bernoulli-equationfluid dynamics
Let's say that I have a duct with the length $ l $ that is connected to a fan, and through the duct there is air flowing with the velocity $ u $. Moreover, let's say we install a manometer right after the fan that measures the gauge pressure $ p $.
I want to understand why the gauge pressure $ p $ in this case is the sum of major and minor losses. I know the formulas, I can do the calculations, but I really want to understand the principles.
Assume steady flow.
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.
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.
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.
Yes. This is an interesting situation. As you said, the change in enthalpy per unit mass of the gas passing through the pipe is zero, and so, for an ideal gas, this means that the temperature change is also zero. But the paradox here is that, even though there is viscous heating, the temperature change is zero. So what is the solution to the paradox? Well mechanistically, there are two things happening: (1) the gas is experiencing viscous heating and (2) the gas is experiencing expansion cooling, caused by the work which each expanding parcel of gas does on its neighbors. For an ideal gas, these two effects exactly cancel one another, so that, even though the flow is adiabatic, it is also isothermal.
Best Answer
By convention, gauge pressure references atmospheric pressure- for "gauge pressure p in this case is the sum of major and minor losses" to be valid, the duct must discharge to atmosphere. This requires that your Minor losses include the discharge loss, where the duct delivers/returns air to atmospheric conditions (and all kinetic energy/dynamic pressure is lost),where $v_{discharge}=0$, $P_{discharge}=P_{atm}$, and $T_{discharge}=T_{atm}$.
The fan increases the pressure above atmospheric pressure enough to induce a desired flow rate. The manometer measures the difference in pressure between two points- in this case: $P_{manometer} - P_{discharge} = \Delta P$, where $P_{manometer} \approx P_{fan}$.
The pressure difference, $\Delta P$ is a physical measure of flow potential between points. For your problem statement to hold, all of the pressure differential is lost. These pressure losses are due to Major (friction) losses and Minor (geometry restriction) losses. Without knowing other factors like duct geometry, roughness, velocity, etc. you cannot determine what the Major and Minor losses are, you can only state:
$$P_{gauge}=\Delta P=Major Loss+Minor Loss$$
Assuming uniform roughness and restrictions throughout the duct, moving the manometer closer to the discharge of the duct results in lower gauge pressures.
(Forgive the pitiful diagram!)
Perhaps a more intuitive way to envision your question: If you wanted to determine losses in a given section of duct, you could place two manometers at chosen locations. Solving for $P_{manometer}$ at both manometers, you can calculate $\Delta P$ between manometers, again, the sum of Major and Minor losses in that section.