Consider an elemental layer at a distance of x having length dx and area dA.

The temperature of the start of layer is thand tc for hot and cold fluid.

Elemental heat transfer for Heat Exchanger is,

dQ=U.dA.θ …(1)

For hot fluid-heat rejcted

dQ=-Ch.dth (Where Ch is heat capacity of hot fluid)

∴dth=-dQCh

For cold fluid-heat absorbed

dQ=Cc.dtc (Where Cc is heat capacity of cold fluid)

∴dtc=-dQCc

∴(dth-dtc)=-dQ(1Ch+1Cc) …(2)

Substitute (1) in (2)

(dth-dtc)=-U.dA.θ(1Ch+1Cc)

dθθ=-U(1Ch+1Cc)dA

Integrating both sides

θ1θ2dθθ=-U(1Ch+1Cc) dA

In (θ2θ1) =-U.A(1Ch+1Cc)

∴θ2θ1=e-U.A.(1Ch+1Cc) …(3)

If, Cc < Ch, Then Effectiveness

ε=Cc(tc2-tc1)Cmin(th1-tc1)

∴ tc2= ε.Cmin.(th1-tc1)Cc+tc1

If, Ch < Cc, Then Effectiveness

ε=Ch(th1-th2)Cmin(th1-tc1)

∴ th2=th1- ε.Cmin.(th1-tc1)Ch

∴(th2-tc2)= th1- ε.Cmin.(th1-tc1)Ch-ε.Cmin.(th1-tc1)Cc-tc1

∴ (th2-tc2)=(th1-tc1)- [1-ε.Cmin(1Ch+1Cc)]

∴θ2θ1=1-ε.Cmin(1Ch+1Cc) ...(4)

Equating (3) and (4)

∴ 1-ε.Cmin(1Ch+1Cc)=e-U.A.(1Ch+1Cc)

∴ε= 1-e-U.A.(1Ch+1Cc)Cmin(1Ch+1Cc)

∴ε= 1-e-U.ACh(1+ChCc)CminCh(1+ChCc)

assumingCh<cc, ch="Cmin" and="" cmax<="" p="">

∴ε= 1-e-NTU(1+C)1+C

In above equation C=CminCmax and NTU=U.ACmin