Heat and Mass Transfer

Question: From the answered problem below, assume 361 K as T_co or C_p = 4.184 kJ/kg K

Water Flowing at a rate of 0.667kg/s enters a countercurrent heat exchanger at 308K and is heated by an oil stream (Cp=1.89(kJ/kgK) at 383K at a rate of 2.85kg/s, the overall U=300W/m2K and A=15.0m2. Calculate: a) heat-transfer rate (q) b) exit water stream temperature (Tco) Given: mH=2.85kg/s Tci=308KCpH=1.89kJ/kgK mc=0.667kg/sU=300W/m2k Assuming: Tco1=370K Cpc=4.192kJ/kgK A=15.0m2 Solution: NTU=UA/Cmin CcCc=2.79kW/KCH=(2.85kg/s)(1.89kJ/kgK)CH=5.39kW/kCmin=2.79kW/k=mcCpc=(0.667kg/s)(4.192kJ/kgK)CmaxCmin=CHCc0.51=0.71qq=min(THITCI)=0.71(2.79KkW)(383308)K=148kW=2790KW(300m2KW)(15.0m2)NTU=1.61 b)q=Cc(TcoTci)Ccq+Tci=TcoTco=2.79KkW+308K148kWTco=361K