Question

An engineer who is working on the heat transfer analysis of a house in English units needs the convection heat transfer coefficient on the outer surface of the house. But the only value he can find from his handbooks is $14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, which is in SI units. The engineer does not have a direct conversion factor between the two unit systems for the convection heat transfer coefficient. Using the conversion factors between \(\mathrm{W}\) and \(\mathrm{Btu} / \mathrm{h}, \mathrm{m}\) and \(\mathrm{ft}\), and \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\), express the given convection heat transfer coefficient in Btu/h \(\mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). Answer: \(2.47\) Btu/h $\cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$

Short answer

Question: Convert the given convection heat transfer coefficient \(14 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\) to \(\mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^2 \cdot { }^{\circ} \mathrm{F}\). Answer: The equivalent convection heat transfer coefficient is approximately \(2.47 \mathrm{Btu/h} \cdot \mathrm{ft}^2{ }^{\circ} \mathrm{F}\).

Step-by-step solution

Write down the given convection heat transfer coefficient and conversion factors

The given convection heat transfer coefficient is \(14 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\). We are given the following conversion factors: 1. Watts (W) to Btu/h: \(1\mathrm{~W} = 3.412\mathrm{~Btu/h}\) 2. Meters (m) to feet (ft): \(1\mathrm{~m} = 3.281\mathrm{~ft}\) 3. Kelvin (K) to Fahrenheit (F): \(1\mathrm{~K} = 1.8 { }^{\circ} \mathrm{F}\)

Multiply the convection heat transfer coefficient value with conversion factors

Using the given conversion factors, we can convert each unit in the convection heat transfer coefficient individually. \(\frac{14\mathrm{~W}}{\mathrm{m}^2 \cdot \mathrm{K}} \times \frac{3.412\mathrm{~Btu/h}}{1\mathrm{~W}} \times \frac{1\mathrm{~m}}{3.281\mathrm{~ft}} \times \frac{1\mathrm{~m}}{3.281\mathrm{~ft}} \times \frac{1.8 { }^{\circ} \mathrm{F}}{1\mathrm{~K}}\)

Simplify the expression

Now, we will cancel out the units and simplify the expression: \(\frac{14\times 3.412\mathrm{~Btu/h}}{3.281^2 \mathrm{ft}^2 \cdot 1.8 { }^{\circ} \mathrm{F}}\)

Calculate the equivalent convection heat transfer coefficient

Performing the calculation, we get: \(\frac{47.768\mathrm{~Btu/h}}{35.7254\mathrm{~ft}^2 \times { }^{\circ} 1.8\mathrm{F}} = \frac{47.768}{64.3057} \mathrm{Btu/h} \cdot \mathrm{ft}^2{ }^{\circ} \mathrm{F} \approx 2.47 \mathrm{Btu/h} \cdot \mathrm{ft}^2{ }^{\circ} \mathrm{F}\) So, the convection heat transfer coefficient in Btu/h \(\cdot \mathrm{ft}^2{ }^{\circ} \mathrm{F}\) is approximately \(2.47\) Btu/h \(\cdot \mathrm{ft}^2{ }^{\circ} \mathrm{F}\).