Effectiveness-NTU analysis of a heat exchanger

The effectiveness-NTU method

Effectiveness $\varepsilon$ is the ratio of the actual heat transfer to the maximum possible:

$$\varepsilon=\frac{Q}{Q_{max}},\qquad Q_{max}=C_{min}\,(T_{h,in}-T_{c,in})$$

where $C = \dot m c_p$ is the heat capacity rate. Number of transfer units $NTU = UA/C_{min}$ and capacity ratio $C_r = C_{min}/C_{max}$. For each flow arrangement, $\varepsilon = f(NTU, C_r)$.

It is preferred when the outlet temperatures are unknown (rating or performance problems): the LMTD method would need trial and error, while the NTU method is direct. LMTD is easier for sizing when all four temperatures are known.

Solution

Heat capacity rates:

$$C_h = 0.8\times2100 = 1680\ \text{W/K},\qquad C_c = 0.6\times4180 = 2508\ \text{W/K}$$

So $C_{min} = C_h = 1680$ W/K (oil), $C_{max} = 2508$ W/K.

$$C_r = \frac{1680}{2508} = 0.670,\qquad NTU = \frac{UA}{C_{min}} = \frac{350\times3}{1680} = 0.625$$

Effectiveness (counterflow):

$$\varepsilon = \frac{1-e^{-0.625(1-0.670)}}{1-0.670\,e^{-0.625(1-0.670)}} = \frac{1-0.8134}{1-0.5449} = \frac{0.1866}{0.4551}= 0.410$$

Heat transfer rate:

$$Q = \varepsilon C_{min}(T_{h,in}-T_{c,in}) = 0.4097\times1680\times(140-25) = 79\,160\ \text{W}\approx79.2\ \text{kW}$$

Outlet temperatures:

$$T_{h,out} = 140 - \frac{79\,160}{1680} = 92.9\ ^\circ\text{C},\qquad T_{c,out} = 25 + \frac{79\,160}{2508} = 56.6\ ^\circ\text{C}$$

(For comparison, a parallel-flow exchanger of the same size would have $\varepsilon = \frac{1-e^{-NTU(1+C_r)}}{1+C_r} = 0.388$.)

Answer: $\varepsilon \approx 0.41$; $Q \approx 79.2$ kW; $T_{h,out} \approx 92.9$ °C; $T_{c,out} \approx 56.6$ °C.