Evaluate the Prandtl number from the following data: \(c_{p}=0.5 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}, k=2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R}, \mu=0.3 \mathrm{lbm} / \mathrm{ft} \cdot \mathrm{s}\).

First, we'll convert the given units to SI units. For \(c_p\), we'll convert Btu/lbm·R to J/kg·K (1 Btu = 1055.06 J, 1 lbm = 0.453592 kg, and (9/5) R = 1 K): $$c_{p}=0.5\frac{\mathrm{Btu}}{\mathrm{lbm}\cdot\mathrm{R}} \times \frac{1055.06\mathrm{J}}{\mathrm{Btu}} \times \frac{0.453592\mathrm{kg}}{\mathrm{lbm}} \times \frac{9}{5}$$ For \(k\), we'll convert Btu/h·ft·R to W/m·K (1 Btu/h·ft·R = 1.73073 W/m·K and (9/5) R = 1 K): $$k=2\frac{\mathrm{Btu}}{\mathrm{h}\cdot\mathrm{ft}\cdot\mathrm{R}} \times \frac{1.73073\mathrm{W}}{\mathrm{m}\cdot\mathrm{K}}$$ For \(\mu\), we'll convert lbm/ft·s to kg/m·s (1 lbm = 0.453592 kg, and 1 ft = 0.3048 m): $$\mu=0.3\frac{\mathrm{lbm}}{\mathrm{ft}\cdot\mathrm{s}} \times \frac{0.453592\mathrm{kg}}{\mathrm{lbm}} \times \frac{1}{0.3048\mathrm{m}}$$

Now we'll calculate the values of \(c_p\), \(k\), and \(\mu\) in SI units: $$c_p = 0.5 \times 1055.06 \times 0.453592 \times \frac{9}{5} = 2140.52\, \mathrm{J/kg\cdot K}$$ $$k = 2 \times 1.73073 = 3.46146\, \mathrm{W/m\cdot K}$$ $$\mu = 0.3 \times 0.453592 \times \frac{1}{0.3048} = 0.4476\, \mathrm{kg/m\cdot s}$$

Using the formula, we can now calculate the Prandtl number: $$Pr = \frac{c_p \cdot \mu}{k} = \frac{2140.52 \cdot 0.4476}{3.46146}$$ Calculate the value of the Prandtl number: $$Pr \approx 276.47$$ The Prandtl number for the given data is approximately 276.47.