Question 9.26: Using the data of Example 9.1, calculate the surface area re......

As in Example 9.1, the heat load = 1672 kW

With reference to Figure 9.71: T_{1} = 360 K, T_{2} = 340 K
and hence: X={\frac{\theta_{2}-\theta_{1}}{T_{1}-\theta_{1}}}={\frac{316-300}{360-300}}=0.267

and: Y={\frac{T_{1}-T_{2}}{\theta_{2}-\theta_{1}}}={\frac{360-340}{316-300}}=1.25

from Figure 9.58:  F = 0.97

and hence: F\theta _{m} = (41.9 x 0.97) = 40.6 K
The heat transfer area is then: A={\frac{1672}{2\times40.6}} = \underline{\underline{20.6\ m^{2}}}