ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER ME 650 Heat and Mass Transfer III YEAR-V SEMESTER NAME :. REG.NO :. BRANCH :... YEAR & SEM :. 1
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 1. State Fourier s Law of conduction. UNIT-1 CONDUCTION Part-A The rate of heat conduction is proportional to the area measured normal to the direction of heat flow and to the temperature gradient in that direction. Q - A dt dx dt Q - KA where A are in m dx dt dx - Temperature gradient in K/m K Thermal conductivity W/mK.. Define Thermal Conductivity. Thermal conductivity is defined as the ability of a substance to conduct heat.. Write down the equation for conduction of heat through a slab or plane wall. Heat transfer Q T overall Where T = T 1 T R L R - Thermal resistance of slab KA L = Thickness of slab, K = Thermal conductivity of slab, A = Area 4. Write down the equation for conduction of heat through a hollow cylinder. Heat transfer Q T overall Where, T = T 1 T R R 1 r in LK r1 thermal resistance of slab L Length of cylinder, K Thermal conductivity, r Outer radius, r 1 inner radius 5. State Newton s law of cooling or convection law. Heat transfer by convection is given by Newton s law of cooling
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Q = ha (T s - T ) Where A Area exposed to heat transfer in m, h - heat transfer coefficient in W/m K T s Temperature of the surface in K, T - Temperature of the fluid in K. 6. Write down the general equation for one dimensional steady state heat transfer in slab or plane wall with and without heat generation. T T T 1 T x y z t T T T q 1 T x y z K t 7. Define overall heat transfer co-efficient. The overall heat transfer by combined modes is usually expressed in terms of an overall conductance or overall heat transfer co-efficient U. Heat transfer Q = UA T. 8. Write down the equation for heat transfer through composite pipes or cylinder. Heat transfer T Q R overall, Where, T = T a T b, r r 1 In In L r1 r L har1 K1 K hbr 1 1 1 R. 9. What is critical radius of insulation (or) critical thickness? Critical radius = r c Critical thickness = r c r 1 Addition of insulating material on a surface does not reduce the amount of heat transfer rate always. In fact under certain circumstances it actually increases the heat loss up to certain thickness of insulation. The radius of insulation for which the heat transfer is maximum is called critical radius of insulation, and the corresponding thickness is called critical thickness. 10. Define fins (or) extended surfaces. It is possible to increase the heat transfer rate by increasing the surface of heat transfer. The surfaces used for increasing heat transfer are called extended surfaces or sometimes known as fins.
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 11. State the applications of fins. The main applications of fins are 1. Cooling of electronic components. Cooling of motor cycle engines.. Cooling of transformers 4. Cooling of small capacity compressors 1. Define Fin efficiency. The efficiency of a fin is defined as the ratio of actual heat transfer by the fin to the maximum possible heat transferred by the fin. fin Q Q fin max 1. Define Fin effectiveness. Fin effectiveness is the ratio of heat transfer with fin to that without fin Fin effectiveness = Q Q with fin without fin Part -B 1. A wall is constructed of several layers. The first layer consists of masonry brick 0 cm. thick of thermal conductivity 0.66 W/mK, the second layer consists of cm thick mortar of thermal conductivity 0.6 W/mK, the third layer consists of 8 cm thick lime stone of thermal conductivity 0.58 W/mK and the outer layer consists of 1. cm thick plaster of thermal conductivity 0.6 W/mK. The heat transfer coefficient on the interior and exterior of the wall are 5.6 W/m K and 11 W/m K respectively. Interior room temperature is C and outside air temperature is -5C. Calculate a) Overall heat transfer coefficient b) Overall thermal resistance c) The rate of heat transfer d) The temperature at the junction between the mortar and the limestone. Given Data 4
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Thickness of masonry L 1 = 0cm = 0.0 m Thermal conductivity K 1 = 0.66 W/mK Thickness of mortar L = cm = 0.0 m Thermal conductivity of mortar K = 0.6 W/mK Thickness of limestone L = 8 cm = 0.08 m Thermal conductivity K = 0.58 W/mK Thickness of Plaster L 4 = 1. cm = 0.01 m Thermal conductivity K 4 = 0.6 W/mK Interior heat transfer coefficient h a = 5.6 W/m K Exterior heat transfer co-efficient h b = 11 W/m K Interior room temperature T a = C + 7 = 95 K Outside air temperature T b = -5C + 7 = 68 K. Solution: Heat flow through composite wall is given by T Q overall [From equation (1)] (or) [HMT Data book page No. 4] R Where, T = T a T b 5
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 1 L L L L 1 1 4 R h a A K 1 A K A K A K 4 A h b A Ta Tb Q 1 L1 L L L4 1 h A K A K A K A K A h A a 1 4 95 68 Q/ A 1 0.0 0.0 0.08 0.01 1 5.6 0.66 0.6 0.58 0.6 11 Heat transfer per unit area Q/A = 4.56 W/m b We know, Heat transfer Q = UA (T a T b ) [From equation (14)] Where U overall heat transfer co-efficient Q U A ( T T ) 4.56 U 95 68 a b Overall heat transfer co - efficient U = 1.8 W/m We know Overall Thermal resistance (R) K 1 L L L L 1 1 4 R h a A K 1 A K A K A K 4 A h b A For unit Area 1 L L L L 1 1 4 R h a K 1 K K K 4 h b 1 0.0 0.0 0.08 0.01 1 = 56 0.66 0.6 0.58 0.6 11 R 0.78 K / W Interface temperature between mortar and the limestone T Interface temperatures relation 6
1 1 1 SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER T T T T T T T T T T T T R R R R R R a 1 1 4 4 5 5 Q Ta T1 Q R 1 4 95-T 1 1 Q = R a 1/ haa haa 95 T1 Q/ A 1/ h 95 T1 4.56 1/ 5.6 T 88.8 K a a T T Q R a b b 88.8 T Q L KA 1 1 1 1 1 R L ka 88.8 T Q/ A L1 K 88.8 T 4.56 0.0 0.66 T 78. K Q = T T R 78. T Q L KA 78. T Q/ A L K 78. T 4.56 0.0 0.6 T 76.5 K 1 R L KA Temperature between Mortar and limestone (T is 76.5 K) 7
