Question

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

Short answer

Question: Calculate the Prandtl number based on the given values of specific heat at constant pressure (\(c_{p}\)), thermal conductivity (k), and dynamic viscosity (\(\mu\)) as 0.5 Btu/lbm·R, 2 Btu/h·ft·R, and 0.3 lbm/ft·s, respectively. Answer: The Prandtl number is 270.

Step-by-step solution

Convert the given values to consistent units

The given values of specific heat, thermal conductivity, and dynamic viscosity need to be in consistent units. Since we have specific heat in Btu/lbm·R and thermal conductivity in Btu/h·ft·R, we can convert dynamic viscosity to lbm/h·ft: 1 hour = 3600 seconds, so multiply the dynamic viscosity by 3600 to convert it from lbm/ft·s to lbm/h·ft. $$\mu = 0.3 \frac{\text{lbm}}{\text{ft} \cdot \text{s}} \times 3600 \, \text{s/h} = 1080 \frac{\text{lbm}}{\text{h} \cdot \text{ft}}$$ Now, we have specific heat in Btu/lbm·R, thermal conductivity in Btu/h·ft·R, and dynamic viscosity in lbm/h·ft, which are consistent units.

Use the formula to calculate the Prandtl number

Now that we have the correct units, we can use the formula for the Prandtl number: $$Pr = \frac{c_{p}\mu}{k}$$ We plug in the given values of specific heat (\(c_{p}\) = 0.5 Btu/lbm·R), thermal conductivity (\(k\) = 2 Btu/h·ft·R), and the dynamic viscosity (\(\mu\) = 1080 lbm/h·ft) to get the Prandtl number: $$Pr = \frac{0.5 \, \text{Btu/lbm} \cdot \text{R} \times 1080 \, \text{lbm/h} \cdot \text{ft}}{2 \, \text{Btu/h} \cdot \text{ft} \cdot \text{R}}$$

Simplify the expression and evaluate the Prandtl number

Now, we can simplify the expression and cancel out the units: $$Pr = \frac{540 \, \text{Btu} \cdot \text{lbm}}{2 \, \text{Btu} \cdot \text{R}} = 270$$ Hence, the Prandtl number is 270.