Chapter 7, Problem 7.1P

Note: In the following problems, you will deal with both the International System of Units (SI) (N, kg, m, s, K) and the English Engineering System (lb, slug, ft, s, $°\text{R}$). Which system to use will be self-evident in each problem. All problems deal with calorically perfect air as the gas, unless otherwise noted. Also, recall that $1\text{atm}=2116{\text{lb/ft}}^{\text{2}}=1.01\times {10}^5{\text{N/m}}^{\text{2}}$.

The temperature and pressure at the stagnation point of a high-speed missile are $\text{934 }°\text{R}$ and 7.8 atm, respectively. Calculate the density at this point.

To determine

The density of missile at the stagnation point.

Answer to Problem 7.1P

The density of missile at stagnation point is ${\rho }_o=0.0103\text{slug}/{\text{ft}}^3$

Explanation of Solution

Given Information:

  $\begin{array}{l}\text{Pressure of missile at stagnation point is, }{P}_o=7.8\text{atm}\\ \text{Temperature of missile at stagnation point is, }{T}_o=934°R\\ \text{The gas constant is Btu is, }R=1716\text{ft}\text{.lb}/\left(\text{slug}\text{.}°\text{R}\right)\end{array}$

Calculation:

From ideal gas equation, the density at the given point can be calculated as,

  $\begin{array}{l}{P}_o={\rho }_{o}R{T}_o\\ {\rho }_o=\frac{P_o}{R{T}_o}\\ \text{Plugging }R=1716\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right),T_o=934°\text{R}\text{\&}{P}_o=7.8\text{atm}\\ {\rho }_o=\frac{7.8\left(\text{atm}\right)}{\left[1716\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\right]\left[934°\text{R}\right]}\left(\frac{2116\text{lb}/{\text{ft}}^2}{1\text{atm}}\right)\\ {\rho }_o=0.0103\text{slug}/{\text{ft}}^3\end{array}$

Hence, the density of missile at the stagnation point is ${\rho }_o=0.0103\text{slug}/{\text{ft}}^3$