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER. A furnace wall made up of 7.5 cm of fire plate and 0.65 cm of mild steel plate. Inside surface exposed to hot gas at 650C and outside air temperature 7C. The convective heat transfer coefficient for inner side is 60 W/m K. The convective heat transfer co-efficient for outer side is 8W/m K. Calculate the heat lost per square meter area of the furnace wall and also find outside surface temperature. Given Data Thickness of fire plate L 1 = 7.5 cm = 0.075 m Thickness of mild steel L = 0.65 cm = 0.0065 m Inside hot gas temperature T a = 650C + 7 = 9 K Outside air temperature T b = 7C + 7 = 00K Convective heat transfer co-efficient for Inner side h a = 60W/m K Convective heat transfer co-efficient for Outer side h b = 8 W/m K. Solution: (i) Heat lost per square meter area (Q/A) Thermal conductivity for fire plate K 1 = 105 10 - W/mK [From HMT data book page No.11] Thermal conductivity for mild steel plate K = 5.6W/mK [From HMT data book page No.1] 8
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Heat flow Q T overall, Where R T = T a T b 1 L L L 1 1 R h a A K 1 A K A K A h b A Ta Tb Q= 1 L1 L L 1 h A K A K A K A h A a 1 b [The term L is not given so neglect that term] Ta Tb Q= 1 L1 L L 1 h A K A K A K A h A a The term L 1 Ta Tb Q = 1 L1 L 1 h A K A K A h A 1 9 00 Q/ A 1 0.075 0.0065 1 60 1.05 5.6 8 Q / A 907.79 W / m is not given so neglect that term] a b b 9
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER (ii) Outside surface temperature T We know that, Interface temperatures relation Ta Tb Ta T1 T1 T T T T Tb Q...( A) R R R R R T T ( A) Q R where R b 1 ha b b T T Q 1 ha b b b a 1 b Q/A = T T 1 h T 907.79 b b 00 1 8 T 66.47 K. A steel tube (K = 4.6 W/mK) of 5.08 cm inner diameter and 7.6 cm outer diameter is covered with.5 cm layer of insulation (K = 0.08 W/mK) the inside surface of the tube receivers heat from a hot gas at the temperature of 16C with heat transfer co-efficient of 8 W/m K. While the outer surface exposed to the ambient air at 0C with heat transfer coefficient of 17 W/m K. Calculate heat loss for m length of the tube. Given Steel tube thermal conductivity K 1 = 4.6 W/mK Inner diameter of steel d 1 = 5.08 cm = 0.0508 m Inner radius r 1 = 0.054 m Outer diameter of steel d = 7.6 cm = 0.076 m Outer radius r = 0.081 m Radius r = r + thickness of insulation Radius r = 0.081 + 0.05 m r = 0.061 m Thermal conductivity of insulation K = 0.08 W/mK Hot gas temperature T a = 16C + 7 = 589 K 10
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Ambient air temperature T b = 0C + 7 = 0 K Heat transfer co-efficient at inner side h a = 8 W/m K Heat transfer co-efficient at outer side h b = 17 W/m K Length L = m Solution : Toverall Heat flow Q [From equation No.(19) or HMT data book Page No.5] R Where T = T a T b 1 1 1 r 1 r 1 r 4 1 R In In In L ha r1 K1 r1 K r K r hb r4 Ta Tb Q = 1 1 1 r 1 r 1 r 4 1 In In In L har1 K1 r1 K r K r hb r4 [The terms K and r 4 are not given, so neglect that terms] Q = T a T 1 1 1 r 1 r 1 In In L har1 K1 r1 K r hb r b Q = 589-0 1 1 1 0.081 1 0.061 1 In + In 8 0.054 4.6 0.054 0.08 0.081 17 0.061 Q 119.4 W Heat loss Q = 119.4 W. 4. Derive an expression of Critical Radius of Insulation For A Cylinder. Consider a cylinder having thermal conductivity K. Let r 1 and r 0 inner and outer radii of insulation. 11
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Heat transfer Ti T Q r 0 In r1 KL [From equation No.()] Considering h be the outside heat transfer co-efficient. Q = Here A T T r 0 In r 1 1 KL A h r L 0 0 Ti T Q r 0 In r 1 1 KL r Lh i 0 0 zero. To find the critical radius of insulation, differentiate Q with respect to r 0 and equate it to 0 (T T ) 1 1 dq KLr hlr dr0 1 r 0 1 In KL r 1 hlr0 since (T T ) 0 i 1 1 0 KLr hlr K r r h 0 c i 0 0 0 0 5. A wire of 6 mm diameter with mm thick insulation (K = 0.11 W/mK). If the convective heat transfer co-efficient between the insulating surface and air is 5 W/m L, find the critical thickness of insulation. And also find the percentage of change in the heat transfer rate if the critical radius is used. 1
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Given Data Solution : d 1 = 6 mm r 1 = mm = 0.00 m r = r 1 + = + = 5 mm = 0.005 m K = 0.11 W/mK h b = 5 W/m K K 1. Critical radius r c [From equation No.(1)] h 0.11 rc 4.4 10 m 5 r c 4.4 10 m Critical thickness = r c r 1 4.410 0.00 Critical thickness t c = 1.4 1.410 m - 10 (or) 1.4 mm. Heat transfer through an insulated wire is given by Ta Tb Q1 r In 1 r 1 1 L K1 hbr 1
ME650 HEAT AND MASS TRNSFER = Q1 = MARKS & 16 MARKS QUESTION AND ANSWER From HMT data book Page No.5 L (T T ) 0.005 In 0.00 1 0.11 5 0.005 L (Ta T b) 1.64 a b Heat flow through an insulated wire when critical radius is used is given by T T Q r r r c In 1 r 1 1 L K1 hbr c a b c = Q = L (T T ) 4.4 10 In 0.00 1 0.11 5 4.4 10 L (Ta T b) 1.57 a b Percentage of increase in heat flow by using Q Q1 Critical radius = 100 Q 1 1 100 1.57 1.64 1 1.64 0.55% 1 14
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 6. An aluminium alloy fin of 7 mm thick and 50 mm long protrudes from a wall, which is maintained at 10C. The ambient air temperature is C. The heat transfer coefficient and conductivity of the fin material are 140 W/m K and 55 W/mK respectively. Determine 1. Temperature at the end of the fin.. Temperature at the middle of the fin.. Total heat dissipated by the fin. Given Thickness t = 7mm = 0.007 m Length L= 50 mm = 0.050 m Base temperature T b = 10C + 7 = 9 K Ambient temperature T = + 7 = 95 K Heat transfer co-efficient h = 140 W/m K Thermal conductivity K = 55 W/mK. Solution : Length of the fin is 50 mm. So, this is short fin type problem. Assume end is insulated. We know Temperature distribution [Short fin, end insulated] T T cos h m [L -x]...(a) T T cos h (ml) b [From HMT data book Page No.41] (i) Temperature at the end of the fin, Put x = L 15
