Question

2) Hot air enters a heat exchanger at 350°C and exits at 155°C. The heat extracted is used to boil 0.283 kg/s of 100°C water (from saturated liquid to saturated steam). The heat exchanger is a single-shell shell-and-tube heat exchanger with two tube passes. The overall heat transfer coefficient for the hot side is 240 W/(m²°C). Assume a constant specific heat for air of Cp = 1.005 kJ/(kg °C). The pressure of the hot air and the boiling water is P = 1.0 atm, Find the following: a. the NTU for the heat exchanger NTU = (4 0.01) b. the effectiveness of the heat exchanger (+0.01) C. the net heat transfer from the hot air. q= _kW (+0.5 kW) d. the required hot-side area. An = m² +0.1 m)

Answer 1

I3 = Ty=1000 Ques@ Solution Ti 350° ma Oz Th 1550 01 co mw BT4 3 100°C m2°C mroci Kg @ For Alates ; mais flow Rate (z 0.283k9led) @Overall HoT coefficient le= 2400 -0.24ku of Hot side For later at 100°c and 1 atm Pressure enthalpy of Vapowrisation [h=2256 k5 for aur ; Specific Heat (Cp = 1005 kJ kq Ć may Flow Rate = lima .? Claculation of LMTD Chogarithmic mean Temperature difference)- 10m = 02-01 22 7-73 0 1 = T2-T4 mais ** l In (or)

; 02 350-100 102 = 250 = 155-100 101=55 Om= 250-55 • Om=128.78710 In (250 *** -55. h mair mair xCpaul Chi E= 0.78 energy Balance in heat exchemores: x Caist (350-155) = mo 0.283 x 2256 195 Cmin = max clairs = 3.27409 Ko e o effectiveness of H.E E=2 E= (AT) ai T - Tz (AT) mare Ti-Ty Ans E=850-155 350-100 NTU=N ; Formula of effectiveness In terms of NTU is E- e-N=0.22 Taking (2n) Both Side In en Ln 0.22 -N = -1-514127 N= 1-5141271 Any © net heat Bansfer from the Hat air dair = (ma CP) CT, - Iz) = (min (TI-T2) = 3.27409 (350-155) Pais = 638.448 kco.) Ang @ Required Hat side air area : - Formula for HoT a = M A Om 638.448 = 0.24xAx(128.787) A- 20.655 m2 l-e-N 0.78= 1-eN In terms of LMTD. Ans