Example(Cont'd Solution. Part 1 350K T aterIn 280K ≈295K,Cn=4181J/KgK p,c q=Ch(Th;-Tn。)=0.5×2090×(375-350 26125W ar T=1n+q1C=280+26125(0.2×4181) 311k For parallel flow,△Tm=95△7om=39 △T △T 95-39 △ out Im PF 63 hn△T/△ ln(95/39) For counter flow,△T=64△T=70 out △T-△T 64-70 Im cF =67 n△Tmn/△ Tout In(64/70) Heat Transfer Su Yongkang School of Mechanical EngineeringHeat Transfer Su Yongkang School of Mechanical Engineering # 12 Example (Cont’d) Solution, Part 1: • For parallel flow, • For counter flow, Toil,in = 375K Toil,in = 350K Twater,in = 280K ? Twater,out = Tc  295K, c p,c = 4181J / Kg.K ( ) / and , , , , c o c i c h h i h o T T q C q C T T = + = − 26125 W 0.5 2090 (375 350) = =   − 311 K 280 26125/(0.2 4181) = = +  63 ln( 95/ 39) 95 39 ln / Tlm,P F = − =    −   = i n out i n out T T T T Tin = 95 Tout = 39 Tin = 64 Tout = 7067 ln( 64 / 70) 64 70 ln / Tlm,CF = − =    −   = i n out i n out T T T T