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER (A) T - T cos h m [L-L] Tb T cos h (ml) T - T 1 Tb T cos h (ml)...(1) where m = hp KA P = Perimeter = L (Approx) = 0.050 P = 0.1 m A Area = Length thickness = 0.050 0.007 A.5 10 m hp m = KA 4 140 0.1 55.5 10 m 6.96 4 (1) T - T 1 T T cos h (6.9 0.050) b T - T 1 T T.05 b T - 95 1 9-95.05 T - 95 = 47.8 T = 4.8 K Temperature at the end of the fin T xl (ii) Temperature of the middle of the fin, Put x = L/ in Equation (A) 4.8 K 16
(A) b b SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER T - T cos hm [L-L/] T T cos h (ml) 0.050 cos h 6.9 0.050 - T - T T T cos h 6.9 (0.050) T- 95 1.4 9-95.049 T - 95 0.605 9-95 T 54.04 K Temperature at the middle of the fin T xl / 54.04 K (iii) Total heat dissipated [From HMT data book Page No.41] 1/ Q = (hpka) (Tb T )tan h (ml) -4 1/ [140 0.1 55.5 10 ] (9 95) tan h (6.9 0.050) Q = 44.4 W 7. A copper plate mm thick is heated up to 400C and quenched into water at 0C. Find the time required for the plate to reach the temperature of 50C. Heat transfer co-efficient is 100 W/m K. Density of copper is 8800 kg/m. Specific heat of copper = 0.6 kj/kg K. Plate dimensions = 0 0 cm. Given Thickness of plate L = mm = 0.00 m Initial temperature T 0 = 400C + 7 = 67 K Final temperature T = 0C + 7 = 0 K Intermediate temperature T = 50C + 7 = K Heat transfer co-efficient h = 100 W/m K Density = 8800 kg/m Specific heat C = 60 J/kg k 17
Plate dimensions = 0 0 cm SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER To find Time required for the plate to reach 50C. [From HMT data book Page No.] Solution: Thermal conductivity of the copper K = 86 W/mK For slab, L Characteristic length Lc 0.00 = L c 0.001 m We know, hlc Biot number Bi K 100 0.001 86 4 B i =.59 10 0.1 Biot number value is less than 0.1. So this is lumped heat analysis type problem. For lumped parameter system, T T T T 0 We know, ha t C V e.(1) [From HMT data book Page No.48] Characteristics length L c = V A 18
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER T-T (1) e T T - 0 e 67-0 t = 9.4 s 0 h t C Lc 100 t 600.0018800 Time required for the plate to reach 50C is 9.4 s. 8. A steel ball (specific heat = 0.46 kj/kgk. and thermal conductivity = 5 W/mK) having 5 cm diameter and initially at a uniform temperature of 450C is suddenly placed in a control environment in which the temperature is maintained at 100C. Calculate the time required for the balls to attained a temperature of 150C. Take h = 10W/m K. Given Specific heat C = 0.46 kj/kg K = 460 J/kg K Thermal conductivity K = 5 W/mK Diameter of the sphere D = 5 cm = 0.05 m Radius of the sphere R = 0.05 m Initial temperature T 0 = 450C + 7 = 7 K Final temperature T = 100C + 7 = 7 K Intermediate temperature T = 150C + 7 = 4 K Heat transfer co-efficient h = 10 W/m K To find Time required for the ball to reach 150C [From HMT data book Page No.1] Solution Density of steel is 78 kg/m 78 kg/m For sphere, R Characteristic Length Lc 19
We know, hlc Biot number Bi K 108.10 5 SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 0.05 = Lc 8. 10 m B i =.8 10 - < 0.1 Biot number value is less than 0.1. So this is lumped heat analysis type problem. For lumped parameter system, T T T T 0 ha t C V e.(1) We know, [From HMT data book Page No.48] Characteristics length L c = V A T-T (1) e T T 0 h t C Lc 10 t 4608.10 78 4-7 e 7-7 4-7 In 10 t 7-7 460 8. 10 78 t = 5840.54 s Time required for the ball to reach 150C is 5840.54 s. 0
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 9. Alloy steel ball of mm diameter heated to 800C is quenched in a bath at 100C. The material properties of the ball are K = 05 kj/m hr K, = 7860 kg/m, C = 0.45 kj/kg K, h = 150 KJ/ hr m K. Determine (i) Temperature of ball after 10 second and (ii) Time for ball to cool to 400C. Given Diameter of the ball D = 1 mm = 0.01 m Radius of the ball R = 0.006m Initial temperature T 0 = 800C + 7 = 107 K Final temperature T = 100C + 7 = 7 K Thermal conductivity K = 05 kj/m hr K 051000J 600 s mk 56.94 W /mk [ J/s = W] Density = 7860 kg/m Specific heat C = 0.45 kj/kg K = 450 J/kg K Heat transfer co-efficient h = 150 kj/hr m K 150 1000J 600 s m K 41.66 W /m K Solution Case (i) Temperature of ball after 10 sec. For sphere, R Characteristic Length Lc 0.006 = We know, hlc Biot number Bi K 41.667 0.00 56.94 L c 0.00 m 1
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER B i = 1.46 10 - < 0.1 Biot number value is less than 0.1. So this is lumped heat analysis type problem. For lumped parameter system, T T T T 0 We know, ha t C V e.(1) [From HMT data book Page No.48] Characteristics length L c = V A 0 h t C Lc T-T (1) e...() T T T - 7 e 107-7 T = 10.95 K 41.667 10 4500.007860 Case (ii) Time for ball to cool to 400C T = 400C + 7 = 67 K 0 h t C Lc T-T () e...() T T 41.667 t 4500.007860 67-7 e 107-7 67-7 41.667 In t 107-7 450 0.00 7860 t = 14.849 s 10. A large steel plate 5 cm thick is initially at a uniform temperature of 400C. It is suddenly exposed on both sides to a surrounding at 60C with convective heat transfer co-efficient of 85
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER W/m K. Calculate the centre line temperature and the temperature inside the plate 1.5 cm from themed plane after minutes. Take K for steel = 4.5 W/mK, for steel = 0.04 m /hr. Given Thickness L = 5 cm = 0.05 m Initial temperature T i = 400C + 7 = 67 K Final temperature T = 60C + 7 = K Distance x = 1.5 mm = 0.015 m Time t = minutes = 180 s Heat transfer co-efficient h = 85 W/m K Thermal diffusivity = 0.04 m /hr = 1.19 10-5 m /s. Thermal conductivity K = 4.5 W/mK. Solution For Plate : L Characteristic Length Lc 0.05 = We know, hlc Biot number Bi K 85 0.05 4.5 Bi 0.1675 Infinite Solids Case (i) L c 0.05 m 0.1 < B i < 100, So this is infinite solid type problem. [To calculate centre line temperature (or) Mid plane temperature for infinite plate, refer HMT data book Page No.59 Heisler chart].
