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Let's assume we have a simple duct with a fan (compressor) situated somewhere along it. The fan runs at the same speed always. At the end of the duct, downstream of the fan, there is a throttle valve. The throttle valve is simply a plate facing normal to the flow in the duct, and can move horizontally (towards and away from the duct). Effectively the plate can completely block the duct outlet, giving the ducted flow no where to exit, or it can be placed sufficiently far away such that the flow "exhausts" to the atmosphere freely.

My question is now: Why does throttling (moving the plate closer to the duct exit) reduce the mass flow rate of the flow in the duct?

Will the same result be observed in a potential (irrotational) flow?

I understand that the throttling process is isenthalpic, which results in a loss of stagnation pressure across it, if that helps from the start.

Edit: I would also like to understand the causality of the process: does the throttling cause a reduced mass flow which means the fan will now achieve a higher head (pressure increase) at the same speed or does the throttling cause an increased pressure which reduces the mass flow?

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This is a great question, and the answer relates intimately to why turbofan engines equipped with afterburners require variable geometry exhaust nozzles. Without increasing the throat area to accommodate the larger volumetric flow rate, lighting the afterburner would back-pressure the fan and very possibly lead to a compressor stall. Similarly, mechanically reducing the downstream area (all other things being equal) will require the flow to have a higher upstream stagnation pressure, which means the fan/pump will be required to work harder.

GE F110-129

Now, as to your question about why the flowrate decreases when the exit area is closed, we need to expound a bit on how fans and compressors operate. The fan speed, massflow rate, and pressure ratio are related in a complex way and are usually represented graphically by a fan map.

enter image description here

For a given rotational speed, there is a single steady-state characteristic relating pressure ratio and massflow rate. The shape of this curve can vary (e.g. compare the 40% Nf line with the 100% Nf line above), but generally speaking the higher the pressure ratio, the lower the massflow rate for a given engine RPM. This makes some intuitive sense because the faster the bladetip velocity compared to the axial velocity, the higher will be the flow turning within the bladerow. Work done and pressure rise are proportional to flow turning within the rotor, so higher pressure ratios are positively correlated with lower massflow rates/axial velocities (up to a point).

To truly understand the causal relationship between massflow rate and back-pressure requires that we abandon steady-state thinking altogether. If the exit area is reduced, unsteady compression waves propagate upstream at the speed of sound, incrementally increasing the static pressure at the exit of the fan. This increased back-pressure means that the entering flow now encounters an adverse streamwise pressure gradient and slows down. This slower flow is then worked harder by the spinning bladerow, which results in larger stagnation pressure and temperature rises.

Remember that the flow always exits the device at atmospheric pressure so long as it is subsonic, precisely because of the information propagated upstream by unsteady pressure waves. Thus, if reducing the exit area means a higher exit Mach number is required to conserve mass, the total-to-static pressure ratio must increase. A fixed exit static pressure and increased total-to-static pressure ratio demands that the upstream stagnation pressure increase, and so the upstream turbomachinery will be affected.

If you are looking to put numbers on things, the isentropic flow function is a useful and straightforward way to determine the massflow rate of a compressible fluid if other of the fluid's basic properties are known. In general, the massflow rate of a fluid through a cross-sectional area $A$ is equal to

$\dot{m}=\rho VA$.

Now, if the fluid is compressible and the Ideal Gas Law applies, then

$\dot{m}=\rho VA=\left(\frac{P}{RT}\right)(M\sqrt{\gamma RT})A=PAM\sqrt{\frac{\gamma}{RT}}$.

Both the stagnation temperature and stagnation pressure are preferred flow variables to their static counterparts, so the above equation can be rewritten as

$\dot{m}=P_0 \left(\frac{P}{P_0}\right)AM\sqrt{\frac{\gamma (T_0/T)}{R(T_0)}}$,

and the stagnation properties (as well as the through-flow area) can be moved to the LHS of the equation:

$\frac{\dot{m}\sqrt{T_0}}{P_0 A}=\left(\frac{P}{P_0}\right)M\sqrt{\frac{\gamma}{R}\left(\frac{T_0}{T}\right)}$

If the flow is isentropic (as we are assuming), we know that

$\frac{P}{P_0}=\left(\frac{P_0}{P}\right)^{-1}=\left(\frac{T_0}{T}\right)^\frac{\gamma}{1-\gamma}$,

which gives us

$\frac{\dot{m}\sqrt{T_0}}{P_0 A}=M\sqrt{\frac{\gamma}{R}}\left(\frac{T_0}{T}\right)^{\frac{1}{2}+\frac{\gamma}{1-\gamma}}=M\sqrt{\frac{\gamma}{R}}\left(\frac{T_0}{T}\right)^{\frac{1+\gamma}{2(1-\gamma)}}$.

Again invoking our assumption of isentropic flow, we know that the stagnation temperature ratio is related to the local Mach number by the following equation:

$\frac{T_0}{T}=1+\frac{\gamma-1}{2}M^2$

which, when plugged into the previously derived expression gives us the isentropic flow function $FF_T$:

$FF_T=\frac{\dot{m}\sqrt{T_0}}{P_0 A}=M\sqrt{\frac{\gamma}{R}}\left(1+\frac{\gamma-1}{2}M^2\right)^{\frac{1+\gamma}{2(1-\gamma)}}$

To compute the massflow rate we simply rearrange the isentropic flow function relation...

$\boxed{\dot{m}=P_0 AM\sqrt{\frac{\gamma}{RT_0}}\left(1+\frac{\gamma-1}{2}M^2\right)^{\frac{1+\gamma}{2(1-\gamma)}}}$.

**Note: The above equation is true at any given section within a compressible flow, but the stagnation properties may change from location to location (or over time) based on the specifics of the exact flow the equation is being applied to.

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  • $\begingroup$ Mathjax tip: Instead of $...$, use $$...$$; then you don't need blank lines buffering the equation, and more importantly it doesn't try to squeeze things vertically to fit inline. (Nice answer by the way :) $\endgroup$ – user10851 Jul 15 '14 at 22:08
  • $\begingroup$ Thanks, this is a great answer, it explains it very well. $\endgroup$ – Dipole Jul 16 '14 at 23:01
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For a given pressure drop (pressure upstream minus pressure downstream), the flow rate is proportional to the 4th power of radius, according to the Hagen–Poiseuille equation.

As you close the valve, flow across the valve will decrease, and pressure between the fan and the valve will increase. Because the pressure on the downstream side of the fan increases, flow across the fan will decrease. A steady state where flow across the valve and across the fan will both have decreased to the same lower value will be reached.

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