Heat Exchanger Tutorial-1 Question-1 The overall heat transfer coefficient of an automotive radiator that is a compact cross flow water to air heat exchanger with both fluids (air and water) unmixed as shown in the following figure is tested. The radiator has 35 tubes of internal diameter 0.5 cm and length 65 cm in a closely spaced plate-finned matrix. Hot water enters the tubes at 90°C at the rate of 0.6 kg/s and leaves at 70°C. Air flows across the radiator through the interfin spaces and is heated from 20°C to 35°C. Determine the overall heat transfer coefficient of this radiator based on the inner surface area of the tubes. [Assumption: the specific heat capacity of water is 4.195 kJ/(kg.C)]. Question-1 Question-1: Solution Using LMTD method, we have Given C p , w = 4.195 kJ Di = 0.005 m L = 65 m Q = UAF ( LMTD ) kg .C T1 = Th ,in − Tc ,out = 55 C T2 = Th ,out − Tc ,in = 50 C T1 − T2 = 52.46 C ln ( T1 T2 ) Th ,in = 90 C LMTD = Th ,out = 70 C A = n DL = 35 0.005 .65 = 0.36 m 2 Tc ,in = 20 C Tc ,out = 35 C t2 − t1 Th ,out − Th ,in P= = = 0.3 T1 − t1 Tc ,in − Th ,in F = 0.985 T −T T −T R = 1 2 = c ,in c ,out = 0.75 t2 − t1 Th ,out − Th ,in t = Tube side T = Shell side Question-1: Solution Solution 4 LMTD = 52.46 C Q = mC p , w (T ) = 0.6 4.195 (90 − 70) = 50.34kW A = 0.36 m 2 Q U= = 2.73 kW m 2 .C AF ( LMTD ) F = 0.985 Question-2 Hot oil is to be cooled by water in a 1-shell-pass and 8 tube passes heat exchanger. The tubes are thin-walled and are made of copper with an internal diameter of 1.4 cm. The length of each tube pass in the heat exchanger is 6 m, and the overall heat transfer coefficient is 300 W/m2.C. Water flows through the tubes at the rate of 0.2 kg/s, and the oil through the shell at a rate of 0.3 kg/s. The water and the oil enter at the temperature of 30֯C and 100֯C, respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperature of the water and the oil. The specific heat capacity of water and oil are assument to be C p , water = 4.195 kJ kg .C and C p ,oil = 2.13 kJ kg .C Question-2: Solution Di = 0.014 m Leach −tube = 6 m U = 300 W m 2 .C mw = 0.2 kg s mo = 0.3 kg s Tw,in = 30 C To ,in = 100 C Given → not enough info for LMTD method -NTU method will be implemented Question-2: Solution kg 4.195 kJ = 0.839 kW kg .C C s kg Ch = moC p ,oil = 0.3 2.13 kJ = 0.639 kW kg .C C s Cmin = Ch = 0.639 kW C C 0.639 c = min = = 0.762 Cmax 0.836 = 0.5 Qmax = Cmin (100 − 30) = 44.73 kW Cc = mwC p , water = 0.2 Q = Qmax → Need to find To find 2 UA 300 W m .C ( n DL ) NTU = = Cmin Cmin 300 W m 2 .C ( 8 0.014 6 ) NTU = = 0.99 639 Question-2: Solution c = 0.764 = 0.5 NTU = 0.99 Q = Qmax = 0.5 44.73 = 22.37 kW Q = mwC p , w ( T ) w = moC p ,o ( T )o Tw,out = 56.7 C To ,out = 65 C Question-3 Steam in the condenser of a power plant is to be considered at a temperature of 30oC with cooling water from a nearby lake, which enters the tubes of the condenser at 14oC and leaves at 22oC. The surface area of the tube is 45m2, and the overall heat transfer coefficient is 1500 W/m2C. Determine the mass flow rate of the cooling water needed and the rate of condensation of the steam in the condenser. Use Q = UA ( LMTD ) and kJ kg For condensation, the heat reject from h fg at condenser = 2431 the process of steam to condensate water can be calculated as multiplication of mass and h fg at condenser (enthalpy difference between phase changes). Question-3: Solution Using LMTD method, we have Q = UA( LMTD ) T1 = Th ,in − Tc ,out = 30 − 22 = 8 C T2 = Th ,out − Tc ,in = 30 − 14 = 16 C LMTD = LMTD = Tm = T1 − T2 = 11.5 C ln ( T1 T2 ) T1 − T2 T ln 1 T 2 where A = 45 m ( Given ) 2 T1 = Th ,in − Tc ,out T2 = Th ,out − Tc ,in Q = UA ( LMTD ) W 2 45 m 11.54 C 2 mC Q = 779.055 kW Q = 1500 Q = mwc p , w ( T ) w = ms h fg 779.055 kW = mw 4.195 mw = 23.2 kg s kJ (22 − 14) K kg .K Question-4 The overall heat transfer coefficient U of a circular fin heat exchanger in which a hot gas flows across the finned tubes and oil flows through the tubes is given by 1 1 1 = + U t hg o g ho i. Explain briefly how the efficiency on the gas side t of the finned tube is determined. ii. For the case of circular finned tube, surface CF-8.7-5/8J, determine the value of for the o flow of oil inside the finned tubes. For surface CF-8.7-5/8J the following expression may be assumed: = Ac A = rh fr The symbols have their usual meanings. The data for surface CF-8.7-5/8J are given in the following figure. Use the plate spacing of 1.848 in. Question-4 Question-4: Solution 1 1 1 = + U t hg o g ho t = efficiency of finned tube or overall surface efficiency t A = ( A − A f ) + ( A f f ) t = ( A − Af ) t = 1 − A Af A ( + 1 − ) ( A f f f A ) Question-4: Solution = = rh o (oil side) = A rh = A /V = = LAfr rh Ac Arh = Afr LA fr 2 ( 5) Arh = V Ac = Afr rh 2 D = 2 4 8 4 Afr = S 1.35 = 1.848 1.35 Ac = rh = flow cross sectional area/wetter perimeter D2 4 =D rh = D 4 D2 4 Ac D in 2 o (oil side) = = = = 0.787 3 Afr rh 1.848 1.35 D 1.848 1.35 in 4 Question-5 Figure shows the core geometry of a circular, finned-tube core structure (CF-7.0-5/8J) for the design of a cross-flow heat exchanger for recovery for heat from the exhaust gas of a micro-turbine. In operation, hot exhaust gas flows across the finned tube while water flows through the tubes. (i) Determine the frontal area, free flow area, and the total heat transfer area on the gas side based on a tube length of 1 inch. (ii) Determine the free flow area, the frontal area and the hydraulic radius for the flow on the water side. The tube outside diameter is 0.645 in or 16.38 mm, fin pitch is 7.0 per inch or 276 per m. The fin thickness is 0.01 inch or 0.254 mm. Question-5: Solution Air Side: S = 1.232" d o = 0.645" Frontal Area = Afr = S 1 = 1.232 in 2 Free flow area = Ac = Agap + Awithin − fins d Agap = S − 2 o 2 Awithin − fins 1.121 − d o − 2 2 1.121 − d o = 2 2 Ac = 0.5537in 2 2 1 = 0.111 in 2 1 − 7 0.01 = 0.4427 in ( ) total fin thickness Question-5: Solution A = total heat transfer area = Afin + Afin −tip + Aunfin Afin = 1.121 ( 4 2 ) − 0.6452 7 no. of fins 2 two surfaces Afin −tip = 1.121 0.01 7 Aunfin = 0.645 (1 − 0.01 7) A = 11.37 in 2 OilSide Ac ,o = Afr ,o 5 2 2 = 0.3068 in 4 8 = 1.35 1.232 = 1.6632 in 2 rh = flow cross sectional area/wetter perimeter D2 5 D 4 = = 8 = 0.15625 in rh = D 4 4 END END