*Originally posted in Thermodynamics.. reposted here

Helloooo everyone! Self proclaimed jet engine designer here~ Don't worry I won't be designing airliner engines anytime soon.

I know when air passes through a diffuser or a duct of progressively larger cross section it gets slowed down, compressed with an increase in temperature. And when the passage is of progressively smaller cross section the opposite would happen.

Now.. a great many online publications for laymen explain the above fact clearly.. sometimes with nice illustrations but they don't go as far to detail the calculations. And I find specialized literature to be rather difficult to understand.

For example what would the outcome be if air is discharged out of a compressor (at 300K 300m/s 1.5 bar) and passes through a diffuser that has an exit area of twice that of inlet area? I don't think it will be so simple because all three properties I've mentioned must change simultaneously..

Could an expert thermodynamitician explain this please? Thanks in advance!


1 Answer 1


I notice a couple of issues with your problem:

  • Underspecified: The ratio of the exit to the entrance areas is not enough information to provide an answer. You need to provide another property, such as the pressure, temperature, or isentropic efficiency of the diffuser.
  • More of a fluids problem than thermodynamics: Thermodynamics is more about energy transfer from heat and work, while your problem seems to be more about fluid speed. That's okay though, since fluids and thermo are tightly connected.

Symbolic solution

Despite the problem being underspecified for a numeric answer, you can still set up the problem, but first you need to state some simplifying assumptions. My solution assumes:

  1. Steady state: the flow does not vary with time.
  2. Inviscid fluid: the air has negligible viscosity, losing negligible energy to friction.
  3. Ideal gas: the air behaves as an ideal gas.
  4. Constant potential energy: you didn't mention a height change, so the inlet and exit are presumably at about the same height, and thus have the same potential energy.


Density can be found from the ideal gas law, which you might remember from chemistry as: $$P \cdot V = n \cdot \bar R \cdot T$$ where $P$ is pressure, $T$ is temperature, $V$ is volume, $n$ is the number of moles of gas, and $\bar R$ is the ideal gas constant in units of Joules per mole per Kelvin. In thermo we can replace the $\bar R$ with a gas constant measured in kilograms that is unique to each gas. For air, $R_{air}=286.9 ~ J \cdot kg^{-1} \cdot K^{-1}$. The gas law then becomes: $$P \cdot V = m \cdot R_{air} \cdot T$$ Recalling the definition of density $\rho$ as the ratio of mass to volume, you can divide both sides by $V$ and then rearrange, producing an equation for density: $$\rho = \frac{P}{R_{air} \cdot T}$$ For the values you gave, $\rho_i = 1.74 ~ kg \cdot m^{-3}$

Conservation of mass

Because the flow is steady, mass flow into the diffuser must equal mass flow exiting: $$\dot m_i=\dot m_e$$ From fluids, mass flow is the product of density, area, and velocity, so the mass flow equation becomes: $$\rho_i \cdot A_i \cdot v_i = \rho_e \cdot A_e \cdot v_e$$ You stated that the exit area is twice that of the inlet area, so this equation becomes: $$\frac{\rho_i \cdot v_i}{2} = \rho_e \cdot v_e$$ For the values you gave, $\rho_e \cdot v_e = 261 ~ kg \cdot m^{-2} \cdot s^{-1}$

Compressible Bernoulli Equation

We have changing pressures and velocities along the streamline of an inviscid fluid, so Bernoulli's equation is the next tool we pull out. We need the compressible form since the air density is likely to change quite a bit between inlet and exit. The compressible Bernoulli equation is: $$\frac{v^2}{2} + \frac{\gamma}{\gamma -1} \cdot \frac{P}{\rho} + g \cdot z = constant$$ $v$ is fluid velocity. $\gamma$ is the gas's ratio of specific heats, which for air is about 1.4. I'll refer to the $\gamma/(\gamma - 1)$ as $\alpha$, because it's annoying to type otherwise. The constant I will refer to as $\beta$ (there's no significance to me choosing $\alpha$ and $\beta$ by the way, I just needed letters and who cares). $g$ is gravitational acceleration and $z$ is height. We can neglect $g \cdot z$ because we are assuming that potential energy is constant. Using these simplifications, we can rewrite the equation into a more manageable form: $$\frac{v^2}{2} + \alpha \cdot \frac{P}{\rho} = \beta$$ We know the properties at the inlet, so we can calculate beta, which comes out to be $\beta = 347,000 ~ m^2 \cdot s^{-2}$ for the values you gave.

Why it's underspecified

At present, we have 2 equations to use: $$\rho_e \cdot v_e = \frac{\rho_i \cdot v_i}{2} = 261 ~ kg \cdot m^{-2} \cdot s^{-1}$$ $$and$$ $$\frac{v_e^2}{2} + \alpha \cdot \frac{P_e}{\rho_e} = \beta = 347,000 ~ m^2 \cdot s^{-2}$$ Unfortunately, we have 3 unknowns: $v_e, P_e,$ and $\rho_e$. Algebra teaches us that this is a craptacular situation, as the number of unknowns must equal the number of equations in order to solve the system. You might think that we could use our gas equation: $$\rho_e = \frac{P_e}{R_{air} \cdot T_e}$$ but unless we define $T_e$ we've just changed our situation to 3 equations with 4 unknowns.

Wrapping up

I hope I've been able to help with understanding some of the math. If you're wondering why I said that this is more a fluid dynamics question than a thermo one, notice that I didn't mention entropy, enthalpy, work, that sort of thing. Perhaps I could have used them, but based on the info you provided I didn't see it (maybe with exit temperature and/or pressure, enthalpy could be used).

Please correct me anywhere you find a mistake!


  1. BS in Mechanical Engineering from Virginia Tech (Go Hokies!)
  2. Wikipedia - Bernoulli equation
  3. Engineering Toolbox - individual gas constant
  • $\begingroup$ Hello! Thank you for the detailed explanation. I'm sure youre a very likeable person in real life as well ^^. $\endgroup$
    – thermoboi
    Sep 22, 2016 at 9:44
  • $\begingroup$ And yes it seems i've omitted a few(..) things to begin with.. but the assumptions that you've made such as it being steady flow etc are what i had in mind but failed to convey on text. Like you said there are only a few leads to what I've been trying to figure out, and what I have is basically a centrifugal impeller that discharges a stream of gas of known exit velocity(300m/s) at ambient static pressure(all static pressure rise occuring in diffuser) out of known exit area A1. $\endgroup$
    – thermoboi
    Sep 22, 2016 at 9:51
  • $\begingroup$ And much to my dismay the geometrical restriction within my little engine design the diffuser exit area A2 can only assume twice that of A1(lets ignore impeller discharge swirl correction for now..). Then assuming isentropic flow of little or no loss what might the static pressure and velocity be at the exit of the diffuser? $\endgroup$
    – thermoboi
    Sep 22, 2016 at 9:56
  • $\begingroup$ I've been having problems solving this.. but real engineers seem to have sorted it out long ago with beautiful engines. Thanks! $\endgroup$
    – thermoboi
    Sep 22, 2016 at 9:57
  • $\begingroup$ I've looked at it long and hard... and i think enthalpy just might have the key to fill in the blank.. I did see some calculations using enthalpy as a parameter in turbine design but it got no further in explaining it in detail.. I'll continue to look.. $\endgroup$
    – thermoboi
    Sep 22, 2016 at 15:26

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