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### Effectiveness for Counterflow heat exchanger

1. :-
2. Flow arrangement and Temperature distribution for counter-flow H.T
3. Logarithmic Mean Temperature Difference (LMTD) for Counter-Flow Considering an elementary area dA of the heat exchanger. The rate of flow of heat through this elementary area is given by :- Deriving this we get, Where, θm=LMTD θ1 = th1 − tc2 θ2 = th2 − tc1 dQ = U ⋅ dA(th − tc) = U ⋅ dA ⋅ Δt …….(i) Q = )UA(θ2 − θ1 )ln(θ2 θ1 Since, Q=U ⋅ A ⋅ θm ∴ θm = θ2 − θ1 )ln(θ2 θ1
4. Effectiveness :- The heat exchange dQ through an area dA of the heat exchanger is given by , …….(ii) From expression (ii),we get, Substituting the value of dQ from expression (i) and rearranging the equation, we get, d(th − tc) = −U ⋅ dA(th − tc) 1 Ch − 1 Cc dQ = −m • hCph dth = m • cCpc dtc = −Chdth = Ccdtc dth = −dQ Ch and dtc = dQ Cc or d(th − tc) = −dQ 1 Ch − 1 Cc
5. d(th − tc) (th − tc) = − U ⋅ dA 1 Ch − 1 Cc ln (th2−tc1) (th1−tc2) = −U ⋅ A 1 Ch − 1 Cc ∴ ln (th2−tc1) (th1−tc2) = UA CC 1 − Cc Ch ∴ (th2 − tc1 ) (th1 − tc2 ) = exp UA CC 1 − Cc Ch Now, From effectiveness equation, ε= Q ⋅ Qmax = Actual H.T Max possible H.T ε = Ch(th1 − th2 Cmin(th1 − tc1 = Cc(tc2 − tc1 Cmin(th1 − tc1 ∴ 𝑡ℎ2 = 𝑡ℎ1 − 𝜀𝐶min(𝑡ℎ1 − 𝑡 𝑐1 𝐶ℎ
6. ∴ tc2 = εCmin(th1 − tc1 Cc + tc1 Substituting the values of tc2 and th2 , we get Since,Cmin Cmax=R and UA/Cmin=NTU ε = 1 − exp −NTU(1 − R) 1 − Rexp −NTU(1 − R) ε = 1 − exp (−U A Cmin) 1 − Cmin Cmax 1 − Cmin Cmax exp (−U A Cmin) 1 − Cmin Cmax
7. Here, we find that effectiveness of parallel flow and counter- flow . ε = 1−exp −NTU(1−R) 1−Rexp − NTU(1−R) …..Counter flow ε = 1−exp −NTU(1+R) 1+R …….(parallel flow)
8. Example:- A Counter-flow heat exchangeris used to cool 0.55 kg/sec with Cp=2.45kJ/kg˚C of Oil from 115˚C to 40˚C by the use of water. The inlet and outlet temp. of cooling water are 15˚C and 75˚C respectively. The Overall heat transfer coefficient is 1450W/m2˚C. Using NTU method, Calculate the following : 1)The mass flow rate of water 2)The effectiveness of Heat exchanger 3)The Surface area required. Given Data :- th1 = 115˚C , th2 = 40˚C , tc1 = 15˚C , tc2 = 75˚C , U= 1450W/m2˚C, mh= 0.55 kg/sec , Cph =2.45kJ/kg˚C , Cpc (water)=4.187kJ/kg˚C
9. i) Q=mCp Δt ∴ m • hCph (th1 − th2 ) = m • cCpc (tc2 − tc1 ) ∴ 0.55 × 2.45 115 − 40 = m • c×4.187(75-15) ∴ m • c= 0.40 kg/sec ii) Ch = m • hCph Cc = m • cCpc =2.45×0.55 =4.187×0.40 =1.3475 =1.6748 Here, Ch<Cc ∴ Cmin = Ch=1.3475 ε = Ch(th1−th2 Cmin(th1−tc1 = )1.3475(115−40 )1.3475(115−15 ∴ ε = 0.75
10. iii) Heat capacity ratio, R = Cmin Cmax = 1.3475 1.6748 = 0.81 ε = 1 − exp −NTU(1 − R) 1 − Rexp − NTU(1 − R) ∴ 0.75 = 1 − exp −NTU(1 − 0.81) 1 − 0.81exp − NTU(1 − 0.81) ∴ NTU = 2.4736 ∴ 2.4736 = 1450 × A 1.3475 NTU = UA Cmin ∴ A = 2.2987 m2
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