Question 9.27: Estimate the heat transfer area required for the system cons......

As in the previous examples,

heat load = 1672 kW
and:
corrected mean temperature difference, F\theta _{m} = 40.6 deg K
In the tubes;
mean water temperature, T = 0.5(360 + 340) = 350 K
Assuming a tube diameter, d = 19 mm or 0.0019 m and a water velocity, u = 1 m/s, then, in equation 9,221:

h=4280(0.00488T-1)u^{0.8}/d^{0.2}              (9.221)

h_{i}=4280((0.00488\times350)-1)1.0^{0.8}/0.0019^{0.2} = 10610 W/m²K or 10.6 kW/m²K

From Table 9.18, an estimate of the shell-side film coefficient is:
h_{o} = 0.5(1700 + 11000) = 6353 W/m²K or 6.35 kW/m²K
For steel tubes of a wall thickness of 1.6 mm, the thermal resistance of the wall, from Table 9.15 is:
x_{w}/k_{w} = 0.025 m²K/kW
and the thermal resistance for treated water, from Table 9.16, is 0.26 m²K/kW for both layers of scale. Thus, in Equation 9.201:

\frac{1}{U A}=\frac{1}{h_{i}A_{i}}+\frac{x_{w}}{k_{w}A_{w}}+\frac{1}{h_{o}A_{o}}+\frac{R_{o}}{A_{o}}+\frac{R_{i}}{A_{i}}                  (9.201)

(1/U) = (1/h_{o}) + (x_{w}/k_{w}) + R_{i} + R_{o} + (1/h_{i})
= (1/6.35) + 0.025 + 0.52 + (1/10.6) = 0.797 m²K/kW
and:
U = 1.25 kW/m²K
The heat transfer area required is then:
A = Q/F\theta _{m}U = 1672/(40.6 x 1.25) = \underline{\underline{32.9\ m^{2}}}