Heat transfer lab manual for Mechanical engineering

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G.H.Raisoni College of Engineering,Nagpur Department of Mechanical Engineering HEAT TRANSFER LAB MANUAL Semester- V Expt -1 HEAT TRANSFER THROUGH COMPOSITE WALL DESCRIPTION: The apparatus consists of a central heater sandwiched between two sheets. Three types of slabs are provided 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 dimmerstat is provided for varying the input to the heater and measurement of input is carried out by a voltmeter, ammeter. Thermocouples are embedded between interfaces of the slabs, to read the temperature at the surface. The experiments can be conducted at various values of input and calculation can be made accordingly. SPECIFICATIONS: 1. Slab assembly arranged symmetrically on both sides of heater. 2. Heater : Nochrome heater wound on mica former and insulation with control unit capacity 300 watt maximum. 3. Heater Control Unit : 0-230V. Ammeter 0-2Amps. Single phase dimmerstat (1No.). 4. Voltmeter 0-100-200V. Ammeter 0-2Amps. o 5. Temperature Indicator (digital type): 0-200 C. Service required – A. C. single phase 230 V. earthed electric supply. EXPERIMENTS TO BE CARRIED OUT: a) To determine total thermal resistance and thermal conductivity of composite wall. b) To plot temperature gradient along composite wall structure. PRECAUTIONS: 1. Keep dimmerstat to zero before start. 2. Increase the voltage slowly. 3. Keep all the assembly undisturbed. 4. Remove air gap between plates by moving hand press gently. 5. While removing the plates do not disturb the thermocouples. 6. Operate selector switch of temperature indicator gently. PROCEDURE : Arrange the plates in proper fashion (symmetrical) on both sides of the heater plates. 1. See that plates are symmetrically arranged on both sides of the heater plates. 2. Operate the hand press properly to ensure perfect contact between the plates. 3. Close the box by cover sheet to achieve steady environmental conditions. 4. Start the supply of heater by varying the dimmerstat; adjust the input at the desired value. 5. Take readings of all the thermocouples at an interval of 10 minutes until fairly steady temperatures are achieved and rate of rise is negligible. 6. Note down the reading in observation table. OBSERVATIONS : COMPOSITE SLABS : 1. Wall thickness : a. Cast iron = b. Hylam = c. Wood = 2. Slab diameter = 300mm. SET I SET II SET III READINGS 1.Voltmeter V (Volts) 2.Ammeter I (Amps) Heat supplied = 0.86 VI (in MKS units) = VI (SI units) 0 Thermocouple Reading C T1 T2 T3 T4 T5 T6 T7 T8 (T1 + T2) Mean Readings : T = - A 2 (T3 + T4) T = - B 2 (T5 + T6) T = - C 2 (T7 + T8) T = - D 2 CALCULATIONS : Read the Heat supplied Q = V x I Watts (In S. I. Units) For calculating the thermal conductivity of composite walls, it is assumed that due to large diameter of the plates, heat flowing through central portion is unidirectional i. e. axial flow. Thus for calculation, central half diameter area where unidirectional flow is assumed is considered. Accordingly, thermocouples are fixed at close to center of the plates. Q 2 Now q = Heat flux = W / m A 2 Where A = π / 4 x d = half dia. of plates. 1. Total thermal resistance of composite slab (T – T ) A D R total = - q 2. Thermal conductivity of composite slab. q x b K composite = (T – T ) A D b = Total thickness of composite slab. 3. To plot thickness of slab material against temperature gradient. REFERENCES : 1. Principles of Engineering Heat Transfer………Warren H. Giedt 2. A text book on Heat Transfer…………………Dr. S. P. Sukhatme 3. Engineering Heat Transfer. ……………C. P. Gupta & R. Prakash. THERMAL CONDUCTIVITY OF INSULATING POWDER DESCRIPTION: The apparatus consists of two thin walled concentric copper spheres. The inner sphere houses the heating coil. The insulating powder (Asbestos powder – Lagging Material) is packed between the two shells. The powder supply to the heating coil is by using a dimmerstat and is measured by Voltmeter and Ammeter. Choromel Alumel thermocouples are use to measure the temperatures. Thermocouples (1) to (4) are embedded on inner sphere and (5) to (10) are as shown in the fig. Temperature readings in turn enable to find out the Thermal Conductivity of the insulating powder as an isotropic material and the value of Thermal Conductivity can be determined. Consider the transfer of heat by conduction through the wall of a hollow sphere formed by the insulating powdered layer packed between two thin copper spheres (Ref. Fig. 1) Let, ri = Radius of inner sphere in meters. ro = Radius of outer sphere in meters. o Ti = Average Temperature of the inner sphere in C o To = Average Temperature of the outer sphere in C T1 + T2 + T3 + T4 Where, Ti = - 4 T5 + T6 + T7 + T8 + T9 + T10 and To = 6 Note that T1 to T10 denote the temperature of thermocouples (1) to (10). Form the experimental values of q, Ti and To the unknown thermal conductivity K cal be determined as … q (ro – ri ) K = - 4 π ri x ro (Ti +To) SPECIFICATIONS : 1. Radius of the inner copper sphere, ri = 50mm 2. Radius of the outer copper sphere, ro = 100mm 3. Voltmeter (0 – 100 – 200 V). 4. Ammeter (0 – 2 Amps.) 0 5. Temperature Indicator 0 – 300 C calibrated for chromel alumel. 6. Dimmerstat 0 – 2A, 0 – 230 V. 7. Heater coil - Strip Heating Element sandwiched between mica sheets – 200 watts. 8. Chromel Alumel Thermocouples – No. (1) to (4) embedded on inner sphere to measure Ti. 9. Chromel Alumel Thermocouples – No. (5) to (10) embedded on outer sphere to measure To. 10. Insulating Powder – Asbestos magnesia commercially available powder and packed between the two spheres. EXPERIMENTAL PROCEDURE : 1. Start main switch of control panel. 2. Increase slowly the input to heater by the dimmerstat starting from zero volt position. 3. Adjust input equal to 40 Watts Max. by Voltmeter and Ammeter. Wattage W = VI 4. See that this input remains constant throughout the experiment. 5. Wait till fairly steady state condition is reached. This can be checked by reading temperatures of thermocouples (1) to (10) and note changes in their readings with time. 6. Note down the readings in the observations table as given below : OBSERVATION TABLE : 1. Voltmeter reading (V) = Volts. 2. Ammeter reading ( I ) = Amps. 3. Heater input (VI) = Watts. INNER SPHERE : Thermocouple 1 2 3 4 No. T1 T2 T3 T4 Mean Temp. Ti T1 + T2 + T3 + T4 Ti = 4 0 Temp. C OUTER SPHERE : Thermocouple 5 6 7 8 9 10 No. T5 T6 T7 T8 T9 T10 Mean Temp. Ti T5 + T6 +…..+ T10 Ti = - 4 0 Temp. C CALCULATION : W = V x I Watts. 0 Ti = Inner sphere mean temp. C 0 To = Outer sphere mean temp. C ri = Radius of inner copper sphere = 50 mm. ro = Radius of outer copper sphere = 100 mm. Using Equation : q = 0.86 W Kcal/hr (In MKS units) 0.86W ( ro – ri ) K = - 4π ri x ro (Ti - To) q = V x I w / m – k (In SI units) q ( ro – ri ) K = - 4π ri x ro (Ti - To) PRECAUTIONS : 1. Keep dimmerstat to zero volt position before and after the experiment. Check this before switching ON the supply. 2. Handle the changeover switch of temperature indicator gently. EXERCISE : 1. Note down the values of Thermal Conductivity of various insulating material from the literature. 2. Compare the value of K for insulating powder obtained experimentally with that reported in the literature. THERMAL CONDUCTIVITY OF METAL ROD INTRODUCTION : Thermal conductivity is the physical property of the material denoting the ease with a particular substance can accomplish the transmission of thermal energy by molecular motion. Thermal conductivity of material is found to depend on the chemical composition of the substance or substance of which it is a composed, the phase (i. e. gas, liquid or solid) in which it exists, its crystalline structure if a solid, the temperature and pressure to which it is subjected, and whether or not it is a homogeneous material. Table 1 lists the values of thermal conductivity for some common metal : METAL THERMAL CONDUCTIVITY STATE 0 kcal / hr – m - c SOLID’S 20 degree Pure Copper 330 Brass 95 - - do - - Steel (0.5%C) 46 - - do - - S. S. 14 - - do Mechanism Of Thermal Energy Conduction In Metals : Thermal energy may be conducted in solids by two modes : 1. Lattice Vibration. 