The throttling process is a constant enthalpy process, in which the pressure decreases. So,

$$h_1 = h_2$$

$$u_1 + p_1v_1 = u_2 + p_2v_2$$

If the pressure is decreasing and if we consider the Joule-Thomson coefficient to be such that the temperature is decreasing too (so internal energy, $u$, will decrease), then the specific volume will increase. If the specific volume increases won't it cause a change in velocity and so the kinetic energy? Why is it said that there is a negligible change in kinetic energy in a throttling process?

Also when we say the enthalpy is conserved in a throttling process do we mean static enthalpy or stagnation enthalpy?

  • $\begingroup$ If you calculate the change in kinetic energy of the gas, it is typically very small. $\endgroup$ May 23, 2020 at 16:56
  • $\begingroup$ @Chet Miller So the volumetric change does cause velocity to increase but not a significant enough change? $\endgroup$
    May 23, 2020 at 17:04
  • $\begingroup$ That's right. Just calculate the change in temperature that the kinetic energy change would translate into in a typical case. $\endgroup$ May 23, 2020 at 17:43
  • $\begingroup$ @ChetMiller What if both temperature and pressures decreases during throttling? Won't the volumetric change be large enough to cause a significant flow velocity change and so a significant change in kinetic energy? $\endgroup$
    May 23, 2020 at 18:01
  • $\begingroup$ Try a sample calculation and see. Certainly if the mass flow rate is low, the kinetic energy and its change will get very low. $\endgroup$ May 23, 2020 at 18:37

1 Answer 1


In adiabatic throttling process ($Q=0$), if there is a drop in the temperature of fluid (i.e. for positive Joule Thomson coefficient) there is decrease in internal energy $u$ (i.e. $u_2<u_1$) which increases specific volume $v$ (i.e. $v_2>v_1$) but the pressure $p$ decreases (i.e. $p_2<p_1$). Decrease in pressure $p$ is less than increase in specific volume $v$ as a result there is an increase in the flow work i.e. $p_2v_2>p_1v_1$ so that the enthalpy $h=u+pv$ remains constant during the (adiabatic) throttling process. Practically, there is negligible increase in kinetic energy of fluid (ideally zero) as follows $$\Delta K.E.=\left(u_2+p_2v_2\right)-\left(u_1+p_1v_1\right)=h_2-h_1\approx 0$$ Therefore the increase in the velocity of fluid undergoing adiabatic throttling process is negligible.

  • $\begingroup$ Yes, that is my confusion. I do understand the derivation. But if the fluid undergoes expansion, say during throttling through a valve, and the diameter of the pipe on both sides of the valve is equal, shouldn't the velocity if the fluid increase on account on the expansion (increase in volume)? What am I missing here... $\endgroup$
    May 23, 2020 at 18:26
  • $\begingroup$ No, expansion means fluid does work to push itself against the surrounding. When a fluid expands in throttling, it does work on the surrounding to maintain flow (i.e. flow energy) & its internal energy decreases however the sum of two i.e. enthalpy remains constant. Ideally, this doesn't affect K.E. or velocity of fluid $\endgroup$ May 23, 2020 at 18:37
  • $\begingroup$ Ok...let us consider a mass of fluid occupying a certain volume going through the throttling valve, (from state1 to state2). As it goes through the throttling valve the volume occupied by this mass of fluid increases, so the same mass of fluid is takin up more space. If the diameter of pipe is equal on both sides of the valve how will the velocity remain unchanged? $\endgroup$
    May 23, 2020 at 18:57
  • $\begingroup$ So my confusion is if a packet of a fluid undergoing throttling takes up more space (as there is an increase in specific volume) or not after the process. $\endgroup$
    May 23, 2020 at 19:25
  • $\begingroup$ Here it says velocity increases during throttling process:engineersedge.com/thermodynamics/throttling_process.htm $\endgroup$
    May 23, 2020 at 20:13

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