Chapter 1, Problem 1.1P

For most gases at standard or near standard conditions, the relationship among pressure, density, and temperature is given by the perfect gas equation of state: $p=\rho RT$, where R is the specific gas constant. For air at near standard conditions, $R=287/\left(kg\cdot K\right)$ in the International System of Units and $R=1716\text{ft}\cdot \text{lb}/\left(\text{slug}\cdot °\text{R}\right)$ in the English Engineering System of Units. (More details on the perfect gas equation of state are given in Chapter 7.) Using the above information, consider the following two cases:

a. At a given point on the wing of a Boeing 727, the pressure and temperature of the air are $1.9\times {10}^4{\text{N/m}}^{\text{2}}$ and 203 K, respectively. Calculate the density at this point.

b. At a point in the test section of a supersonic wind tunnel, the pressure and density of the air are $1058{\text{lb/ft}}^{\text{2}}$ and $1.23\times {10}^{-3}{\text{slug/ft}}^{\text{3}}$, respectively. Calculate the temperature at this point.

To determine

(a)

The density at a given point on the wing of a Boeing 727.

Answer to Problem 1.1P

$\rho \text{ = 0}\text{.326}{\text{kg/m}}^3$

Explanation of Solution

Given:

Pressure $p=1.9\times {10}^4{\text{Ν/m}}^{\text{2}}$

Temperature $T=203{\rm K}$

Gas constant $R=287\text{J/}\left(\text{Kg}\text{.}\text{K}\right)$

Calculation:

The perfect gas equation is $p=\rho RT$

Here, P is the pressure, ρ is the density, R is the specific gas constant and T is the temperature of air.

Density, $\rho =\frac{p}{RT}$

$\begin{array}{l}\rho =\frac{1.9\times {10}^4}{287\times 203}\\ \rho =0.326{\text{Kg/m}}^{\text{3}}\end{array}$

Conclusion:

Thus, the density of air at given condition is $\rho \text{ = 0}\text{.326}{\text{kg/m}}^3.$

To determine

(b)

The temperature at a given point on the wing of a Boeing 727.

Answer to Problem 1.1P

${\text{T= 501}}^{°}\text{R}$

Explanation of Solution

Given:

Pressure $p=1058{\text{lb/ft}}^{\text{2}}$

Density $\rho =1.23\times {10}^{-3}{\text{slug/ft}}^{\text{3}}$

Gas constant $R=1716\text{ft}\text{.}\text{lb/}\left(\text{slug}\text{.}{}^{\text{°}}\text{R}\right)$

Calculation:

The perfect gas equation is $p=\rho RT$

Here, P is the pressure, ρ is the density, R is the specific gas constant and T is the temperature of air.

Temperature, $T=\frac{p}{\rho R}$

$\begin{array}{l}T=\frac{1058}{\left(1.23\times {10}^{-3}\right)\times 1716}\\ T={501}^{°}R\end{array}$

Conclusion:

Thus, the temperature of air at given condition is ${\text{T= 501}}^{°}\text{R}.$