Chapter 11, Problem 11.1P

Consider a subsonic compressible flow in cartesian coordinates where the velocity potential is given by $\varphi \left(x,y\right)=V_{\infty }x+\frac{70\sin \left(2\pi x\right)}{\sqrt{1-M_{{}^{\infty }}^2}}{e}^{-2\pi y\sqrt{1-M^{{}_{\infty }^2}}}$

If the freestream properties are given by $V_{\infty }=700\text{ft/s},p_{\infty }=1\text{atm}$, and $T_{\infty }=\text{519}°\text{R}$, calculate the following properties at the location $\left(x,y\right)=\left(0.2\text{ft},0.2\text{ft}\right):M,p$, and T.

To determine

The Mach number of the subsonic compressible flow.

The temperature of the subsonic compressible flow.

The pressure of the subsonic compressible flow.

Answer to Problem 11.1P

The Mach number for the fluid at the given point is $M=0.7067$ .

The pressure for the fluid at the given point is $p=0.933\text{atm}$ .

The temperature for the fluid at the given point is $T=508.93°\text{R}$ .

Explanation of Solution

Given:

The freestream velocity of the compressible flow is $V_{\infty }=700\text{ft}/\text{s}$ .

The pressure of the compressible flow is $p_{\infty }=1\text{atm}$ .

The temperature of the compressible flow is $T_{\infty }=519°\text{R}$ .

Formula used:

The velocity component in the x-direction is given as,

  $u=\frac{\partial \varphi }{\partial x}$

The velocity component in the y-direction is given as,

  $v=\frac{\partial \varphi }{\partial y}$

The expression for the Mach number is given as,

  $M=\frac{V}{\sqrt{\gamma RT}}$

Here, $\gamma$ is the ratio of specific heat, $R$ is the specific gas constant and $V$ is velocity of the flow.

The velocity of the flow is given as,

  $V=\sqrt{u^2+v^2}$

The expression for the temperature for the given point is given as,

  $C_P{T}_0=C_{P}T+\frac{V^2}{2}$

The expression for the pressure is given as,

  $\frac{p_0}{p}={\left(1+\frac{\gamma -1}{\gamma }{M}^2\right)}^{\frac{\gamma }{\gamma -1}}$

Calculation:

The “Properties of air” is given as,

  $\begin{array}{l}\gamma =1.4\\ R=1716\text{ft}\cdot \text{lb}/\text{slug°R}\\ C_P=6006\text{ft}\cdot \text{lb}/\text{slug°R}\end{array}$

The velocity component of the fluid in x-direction can be calculated as,

  $\begin{array}{c}u=\frac{\partial \varphi }{\partial x}\\ u_{\left(0.2\text{ft,}0.2\text{ft}\right)}=\frac{\partial (V_{\infty }x+\frac{70\sin \left(2\pi x\right)}{\sqrt{1-M_{\infty }^2}}{e}^{-2\pi y\sqrt{1-m_{\infty }^2}}}{\partial x}\\ u=765.55\text{ft}/\text{s}\end{array}$

The velocity component of the fluid in the y-direction can be calculated as,

  $\begin{array}{c}v=\frac{\partial \varphi }{\partial y}\\ v_{\left(0.2\text{ft,}0.2\text{ft}\right)}=\frac{\partial (V_{\infty }x+\frac{70\sin \left(2\pi x\right)}{\sqrt{1-M_{\infty }^2}}{e}^{-2\pi y\sqrt{1-m_{\infty }^2}}}{\partial x}\\ v=-157.23\text{ft}/\text{s}\end{array}$

The resultant velocity of the flow can be calculated as,

  $\begin{array}{l}V=\sqrt{u^2+v^2}\\ V=\sqrt{{\left(765.55\text{ft}/\text{s}\right)}^2+{\left(-157.23\text{ft}/\text{s}\right)}^2}\\ V=781.53\text{ft}/\text{s}\end{array}$

The Mach number of the supersonic flow can be calculated as,

  $\begin{array}{l}{M}_{\infty }=\frac{V_{\infty }}{\sqrt{\gamma R{T}_{\infty }}}\\ M_{\infty }=\frac{700\text{ft}/\text{s}}{\sqrt{1.4\times 1716\text{ft}\text{lb}/\text{slug°R}\times 519°R}}\\ M_{\infty }=0.6268\end{array}$

For $M_{\infty }=0.6268$ , the temperature at start can be calculated as,

  $\begin{array}{c}\frac{T_0}{T\infty }=1+\frac{\gamma -1}{2}{M}_{\infty }^2\\ \frac{T_0}{519°R}=1+\frac{1.4-1}{2}{\left(0.6268\right)}^2\\ T_0=559.78°R\end{array}$

The pressure at starts can be calculated as,

  $\begin{array}{c}\frac{p_0}{p_{\infty }}={\left(1+\frac{\gamma -1}{2}{M}_{\infty }^2\right)}^{\frac{\gamma }{\gamma -1}}\\ \frac{p_0}{1\text{atm}}={\left(1+\frac{1.4-1}{2}\times {0.6268}^2\right)}^{\frac{1.4}{1.4-1}}\\ p_0=1.303\text{atm}\end{array}$

The required temperature can be calculated by the energy equation as,

  $\begin{array}{c}{C}_P{T}_0=C_{P}T+\frac{V^2}{2}\\ T=T_0-\frac{V^2}{2C_P}\\ T=559.78°R-\frac{{\left(781.53\text{ft}/\text{s}\right)}^2}{2\times 6006\text{ft}\text{lb}/\text{slug°R}}\\ T=508.93°R\end{array}$

The required Mach number can be calculated as,

  $\begin{array}{l}M=\frac{V}{\sqrt{\gamma RT}}\\ M=\frac{781.53\text{ft}/\text{s}}{\sqrt{1.4\times 1716\text{ft}\text{lb}/\text{slug°R}\times 508.93°R}}\\ M=0.7067\end{array}$

The required pressure for the flow can be calculated as,

  $\begin{array}{c}\frac{p_0}{p}={\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{\frac{\gamma }{\gamma -1}}\\ \frac{1.303\text{atm}}{p}={\left(1+\frac{1.4-1}{2}\times {\left(0.7067\right)}^2\right)}^{\frac{1.4}{1.4-1}}\\ p=0.933\text{atm}\end{array}$

Conclusion:

Therefore, the Mach number for the fluid at the given point is $M=0.7067$ .

Therefore, the pressure for the fluid at the given point is $p=0.933\text{atm}$ .

Therefore, the temperature for the fluid at the given point is $T=508.93°R$ .