Question 11.28: Using Table 11.6 for the appropriate fouling factors, calcul...
Using Table 11.6 for the appropriate fouling factors, calculate the surface required for Illustrative Problem 11.27.
| Fouling Resistance | ||
| Types of Fluid | \left( h \cdot{ }^{\circ} F \cdot ft .^{2} / Btu \right)^{ a } | (m^{2}·°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 |
For the oil side, a resistance (fouling factor) of 0.005 h·°F·ft .^{2}/Btu can be used, and for the water side, a fouling factor of 0.001 h·°F·ft .^{2}/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·ft .^{2}·°F). The overall resistance and the overall heat transfer coefficient are obtained as
\begin{array}{rlr}\text { oil, } R & = & 0.005 \\\text { water, } R & = & 0.001 \\\text { clean unit, } R & =1 / 40 & 0.025 \\\hline R_{\text {overall }} & = & 0.031 \\U_{\text {overall }} & =\frac{1}{0.031}=32.3 \frac{ Btu }{ h \cdot ft .{ }^{2} \cdot{ }^{\circ} F }\end{array}Because all other parameters are the same, the surface required will vary inversely as U. Therefore, A = 569 × 40/32.3 = 705 ft .^{2} or an increase in surface required of approximately 24% due to fouling. This obviously represents an important consideration in the design of industrial equipment.