3
$\begingroup$

I have some questions about the enthalpies between two points during an isentropic process in a nozzle. By doing the energy balance between the two points, one gets:

$h_1 + \frac{v_1^2}{2}+gz_1=h_2+\frac{v_2^2}{2}+gz_2$

If one neglects the potential energy and assuming that the velocity at the inlet is 0, one ends up with this:

$h_1 =h_2+\frac{v_2^2}{2}$

This is the part that raises some questions. If the process is isentropic, it means that the entropy doesn't change and therefore there is no heat added (or vice-versa) to the flow. But wouldn't that make $h_1 = h_2$?

My guess is that because $h = u + Pv$, the Pv term is the one that makes them different, considering that the flow energy (Pv) is different. But this would make $u_1 = u_2$. Is this true?

And if this is true, why don't we just use the following relation:

$P_1v_1=P_2v_2+\frac{v_2^2}{2}$

At this point, one can't really define $Cp$ or $k=Cp/Cv$, which would lead to the well-known isentropic flow equations. Am I getting something wrong? Can anyone help me to understand this? Thank you!

$\endgroup$
4
$\begingroup$

Constant entropy doesn't mean constant temperature, the gas flow through nozzle is subject to change in pressure and velocity (continuity equation) due to the variation in cross section, and being compressible the change in pressure will be reflected as change in temperature and density.

That being said, since the flow is isentropic and there will be no energy dissipation due to friction and by applying the first law of thermodynamics to the nozzle you will (naturally!) find that the sum of energy across the nozzle is constant, meaning that the stagnation enthalpy ($h_1$) will equal the sum of enthalpy and kinetic energy at every other point in the nozzle (any increase in kinetic energy will be reflected as decrease in enthalpy and vice versa).

So, to answer your question: $h_1 \neq h_2$ and also the sensible temperature $T$ reflected in internal energy is not necessarily the same for the two point in comparison, meaning that also $u_1 \neq u_2$, you have to look at the big picture where the change in cross section will affect the velocity, pressure, temperature and density (that's why the analysis of compressible flow is far more complex than incompressible flow).

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.