To determine a specific form of the effectiveness–NTU relation

ε=$f(NTU,\frac{C_min}{C_max} )$

consider a parallel-flow heat exchanger for which $C_min$=$C_h$

ε=$\frac{(T_(h,i)-T_(h,o))}{(T_(h,i)-T_(c,i) )}$

But heat transfer rate would be $\frac{C_min}{C_max}$ =$\frac{(m_h C_(p,h))}{(m_c C_(p,c) )}$=$\frac{(T_(c,o)-T_(c,i))}{(T_(h,i)-T_(h,o) )}$

Now returning to derivation of the LMTD we know that for a parallel flow hxgr

$ln\frac{(∆T_2)}{∆T_1 }$=$-UA(\frac{1}{C_h} +\frac{1}{C_c} )$=-$\frac{UA}{C_h}$ (1+$\frac{C_h}{C_c}$ )=-$\frac{UA}{C_min}$ (1+C_r)

$ln\frac{⁡(T_ho-T_co)}{(T_hi-T_ci )}$=-NTU(1+$C_r$)

And

$\frac{((T_ho-T_co)}{(T_hi-T_ci ))}$=exp(-NTU(1+$C_r$))

Now we have three expression for temperature that relates to ε,NTU and $C_r$ and we can eliminate temperature altogether to find ε only in terms of NTU and $C_r$ The derivation is as follows:

$T_∞=C_r(T_hi-T_ho)+T_ci$

$\frac{T_ho-T_∞}{T_hi-T-ci}=\frac{(T_ho-T_hi)+(T_hi-T_co)}{T_hi-T-ci}$

$\frac{T_ho-T_co}{T_hi-T-ci} =\frac{T_ho-T_hi}{T_hi-T-ci} +\frac{T_hi-T-ci}{T_hi-T-ci} - C_r\frac{T_hi-T-ho}{T_hi-T-ci}$

$\frac{T_ho-T_∞}{T_hi-T-ci}=(-s)+(1)-C_r(s)=exp(-NTU(1+C_r))$

$\epsilon$=$\frac{1-exp(-NTU(1+C_r))}{1+C_r}$

The above equation is the effectiveness of parallel flow heat exchanger by ntu method.