# 11.4.3.1: The pressure Mach number relationship

Before going further in the mathematical derivations it is worth looking at the physical meaning of equation (42). The term $${\rho\, U^{2} / A}$$ is always positive (because all the three terms can be only positive). Now, it can be observed that $$dP$$ can be positive or negative depending on the $$dA$$ and Mach number. The meaning of the sign change for the pressure differential is that the pressure can increase or decrease. It can be observed that the critical Mach number is one. If the Mach number is larger than one than $$dP$$ has opposite sign of $$dA$$. If Mach number is smaller than one $$dP$$ and $$dA$$ have the same sign. For the subsonic branch $$M<1$$ the term $$1 /(1 -M^2)$$ is positive hence
$dA > 0 \Longrightarrow dP> 0 \\ dA < 0 \Longrightarrow dP< 0$
From these observations the trends are similar to those in incompressible fluid. An increase in area results in an increase of the static pressure (converting the dynamic pressure to a static pressure). Conversely, if the area decreases (as a function of $$x$$) the pressure decreases. Note that the pressure decrease is larger in compressible flow compared to incompressible flow. For the supersonic branch $$M> 1$$, the phenomenon is different. For $$M > 1$$ the term $$1 / 1 - M^{2}$$ is negative and change the character of the equation.
\begin{align*}
dA > 0 \Rightarrow dP< 0 \\
dA < 0 \Rightarrow dP> 0
\end{align*}
This behavior is opposite to incompressible flow behavior. For the special case of $$M=1$$ (sonic flow) the value of the term $$1 - M^{2}= 0$$ thus mathematically $$dP \rightarrow \infty$$ or $$dA = 0$$. Since physically $$dP$$ can increase only in a finite amount it must that $$dA = 0$$. It must also be noted that when  $$M=1$$ occurs only when $$dA = 0$$. However, the opposite, not necessarily means that when $$dA = 0$$ that $$M=1$$. In that case, it is possible that $$dM=0$$ thus the diverging side is in the subsonic branch and the flow isn't choked. The relationship between the velocity and the pressure can be observed from equation (36) by solving it for $$dU$$.
$dU = - \dfrac{ dP }{ P \,U} \label{gd:iso:eq:relationshipU-rho} \tag{43}$
From equation (43) it is obvious that $$dU$$ has an opposite sign to $$dP$$ (since the term $$PU$$ is positive). Hence the pressure increases when the velocity decreases and vice versa. From the speed of sound, one can observe that the density, $$\rho$$, increases with pressure and vice versa (see equation (44)).
$d\rho = \dfrac{ 1 }{ c^2} dP \label{gd:iso:eq:relationshipP-rho} \tag{44}$
It can be noted that in the derivations of the above equations (??) - (??), the equation of state was not used. Thus, the equations are applicable for any gas (perfect or imperfect gas). The second law (isentropic relationship) dictates that $$ds=0$$ and from thermodynamics $$ds = 0 = C_p \,\dfrac{dT }{ T} - R \,\dfrac{dP }{ P}$$ and for perfect gas
$\dfrac{dT }{ T} = \dfrac{k - 1 }{ k } \; \dfrac{dP }{ P} \label{gd:iso:eq:isentropicDiff} \tag{45}$
Thus, the temperature varies in the same way that pressure does. The relationship between the Mach number and the temperature can be obtained by utilizing the fact that the process is assumed to be adiabatic $$dT_0 = 0$$. Differentiation of equation (27), the relationship between the temperature and the stagnation temperature becomes
$dT_0 = 0 = dT \left( 1 + \dfrac{ k - 1 }{ 2 } M^2 \right) + T (k -1) M dM \label{gd:iso:eq:temp-M1} \tag{46}$
and simplifying equation (46) yields
$\dfrac{ dT }{ T } = - \dfrac{(k -1) \,M\, dM }{ 1 + \dfrac{ k - 1 }{ 2 }\, M^{2} } \label{gd:iso:eq:temp-M} \tag{47}$

### Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.