Chapter 7, Problem 7.2P

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 $1\text{atm}=2116{\text{lb/ft}}^{\text{2}}=1.01\times {10}^5{\text{N/m}}^{\text{2}}$.

Calculate $c_p,c_v,\text{}\text{}e,$ and h for

a. The stagnation point conditions given in Problem 7.1

b. Air at standard sea level conditions

(If you do not remember what standard sea level conditions are, find them in an appropriate reference, such as Reference 2.)

To determine

Specific heat at constant pressure and constant volume, specific internal energy &enthalpy at the given stagnation point.

Answer to Problem 7.2P

Specific heat capacity at constant pressure at stagnation point is, $c_p=6006\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)$

Specific heat capacity at constant volume at stagnation point is, $c_v=4290\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)$

Specific internal energy at the stagnation point is $e=4.01\times {10}^6\text{ft}\text{.lb}/\text{slug}$

Specific enthalpy energy at the stagnation point is $h=5.61\times {10}^6\text{ft}\text{.lb}/\text{slug}$

Explanation of Solution

Given:

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

Calculation:

Specific heat constant pressure can be given as,

  $\begin{array}{l}{c}_p=\frac{\gamma R}{\gamma -1}\\ \text{Plugging }\gamma \text{=1}\text{.4, }R=1716\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\\ c_p=\frac{1.4\times 1716}{1.4-1}\\ c_p=6006\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\end{array}$

Hence, the specific heat at constant pressure is $c_p=6006\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)$

Specific heat constant volume can be given as,

  $\begin{array}{l}{c}_v=\frac{R}{\gamma -1}\\ \text{Plugging }\gamma \text{=1}\text{.4, }R=1716\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\\ c_v=\frac{1716}{1.4-1}\\ c_v=4290\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\end{array}$

Hence, the specific heat at constant volume is $c_v=4290\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)$

Specific internal energy can be given as,

  $\begin{array}{l}e=c_v{T}_o\\ \text{Plugging }{c}_v=4290\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\&T_o=934°\text{R}\\ e=4290\times 934\\ e=4.01\times {10}^6\text{ft}\text{.lb}/\text{slug}\end{array}$

Hence, the specific internal energy is $e=4.01\times {10}^6\text{ft}\text{.lb}/\text{slug}$

Specific enthalpy can be given as,

  $\begin{array}{l}h=c_p{T}_o\\ \text{Plugging }{c}_v=6006\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\&T_o=934°\text{R}\\ h=6006\times 934\\ h=5.61\times {10}^6\text{ft}\text{.lb}/\text{slug}\end{array}$

Hence, the specific enthalpy energy is $h=5.61\times {10}^6\text{ft}\text{.lb}/\text{slug}$

To determine

Specific heat at constant pressure and constant volume, specific enthalpy and specific internal energy at the sea level.

Answer to Problem 7.2P

Specific heat capacity at constant pressure at sea level is, $c_p=6006\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)$

Specific heat capacity at constant volume at sea level is, $c_v=4290\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)$

Specific internal energy at sea level is $e=2.23\times {10}^6\text{ft}\text{.lb}/\text{slug}$

Specific enthalpy energy at sea level is $h=3.12\times {10}^6\text{ft}\text{.lb}/\text{slug}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Pressure of missile at stagnation point is, }{P}_o=7.8\text{atm}\\ \text{Temperature of missile at sea level is, }{T}_o=519°R\\ \text{The gas constant is Btu is, }R=1716\text{ft}\text{.lb}/\left(\text{slug}\text{.}°\text{R}\right)\end{array}$

Calculation:

Specific heat constant pressure can be given as,

  $\begin{array}{l}{c}_p=\frac{\gamma R}{\gamma -1}\\ \text{Plugging }\gamma \text{=1}\text{.4, }R=1716\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\\ c_p=\frac{1.4\times 1716}{1.4-1}\\ c_p=6006\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\end{array}$

Hence, the specific heat at constant pressure is $c_p=6006\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)$

Specific heat constant volume can be given as,

  $\begin{array}{l}{c}_v=\frac{R}{\gamma -1}\\ \text{Plugging }\gamma \text{=1}\text{.4, }R=1716\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\\ c_v=\frac{1716}{1.4-1}\\ c_v=4290\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\end{array}$

Hence, the specific heat at constant volume is $c_v=4290\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)$

Specific internal energy can be given as,

  $\begin{array}{l}e=c_v{T}_o\\ \text{Plugging }{c}_v=4290\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\&T_o=519°\text{R}\\ e=4290\times 519\\ e=2.23\times {10}^6\text{ft}\text{.lb}/\text{slug}\end{array}$

Hence, the specific internal energy is $e=2.23\times {10}^6\text{ft}\text{.lb}/\text{slug}$

Specific enthalpy can be given as,

  $\begin{array}{l}h=c_p{T}_o\\ \text{Plugging }{c}_v=6006\text{ft}\text{.lb}/\left(\text{slug}\cdot °\text{R}\right)\&T_o=519°\text{R}\\ h=6006\times 519\\ h=3.12\times {10}^6\text{ft}\text{.lb}/\text{slug}\end{array}$

Hence, the specific enthalpy energy is $h=3.12\times {10}^6\text{ft}\text{.lb}/\text{slug}$