Question 11.26: A counterflow heat exchanger is used to cool a flow of 400 l...
A counterflow heat exchanger is used to cool a flow of 400 lb./min of lubricating oil. Hot oil enters at 215°F and leaves at 125°F. The specific heat of the oil is 0.85 Btu/lb.·°F, and the overall coefficient of heat transfer of the unit is 40 Btu/(h·ft .^{2}·°F) (of the outside tube surface). Water enters the unit at 60°F and leaves at 90°F. Determine the outside tube surface required.
From Figure 11.30, \theta_{A} = 215 − 90 = 125°F, and \theta_{B} = 125 − 60 = 65°F. Therefore,
(\Delta t)_{m}=\frac{\theta_{A}-\theta_{B}}{\ln \left(\theta_{A} / \theta_{B}\right)}=\frac{125-65}{\ln (125 / 65)}=92^{\circ} FFrom the oil data, the heat transfer \dot{Q}=\dot{m} c_{p} \Delta t= 400 × 60 × 0.85(215 – 125) = 1836000
Btu/h, and from the heat transfer equations, \dot{Q}=U A(\Delta t)_{m} =1836000 = × 40 A × 92. Therefore, A = 499 ft .^{2} of the outside surface is required.
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