2. Transport by free electrons. In good electrical conductors a rather large number of free electrons move about in the lattice structure of the material. Just as these electrons may transport electric charge, they may also carry thermal energy from a high temperature region to a low temperature region. In fact, these electrons are frequently referred as the electron gas. Energy may also be transmitted as vibrational energy in the lattice structure of the material. In general, however, this latter mode of energy transfer is not as large as the electrons transport and it is for this reason that good electrical conductors are almost always good heat conductor viz. Copper, Aluminium and silver. With increase in the temperature, however the increased lattice vibrations come in the way of the transport by free electrons for most of the pure metals the thermal conductivity decreases with increase in the temperature. Fig. 1 shows the trend of vibration of thermal conductivity with temperature for some metals. APPARATUS: The experimental set up consists of the metal bar, one end of which is heated by an electric heater while the other end of the bar projects inside the cooling water jacket. The middle portion of the bar is surrounded by a cylindrical shell filled with the asbestos insulating powder. The temperature of the bar is measured at eight different sections Fig. 2 (1) to (4) while the radial temperature distribution is measured by separate thermocouples at two different sections in the insulating shell. The heater is provided with a dimmerstat for controlling the heat input. Water under constant heat condition is circulated through the jacket and its flow rate and temperature rise are noted. SPECIFICATION: 1. Length of the metal bar (total) : 410 mm 2. Size of the metal bar (diameter) : 25 mm 3. Test length of the bar : 200mm 4. No. of thermocouple mounted on the Bar (Positions are shown fig. 2) : 9 5. No. of thermocouples in the insulation shell (shown in fig. 2) : 2 6. Heater coil (Bald type ) : Nichrome. 7. Water jacket diameter : 80mm 8. Temperature indicator, 13 channel : 200 Degree. 9. Dimmerstat for heater coil : 2A / 230 V. 10. Voltmeter 0 to 300 volts. 11. Ammeter 0 to 2 Amps. 12. Measuring flask for water flow rate. 13. Stop clock. THEORY : The heater will heat the bar at its end and heat will be conducted through the bar to other end. After attaining the steady state Heat flowing out of bar. Heat flowing out of bar = Heat gained by water Q = m x Cp x (T – T ) = m Cp ( ∆T ) = m Cp (T – T ) w w w out in w w w w w out in Where, m : Mass flow rate of the cooling water In Kg / hr w Cp : Specific Heat of water (Given 1) T : (T – T ) for water out in Thermal Conductivity of Bar 1. Heat Conducted through the Bar (Q) 2 π KL (To – T I) Q = Qw + - Log e ro / ri Where, Qw : Heat conducted through water K : Thermal conductivity of Asbestos powder is 0.3 Kcal / hr – m- degree ro & ri : Radial distance of thermocouple in insulating shell. 2. Thermal conductivity of Bar (K) Q = K dt / dx x A Where, dt : Change in temperature. (T1 – T9) dx : Length across temperature. (0.2) A : Area of the bar (π / 4 x d2). – 4 2 π / 4 x (0.025)2 = 4.9 x 10 m PROCEDURE : 1. Start the electric supply. 