When using the general Bernoulli equation along a streamline one of the assumptions that you have to make is that the fluid is incompressible. Making this assumption is reasonable when you are analyzing a liquid. However, what if you have a compressible flow such as for a gas. If you are analyzing a gas making this assumption will cause inaccuracies in your result. To remedy this problem you can make modifications to Bernoulli equation to analyze a compressible fluid.
First, before modifying Bernoulli equation for a gas you should analyze the static and stagnation pressure. For an incompressible fluid the pressure change between the static pressure and stagnation pressure is represented by the following equation.
(Eq 1) $Δp = p_{stag}-p_{static}=ρv^2/2$
$ρ$ = fluid density
$v$ = fluid velocity
The above equation states that the stagnation pressure minus the static pressure will equal the dynamic pressure. This statement can still be made for a gas if the following is true. The dynamic pressure can’t be much larger than your static pressure. Basically if the static pressure is more dominant than the dynamic pressure than the density change will be low enough between the two points of interest that it can be considered negligible. However, as the dynamic pressure increases the density change between the two points will increase causing inaccuracies. To compensate for the density change, the Bernoulli equation will become the following.
(Eq 2) $\int\frac{dp}{ρ}+\frac{1}{2}v^2 + gz = constant$
$g$ = gravitational constant
$z$ = fluid height
Isothermal Flow
When you are analyzing a compressible flow you will need to take a look at its thermodynamic properties. First, let’s take a look at an isothermal flow. For an isothermal flow to occur the temperature of the gas must remain constant along the streamline. If this is the case you can take advantage of the ideal gas law. Recall that the ideal gas law states $p=ρRT$. As a result, for an isothermal flow, Bernoulli equation will become the following.
(Eq 3) $RT\int\frac{dp}{}+\frac{1}{2}v^2 + gz = constant$
$R$ = gas constant
$T$ = absolute temperature
Finally, since $ρ=p/RT$, the pressure term in equation 3 can be easily integrated. This will result in the following equation.
(Eq 4) $\frac{v_1^2}{2g}+z_1+\frac{RT}{g}ln\left(\frac{p_1}{p_2}\right)=\frac{v_2^2}{2g}+z_2$
Remember, even though the above equation will allow you to use Bernoulli equation for a compressible isothermal flow you will still need to make the following assumption. First, the flow must be inviscid and second, the flow must be steady. If these assumptions cannot be satisfied there will be inaccuracies in your results.
Isentropic Flow
Instead of an isothermal flow, what if the flow was isentropic? As with an isothermal flow an isentropic flow is a compressible flow. Which means the density of the gas will change. However, for the flow to be isentropic the entropy of the gas must be constant. This type flow is a reversible adiabatic process. Which means that there will be no friction and heat transfer cannot occur. As a result the pressure and density of the gas can be related by using the specific heat ratio. As a result $p/ρ^k=C$ where C is a constant and k is the specific heat ratio. This can than be used to modify equation 2 resulting in the following equation.
(Eq 5) $C^{1/k}\int{}p^{-1/k}dp+\frac{1}{2}v^2+gz = constant$
Next you will need to integrate the pressure term between the two points of interest along the streamline. This will result in the following.
$C^{1/k}\int{^{p_2}_{p_1}}p^{-1/k}dp = \left(\frac{k}{k-1}\right)\left(\frac{p_2}{ρ_2}-\frac{p_1}{ρ_1}\right)$
As a result equation 5 will become the following equation for an isentropic flow.
(Eq 6) $\left(\frac{k}{k-1}\right)\frac{p_1}{ρ_1}+\frac{v_1^2}{2}+gz_1 = \left(\frac{k}{k-1}\right)\frac{p_2}{ρ_2}+\frac{v_2^2}{2}+gz_2$