X axis SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER c = c X axis Fourier number =.4 hl Curve K 85 0.05 0.167 4.5 hl K c Curve 0.167 t Fourier number = L -5 1.19 10 180 (0.05) X axis value is.4, curve value is 0.167, corresponding Y axis value is 0.64 T T T T 0 Y axis = 0.64 T0 T T T i 0.64 i 0 T T T T i 0.64 T0 0.64 67 T 0 550.6 K Center line temperature T 0 550.6 K Case (ii) Temperature (T x ) at a distance of 0.015 m from mid plane [Refer HMT data book Page No.60, Heisler chart] 4
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER hl K x 0.015 Curve 0.5 L 0.05 c X axis Biot number Bi 0.167 c X axis value is 0.167, curve value is 0.5, corresponding Y axis value is 0.97. T T x 0 T T 0.97 x Y axis = 0.97 T T x 0 x T T T T 0 T T 0.97 Tx 0.97 550.6 T 544 K Temperature inside the plate 1.5 cm from the mid plane is 544 K. 5
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 1. Define convection. UNIT- CONVECTION Part-A Convection is a process of heat transfer that will occur between a solid surface and a fluid medium when they are at different temperatures.. Define Reynolds number (Re) & Prandtl number (Pr). Reynolds number is defined as the ratio of inertia force to viscous force. Re Inertia force Viscous force Prandtl number is the ratio of the momentum diffusivity of the thermal diffusivity. Momentum diffusivity Pr Thermal diffusivity. Define Nusselt number (Nu). It is defined as the ratio of the heat flow by convection process under an unit temperature gradient to the heat flow rate by conduction under an unit temperature gradient through a stationary thickness (L) of metre. Nusselt number (Nu) = Q conv. Q cond 4. Define Grash of number (Gr) & Stanton number (St). It is defined as the ratio of product of inertia force and buoyancy force to the square of viscous force. Inertia force Buyoyancy force Gr (Viscous force) Stanton number is the ratio of nusselt number to the product of Reynolds number and prandtl number. 6
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Nu St Re Pr 5. What is meant by Newtonian and non Newtonian fluids? The fluids which obey the Newton s Law of viscosity are called Newtonian fluids and those which do not obey are called non Newtonian fluids. 6. What is meant by laminar flow and turbulent flow? Laminar flow: Laminar flow is sometimes called stream line flow. In this type of flow, the fluid moves in layers and each fluid particle follows a smooth continuous path. The fluid particles in each layer remain in an orderly sequence without mixing with each other. Turbulent flow: In addition to the laminar type of flow, a distinct irregular flow is frequency observed in nature. This type of flow is called turbulent flow. The path of any individual particle is zig zag and irregular. Fig. shows the instantaneous velocity in laminar and turbulent flow. 7. What is meant by free or natural convection & forced convection? If the fluid motion is produced due to change in density resulting from temperature gradients, the mode of heat transfer is said to be free or natural convection. If the fluid motion is artificially created by means of an external force like a blower or fan, that type of heat transfer is known as forced convection. 8. Define boundary layer thickness. The thickness of the boundary layer has been defined as the distance from the surface at which the local velocity or temperature reaches 99% of the external velocity or temperature. 9. What is the form of equation used to calculate heat transfer for flow through cylindrical pipes? Nu = 0.0 (Re) 0.8 (Pr) n n = 0.4 for heating of fluids n = 0. for cooling of fluids 7
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 10. What is meant by Newtonian and non Newtonian fluids? The fluids which obey the Newton s Law of viscosity are called Newtonian fluids and those which do not obey are called non Newtonian fluids. Part-B 1. Air at 0C, at a pressure of 1 bar is flowing over a flat plate at a velocity of m/s. if the plate maintained at 60C, calculate the heat transfer per unit width of the plate. Assuming the length of the plate along the flow of air is m. Given : Fluid temperature T = 0C, Pressure p = 1 bar, Velocity U = m/s, Plate surface temperature T w = 60C, Width W = 1 m, Length L = m. Solution : We know, Film temperature 60 0 T 40C f T f T w T Properties of air at 40C: Density = 1.19 Kg/m Thermal conductivity K = 6.56 10 W /mk, 6 Kinematic viscosity v = 16.96 10 m / s. Prandtl number Pr = 0.699 We know, Reynolds number Re = UL v 16.96 10 5.77 10 6 4 8
4 5 Re 5.77 10 5 10 SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Reynolds number value is less than 5 10 5, so this is laminar flow. For flat plate, Laminar flow, Local Nusselt Number Nu x = 0. (Re) 0.5 (Pr) 0. Nu 0. (5.77 10 ) (0.699) x Nu 175.7 x We know that, Local Nusselt Number Nu h s 175.7 6.56 10 4 0.5 0. x hs L K Local heat transfer coefficient h x =.7 W/m K We know, Average heat transfer coefficient h = h x h.7 h = 4.65 W/m K Heat transfer Q = h A (T w - T ) 4.65 (60 0) [ Area widthlength 1 ] Q 7 Watts.. Air at 0C at atmospheric pressure flows over a flat plate at a velocity of m/s. if the plate is 1 m wide and 80C, calculate the following at x = 00 mm. 1. Hydrodynamic boundary layer thickness,. Thermal boundary layer thickness,. Local friction coefficient, 4. Average friction coefficient, 9
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 5. Local heat transfer coefficient 6. Average heat transfer coefficient, 7. Heat transfer. Given: Fluid temperature T = 0C Velocity U = m/s Wide W = 1 m Surface temperature Tw = 80C Distance x = 00 mm = 0. m Solution: We know, Film temperature Tw T Tf 80 0 T 50C f Properties of air at 50C Density = 1.09 kg/m Kinematic viscosity v = 17.95 Pr andt l number Pr =0.698 Thermal conductivity K = 8.6 We know, Reynolds number Re = UL v 0. 6 17.9510 Re 5.0110 510 4 5 Since Re < 5 10 5, flow is laminar For Flat plate, laminar flow, -6 10 m / s - 10 W / mk 1. Hydrodynamic boundary layer thickness: 0
hx hx 5 x (Re) 6.710 m SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 0.5 = 5 0. (5.0110 ) 4 0.5. Thermal boundary layer thickness: TX (Pr) hx TX TX 0. 6.7 10 (0.698) 7.5 10 m 0.. Local Friction coefficient: C fx fx 0.664(Re) - 0.5 = 0.664 (5.0110 ) C =.96 10 4. Average friction coefficient: 4 0.5 C fl fl 1.8 (Re) - -0.5 = 1.8 (5.0110 ) = 5.9 10 C 5.9 10 4 0.5 5. Local heat transfer coefficient (h x ): Local Nusselt Number Nu x = 0. (Re) 0.5 (Pr) 0. 0. (5.0110 ) (0.698) Nu 65.9 x We know Local Nusselt Number 4 0. 1