2. Adjust the temperature in the temperature indicator by means of rotating the knob for compensation of temperature equal to room temperature. (Normally this is per adjusted) 3. Give input to the heater by slowly rotating the dimmerstat and adjust it to voltage equal to 80 V, 120 V etc. 4. Start the cooling water supply through the jacket and adjust it about 350cc per minute. 5. Go on checking the temperature at some specified time interval say 5 minute and continue this till a satisfactory steady state condition is reached. 6. Note the temperature reading 1 to 13. 7. Note the mass flow rate of water in Kg/minute and temperature rise in it. Observation Table : Sr. No. Mass Flow Rate in Kg/Min Temperature in Degree Centigrade T1, T2, T3, T4,………………………T13 1. 2. 3. 4. 5. Observations: Mass flow rate of water (m) : Kg/min Water inlet temperature (T12) : Degree Centigrade Water outlet temperature (T13) : Degree Centigrade Rod Temperature (T1 to T9) : Degree Centigrade Radial distance of Thermocouples (ro) : 40mm in insulating shell. (ri) : 25mm 0 Specific heat of water (Cp) : 1 Kcal/Kg K = 4.186 KJ/KgK 0 Thermal conductivity of Asbestos powder (K) : 0.3 Kcal/hr-m- C 0.3 x 4.18 KJ/KgK Length of bar (L) : 200mm Demeter of bar (d) : 50mm – 4 2 Area of the bar (A) : 4.9 x 10 m Plot the temperature distribution along the length of the bar using observed values. Calculations: 1. Heat flowing out of bar. Qbar = Qw Qw = m x Cp x (∆Tw) (Kcal/hr) Where, m : Mass flow rate of the cooling water In Kg/hr Cp :Specific Heat of water (Given 1) ∆Tw : (Tout – Tin ) for water 2. Heat conducted through the Bar (Q) 2n KL (T10 – T11) Q = Qw + - (Kcal / Hr) Log e ro / ri Where, Qw : Heat conducted through water K : Thermal conductivity of Asbestos powder is 0.3 Kcal/ hr-m-degree. ro & ri : Radial distance of thermocouple in insulating shell. 3. Thermal conductivity of Bar (K) 0 Q = K dt/dx x A (Kcal/Hr-m- C) Where, dt : Change in temperature. (T1 – T9) dx : Length Across temperature. (0.2) A : Area of the bar (n/4 x d2). - 4 2 n/4 x (0.025)2 = 4.9 x 10 m References : 1. Engineering Heat Transfer Gputa & Prakash 2. Experimental method for Engineers Tata Mc Grew Hill Company THERMAL CONDUCTIVITY OF METAL ROD INTRODUCTION : Thermal conductivity is the physical property of the material denoting the ease with a particular substance can accomplish the transmission of thermal energy by molecular motion. Thermal conductivity of material is found to depend on the chemical composition of the substance or substance of which it is a composed, the phase (i. e. gas, liquid or solid) in which it exists, its crystalline structure if a solid, the temperature and pressure to which it is subjected, and whether or not it is a homogeneous material. Table 1 lists the values of thermal conductivity for some common metal : METAL THERMAL CONDUCTIVITY STATE 0 kcal / hr – m - c SOLID’S Pure Copper 330 20 degree Brass 95 - - do - - 46 - - do - - Steel (0.5%C) S. S. 14 - - do Mechanism Of Thermal Energy Conduction In Metals : Thermal energy may be conducted in solids by two modes : 3. Lattice Vibration. 4. Transport by free electrons. In good electrical conductors a rather large number of free electrons move about in the lattice structure of the material. Just as these electrons may transport electric charge, they may also carry thermal energy from a high temperature region to a low temperature region. In fact, these electrons are frequently referred as the electron gas. Energy may also be transmitted as vibrational energy in the lattice structure of the material. In general, however, this latter mode of energy transfer is not as large as the electrons transport and it is for this reason that good electrical conductors are almost always good heat conductor viz. Copper, Aluminium and silver. With increase in the temperature, however the increased lattice vibrations come in the way of the transport by free electrons for most of the pure metals the thermal conductivity decreases with