The principle behind throttling valves

[Chestermiller]

Hi
On what you term the first question, you are correct but it is generally not necessary to delve into the tensors of Reynolds and normal stresses when control volumes are considered. I am quite familiar with the dissipation term in the differential energy equation, actually spent quite a while working with it. The above result that propose from the textbook I expect is easily found from the first law Q-W=d(H)+d(KE)+d(PE). For throttle everything except d(H) is assumed to go to zero. You may then expand out d(H) and show that viscous dissipation results in a temperature change but then this would contradict the constant h across the throttle assumption (which must be true since it is simply energy conservation afterall). So I think one can view it from two perspectives,

Allow the frictional work to show up in the energy equation giving W(viscous) = d(H) but this implies the the control volume must be drawn within the throttle rather than around it.

Or allow h to be constant when control volume is drawn around the throttle and then split the enthalphy into two terms, u+PV, the u term can then be assigned to equal the viscous dissipation (Cv(dT)) and hence giving a temperature rise. This initially seems in conflict with the constant enthalpy requirement as it indicates a temperature rise but this temperature rise is compensation for by the (V−T∂V∂T)dP using your equation above. This way the enthalpy can remain constant even though there is a pressure rise.

Both ways to think about it are probably reasonable and I think correct.

I looked over Chapter 11 in Bird, Stewart, and Lightfoot last night, and I'm very confident you are going to find the answers to all your questions in that chapter. Unfortunately, I have not been able to articulate the essence of that development in a way that satisfies your doubts.

I also am sending you a private message in which I am proposing that we work together to apply the equations in BSL to solve a specific problem, in order to develop a practical understanding of exactly what is going on. I hope you are interested in participating.

Chet

 

[Chestermiller]

Several weeks ago, in posts to this thread, Toolbox13 and I had been puzzling over some of the thermodynamics fundamentals of throttle values, particularly with regard to the manner in which viscous heating comes into play, and its influence on the temperature of the fluid exiting from the valve. As a result of these interactions, in private messages, we resolved to collaborate in analyzing the system in much greater detail, with the objective of removing the gaps in our understanding. We are now reporting back to the thread on the results of our deliberations, with which we are now in full agreement.

In order to keep things simple, we decided that, rather than deal with the geometric complexities of a throttle valve, we would analyze the simpler situation involved in the thermodynamics of adiabatic laminar fluid flow through a long straight section of pipe. The qualitative results we obtained for this case are fully compatible with those for a throttle valve.

We considered three types of fluids: Ideal (compressible) gases, ideal incompressible liquids, and real compressible gases. In all three cases, viscous dissipation was included in our analyses.

We also included three versions of energy balance equations:
1. Mechanical energy balance equation (obtained by dotting the differential equation of motion with the velocity vector)
2. Overall energy balance equation, including both mechanical and thermal effects.
3. Thermal energy balance equation, obtained by subtracting the mechanical energy balance equation from the overall energy balance equation. This focuses mainly on how the thermodynamic functions vary with position and time.

We considered the differential forms of these energy balance equations as well as the forms of the equations obtained by integrating over a fixed control volume, such as a pipe or a throttle valve. We also considered the predictions for specific thermodynamic functions, such as internal energy or enthalpy.

The following is a summery of our findings:
1. The change in mixing cup average enthalpy between inlet and exit of the pipe section (or throttle valve) is zero.
2. Even though the change in mixing cup average enthalpy is zero, there are non-uniform enthalpy variations over the exit cross section of the pipe. In the case of an ideal gas, this translates into radial variations in temperature at the exit; however, the mixing cup average temperature of the gas at the exit is the same as at the entrance.
3. In the case of an ideal gas, there are two competing physical mechanisms affecting the internal energy change and the temperature change from inlet to exit. These are: viscous heating and expansion cooling. These two effects exactly cancel one another for an ideal gas, such that there is no change in mixing cup average temperature.
4. For incompressible liquids, even though the mixing cup enthalpy change is zero, there is no expansion cooling, and the viscous heating effect brings about a temperature rise.
5. For real gases, even though the mixing cup enthalpy change is zero, the effect of pressure on enthalpy leads to Joule-Thompson heating or cooling.

 

[Dennis C]

For Tony:
"I don't understand why pressure will drop when the fluid pass through the valve but not "Why does the pressure not recover?"
Could you please help again? I'm such an idiot..."
You don't want it to recover, you are controlling the valve (opening and closing it) depending on the downstream load.

Tony, minus all the engineering, think of trying to regulate pressure downstream of a MANUAL valve with a varying downstream load.
You tie a 1" water line into a 6" main header that is at 100 psi. you put a manual gate valve on it and it is in the closed position.
On the downstream end of the manual gate valve you have a garden hose (for simplicity) and another valve, your "load" valve. Now you open your load valve wide to atmosphere, then you throttle open the manual gate valve 1/2 way, the upstream line will still be at 100 psi, and if you have a gauge installed downstream of your throttle valve it may well read zero because we are open to atmosphere and can move the water without line losses. It will never recover because the load is there (open to atmosphere). Now, close your load valve to 90 percent, (Your downstream load is reduced) your original throttle valve still open 50% will now not produce the drop you want, if fact it may very well "recover" because the load is so low and the valve is still at 50%. When they are talking about throttling, the downstream load has to be there for a pressure drop.

Let's say you want to drop pressure in a steam line from 200 psi to 80 psi
the control valve will vary its position depending on downstream load, that is why it won't "recover". It is not suppose to. If the downstream load on the steam main drops 90 percent (like when you closed the load valve 90% on the hose) because a process or a heat exchanger was shut down,the throttle valve (control valve) will also close off a huge percentage to keep the downstream press at your predetermined pressure, 80 psi. It is not suppose to recover!
Back to the original manual valve, you can not throttle pressure with a manual valve if the downstream load varies! if the downstream load is constant and the fluid and upstream pressure is constant, you can.
Hope this helps
Dennis

 

[PepperAngus]

Can anybody explain why the pressure decreases across a throttling valve using the equation U1 + P1V1 = U2 + P2V2? Furthermore, for gases flowing through a throttling valve, is it necessary for the outlet pipe to be greater than the inlet? If not, how can there be an "expansion" or increase in volume?

 

[Chestermiller]

Can anybody explain why the pressure decreases across a throttling valve using the equation U1 + P1V1 = U2 + P2V2?

The pressure decreases across a throttling valve, not because of conservation of energy between inlet and outlet, but because of viscous drag forces acting on the fluid in passing through the valve, which converts mechanical energy to thermal energy. If you had flow through a porous plug instead of a valve, would you really feel that the inlet and outlet pressures should be the same?

Furthermore, for gases flowing through a throttling valve, is it necessary for the outlet pipe to be greater than the inlet?

No.

If not, how can there be an "expansion" or increase in volume?

The velocity of the gas increases.

Chet

 

[piyush007]

so that means if i have a throtlling valve which decreases the pressure to 0.2 bar from 23 bar of steam, and i am planning to have an arrangement that will give 0.2 bar from 12 bar of steam, which is more beneficial and why, if possible prove by calculation.

 

[Chestermiller]

so that means if i have a throtlling valve which decreases the pressure to 0.2 bar from 23 bar of steam, and i am planning to have an arrangement that will give 0.2 bar from 12 bar of steam, which is more beneficial and why, if possible prove by calculation.

Exactly what do you mean by "more beneficial?"