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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?

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    $\begingroup$ a throttle valve is inline with a pipe and thus the total velocity is the same. It's just mass flow rate over an area, the same both in and out. A nozzle is defined by the change in cross-section. $\endgroup$
    – Tiger Guy
    Jul 31 at 23:38

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