increase in the temperature. Fig. 1 shows the trend of vibration of thermal conductivity with temperature for some metals. APPARATUS: The experimental set up consists of the metal bar, one end of which is heated by an electric heater while the other end of the bar projects inside the cooling water jacket. The middle portion of the bar is surrounded by a cylindrical shell filled with the asbestos insulating powder. The temperature of the bar is measured at eight different sections Fig. 2 (1) to (4) while the radial temperature distribution is measured by separate thermocouples at two different sections in the insulating shell. The heater is provided with a dimmerstat for controlling the heat input. Water under constant heat condition is circulated through the jacket and its flow rate and temperature rise are noted. SPECIFICATION: 14. Length of the metal bar (total) : 410 mm 15. Size of the metal bar (diameter) : 25 mm 16. Test length of the bar : 200mm 17. No. of thermocouple mounted on the Bar (Positions are shown fig. 2) : 9 18. No. of thermocouples in the insulation shell (shown in fig. 2) : 2 19. Heater coil (Bald type ) : Nichrome. 20. Water jacket diameter : 80mm 21. Temperature indicator, 13 channel : 200 Degree. 22. Dimmerstat for heater coil : 2A / 230 V. 23. Voltmeter 0 to 300 volts. 24. Ammeter 0 to 2 Amps. 25. Measuring flask for water flow rate. 26. Stop clock. THEORY : The heater will heat the bar at its end and heat will be conducted through the bar to other end. After attaining the steady state Heat flowing out of bar. Heat flowing out of bar = Heat gained by water Q = m x Cp x (T – T ) = m Cp ( ∆T ) = m Cp (T – T ) w w w out in w w w w w out in Where, m : Mass flow rate of the cooling water In Kg / hr w Cp : Specific Heat of water (Given 1) T : (T – T ) for water out in Thermal Conductivity of Bar 3. Heat Conducted through the Bar (Q) 2π KL (To – T I) Q = Qw + - Log e ro / ri Where, Qw : Heat conducted through water K : Thermal conductivity of Asbestos powder is 0.3 Kcal / hr – m- degree ro & ri : Radial distance of thermocouple in insulating shell. 4. Thermal conductivity of Bar (K) Q = K dt / dx x A Where, dt : Change in temperature. (T1 – T9) dx : Length across temperature. (0.2) A : Area of the bar (π / 4 x d2). – 4 2 π / 4 x (0.025)2 = 4.9 x 10 m PROCEDURE : 8. Start the electric supply. 9. Adjust the temperature in the temperature indicator by means of rotating the knob for compensation of temperature equal to room temperature. (Normally this is per adjusted) 10. Give input to the heater by slowly rotating the dimmerstat and adjust it to voltage equal to 80 V, 120 V etc. 11. Start the cooling water supply through the jacket and adjust it about 350cc per minute. 12. Go on checking the temperature at some specified time interval say 5 minute and continue this till a satisfactory steady state condition is reached. 13. Note the temperature reading 1 to 13. 14. Note the mass flow rate of water in Kg/minute and temperature rise in it. Observation Table : Sr. No. Mass Flow Rate in Kg/Min Temperature in Degree Centigrade T1, T2, T3, T4,………………………T13 1. 2. 3. 4. 5. Observations: Mass flow rate of water (m) : Kg/min Water inlet temperature (T12) : Degree Centigrade Water outlet temperature (T13) : Degree Centigrade Rod Temperature (T1 to T9) : Degree Centigrade Radial distance of Thermocouples (ro) : 40mm in insulating shell. (ri) : 25mm 0 Specific heat of water (Cp) : 1 Kcal/Kg K = 4.186 KJ/KgK 0 Thermal conductivity of Asbestos powder (K) : 0.3 Kcal/hr-m- C 0.3 x 4.18 KJ/KgK Length of bar (L) : 200mm Demeter of bar (d) : 50mm – 4 2 Area of the bar (A) : 4.9 x 10 m Plot the temperature distribution along the length of the bar using observed values. Calculations: 4. Heat flowing out of bar. Qbar = Qw Qw = m x Cp x (∆Tw) (Kcal/hr) Where, m : Mass flow rate of the cooling water In Kg/hr Cp :Specific Heat of water (Given 1) ∆Tw : (Tout – Tin ) for water 5. Heat conducted through the Bar (Q) 2n KL (T10 – T11) Q = Qw + - (Kcal / Hr) Log e ro / ri Where, Qw : Heat conducted through water K : Thermal conductivity of Asbestos powder is 0.3 Kcal/ hr-m-degree. ro & ri : Radial distance of thermocouple in insulating shell. 