Problem 3.1

Given:
\(dV/V=\) 1% = 0.01, isentropic flow.

To plot: \( dp/p, dT/T, d\rho/\rho \) for M=[0.2, 2].

Using the conservation of momentum equation (Euler equation),

\[ \frac{dp}{\rho}+V\,dV=0 \]

dividing by \( V^{2} \),

\[ \frac{p}{\rho V^{2}}\,\frac{dp}{p}+\frac{dV}{V}=0\ . \]

Using the equation for speed of sound, \( a^{2}=\gamma\,p/\rho\implies p/\rho=a^{2}/\gamma \), the above equation may be written as,

\[ \frac{a^{2}}{\gamma\,V^{2}}\,\frac{dp}{p}+\frac{dV}{V}=0\ , \] \[ \frac{1}{\gamma\,M^{2}}\,\frac{dp}{p}+\frac{dV}{V}=0\ , \] \[ \frac{dp}{p}=-\gamma\,M^{2}\frac{dV}{V}\ . \]

Substituting the value of \( \gamma= \) 1.4, for air, and \( dV/V= \) 0.01, the above equation becomes,

\begin{equation} \boxed{\frac{dp}{p}=-0.014\,M^{2}}\ .\label{eq:dp_p} \end{equation}

Using the equation for conservation of energy for an isentropic flow,

\[ c_{p}dT+V\,dV=0 \]

Dividing the equation by \( V^{2} \), we obtain,

\[ \frac{c_{p}T}{V^{2}}\,\frac{dT}{T}+\frac{dV}{V}=0\ . \]

The value of specific heat \( c_{p} \) can be obtained using the equation, \( c_{p}=\gamma R/\left(\gamma-1\right) \), which results in,

\[ \frac{1}{\gamma-1}\,\frac{\gamma\,R\,T}{V^{2}}\,\frac{dT}{T}+\frac{dV}{V}=0\ . \]

Using the equation for speed of sound, \( a^{2}=\gamma\,R\,T \), we can write,

\[ \frac{1}{\gamma-1}\,\frac{a^{2}}{V^{2}}\,\frac{dT}{T}+\frac{dV}{V}=0\, \] \[ \frac{1}{\left(\gamma-1\right)M^{2}}\,\frac{dT}{T}+\frac{dV}{V}=0\ , \] \[ \frac{dT}{T}=-\left(\gamma-1\right)M^{2}\,\frac{dV}{V}\ . \]

Substituting the value of \( \gamma = \) 1.4, for air, and \( dV/V \) = 0.01, the above equation becomes,

\begin{equation} \boxed{\frac{dT}{T}=-0.004\,M^{2}}\ .\label{eq:dT_T} \end{equation}

Using the differential form of the equation of state for an ideal gas,

\[ \frac{dp}{p}-\frac{d\rho}{\rho}-\frac{dT}{T}=0\ , \]

Substituting the derived formula for \( dp/p \) and \( dT/T \) in the above equation,

\[ -\gamma\,M^{2}\frac{dV}{V}-\frac{d\rho}{\rho}+\left(\gamma-1\right)M^{2}\,\frac{dV}{V}=0\ . \]

Which can be solved for \( d\rho/\rho \) to be,

\[ \frac{d\rho}{\rho}=-M^{2}\,\frac{dV}{V}\ . \]

Substituting the value of \( dV/V= \) 0.01, the above equation becomes,

\begin{equation} \boxed{\frac{d\rho}{\rho}=-0.01\,M^{2}}\ .\label{eq:drho_rho} \end{equation}

The equations (\ref{eq:dp_p}), (\ref{eq:dT_T}) and (\ref{eq:drho_rho}) are plotted in Fig. 1, with varying Mach number, \( M \) in range [0.2, 2].

Fig. 1: Percentage changes in pressure, temperature and density as a function of Mach number for an isentropic flow

Problem 3.1:

Solution: