Can a Wide Angle in a Diffuser Cause Total Pressure Loss and Affect Gas Flow?

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SUMMARY

The discussion focuses on the impact of wide angles in a diffuser on gas flow and total pressure loss. Key equations include the Steady Flow Energy equation and the ideal gas law, specifically P * V = m R T, which relates pressure, volume, mass, and temperature. The conversation highlights the importance of considering compressible flow and the effects of area expansion on velocity and pressure changes. It concludes that an excessively wide angle can lead to flow separation and a loss of isentropic conditions, invalidating Bernoulli's principle.

PREREQUISITES
  • Understanding of the Steady Flow Energy equation
  • Familiarity with the ideal gas law (P * V = m R T)
  • Knowledge of compressible flow dynamics
  • Basic principles of thermodynamics and fluid mechanics
NEXT STEPS
  • Research the application of Bernoulli's equation in compressible flow scenarios
  • Study the effects of area expansion on gas velocity and pressure in diffusers
  • Explore isentropic processes and their implications in gas dynamics
  • Learn about flow separation and its impact on total pressure loss in fluid systems
USEFUL FOR

Engineers, fluid dynamics researchers, and students studying thermodynamics and gas flow mechanics will benefit from this discussion.

airrocket
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Change in gas velocity and pressure entering and exiting a duct or pipe expansion section (diffuser). Looking for a formula to determine velocity, pressure and density of a gas as it exits a duct or pipe diffuser section based on the entrance area to exit area ratio. Ambient temperature, subsonic, subcritical flow.
 
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Im pretty sure this has to do with the Steady Flow Energy equation from thermodynamics.

This equation can equate velocities, pressures, mass flow rates, heat input/output and work input/output, enthalpy's and physical changes in height.

for enthalpy use,
h = c(p) dt

For the density of the gas, assuming it is a perfect gas use
P(pressure)*Velocity=m(mass)*R*T(temp in kelvin)

where R=(Universal Gas constant = 8.315)/M ... (M is the molecular mass of the gas)
keep in mind the units of universal gas constant its in "k" or thousand

good luck
 
Sorry I've made a mistake ..

that second equation is meant to be

P * V(volume) = m R T

note that m /v is density so P/RT = density

hope that helps
 
Thanks for the tips and equations. Yes I believe that maybe correct also related to Bernoulli's "conservation energy equation only modified for compressible flow. Now I have to figure out how to apply them to solve for the ratio of area expansion to change in velocity or pressure. Gravity does not appear to be a factor so I can ignore it. Once I get one the rest should all follow.
 
airrocket said:
Thanks for the tips and equations. Yes I believe that maybe correct also related to Bernoulli's "conservation energy equation only modified for compressible flow. Now I have to figure out how to apply them to solve for the ratio of area expansion to change in velocity or pressure. Gravity does not appear to be a factor so I can ignore it. Once I get one the rest should all follow.

Gravity can only be ignored if u have a single datum line. This means that the mass flow rate coming into the diffuser is leaving it at the same physical level(height) with respect to a reference plane. If the datum line changes(ie. change in distance from reference plane to the centroid of the diffuser inlet/outlet) between input and output, gravity does become a factor in the steady flow energy equation.
 
Expansion of an incompressible fluid through a diffuser can be modeled as Ksum mentions, simply using Bernoulli’s equation (less the ideal gas equation). However, for a gas, it’s not quite so simple. The gas is expanding isentropically, so you have to account for this. See attached.
 

Attachments

Yes, for now the flow is subsonic yet above M.3 so it is compressible. In the future it maybe supersonic in which case as I see diffusing/converging have opposite effects. Sounds simple yet gets very complex. Appreciate the assistance very helpful seems to confirm what I was thinking.
 
If the angle is to wide the flow will separate and you will have a total pressure loss hence the flow is not isentropic anymore and bernoulli does not nessesarily aply anymore.
 

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