6. Thermal conductivity of Bar (K) 0 Q = K dt/dx x A (Kcal/Hr-m- C) Where, dt : Change in temperature. (T1 – T9) dx : Length Across temperature. (0.2) A : Area of the bar (n/4 x d2). - 4 2 n/4 x (0.025)2 = 4.9 x 10 m References : 1. Engineering Heat Transfer Gputa & Prakash 2. Experimental method for Engineers Tata Mc Grew Hill Company HEAT TRANSFER FROM A PIN-FIN APPARATUS INTRODUCTION: Extended surfaces of fins are used to increase the heat transfer rate from a surface to a fluid wherever it is not possible to increase the value of the surface heat transfer coefficient or the temperature difference between the surface and the fluid. The use of this is variety of shapes (refer fig. 1). Circumferential fins around the cylinder of a motor cycle engine and fins attached to condenser tubes of a refrigerator are a few familiar examples. It is obvious that a fin surface sticks out from the primary heat transfer surface. The temperature difference with surrounding fluid will steadily diminish as one moves out along the fin. The design of the fins therefore required a knowledge of the temperature distribution in the fin. The main objective of this experimental set up is to study temperature distribution in a simple pin fin. APPARATUS: A brass fin of circular cross section in fitted across a long rectangular duct. The other end of the duct is connected to the suction side of a blower and the air flows past the fin perpendicular to the axis. One end of the fin projects outside the duct and is heated by a heater. Temperature at five points along the length of the fin. The air flow rate is measured by an orifice meter fitted on the delivery side of the blower. Schematic diagram of the set up is shown in fig. 2, while the details of the pin fin are as per fig. 3. SPECIFICATIONS: 1. Duct size = 150mm x 100mm. 2. Diameter of the fin = 12.7mm. 3. Diameter of the orifice = 18mm. 4. Diameter of the delivery pipe = 42mm. 5. Coefficient of discharge (or orifice meter) Cd = 0.64. 6. Centrifugal Blower 1 HP single-phase motor. 7. No. of thermocouples on fin = 5. (1) to (5) as shown in fig. 3 and indicated on temperature indicator. 8. Thermocouple (6) reads ambient temperature inside of the duct. 0 9. Thermal conductivity of fin material (Brass) = 110w/m C. 0 10. Temperature indicator = 0 – 300 C with compensation of ambient temperature up 0 to 50 C. 11. Dimmerstat for heat input control 230V, 2 Amps. 12. Heater suitable for mounting at the fin end outside the duct = 400 watts (Band type). 13. Voltmeter = 0 – 100/200 V. 14. Ammeter = 0 – 2 Amps. THEORY: Consider the fin connected at its bast to a heated wall and transferring heat to the surroundings. (Refer fig. 4) Let, A = Cross section area of the fin. C = Circumference of the fin. L = Length of the fin. T = Temp. of the fin at the beginning. 1 T = Duct fluid temperatures. f ∅ = (T – T ) = Rise in temperature. f The heat is conducted along the rod and also lost to the surrounding fluid by convection. Let, h = Heat Transfer coefficient. K = Thermal conductivity of the fin material. Applying the first law of thermodynamics to a controlled volume along the length of the fin at X, the resulting equation of heat balance appears as: 2 d ∅ h. c - - - ∅ = 0 ……………..(1) 2 d x K. A and the general solution of equation (1) is mx 2 -mx ∅ = C . e + C . e …………………………..(2) 1 h. c Where, m = - K. A With the boundary conditions of ∅ = ∅ at x = 0 1 Where, ∅ = T – T and assuming the fin tip to be insulated. 