Question

In an oil cooler, oil flows steadily through a bundle of metal tubes submerged in steady stream of cooling water. Under steady conditions oil enters at ${90}^o$C and leaves at ${30}^o$C. Water enters at ${25}^o$C and leaves at ${70}^o$C. Enthalpy of oil, $h=1.68t+10.5\times {10}^{-4}{t}^2$ kJ/kg (temperature is taken in ${}^o$C). The cooling water required for cooling 2.78 kg/s of oil is _________ $kg/s$.

• A. 1
• B. 2.6
• C. 1.6
• D. 2

Answer 1

The correct option is C 1.6

Now, $\Delta H_{oil}=\Delta H_{water}$

$\Rightarrow$ ${\dot{m}}_{oil}\left[1.68\right(T_1-T_2)+10.5\times {10}^{-4}\times (T_1^2-T_2^2\left)\right]={\dot{m}}_{water}\times C_{P_w}[T_1-T_2{]}_{water}$

$\Rightarrow$ $2.78\left[1.68\right(90-30)+10.5\times {10}^{-4}\times ({90}^2-{30}^2\left)\right]={\dot{m}}_{water}\times 4.18\times (70-25)$

$\Rightarrow$ ${\dot{m}}_{water}=1.6kg/s$