# Why is a throttling valve isenthalpic whilst a nozzle is not?

From my understanding, Nozzle's and throttling valve are devices which both trade pressure for velocity, so they increase velocity at the expense of having to decrease the pressure. If we examine these devices using the first law for control volumes (assuming no height changes and steady state conditions), we get that $$\dot{E}_{in} =\dot{E}_{out}$$ $$\dot{m}(h_{in}+\frac{V^2_{in}}{2}+gz_{in}) = \dot{m}(h_{out}+\frac{V^2_{out}}{2}+gz_{out})$$ $$\Rightarrow (h_{in}+\frac{V^2_{in}}{2}) = (h_{out}+\frac{V^2_{out}}{2})$$ $$\Rightarrow h_{in} = h_{out}+\frac{V^2_{out}}{2}-\frac{V^2_{in}}{2}$$ The above equation is true for both devices. Now my textbook (Cengel and Boles Thermodynamics) explains , without much justification, that for throttling valves we have that $$V_{out}\approx V_{in}$$. This means that $$h_{in}\approx h_{out}$$ implying that throttling valves reduce pressure non-trivially but do not appreciably alter velocity.

My problem is that the author claims that for nozzles, the approximation $$V_{in}\approx V_{out}$$ does not hold. Thus nozzles are not isenthalpic but throttling valves are. Why is this the case? Why do nozzles expend pressure to increase velocity while throttling valves expend pressure but do not appreciably gain velocity? If I imagine individual fluid particle travelling into a nozzle, I can visualize how they must speed up to ensure a constant mass flow rate. Similarly, if I imagine fluid particles travelling through a throttling valve (i.e a porous plug so the particles must flow through a perforated obstacle), I can also visualize how they must speed up when they flow through the perforated porous plug to ensure that the liquid mass flow rate remains constant. So ultimately my question is this : Why are throttling valves able to decrease the pressure non-trivially without appreciably altering velocity in steady flow conditions whilst a nozzle that causes a non-trivial decrease in pressure must appreciably alter the velocity?

The internal design of a nozzle is very different from the internal design of the throttling valve or porous plug. The nozzle is designed with smooth gradually varying cross section such that the flow is nearly isentropic, with dh=vdP. The throttling valve or porous plug features non-smooth tortuous cross sections which results in a highly irreversible flow, and lots of viscous dissipation. So, in the case of a nozzle, we expect substantial temperature changes due of isentropic expansion and compression, while, in the throttle or porous plug, at least for an ideal gas, the drop in pressure does not lead to a decrease in temperature.

In addition, the mass flow rate through a throttle valve or porous plug is much lower than an equivalent nozzle, such that the inlet and outlet kinetic energies of the gas are both much lower than an equivalent nozzle, and the change in kinetic energy is negligible compared to the case of a nozzle.

• Thanks for the response. Just a question I asked in the other response: Suppose we have a perfectly incompressible fluid flow through a throttle valve so that $V_{out}=V_{in}$. In this case we have that $u_f+P_fv=u_i+P_iv$ where $v$ is the constant specific volume. Throttle valves decrease the pressure ($P_f<P_i$) which implies that $\Delta u=u_f−u_i=(P_i−P_f)v>0$ (i.e the fluids internal energy increases as it passes through the valve). Surely this means that the fluid temperature will always increase in a throttle valve if we have an incompressible fluid (i.e any liquid) flow through it? Aug 1 '21 at 13:27
• Sure. This is the result of "viscous heating." Aug 1 '21 at 14:32
• Okay thanks. My issue now is that in the first law applied to control volumes, we assume that $\dot{Q}=\dot{W}=0$ for throttle valves so that no work is done on (or by) the system and no heat transfer occurs. If this is the case, then the increase in internal energy of the gas seems perplexing since no heat or work is being done on the mass within the control volume yet "viscous heating" still occurs and results in an increased internal energy. How can this be so? (I realize this might be better placed as another post on its own) Aug 1 '21 at 15:21
• No shaft work is being done. Work is still being done (lumped in with the vP term of the enthalpy) to force material into and out of the control volume. You need to review the derivation of the open system (control volume) version of the first law to see that the total work is partitioned into shaft work and in-out work. Aug 1 '21 at 15:29
• Okay I think I understand. The higher upstream pressure does positive work $W_{upstream}=P_{up}V$ on a mass particle flowing into the control volume whilst the lower downstream pressure does negative work $W_{downstream}=-P_{down}V$ and the sum of these is positive so that we can have a positive $\Delta u$. My issue was in my interpretation of the "flow work" which I sometimes see described as "pressure energy" and other times as "flow work". Aug 1 '21 at 15:39

As an example, assume a 8 inch diameter pipe. Flanges are installed at the point where a throttling valve is to be installed, the throttling valve is installed between those flanges, and the throttling valve is normally the same size as the pipe that it is installed in, with appropriate valve trim to provide the required pressure drop for the given application. When fluid flows through the valve, the fluid velocity increases substantially as fluid goes through the valve trim, but that fluid exits back into the 8 inch line downstream of the throttling valve. This means that if you take a control volume across the whole throttling valve, the exit velocity equals the entrance velocity for incompressible flow, and the exit velocity is a bit higher than the entrance velocity for compressible flow.

A similar analysis for a nozzle clearly shows that the exit velocity from a nozzle is substantially higher than the entrance velocity to the same nozzle, both for compressible and incompressible flows. The same conclusion would be drawn if you took a control volume from the upstream side of a throttling valve and extended that control volume only to the valve trim, but that particular control volume is seldom (if ever) selected. This leads to what the author published regarding entrance and exit velocities of both devices.

• Thanks. Just a quick question: suppose we have a throttle valve set up in a 8'' pipe as you describe with a perfectly incompressible fluid flow through it so that $V_{out}=V_{in}$. In this case we have that $u_f+P_fv=u_i +P_iv$ where $v$ is the constant specific volume. Throttle valves decrease the pressure ($P_f<_P_i$) which implies that $\Delta u = u_f-u_i=(P_i-P_f)v > 0$ (i.e the fluids internal energy increases as it passes through the valve). Surely this means that the fluid temperature will always increase in a throttle valve if we have an incompressible fluid (i.e any liquid)? Jul 31 '21 at 18:59
• @SalahTheGoat, it's not done that way. What you are describing is an eductor, which "pulls a vacuum on a side stream" in order to entrain liquid from a lower pressure source. Also, note my previous reply ... the conclusions of the author are based on the standard way of drawing control volumes for the given devices, but that's not the only way to select a control volume. The selection of the control volume will influence your answer. Jul 31 '21 at 22:25

Good morning:

I might be late for my answer, but here follows:

The reason why certain fluids cools while expanding is internally driven by their molecules interaction.

An ideal throttling / expansion leads to another pressure level while maintaining same enthalpy and, more importantly, same temperature.

However, when a real throttling / expansion takes place, the collisions between the particles could be weaker, leading to an average kinetic energy that is lower in value. The fluid is undergoing the so called negative Joule-Thomson effect (negative deviation from ideal throttling).

This is why you might see frost near throttling valves at chiller cycles. I have grabbed one of them with my hand, and feels cold indeed!