hx L Nux K h 0..6 10 h SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER x 65.9 x = L = 0.m x 6.0 W/m K Local heat transfer coefficient h 6. Average heat transfer coefficient (h): h h h x 6.0 7. Heat transfer: 1.41 W /m K x 6.0 W / mk We know that, Q h A(T T ) w = 1.41 (10.) (80-0) Q =.8 Watts. Air at 0C flows over a flat plate at a velocity of m/s. The plate is m long and 1.5 m wide. Calculate the following: To find: 1. Boundary layer thickness. Total drag force.. Total mass flow rate through the boundary layer between x = 40 cm and x = 85 cm. Given: Fluid temperature T = 0C Velocity Length Wide W U = m/s L = m W = 1.5 m
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Solution: Properties of air at 0C 1.165 kg/m 6 v 16 10 m / s Pr 0.701 K 6.75 10 W /mk We know, UL Reynolds numberre v 6 16 10 Re.5 10 5 10 Since Re<5 5 5 5 10,flow is laminar For flat plate, laminar flow, [from HMT data book, Page No.99] Hydrodynamic boundary layer thickness hx hx 5 x (Re) 0.5 = 5 (.510 ) 0.0 m 5 0.5 Thermal boundary layer thickness,
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 0. tx hx (Pr) TX =0.0 (0.701) 0.05 m -0. We know, Average friction coefficient, C fl fl 1.8 (Re) 0.5 = 1.8 (.5 10 ) C.65 10-5 0.5 We know C t U t 1.165 () Average shear stress t = 6.110 N/ m fl -.65 10 - - Drag force = Area Average shear stress = 1.5 6.110 Drag force = 0.018 N Drag force on two sides of the plate = 0.018 = 0.06 N Total mass flow rate between x = 40 cm and x= 85 cm. 4
SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 5 m U 85 40 hx hx 8 Hydrodynamic boundary layer thickness hx0.5 5 x (Re) 0.5 U x = 5 0.85 v 0.5 0.5 0.85 5 0.85 6 HX0.85 hx=0.40 0.5-0.5 0.5 0.40 5 0.40 6 HX0.40 16 10 0.010 m = 5 x (Re) U x 5 0.40 v 16 10 8.9 10 m 5 (1) m= 1.165 0.010 8.9 10 8 - m = 5.97 10 Kg/ s, 4. Air at 90C flows over a flat plate at a velocity of 6 m/s. The plate is 1m long and 0.5 m wide. The pressure of the air is 6 kn/m. If the plate is maintained at a temperature of 70C, estimate the rate of heat removed from the plate. Given : Fluid temperature T = 90C Velocity U = 6 m/s. Length L = 1 m Wide W = 0.5 m Pressure of air P = 6 kn/m 6 10 N/m Plate surface temperature T w = 70C 5
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER To find: Heat removed from the plate Tw T Solution: We know, Film temperature Tf 70 90 T 180C f Properties of air at 180C (At atmospheric pressure) 0.799 Kg/m -6 =.49 10 m / s Pr 0.681 K - 7.80 10 W/mK Note: Pressure other than atmospheric pressure is given, so kinematic viscosity will vary with pressure. Pr, K, C p are same for all pressures. P Kinematic viscosity atm P atm given 6.49 10 6 10 N/ m 6.49 10 1 bar Atmospheric pressure = 1 bar 5 10 N/ m 6 10 N/ m 5 1 bar = 110 N/m -4 Kinematic viscosity v = 5.145 10 m / s. We know, Reynolds number UL Re v 6
61 4 5.145 10 Re 1.10 10 5 10 SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 4 5 5 Since Re< 5 10, flow is laminar For plate, laminar flow, Local nusselt number NU 0. (Re) (Pr) x x 0.5 0. 0. (1.10 10 ) (0.681) NU 0.6 4 0.5 0. hl x We know NU x = K h 1 7.80 10 x 0.6 [ L = 1 m] Local heat transfer coefficient h x 1.15 W/m K We know Average heat transfer coefficient h = h x h 1.15 h.1 W/m K We know Heat transferred Q h A (T T ).1 (1 0.5) (56 4) Q 54.1 W Heat transfer from both side of the plate = 54.1 w = 508. W. 7
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 5. Air at 40C flows over a flat plate, 0.8 m long at a velocity of 50 m/s. The plate surface is maintained at 00C. Determine the heat transferred from the entire plate length to air taking into consideration both laminar and turbulent portion of the boundary layer. Also calculate the percentage error if the boundary layer is assumed to be turbulent nature from the very leading edge of the plate. Given : Fluid temperature T = 40C, Length L = 0.8 m, Velocity U = 50 m/s, Plate surface temperature T w = 00C To find : 1. Heat transferred for: i. Entire plate is considered as combination of both laminar and turbulent flow. ii. Entire plate is considered as turbulent flow.. Percentage error. Solution: We know Film temperature T f T w T T 00 40 44 K T 170C f Pr operties of air at 170C: = 0.790 Kg/m 6 1.10 10 m / s Pr 0.6815 K 7 10 W/mK We know UL Reynolds number Re= v 50 0.8 1.6 10 6 1.10 10 6 5 Re = 1.6 10 5 10 Re 5 5 10,so this is turbulent flow 6 8
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Case (i): Laminar turbulent combined. [It means, flow is laminar upto Reynolds number value is 5 10 5, after that flow is turbulent] Average nusselt number = Nu = (Pr) 0. (Re) 0.8 871 Nu = (0.6815) 0. [0.07 (1.6 10 6 ) 0.8 871 Average nusselt number Nu = 1705. hl We know Nu = K h 0.8 1705. 7 10 h 78.8 W / m K Average heat transfer coefficient h=78.8 W/m K Head transfer Q h A (T T ) 1 1 w hl W (T T ) = 78.8 0.8 1 (00-40) Q 1690.4 W w Case (ii) : Entire plate is turbulent flow: Local nusselt number} Nux = 0.096 (Re) 0.8 (Pr) 0. NU x = 0.096 (1.6 10 6 ) 0.8 (0.6815) 0. NU x = 1977.57 We know NU x hx L K h 0.8 x 1977.57 7 10 hx 91.46 W/m K 9
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Local heat transfer coefficient h x = 91.46 W/m K Average heat transfer coefficient (for turbulent flow) h = 1.4 h x = 1.4 91.46 Average heat transfer coefficient} h = 11.41 W/m K We know Heat transfer Q = h A (T w + T ) = h L W (T w + T ) = 11.41 0.8 1 (00 40) Q = 589. W Q Q1. Percentage error = Q 589. - 1690.4 = 100 1690.4 = 4.9% 1 6. 50 Kg/hr of air are cooled from 100C to 0C by flowing through a.5 cm inner diameter pipe coil bent in to a helix of 0.6 m diameter. Calculate the value of air side heat transfer coefficient if the properties of air at 65C are K = 0.098 W/mK = 0.00 Kg/hr m Pr = 0.7 = 1.044 Kg/m 40
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Given : Mass flow rate in = 05 kg/hr 05 Kg/ s in = 0.056 Kg/s 600 Inlet temperature of air T mi = 100C Outlet temperature of air T mo = 0C Diameter D =.5 cm = 0.05 m Tmi Tmo Mean temperature Tm 65 C To find: Heat transfer coefficient (h) Solution: Reynolds Number Re = UD Kinematic viscosity 0.00 Kg/ s m 600 1.044 Kg/m 7 v 7.98 10 m / s Mass flow rate in = A U 41
