Question

P11-104. Cold water (Cp = 4180 J/kg-K) leading to a shower enters a thin-walled double-pipe counterflow heat exchanger at 15°C at a rate of 0.25 kg/s and is heated to 45°C by hot water (cp=4190 J/kg K) that enters at 100°C at a rate of 3 kg/s. If the overall heat transfer coefficient is 950 W/m2.K, determine the rate of heat transfer and the heat surface area of the heat exchanger using the effectiveness-NTU method.

Answer 1

Lov°c in- 3 kg ls. h = 4190 I lkgoc in = 0.25 kg/s. Cpc = 4180 71 kg°C 45°c Cold flued. Heat loss by hot fluid = Heat gain by in Can Thi The) =ine Coc (To-T) → 3x4190x1108 - The) - 0.25% 4180 ( 45-15) → The = 37.505°C] man innan = 3x4190 = 12570 W/oc c = 0.25x4180= 1045 w/oc. min = me R= (min - Smar 1045 = 0.0831. 12570 (effectiveness) E = 45-15 = 0.352 4 (To-To) - mon (The Ti loo - 15 6 Heat transfer (9) mc Coc (Tco-Ti = 0.25x4180 (45-15) = 31350 h

6 for counterflow heat exchanger. effectiveness (E) = 1- exp ENTUCI-R) 1-Rene [NTU (1-R)] let expENTU (1-P)] = P - 1-P 1-RP = 0.352 = 1- 0.0831P 0.352-0.03p = 1 -P 0.97 p = 0.648 P = 0.668 exp ENTU (1-P)] = 0.668. INTU (1-R) = +0.4034 NTU= 0.43gg. UH = NTU Anin 0.4399X 1045 7 H = NYUK (min 950 > A = 0.483 m?!