Topic: Heat Transfer
What is the Prandtl number for a gas at $27^{\circ}C$ and $1\times10^{5}Pa.$ with kinematic viscosity equal to $1.8\times10^{-5}m^{2}/s.$, specific heat capacity at constant pressure $=700~J/kg-K$ and thermal conductivity of 0.02 W/m-K? (Take Gas constant 300 J/kg-K)
- A. 0.52
- B. 0.62
- C. 0.70
- D. 0.78
Answer: C. 0.70
Explanation: Given $T=27^{\circ}C=300~K$, $P=1\times10^{5}~Pa$, $R=300~J/kg-K$. Density $\rho = \frac{P}{RT} = \frac{100000}{300 \times 300} = \frac{10}{9}~kg/m^{3}$. Prandtl number $Pr = \frac{\mu C_{p}}{k} = \frac{\rho v C_{p}}{k}$. Substituting the values: $Pr = \frac{(10/9) \times 1.8\times10^{-5} \times 700}{0.02} = \frac{2\times10^{-5} \times 700}{0.02} = \frac{0.014}{0.02} = 0.70$.