0.056 1.044 D U SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 4 4 0.056 1.044 (0.05) U U = 55.7 m/s UD (1) Re = 55.7 0.05 = -7 7.98 10 6 Re =.44 10 Since Re > 00, flow is turbulent For turbulent flow, general equation is (Re > 10000) Nu 0.0 (Re) (Pr) 0.8 0. This is cooling process, so n = 0. Nu = 0.0 (.44 10 ) (0.7) Nu 661.7 6 0.8 0. We know that, hd Nu K h0.05 661.7 0.098 Heat transfer coefficient h = 66. W/m K 4
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 7. In a long annulus (.15 cm ID and 5 cm OD) the air is heated by maintaining the temperature of the outer surface of inner tube at 50C. The air enters at 16C and leaves at C. Its flow rate is 0 m/s. Estimate the heat transfer coefficient between air and the inner tube. Given : Inner diameter D i =.15 cm = 0.015 m Outer diameter D o = 5 cm = 0.05 m Tube wall temperature T w = 50C Inner temperature of air T mi = 16C Outer temperature of air t mo = C Flow rate U = 0 m/s To find: Heat transfer coefficient (h) Solution: Tmi T Mean temperature T m = 16 T 4C m Properties of air at 4C: = 1.614 Kg/m -6 = 15.9 10 m / s Pr = 0.707 K = 6. - 10 W / mk We know, 4 mo
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Hydraulic or equivalent diameter D P Do Di D D D D h D 4 D D 4A i 4 o o i o i D (D D ) = 0.05 0.015 D h = 0.01875 m i o i Reynolds number Re = 0 0.01875 6 15.910 UD h Re = 5. 10-6 Since Re > 00, flow is turbulent For turbulent flow, general equation is (Re > 10000) Nu = 0.0 (Re) 0.8 (Pr) n This is heating process. So n = 0.4 Nu = 0.0 (5. 10 ) (0.707) Nu 87.19 0.8 0.4 44
hdh We know Nu = K h 0.01875 87.19= 6. 10 h = 1. W/m K SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER - 8. Engine oil flows through a 50 mm diameter tube at an average temperature of 147C. The flow velocity is 80 cm/s. Calculate the average heat transfer coefficient if the tube wall is maintained at a temperature of 00C and it is m long. Given : Diameter D = 50 mm = 0.050 m Average temperature T m = 147C Velocity U = 80 cm/s = 0.80 m/s Tube wall temperature T w = 00C Length L = m To find: Average heat transfer coefficient (h) Solution : Properties of engine oil at 147C = 816 Kg/m -6 = 7 10 m / s Pr = 116 K = 1.8 We know - 10 W/mK 45
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Reynolds number Re = 0.8 0.05 6 7 10 Re = 5714. UD Since Re < 00 flow is turbulent L 40 D 0.050 L 10 400 D For turbulent flow, (Re < 10000) Nusselt number Nu = 0.06 (Re) (Pr) 0.8 0. D L 0.055 0.8 0. 0.050 Nu 0.06 (5714.) (116) Nu 14.8 hd We know Nu = K 0.055 h 0.050 14.8 = 1.8 10 h = 8. W/m K - 9. A large vertical plate 4 m height is maintained at 606C and exposed to atmospheric air at 106C. Calculate the heat transfer is the plate is 10 m wide. Given : Vertical plate length (or) Height L = 4 m Wall temperature T w = 606C 46
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Air temperature T = 106C Wide W = 10 m To find: Heat transfer (Q) Solution: T Film temperature T 606 106 T 56C f -6 55.46 10 m / s - 10 W/mK f w T Properties of air at 56C = 50C = 0.566 Kg/m Pr = 0.676 K = 49.08 1 Coefficient of thermal expansion} = T in K 1 1 56 7 69-1 = 1.58 10 K Grashof number Gr = Gr = g L T - 9.81.4 10 (4) (606 106) v (55.46 10 ) 6 f Gr = 1.61 10 11 47
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Gr Pr = 1.61 10 11 0.676 Gr Pr = 1.08 10 11 Since Gr Pr > 10 9, flow is turbulent For turbulent flow, Nusselt number Nu = 0.10 [Gr Pr] 0. Nu = 0.10 [1.08 10 ] Nu = 471.0 11 0. We know that, Nusselt number hl Nu K 47.0 = h 4 49.08 10 - Heat transfer coefficient h = 5.78 W/m K Heat transfer Q = h A T h W L (T T ) 5.78 10 4 (606 106) Q 115600 W w Q = 115.6 10 W 48
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 10. A thin 100 cm long and 10 cm wide horizontal plate is maintained at a uniform temperature of 150C in a large tank full of water at 75C. Estimate the rate of heat to be supplied to the plate to maintain constant plate temperature as heat is dissipated from either side of plate. Given : Length of horizontal plate L = 100 cm = 1m Wide W = 10 cm = 0.10 m Plate temperature T w = 150C Fluid temperature T = 75C To find: Heat loss (Q) from either side of plate Solution: T Film temperature T 150 75 T 11.5C f -6 = 0.64 10 m / s K = 68 10 W/mK f w T Properties of water at 11.5C = 951 Kg/m Pr = 1.55 49
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 1 1 Coefficient of thermal expansion} = T in K 11.5 7 f 1.5910 K Grashof Number Gr = For horizontal plate, Characteristic length L L c = 0.05 m (1) Gr = Gr =.4110 9 Gr Pr =.4110 1.55 g L T c v W 0.10-9.81.59 10 (0.05) (150 75) 9 (0.6410 ) 6 Gr Pr = 5.9 10 9 Gr Pr value is in between 8 10 6 and 10 11 i.e., 8 10 6 < Gr Pr < 10 11 For horizontal plate, upper surface heated: Nusselt number Nu = 0.15 (Gr Pr) 0. Nu = 0.15 [5.9 10 ] Nu = 59.41 9 0. We know that, Nusselt number Nu = h 0.05 u 59.41 68 10 h u = 54.6 W/m K hl u K c 50
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Upper surface heated, heat transfer coefficient h u = 54.6 W/m K For horizontal plate, lower surface heated: Nusselt number Nu = 0.7 [Gr Pr] 0.5 Nu = 0.7 [5.9 10 ] 9 0.5 Nu = 7.8 We know that, hl Nusselt number Nu = K hl 1 c 7.8 K h1 0.05 7.8 6810 h 1 994.6 W/m K 1 c Lower surface heated, heat transfer coefficient h 1 = 994.6 W/m K Total heat transfer Q = (h u + h 1 ) A T = (h u + h 1 ) W L (T w - T ) = (54.6 + 994.6) 0.10 (150 75) Q = 406.5 W 51
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER UNIT- BOILING AND CONDENSATION,HEAT EXCHANGER PART - A 1. What is meant by Boiling and condensation? The change of phase from liquid to vapour state is known as boiling. The change of phase from vapour to liquid state is known as condensation.. Give the applications of boiling and condensation. Boiling and condensation process finds wide applications as mentioned below. 1. Thermal and nuclear power plant.. Refrigerating systems. Process of heating and cooling 4. Air conditioning systems. What is meant by pool boiling? If heat is added to a liquid from a submerged solid surface, the boiling process referred to as pool boiling. In this case the liquid above the hot surface is essentially stagnant and its motion near the surface is due to free convection and mixing induced by bubble growth and detachment. 4. What is meant by Film wise and Drop wise condensation? The liquid condensate wets the solid surface, spreads out and forms a continuous film over the entire surface is known as film wise condensation. In drop wise condensation the vapour condenses into small liquid droplets of various sizes which fall down the surface in a random fashion. 5. Give the merits of drop wise condensation? In drop wise condensation, a large portion of the area of the plate is directly exposed to vapour. The heat transfer rate in drop wise condensation is 10 times higher than in film condensation. 5
