# How is the kinetic energy change during a throttling process negligible?

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?

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

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. $$u_2) which increases specific volume $$v$$ (i.e. $$v_2>v_1$$) but the pressure $$p$$ decreases (i.e. $$p_2). 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.