-1
$\begingroup$

Why the gas temperature and pressure decreases while passing through a nozzle?

I have read that in a nozzle velocity increases at the expanse of pressure drop.But how temperature varies with that? I didn't get that.

$\endgroup$
2
$\begingroup$

For quasi-one-dimensional compressible flow, the Mach number ($M$) of the gas in a nozzle has the following relationship with the area ratio ($A/A^*$) of that nozzle.

$$ \frac{A}{A^*} = \frac{1}{M}\left[\frac{2}{\gamma+1} \frac{\gamma-1}{2}M^2\right]^{\frac{\gamma+1}{2(\gamma-1)}} $$

where $\gamma$ is the ratio of specific heats of the gas. This equation has a subsonic and supersonic solution for a given area ratio $A/A^*$, which is represented in the figure below.

enter image description here

Notice the Mach number increases with decreasing nozzle area ratio if the flow is subsonic. At the throat condition $A/A^* = 1$, we reach the sonic flow condition where $M = 1$. Downstream of the throat the Mach number increases with increasing nozzle area ratio. Now to actually meet the choking condition at the throat, we need to have a pressure ratio of $p/p_0 = 0.528$ ($\gamma$ = 1.4) at the nozzle throat. Essentially, the chamber pressure must be at least 1.89 times that at the nozzle throat, or we won't reach the supersonic solution common to choked nozzle flows.

Now with all of this defined, we can address your question properly. The pressure and temperature vary subject to the isentropic flow relations in a nozzle (at least as a very good approximation). These relations go as follows:

$$ p = p_0 \left(1 + \frac{\gamma-1}{2} M^2 \right)^{-\frac{\gamma}{\gamma-1}} $$

$$ T = T_0 \left(1 + \frac{\gamma-1}{2} M^2 \right)^{-1} $$

where $p_0$ and $T_0$ are the chamber pressure and temperature, respectively. I hope it is clear from these equation that with increasing Mach number (as one would get through a converging-diverging nozzle) the pressure and temperature decrease.

A common diagram in rocket propulsion is the following:

enter image description here

$\endgroup$
7
  • $\begingroup$ The assumption of isentropic flow through an orifice/nozzle indeed well known. And isentropic implies adiabatic, so no loss or gain of heat. But real nozzles loose heat to the flow stream from the nozzle throat. So the gas temperature in the real nozzle is actually increasing downstream of the throat right? $\endgroup$
    – docscience
    Jan 11 '17 at 20:21
  • $\begingroup$ The temperature continually drops downstream of the thrust chamber. Most of the enthalpy in the thrust chamber is being converted into kinetic energy. As an example, the Space Shuttle Main Engine (SSME) has a chamber temperature of about 3,600 K and a nozzle exit temperature of roughly 1,230 K, with an exit Mach number of about 4.6. It is true that heat is lost through the walls of the nozzle, as they are not perfectly adiabatic, however the temperature will always decrease downstream of the thrust chamber. $\endgroup$
    – TRF
    Jan 11 '17 at 22:12
  • $\begingroup$ You are talking rocket engines. I'm talking expansion nozzles like the valve on a SCUBA tank for example. It's not adibatic, not isentropic, definitely sonic. The valve body turns extremely cold to the point ice accumulates. $\endgroup$
    – docscience
    Jan 11 '17 at 22:20
  • $\begingroup$ If the flow is choked, i.e. sonic orifice, than the temperature is dropping moving through the valve. In which case, the valve acts as a subsonic nozzle. If the flow is not choked, then the valve acts as a subsonic diffuser and the temperature will rise. $\endgroup$
    – TRF
    Jan 11 '17 at 22:36
  • $\begingroup$ The SCUBA valve definitely sonic. Since the valve freezes it is losing heat energy. The heat presumably flows from the valve to the gas stream not to the surrounding air. So shouldn't the gas stream increase in temperature? I recall putting my fingers near the gas, it is hot! The valve near the nozzle cold, the expanding gas hot. No theory, experience. $\endgroup$
    – docscience
    Jan 11 '17 at 22:45
1
$\begingroup$

Consider a pressurised aerosol cannister.

The portion of the gas outside the nozzle, that has already been released, is expanding into the atmosphere, as the atmosphere is at a lower pressure than the inside of the cannister. As the gas expands, it occupies a larger volume, so its pressure must decrease.

The kinetic energy required to expand into the empty space must come from somewhere, and it can only come from the heat energy that gives the gas temperature, so its temperature also must decrease.

The whole fluid/gas is continous, and pressure and temperature are defined continuously, so these must change gradually. This is also why the cannister gets colder.

Pressure or temperature variations within the cannister are very quickly eliminated, due to the large numbers of particles involved.

Similarly for the gas outside the cannister.

So, the only place left where they can change is the nozzle.

$\endgroup$
1
  • $\begingroup$ Ok, I get that but how pressure and temperature varies within nozzle?How the pressure varies in compressible and incompressible flow? Does the gas expands while passing through a convergent nozzle? $\endgroup$
    – Max
    Jan 12 '17 at 6:36
0
$\begingroup$

SUppose it happens in ideal gas, then:

$$pv=kT\\pv^n=C_1$$

where $k$ is Boltzmann's constant and $n$ is polytropic index and $C_i$ are constants. For adiabatic $n=\varkappa=1.4$, Poisson constant.

We can combine those equations and we get: $$p^{\frac{n-1}n}T=C_2$$

Then: $$T=C\cdot p^{\frac{n-1}n}$$

If the polytropic index is constant and you know the pressure along the nozzle, you can calculate the temperature.

$\endgroup$

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.