Question

A counterflow heat exchanger is stated to have an overall heat transfer coefficient based on outside tube area of $50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}$ when operating at design and clean conditions. After a period of use, scale buildup in the heat exchanger gives a fouling factor of $0.002 \mathrm{~h} \cdot \mathrm{ft}^{2},{ }^{\circ} \mathrm{F} /\( Btu. Determine \)(a)$ the overall heat transfer coefficient of the heat exchanger and \((b)\) the percentage change in the overall heat transfer coefficient due to the scale buildup.

Short answer

Answer: The overall heat transfer coefficient with the fouling factor is 48.54 Btu/h·ft²·°F, and the percentage change in the overall heat transfer coefficient due to the scale buildup is -2.92%.

Step-by-step solution

Given values.

The given values are \(U_{clean} = 50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^2 \cdot{ }^{\circ} \mathrm{F}\) and fouling factor \(R_f = 0.002 \mathrm{~h} \cdot \mathrm{ft}^2,{ }^{\circ} \mathrm{F} /\) Btu.

Calculate the overall heat transfer coefficient with the fouling factor.

First, we need to calculate the overall heat transfer coefficient with the fouling factor using the formula: $$\frac{1}{U_{fouled}} = \frac{1}{U_{clean}} + R_f$$ Plugging in the values, we get: $$\frac{1}{U_{fouled}} = \frac{1}{50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^2 \cdot{ }^{\circ} \mathrm{F}} + 0.002 \mathrm{~h} \cdot \mathrm{ft}^2,{ }^{\circ} \mathrm{F} / \mathrm{Btu}$$ Solving for \(U_{fouled}\), we get: $$U_{fouled} = \frac{1}{\frac{1}{50} + 0.002} = 48.54 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^2 \cdot{ }^{\circ} \mathrm{F}$$

Calculate the percentage change in the overall heat transfer coefficient.

Now, we need to calculate the percentage change in the overall heat transfer coefficient using the formula: $$\% \text{ Change} = \frac{U_{fouled} - U_{clean}}{U_{clean}} \times 100$$ Plugging in the values, we get: $$\% \text{ Change} = \frac{48.54 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^2 \cdot{ }^{\circ} \mathrm{F} - 50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^2 \cdot{ }^{\circ} \mathrm{F}}{50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^2 \cdot{ }^{\circ} \mathrm{F}} \times 100$$ $$\% \text{ Change} = -2.92 \%$$ The overall heat transfer coefficient with the fouling factor is \(48.54 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^2 \cdot{ }^{\circ} \mathrm{F}\), and the percentage change in the overall heat transfer coefficient due to the scale buildup is \(-2.92 \%\).