6. What is heat exchanger? SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER A heat exchanger is defined as an equipment which transfers the heat from a hot fluid to a cold fluid. 7. What are the types of heat exchangers? The types of heat exchangers are as follows 1. Direct contact heat exchangers. Indirect contact heat exchangers. Surface heat exchangers 4. Parallel flow heat exchangers 5. Counter flow heat exchangers 6. Cross flow heat exchangers 7. Shell and tube heat exchangers 8. Compact heat exchangers. 8. What is meant by Direct heat exchanger (or) open heat exchanger? In direct contact heat exchanger, the heat exchange takes place by direct mixing of hot and cold fluids. 9. What is meant by Indirect contact heat exchanger? In this type of heat exchangers, the transfer of heat between two fluids could be carried out by transmission through a wall which separates the two fluids. 10. What is meant by Regenerators? In this type of heat exchangers, hot and cold fluids flow alternately through the same space. Examples: IC engines, gas turbines. 11. What is meant by Recuperater (or) surface heat exchangers? This is the most common type of heat exchangers in which the hot and cold fluid do not come into direct contact with each other but are separated by a tube wall or a surface. 1. What is meant by parallel flow and counter flow heat exchanger? In this type of heat exchanger, hot and cold fluids move in the same direction. In this type of heat exchanger hot and cold fluids move in parallel but opposite directions. 5
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 1. What is meant by shell and tube heat exchanger? In this type of heat exchanger, one of the fluids move through a bundle of tubes enclosed by a shell. The other fluid is forced through the shell and it moves over the outside surface of the tubes. 14. What is meant by compact heat exchangers? There are many special purpose heat exchangers called compact heat exchangers. They are generally employed when convective heat transfer coefficient associated with one of the fluids is much smaller than that associated with the other fluid. 15. What is meant by LMTD? We know that the temperature difference between the hot and cold fluids in the heat exchanger varies from point in addition various modes of heat transfer are involved. Therefore based on concept of appropriate mean temperature difference, also called logarithmic mean temperature difference, also called logarithmic mean temperature difference, the total heat transfer rate in the heat exchanger is expressed as Q = U A (T)m Where U Overall heat transfer coefficient W/m K A Area m (T) m Logarithmic mean temperature difference. 16. What is meant by Fouling factor? We know the surfaces of a heat exchangers do not remain clean after it has been in use for some time. The surfaces become fouled with scaling or deposits. The effect of these deposits the value of overall heat transfer coefficient. This effect is taken care of by introducing an additional thermal resistance called the fouling resistance. 17. What is meant by effectiveness? The heat exchanger effectiveness is defined as the ratio of actual heat transfer to the maximum possible heat transfer. Actual heat transfer Q Effectiveness = Maximum possible heat transfer Q 54 max
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Part-B 1. Water is boiled at the rate of 4 kg/h in a polished copper pan, 00 mm in diameter, at atmospheric pressure. Assuming nucleate boiling conditions calculate the temperature of the bottom surface of the pan. Given : m = 4 kg / h 4 kg 600 s m 6.6 10 kg/ s d = 00 mm =.m Solution: We know saturation temperature of water is 100C i.e. T sat = 100C Properties of water at 100C From HMT data book Page No.1 55
Density l = 961 kg/m SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER -6 Kinematic viscosity v = 0.9 10 m / s Pr andtl number P 1.740 r Specific heat Cpl = 4.16 kj/kg K = 416 j/kg K Dynamic viscosity l = l v = 9610.9 10 L 81.57 10 Ns/m 6-6 From steam table (R.S. Khumi Steam table Page No.4) At 100C Enthalpy of evaporation hfg = 56.9 kj/kg hfg 56.9 10 j/kg Specific volume of vapour Vg = 1.67 m /kg Density of vapour 1 v vg 1 1.67 v 0.597 kg/m For nucleate boiling 56
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Q g ( l v) CplT Heat flux lhfg A Csf hfgp We know transferred Q = mhfg Heat transferred Q = m hfg. 1.7 r Q mhg A A Q 6.6 10 56.9 10 A d 4 6.6 10 56.9 10 = (.) 4 Q 10 10 w / m A - surface tension for liquid vapour interface At 100C (From HMT data book Page No.147) 58.8 10 N/ m For water copper Csf = Surface fluid constant = 01 C sf.01 (From HMT data book Page No.145) Q Substitute, l, h fg, l, v,, Cpl, hfg, and P r values in Equation (1) A 57
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER (1) 10 10 81.57 10 56.9 10 9.81961597 58.8 10 6 0.5 416 T.01 56.9 10 (1.74) 1.7 416 T 0.85 759.7 w T(.56).85 T.056 = 0.97 T - 16.7 We know that Excess temperature T = T T 116.7C w w T sat 16.7 = T 100C.. A nickel wire carrying electric current of 1.5 mm diameter and 50 cm long, is submerged in a water bath which is open to atmospheric pressure. Calculate the voltage at the burn out point, if at this point the wire carries a current of 00A. Given : D = 1.5mm = 1.5 10 - m; L = 50 cm = 0.50m; Current I = 00A Solution We know saturation temperature of water is 100C 58
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER i.e. T sat = 100C Properties of water at 100C (From HMT data book Page No.11) l = 961 kg/m 6 v 0.9 10 m / s P - 1.740 r Cpl = 4.16 kj/kg K = 416 j/kg K l= l v 961 0.910 l 81.57 10 Ns/m 6 6 From steam Table at 100C R.S. Khurmi Steam table Page No.4 hfg 56.9 kj/kg hfg = 56.9 v g 1.67m / kg g 10 j/kg 1 1 v 0.597 kg/m 1.67 = Surface tension for liquid vapour interface At 100C 58.8 10 N/m (From HMT data book Page No.147) 59
