Tmf = mean film temperature = o C 3. Discharge of air, )(2 a w md hgACQ ρ ρ ×××××= m3 /sec Where, Cd = Co-efficient of discharge of air = 0.64 A = Cross sectional area of the orifice = 2 4 d π in m2 g = Acceleration due to gravity = 9.81m/s2 hm = Difference in manometer reading in metre ρw = Density of water = 1000kg/m3 ρa = Diameter of the orifice = 1.8×10-2 metre 4. Velocity of air in duct (VD) at T6 o C D D A Q V = m/sec Q = Discharge of air in m3 /sec A = Duct area = 0.15 × 0.10 m2 5. Velocity of air (Vmf) at mean film temperature Tmf o C 273 273 6 × × ×= T T VV mf Dmf m/sec 6. Heat transfer coefficient, h = Where, d = Diameter of the fin, 1.27 10-2 m Nu = 0.615 × (Re)0.466 when 40<Re<4000 7. AK Ph m × × = m-1 Where, h = heat transfer coefficient, W/m2 K P = Circumference of the fin = π × d, m 42
  43. 43. K = Thermal conductivity of the material (Given specimen) A = Cross sectional area of the fin = 2 4 d π in m2 43
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  45. 45. 8. The rate of heat transfer from the fin, )tanh()( 21 mlTTKPhQ ×−×××= Nu = 0.174(Rc)0.618 when 40 < Rc >40000 Re = Reynold’s number = d = Diameter of the orifice= 1.8 x 10-2 m γ = Kinematic viscosity (m2 /sec) at Tmf 9. Efficiency of the fin = Where, L=Length of = 1.5 cm 10. Predicted temperature at each point At starting point T1, T1=15 T2 (Pr) = + T6 T3 (Pr) = + T6 T4 (Pr) = + T6 T5 (Pr) = + T6 Where, X2 = 2.5 cm, X3 = 5 cm, X4 = 7.5 cm, X5 = 10 cm, L = 15cm 45 Length of the fin Temperatureino C Predicted Observed
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  47. 47. Temperature Distribution Result Thus the temperature distribution along the length of a pinfin in natural and forced convention is determined and effectiveness is also determined. 47