Using Table 11.6 for the appropriate fouling factors, calculate the surface required for Illustrative Problem 11.27.

Question

	Fouling Resistance
Types of Fluid	[latex]\left( h \cdot{ }^{\circ} F \cdot ft .^{2} / Btu \right)^{ a }[/latex]	([latex]m^{2}[/latex]·°C/W)
Seawater below 50°C (122°F)	0.0005	0.0001
Seawater above 50°C (122°F)	0.001	0.0002
Treated boiler feedwater	0.001	0.0002
River water	0.001–0.006	0.0002–0.001
Fuel oil	0.005	0.0009
Vapor refrigerants	0.002	0.0004
Alcohol vapors	0.0005	0.0001
Steam, non-oil-bearing	0.0005	0.0001
Industrial air	0.002	0.0004
Refrigerating liquid	0.001	0.0002

Accepted Answer

For the oil side, a resistance (fouling factor) of 0.005 h·°F·[latex]ft .^{2}[/latex]/Btu can be used, and for the water side, a fouling factor of 0.001 h·°F·[latex]ft .^{2}[/latex]/Btu can be used. Because of the approximate nature of these resistances, we shall not correct them for inside or outside reference areas, and we will assume that they can be used directly with the value of U of 40 Btu/(h·[latex]ft .^{2}[/latex]·°F). The overall resistance and the overall heat transfer coefficient are obtained as

[latex]\begin{array}{rlr} ext { oil, } R & = & 0.005 \ ext { water, } R & = & 0.001 \ ext { clean unit, } R & =1 / 40 & 0.025 \\hline R_{ ext {overall }} & = & 0.031 \U_{ ext {overall }} & =\frac{1}{0.031}=32.3 \frac{ Btu }{ h \cdot ft .{ }^{2} \cdot{ }^{\circ} F }\end{array}[/latex]

Because all other parameters are the same, the surface required will vary inversely as U. Therefore, A = 569 × 40/32.3 = 705 [latex]ft .^{2}[/latex] or an increase in surface required of approximately 24% due to fouling. This obviously represents an important consideration in the design of industrial equipment.

Question 11.28: Using Table 11.6 for the appropriate fouling factors, calcul...

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