1 1 F d∅ n = 0 at x = L results in obtaining eq (2) in the form: dx ∅ T – T Cosh m (L – x) F - x - = - ……………(3) ∅ T – T Cosh mL 1 1 F This is the equation for the temperature distribution along the length of the fin. It is seen from the equation that for a fin of given geometry with uniform cross section, the temperature at any point can be calculated by knowing the values of T , T and X. 1 F Temperature T and T will be known for a given situation and the value of h depends on 1 Fwhether the heat is lost to the surrounding by free convection or forced convection and can be obtained by using the correlation as given below: 1. For free convection condition, 1/6 -1 4 Nu = 1.1 (Gr. Pr) . . 10 Gr. Pr. 10 1/4 4 9 Nu = 0.53 (Gr. Pr) . . 10 Gr. Pr. 10 4 1/4 9 12 Nu = 0.13 (Gr. Pr) . . 10 Gr. Pr. 10 2. For forced convection, 0.466 Nu = 0.615 (R ) . . 40 R 4000 e e 0.618 Nu = 0.174 (R ) . . 4000 R 40000 e e h. D Where, Nu = - A Air ρ VD R = - = Reynold’s Number. e ν 3 g. β. D ∆T G = - = Grashoff Number. r 2 ν Cp . μ P = - = Prandt 1 Number r K Air All the properties are to be evaluated at the mean film temperature. The mean film temperature is to arithmetic average of the fin temperature and air temperature. Nomenclature: 3 ρ = Density of air, Kg / m D = Diameter of pin-fin, m 2 μ = Dynamic viscosity, N.sec/m C = Specific heat, KJ/Kg.k p ν = Kinematic viscosity, m2/Sec 0 K = Thermal conductivity of air, W/m C 2 g = Acceleration due to gravity, 9.81m/sec Tm = Average fin temperature (T1 + T2 + T4 + T5) = - 5 ∆T = T – T m F T + T m F T = mF 2 β = Coefficient of thermal expansion 1 = T + 273 mF ν = Velocity of air in the duct. The velocity of air can be obtained by calculating the volume flow rate through the duct. 3 π ρW m 2 Q = Cd - x d x 2g (H. -) - 4 ρ Sec a Where, H = Difference of levels in manometer, M 3 ρW = Density of water 1000 Kg/m ρ = Density of air at T a f Cd = 0.64 d = Diameter of the orifice = 18mm. Q Velocity of air at T = - = m/sec f Duct c/s Area Use this velocity in the calculation of R e. The rate of heat transfer from the fin can be calculated as, Q = h. c. k. A x (T1 – Tf) tanh mL ………………(6) And the effectiveness of the fin can also be calculated as, tanh mL η = …………………..(7) mL EXPERIMENTAL PROCEDURE: To study the temperature distribution along the length of a pin fin natural and forced convection, the procedure is as under (I) NATURAL CONVECTION: 1. Start heating the fin by switching ON the heater element and adjust the voltage on dimmerstat to say 80 volt (Increase slowly from 0 to onwards) 2. Note down the thermocouple reading 1 to 5. 3. When steady state is reached, record the final readings 1 to 5 and also record the ambient temperature reading 6. 4. Repeat the same experiment with voltage 100 volts and 120 volts. PRECAUTIONS: 1. See that throughout the experiment, the blower is OFF. (II) FORCED CONVECTION: 1. Start heating the fin by switching ON the heater and adjust dimmerstat voltage equal to 100 volts. 2. Start the blower and adjust the difference of level in the manometer with the help of gate valve. 3. Note down the thermocouple readings (1) to (5) at a time interval of 5 minutes. 4. When the steady state is reached, record the final reading (1) to (5) and also record the ambient temperature reading (6). 5. Repeat the same experiment with different manometer readings. PRECAUTIONS: 1. See that the dimmerstat is at zero position before switching ON the heater. 2. Operate the changeover switch of temperature indicator, gently. 3. Be sure that the steady state is reached before taking the final reading. OBSERVATION TABLE: I) NATURAL CONVECTION: Sr. V I Fin Temperatures Ambient Temp. 0 0 0 0 0 0 No. Volts Amps T ( C) T ( C) T ( C) T ( C) T ( C) T = T ( C) 1 2 2 4 5 6 f II) FORCED CONVECTION: Sr. V I Manometer Fin Temperatures Ambient Temp 0 No Volts Amps reading T T T T T T = T ( C) 1 2 2 4 5 6 f 0 0 0 0 0 (Cm.) ( C) ( C) ( C) ( C) ( C)