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER For nucleate pool boiling critical heat flux (AT burn out) 0.5 Q g ( l - v) 0.18hfg v 1 A v (From HMT data book Page No.14) Substitute h fg, l, v, values in Equation (1) Q A (1) 0.18 56.9 10 0.597 58.8 10 9.81 (961.597).597 Q 1.5 10 W/m A 6 0.5 We know Heat transferred Q = V I Q V I A A V 00 dl 6 V 00 1.510 = 6 1.510 A = dl V 17.9 volts - 1.5 10.50. Water is boiling on a horizontal tube whose wall temperature is maintained ct 15C above the saturation temperature of water. Calculate the nucleate boiling heat transfer coefficient. Assume 60
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER the water to be at a pressure of 0 atm. And also find the change in value of heat transfer coefficient when 1. The temperature difference is increased to 0C at a pressure of 10 atm.. The pressure is raised to 0 atm at T = 15C Given : Wall temperature is maintained at 15C above the saturation temperature. Tw 115C. Tsat 100C Tw 100 15 115 C = p = 10 atm = 10 bar case (i) T 0C; p 10 atm = 10 bar case (ii) p = 0 atm = 0 bar; T - 15C Solution: We know that for horizontal surface, heat transfer coefficient h = 5.56 (T) From HMT data book Page No.18 h = 5.56 (T w T sat ) = 5.56 (115 100) 61
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER h 18765 w/m K Heat transfer coefficient other than atmospheric pressure h p = hp 0.4 From HMT data book Page No.144 = 18765 10 0.4 Heat transfer coefficient hp 47.1 10 W / m K Case (i) P = 100 bar T = 0C From HMT data book Page No.144 Heat transfer coefficient h 5.56 ( T) = 5.56(0) h 150 10 W /m K Heat transfer coefficient other than atmospheric pressure h p = hp 0.4 6
15010 (10) h p 0.4 7710 W /m K SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Case (ii) P = 0 bar; T = 15C Heat transfer coefficient h = 5.56 (T) = 5.56 (15) h 18765 W/m K Heat transfer coefficient other than atmospheric pressure h p = hp 0.4 = 18765 (0) 0.4 hp 6.19 10 W/m K 4. A vertical flat plate in the form of fin is 500m in height and is exposed to steam at atmospheric pressure. If surface of the plate is maintained at 60C. calculate the following. 1. The film thickness at the trailing edge. Overall heat transfer coefficient. Heat transfer rate 4. The condensate mass flow rate. Assume laminar flow conditions and unit width of the plate. 6
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Given : Height ore length L = 500 mm = 5m Surface temperature T w = 60C Solution We know saturation temperature of water is 100C i.e. T sat = 100C (From R.S. Khurmi steam table Page No.4 h fg = 56.9kj/kg h fg = 56.9 10 j/kg We know Film temperature T 60 100 T f 80C f T w T sat Properties of saturated water at 80C (From HMT data book Page No.1) 64
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER - 974 kg/m 6 v 0.64 10 m / s - k = 668.710 W/mk = p v= 974 0.64 10-6 54.5 10 Ns/ m 6 1. Film thickness x We know for vertical plate Film thickness 4K x (Tsat T w ) x ghfg Where X = L = 0.5 m 0.5 6 4 54.510 668.710 0.5100 60 x x 4 1.710 m 9.81 56.910 974. Average heat transfer coefficient (h) 65
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER For vertical surface Laminar flow k gh fg h 0.94 L Tsat Tw 0.5 The factor 0.94 may be replace by 1.1 for more accurate result as suggested by Mc Adams (668.710 ) (974) 9.81 56.910 1.1 6 54.510 1.5100 60 h 6164. W/m k. 0.5. Heat transfer rate Q We know Q ha(t T ) sat = hl W (T T ) w sat = 6164. 0.5 1100-60 Q = 186 W w 4. Condensate mass flow rate m We know 66
Q mh Q m h fg fg 1..86 m 56.9 10 m 0.054 kg/s SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 10. Steam at 0.080 bar is arranged to condense over a 50 cm square vertical plate. The surface temperature is maintained at 0C. Calculate the following. a. Film thickness at a distance of 5 cm from the top of the plate. b. Local heat transfer coefficient at a distance of 5 cm from the top of the plate. c. Average heat transfer coefficient. d. Total heat transfer e. Total steam condensation rate. f. What would be the heat transfer coefficient if the plate is inclined at 0C with horizontal plane. Given : Pressure P = 0.080 bar Area A = 50 cm 50 cm = 50 050 = 0.5 m Surface temperature T w = 0C Distance x = 5 cm =.5 m Solution Properties of steam at 0.080 bar (From R.S. Khurmi steam table Page no.7) 67
T satj / kg SYED AMMAL ENGINEERING COLLEGE ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 41.5C h fg 40.kj/kg = 40.10 j/ kg We know Film temperature T 0+41.5 = T f 0.76C f T w T sat Properties of saturated water at 0.76C = 0C From HMT data book Page No.1 997 kg/m -6 0.8 10 m / s - k 6110 W /mk p v 997 0.8 10 87.5110 Ns / m 6 6 a. Film thickness We know for vertical surfaces 68
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 0.5 4K x (Tsat T w ) x g hfg (From HMT data book Page No.150) 6 4 87.5110 6110.5 (41.5 0)100 x x 4 1.40 10 m 9.81 40.10 997 b. Local heat transfer coefficient h x Assuming Laminar flow h h x k x 6110 1.46 10 x 4 hx 4,191 W/m K c. Average heat transfer coefficient h (Assuming laminar flow) 0.5 k gh fg L Tsat Tw h 0.94 The factor 0.94 may be replaced by 1.1 for more accurate result as suggested by Mc adams 0.5 k g h fg L Tsat Tw h 0.94 69
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER Where L = 50 cm =.5 m (6110 ) (997) 9.81 40.10 h 1.1 h 5599.6 W/m k 6 87.5110.5 41.5 0 0.5 d. Heat transfer (Q) We know Q = ha(t sat T w ) h A (T T ) sat w 5599.6 0.5 (41.5 0 Q 0.19.8 W e. Total steam condensation rate (m) We know Heat transfer Q mh Q m h fg fg 70
ME650 HEAT AND MASS TRNSFER MARKS & 16 MARKS QUESTION AND ANSWER 0.19.8 m 40. 10 m 0.015 kg/ s f. If the plate is inclined at with horizontal h h sin inclined inclined inclined inclined vertical vertical 1/ 4 h h (sin0) h 5599.6 h 1 1/ 4 4.708.6 W/m k 1/ 4 Let us check the assumption of laminar film condensation We know Reynolds Number R e 6 e e 4m w where W = width of the plate = 50cm =.50m 4.015 R 0.50 87.5110 R 10.8 1800 So our assumption laminar flow is correct. 5. A condenser is to designed to condense 600 kg/h of dry saturated steam at a pressure of 0.1 bar. A square array of 400 tubes, each of 8 mm diameter is to be used. The tube surface is maintained at 0C. Calculate the heat transfer coefficient and the length of each tube. 71