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Hot answers tagged compressible-flow

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Two factors contribute to the incompressibility of liquids. The first is the electrostatic repulsion between the electron fields of neighboring molecules keeping the molecules from getting closer to each other without increasing force. The second is that intermolecular spacing is similar to that of solids made from the same material, due to intermolecular ...

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The short answer is that they don't. Choking is not a bad thing. For a gas flow regulator, it's actually a good thing, because under choking conditions, the mass-flow rate is no longer dependent on the downstream pressure. The mass-flow rate is still dependent on the upstream pressure and the size of the aperture connecting the high and low-pressure sides. ...

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Constant entropy doesn't mean constant temperature, the gas flow through nozzle is subject to change in pressure and velocity (continuity equation) due to the variation in cross section, and being compressible the change in pressure will be reflected as change in temperature and density. That being said, since the flow is isentropic and there will be no ...

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How do I explain the fact that the flow chokes only at the nozzle? To explain why the flow only chokes at a single point, it is important to remember that for a system at steady state, there can only be a single flow. All of the fluid entering the system must leave. Since there can only be a single flow rate through the system, you must then consider ...

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Compressors are a combination of rotors and stators along a structure with reducing cross-section area. The rotors spin and push air down the compressor, but due to their spin there arises a circumferential velocity component to the air flow. The stators don't revolve (hence the name) and they "straighten out" the flow, so as to direct the airflow ...

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You can probably assume the natural gas is an ideal gas, then apply the corresponding equation of state on any arbitrary control volume of the flow: $$pV=mRT$$ where $p$ is pressure, $V$ is volume, $m$ is mass, $R$ is the specific gas constant, and $T$ is temperature. $R$ for methane is $518.3 \frac{J}{kg \cdot K}$, which might be close enough. You ...

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You are correct that the Bernoulli equation will not work here because you're in the compressible flow regime. As you already learned, the flow through the orifice will become sonic when the pressure in the bottle exceeds about 30 psi. What you have here is probably mostly adiabatic expansion of the air in the bottle (no heat transfer), however the problem ...

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The text book answer is correct. I think you missed to calculate the density of the downstream of the shock. the equation you entered is valid only in the region 1 or in region 2. means the isentropic relations are not valid across the shock. Across the shock you have to use normal shock relations. Then you would get the right answer

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To reiterate my edit and close the question: Looking at the link suggested and a few textbooks, I found Modern Compressible Flow: with Historical Perspective by John D. Anderson explained conical flow very clearly. To summarise, the ODE to solve is the following: Finding the solution of the flow involves being given either the cone angle and surface Mach ...

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As far as I understand, especially assuming steady conditions, your mass flow rate is held constant. The thing that is normally changing is the mass flow rate per unit area, or mass flux. Knowing that maximum mass flux occurs under stagnation property conditions, you can actually use simple relationships like Fleigner's formula to solve for your mass flux ...

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These are good questions to ask yourself. The most succinct answer to your questions is: Mass Balance and Energy Balance. Wall friction will actually cause a subsonic flow of gas through a pipe to accelerate up to M = 1 (this is still counter-intuitive to me). Gas bulk velocity increases but mass flow rate remains the same. In a gas transmission ...

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Flow rate of the gas into the cylinder will change according to the pressure drop. So first you have to model flow rate through the pipe to the cylinder, which you can probably model as a plenum. I would assume this is a differential equation version of head loss flow. So as the mass of air enters the cylinder you have the pressure increase in the ...

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In, compressible flow, the mass flow rate remains constant throughout the flow. But volume flow rate is changed according to the local velocity and area of cross-section. So one can calculate the volume flow rate locally by diving the mass flow rate by local density. Let say, If you need volume flow rate at the exit, At the exit, the pressure is 1 bar and ...

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Fluids includes gas, liquids and some solids which flows, as you know gas is compressible, in some case fluid flow is considered as compressible like in the case of a rocket design, aircraft etc...

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Use flow (which is a product of pressure). If a single solid will be entrained at the flow rate then it will plug. If the bed has a packing factor of 50% then flow velocity in the bed is twice pipe flow velocity. As the bed fluidizes the packing factor goes down as does flow velocity. VelocityPacking = VelocityPipe / (1 - packing) The smaller ...

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I think not really. You determine power by your choice of motor size, or if you use an FI, rpm. Motor power is propoertional to the third power of rpm (All else equal). So you really don't have value pairs of power and speed to work with. Consider thie field of PV curves: The solid curves correspond to a specific speed each, the dashed to an (approximate!)...

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It is not entirely clear from the question, where the instrument readings take place. The following equations might need to be adapted in order to reflect the sensor positions. The basic equation for the mass-flow $\dot{m}$ is: $\dot{m} = A \cdot \rho \cdot v$ $A$, $\rho$, and $v$ need to be evaluated or measured at the same location. The ideal gas law ...

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I believe that the Borda-Carnot equation is only valid for incompressible flows, while you mention that the injector orifice is choked. While the flow becomes incompressible pretty quickly after the orifice, at this precise point, it is not. In this case, I think you should use 1D isentropic nozzle theory (wikipedia offers a good start on the topic). If ...

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