Exercise 10 Solutions – Heat Exchangers

Exercise Ten Solutions – Heat Exchangers 1. A shell-and-tube heat exchanger with 2-shell passes and 8-tube passes is used to heat ethyl alcohol (cp = 2670 J/kg⋅oC) in the tubes from 25oC to 70oC at a rate of 2.1 kg/s. The heating is to be done by water (cp = 4190 J/kg⋅oC) that enters the shell side at 95oC and leaves at 45oC. If the overall heat transfer coefficient is 950 W/m2⋅oC, determine the heat transfer and surface area of the heat exchanger. (Problem and figure P9.63 from Çengel, Heat and Mass Transfer.) We can solve this problem by finding the heat transfer from the first law energy balance on ethyl alcohol. We have sufficient data to do so; we cannot do this for water because we do not have the mass flow rate for water. If we wanted to, we could find it from the heat transfer. We will then use the other given data to compute the log-mean temperature difference and the correction factor for the 2-shell-pass-8-tube-pass heat exchanger. Once we have these items we can compute the surface area from the equation Q = UAF∆Tlm . For the ethyl alcohol, the cold fluid, our energy balance gives. ( ) 2.1 kg 2650 J 0.001 kW ⋅ s = 252.3 kW 70 o C − 25o C Q = m c c p , c (Tc, out − Tc, in ) = o s kg ⋅ C J We compute the log-mean temperature difference for a counter-flow heat exchanger (the basis for the correction factor method) as follows. ∆Tln = (Th,out − Tc,in ) − (Th,in − Tc,out )  Th,out − Tc,in ln T −T c ,out  h,in     = (45 C − 25 C )− (95 C − 70 C ) = 22.4 C o o o o  45 o C − 25 o C   ln o  95 C − 70 o C    We have to compute the ratios R and P to determine the correction factor. P= R= Ttube,out − Ttube,in Tshell ,in − Ttube,in Tshell ,in − Ttube,in Ttube,out − Ttube,in t 2 − t1 70 o F − 25 o F = = 0.64 T1 − t1 95 o F − 25 o F T −T 95 o F − 45 o F = 1 2 = o = 1.1 t 2 − t1 70 F − 25 o F = o Exercise ten solutions Page 2 From Figure 11.18(8) for two shell passes and any multiple of 4 tube passes we find F = 0.82 for these values of R and P. Solving Q = UAF∆Tlm for the area and substituting the given data and intermediate results gives the desired area. 1000 W ( 252.3 kW )  Q kW A= = = 14.5 m2 950 W UF∆Tlm o (0.8) 22.4 C m 2 ⋅o C ( ) 2. Water (cp = 4180 J/kg⋅oC) is to be heated by solar-heated hot air (cp =1010 J/kg⋅oC) in a double-pipe counterflow heat exchanger. Air enters the heat exchanger at 90oC at a rate of 0.3 kg/s, while water enters at 22oC at a rate of 0.1 kg/s. The overall heat transfer coefficient based on the inner side of the tube is given to be 80 W/m2⋅oC. The length of the tube is 12 m and the internal diameter of the tube is 1.2 cm. Determine the outlet temperatures of the water and the air. Since we only know the inlet temperatures here, we have to apply the effectiveness-NTU method. We start by computing the mass-flow-times-heat-capacity products and determine which of the two is smaller. C h = m h c p ,h = C c = m c c p ,c = 0.3 kg 1010 kJ 1 kW ⋅ s 303 W = o s kg ⋅o C kJ C 0.1 kg 4180 kJ 1 kW ⋅ s 418 W = o s kJ kg ⋅o C C So Cmin = Cc = 303 W/oC. We use this to compute the maximum heat transfer. ( ) 303 W Q max = C min (Th,in − Tc,in ) = o 90 o C − 22 o C = 2.060 x10 4 W C In order to find the heat transfer we have to find the heat exchanger effectiveness. We do this by computing the NTU and using the charts that give the effectiveness as a function of NTU and the ratio of Cmin/Cmax. The surface area is required to compute the NTU. For this simple, counterflow heat exchanger, the total heat transfer surface area is found as follows. As = πDL = 12π(0.012 m )(12 m ) = 0.45 m 2 (0.45 m ) C 80 W NTU = UAs = m ⋅ 303 W C min o C 2 o 303 W 2 = 0.119 c= o C min C = 0.725 = C max 418 W o C For these values of NTU and c = Cmin./Cmax, we can use the equation for effectiveness of counterflow heat exchangers to find ε. 1 − e − NTU (1−c ) 1 − e −0.119(1−0.725) ε= = = 0.108 1 − ce − NTU (1−c ) 1 − 0.119e −0.119(1−0.725) The actual heat transfer is the product of the maximum heat transfer and the heat exchanger effectiveness. Exercise ten solutions Page 3 ( ) Q = εQ max = (0.108) 2.060 x10 4 W = = 2225 W We find the outlet temperatures from the first law mass balances. Q = C c (Tc,out − Tc,in ) ⇒ Tc,out = Tc,in + Q Cc = 22 o C + 2225 W = 27.3oC 418 W o C 2225 W Q = 82.7oC = 90 o C + Q = C h (Th,in − Th,out ) ⇒ Th,out = Th,in − 303 W Cc o C