# How is the equation of Mach number derived?

Wikipedia states that for a pitot-static tachometer, the mach number for subsonic flow equates to

$$M = \sqrt{5\left[\left(\frac{p_t}{p_s}\right)^\frac{2}{7}-1\right]}.$$

How did they get to that result? Is there a derivation, or is it just from a polynomial fit of a tabulated set of data?

Update

I accepted J.G's answer after glancing at the referenced flight test document (a treasure in itself) and realising that $\frac {7}{5}$ is the same as 1.4, but there remains an issue.

Sadly I don't have my uni books anymore with Bernoulli's equation for compressible flow. The issue is with dynamic pressure: for incompressible flows we can take $p_d = \frac {1}{2} \cdot \rho \cdot V^2$, for compressible flow this is $p_d = \frac {1}{2} \cdot \gamma \cdot p_{static} \cdot M^2$.

Right? If I substitute this I don't get to the equation above. So the answer is unfortunately not accepted anymore.

• To be clear: the Mach in Mach number is a reference to Ernst Mach and it should always be capitalized. – Emilio Pisanty Jul 9 '17 at 13:17
• How many newtons per ampere? – Koyovis Jul 9 '17 at 13:20
• Those are units, and the Mach number is not, so the two are not comparable. If you can show that a significant fraction of the literature uses it uncapitalized, then this can change, but until you do, the correct, capitalized usage will remain. – Emilio Pisanty Jul 9 '17 at 13:23
• Ah yeah, found a reference.. Thx. – Koyovis Jul 9 '17 at 13:34
• No worries. (For future reference, rolling back edits without a consensus in the comments is frowned upon on this site.) Welcome, though, and hope you'll stick around. – Emilio Pisanty Jul 9 '17 at 13:44

The factors are not obvious, I agree.

For instance, for a polytrope index, $\gamma$, of 7/5 the exponent of 2/7 corresponds to a term of the form $\left( \tfrac{\gamma - 1}{\gamma} \right)$, which is our first hint. The second hint is that the pitot tube system can be applied to a Bernoulli system. The third thing to note is that for subsonic speeds, which is where a pitot tube actually functions, one can get away with assuming incompressible flow (I know it seems odd since things obviously do compress a little, but the effects can be considered secondary for most intents and purposes).

For a polytropic ideal gas, we know that $P \propto \rho^{\gamma}$. Thus, we can say that: $$P = \kappa \ P_{s} \ \rho^{\gamma} \tag{1}$$ where $P_{s}$ is the static pressure (also can be considered the pressure at infinity). We can rewrite this equation in terms of density to find: $$\rho = \kappa^{-\frac{1}{\gamma}} \ \left( \frac{P}{P_{s}} \right)^{\frac{1}{\gamma}} \tag{2}$$

The differential form of Bernoulli's equation can be given as: $$u \ du + \frac{ 1 }{ \rho } \ \frac{ d P }{ d \rho } \ d \rho = 0 \tag{3}$$ and we know that the speed of sound is given by: \begin{align} C_{s}^{2} & = \frac{ \partial P }{ \partial \rho } \tag{4a} \\ & = \gamma \ \kappa \ P_{s} \ \rho^{\gamma - 1} \tag{4b} \\ & = \frac{ \gamma \ P }{ \rho } \tag{4c} \end{align}

If we replace the $\rho$ in Equation 4b with the form shown in Equation 2, one can show that the 2nd term in Equation 3 can be rewritten as: \begin{align} \frac{ 1 }{ \rho } \ \frac{ d P }{ d \rho } \ d \rho & = \frac{ \gamma \ \kappa \ P_{s} }{ \rho } \ \rho^{\gamma - 1} \ d \rho \tag{5a} \\ & = \frac{ \gamma \ \kappa \ P_{s} }{ \rho } \ \kappa^{-\frac{ \gamma - 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{\frac{ \gamma - 1 }{ \gamma }} \ d \rho \tag{5b} \\ & = \frac{ \gamma \ \ P_{s} }{ \rho } \ \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{\frac{ \gamma - 1 }{ \gamma }} \ d \rho \tag{5c} \end{align}

If we differentiate Equation 2, we find: $$d \rho = \left( \frac{ \rho }{ \gamma \ P_{s} } \right) \ \left( \frac{ P }{ P_{s} } \right)^{-1} \ dP \tag{6}$$ so that Equation 5c can be rewritten as: $$\frac{ \gamma \ \ P_{s} }{ \rho } \ \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{\frac{ \gamma - 1 }{ \gamma }} \ d \rho = \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{-\frac{ 1 }{ \gamma }} \ dP \tag{7}$$

We define $u \ du \rightarrow C_{s}^{2} \ M \ dM$, thus we rewrite Equation 3 as: $$C_{s}^{2} \ M \ dM + \kappa^{\frac{ 1 }{ \gamma }} \ \left( \frac{ P }{ P_{s} } \right)^{-\frac{ 1 }{ \gamma }} \ dP = 0 \tag{8}$$

We also define $\alpha = \tfrac{ P }{ P_{s} }$ so that $dP \rightarrow P_{s} \ d\alpha$. If we integrate Equation 8 with the limits ranging from $P_{s}$ to $P$, the change of variables makes the 2nd term go to: $$\kappa^{\frac{ 1 }{ \gamma }} \ P_{s} \int_{\alpha}^{1} \ d\alpha \ \alpha^{-\frac{ 1 }{ \gamma }} = \left[ \frac{ \gamma \ \kappa^{\frac{ 1 }{ \gamma }} \ P_{s} }{ \gamma - 1 } \ \alpha^{\frac{ \gamma - 1 }{ \gamma }} \right]_{\alpha}^{1} \tag{9}$$

Thus, Equation 8 can be rewritten as: $$0 = \frac{ 1 }{ 2 } C_{s}^{2} \ M^{2} - \frac{ \gamma \ \kappa^{\frac{ 1 }{ \gamma }} \ P_{s} }{ \gamma - 1 } \left[ \alpha^{\frac{ \gamma - 1 }{ \gamma }} - 1\right] \tag{10}$$ which after some algebra reduces to: $$M^{2} = \frac{ 2 }{ \gamma - 1 } \left[ \alpha^{\frac{ \gamma - 1 }{ \gamma }} - 1\right] \tag{11}$$

As stated above, for $\gamma$ = 7/5, this results in the form about which you are concerned.

• Thank you much for your detailed answer, I appreciate it and now I know where the wiki equation originates. I've linked a couple of questions on the Aviation site to your answer. I am surprised that we have to assume that the flow is incompressible, is it possible to quantify what the error is at higher Mach, let's say M 0.9? – Koyovis Jul 14 '17 at 10:34
• @Koyovis - I am not sure "have to" is appropriate. It's rather that this type of instrument works for subsonic flow. For subsonic flows, the air is very close to an incompressible fluid largely because so many particles can respond faster than the flow (i.e., the rms thermal speed is only slightly smaller than the sound speed and it represents the average speed of most particles). – honeste_vivere Jul 14 '17 at 13:25

If you want to know more about calculating a Mach number, it helps to read Wikipedia's article on Mach number. As explained here, the Mach number for subsonic compressible flow is obtainable from Bernoulli's equation (Wikipedia cites this source). The result you cited then follows from $\gamma=\frac{7}{5},\,p_t=q_e+p$.

$M$ is not the speed of sound. It is the Mach number -- the ratio of speed of the aircraft to the speed of sound. The equation is derived from Bernoulli's equation together with a suitable choice of $\gamma=C_P/C_V$ for air.