Example(Cont'd Solution. Part 2 T≈295K,Cn。=4181J/KgK p, c To determine the minimum heat capacity rate C=0.5×2090=1045W/K Cc=0.2×4181=8362W/K=Cm Then max mIn =8362×(375-280)=79440 h T)=26125 1,O The effectiveness is E=q/qm=26125/79440=0.33 Heat Transfer Su Yongkang School of Mechanical EngineeringHeat Transfer Su Yongkang School of Mechanical Engineering # 14 Example (Cont’d) Solution, Part 2: • To determine the minimum heat capacity rate, • Then • The effectiveness is Tc  295K, c p,c = 4181J / Kg.K Ch = 0.52090 =1045 W / K min CC = 0.24181= 836.2 W / K = C 836.2 (375 280) 79440 ( ) max min , , =  − = q = C Th i −Tc i q = Ch (Th,i −Th,o ) = 26125  = q / qmax = 26125/ 79440 = 0.33