For a throttling valve, why is it that temperature remains the same?

[Carbon273]

For a throttling valve C.V analysis, I am wondering why is it known that temperature remains the same for ideal gases. I understand that using the energy balance equation, I end up with h1=h2. Pretty simple so far. By breaking down enthalpy into its components u1+P1v1 = u2+P2v2 I realized that we have an equation set with 1 DOF. If I were to increase internal energy from inlet to outlet, a corresponding decrease in flow energy would occur and vice versa. If I were to increase flow energy, It would lead into a decrease of internal energy, which is usually accompanied by a decrease in temperature (measured by the joule thomson coeeficient) from my understanding. However, it is known that for ideal gases, temperature remains the same as it is constrained (or is a function according my textbook) by enthalpy h=h(T). The book did not seem go into much detail so here I am.

Thanks in advance.

 

[anorlunda]

Ping @Chestermiller

 

[Chestermiller]

For a throttling valve C.V analysis, I am wondering why is it known that temperature remains the same for ideal gases. I understand that using the energy balance equation, I end up with h1=h2. Pretty simple so far. By breaking down enthalpy into its components u1+P1v1 = u2+P2v2 I realized that we have an equation set with 1 DOF. If I were to increase internal energy from inlet to outlet, a corresponding decrease in flow energy would occur and vice versa. If I were to increase flow energy, It would lead into a decrease of internal energy, which is usually accompanied by a decrease in temperature (measured by the joule thomson coeeficient) from my understanding. However, it is known that for ideal gases, temperature remains the same as it is constrained (or is a function according my textbook) by enthalpy h=h(T). The book did not seem go into much detail so here I am.

Thanks in advance.

There are two things happening. First, the pressure of the gas decreases in passing through the throttle valve, so there is a tendency for it to experience expansion cooling. Second, (viscous) friction is responsible for the pressure drop, so there is viscous heating occurring within the throttle. For an ideal gas, these two effects exactly cancel out. (On the other hand, for an incompressible fluid, the expansion cooling is not occurring, so the fluid experiences a temperature rise due to viscous heating).

 

[Carbon273]

There are two things happening. First, the pressure of the gas decreases in passing through the throttle valve, so there is a tendency for it to experience expansion cooling. Second, (viscous) friction is responsible for the pressure drop, so there is viscous heating occurring within the throttle. For an ideal gas, these two effects exactly cancel out. (On the other hand, for an incompressible fluid, the expansion cooling is not occurring, so the fluid experiences a temperature rise due to viscous heating).

You know that is really cool. That’s really interesting. Thanks for that.