4. A shell and tube heat exchanger with single shell and tube passes in counter flow is used to cool the oil of a large marine engine. Lake water (Shell-side fluid) enters the heat exchanger at 2.0 kg/s and 15C, while the oil enters at 1.0 kg/s and 100C. The oil flow 100 brass tubes, each 500 mm long and having inner and outer diameters of 6 mm and 8 mm, respectively. The shell- side heat transfer coefficient is 500 W/mK. Determine the oil and water outlet temperatures (in C) using the LMTD method. (25 points) For water assuming c = 4.182 kJ/kgK For oil assuming k = 0.137 W/mK, Pr = 300, Cp = 2.206 kJ/kgK, p=0.0186 Ns/m For brass, k = 110 W/mK, and pu(T wall) = 0.0836 Ns/m? Where the temperature of the wall is assumed to be 330K The overall heat transfer coefficient with no fouling and no fins is given as In (D/D) UA hA 2kLN h, The Seider-Tate correlation and Reynolds number are given as hD/k = Nu = 1.86 (Re Pr D/L) 3 (www).14 and Re-PVD /u = 4m/(XUDN) PVD 4r = Nu = 1.86(RePrD/L)1/3(u/w) 0.14 Re = 1 1 1 + + hD k TUDN watch Tein = 15C hij=500w/mk mw=2kg/s koil N-100 tubes TH,w=loutc La som mo=1kgls Di = 0.006m 0 0.038m 4. A shell and tube heat exchanger with single shell and tube passes in counter flow is used to cool the oil of a large marine engine. Lake water (Shell-side fluid) enters the heat exchanger at 2.0 kg/s and 15C, while the oil enters at 1.0 kg/s and 100C. The oil flow 100 brass tubes, each 500 mm long and having inner and outer diameters of 6 mm and 8 mm, respectively. The shell- side heat transfer coefficient is 500 W/mK. Determine the oil and water outlet temperatures (in C) using the LMTD method. (25 points) For water assuming c = 4.182 kJ/kgK For oil assuming k = 0.137 W/mK, Pr = 300, Cp = 2.206 kJ/kgK, p=0.0186 Ns/m For brass, k = 110 W/mK, and pu(T wall) = 0.0836 Ns/m? Where the temperature of the wall is assumed to be 330K The overall heat transfer coefficient with no fouling and no fins is given as In (D/D) UA hA 2kLN h, The Seider-Tate correlation and Reynolds number are given as hD/k = Nu = 1.86 (Re Pr D/L) 3 (www).14 and Re-PVD /u = 4m/(XUDN) PVD 4r = Nu = 1.86(RePrD/L)1/3(u/w) 0.14 Re = 1 1 1 + + hD k TUDN watch Tein = 15C hij=500w/mk mw=2kg/s koil N-100 tubes TH,w=loutc La som mo=1kgls Di = 0.006m 0 0.038m