Oil cooler area by LMTD method

(a) Heat duty

$$Q = \dot m_hc_h(T_{h1}-T_{h2}) = 2\times2100\times(100-60) = 168\,000\ \text{W} = 168\ \text{kW}$$

(b) Water outlet

$$T_{c2} = T_{c1}+\frac{Q}{\dot m_cc_c} = 20 + \frac{168\,000}{1.2\times4180} = 20 + 33.49 = 53.5\ ^\circ\text{C}$$

(c) Counterflow LMTD and area

 Oil   100 ----------> 60
 Water  53.5 <---------- 20

$\Delta T_1 = T_{h1}-T_{c2} = 100 - 53.5 = 46.5$ K, $\Delta T_2 = T_{h2}-T_{c1} = 60 - 20 = 40$ K.

$$\Delta T_{lm} = \frac{46.5-40}{\ln(46.5/40)} = \frac{6.51}{0.1505} = 43.2\ \text{K}$$ $$A = \frac{Q}{U\,\Delta T_{lm}} = \frac{168\,000}{400\times43.17} = 9.73\ \text{m}^2$$

(d) Parallel flow

Both fluids enter at the same end: $\Delta T_1 = 100 - 20 = 80$ K, $\Delta T_2 = 60 - 53.5 = 6.5$ K.

$$\Delta T_{lm,p} = \frac{80-6.5}{\ln(80/6.5)} = \frac{73.5}{2.510} = 29.3\ \text{K}$$ $$A_p = \frac{168\,000}{400\times29.29} = 14.3\ \text{m}^2$$

Parallel flow needs about 47 percent more area for the same duty, showing why counterflow is preferred.

Answer: $Q = 168$ kW; $T_{c2} = 53.5$ °C; $\Delta T_{lm} = 43.2$ K and $A = 9.73$ m$^2$ (counterflow); $A = 14.3$ m$^2$ (parallel flow).