Pressure drop across an orifice (Orifice pressure drop in meters?)

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SUMMARY

The discussion centers on calculating the pressure drop across an orifice and converting that pressure drop from Pascals (Pa) to meters of water column. The formula used is ΔP = 1000 x 9.81 x (Orifice pressure drop in m), with a given pressure drop of 470.72 Pa. The conversion to meters is achieved using the equation Orifice pressure drop in meters = (Pa)/(ρg), where ρ is the density of air at 312 K (1.1333 kg/m³) and g is the acceleration due to gravity (9.81 m/s²). The calculated orifice pressure drop in meters is 42.34 meters, which raises concerns about its validity due to the nature of compressible gases.

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  • Understanding of fluid mechanics principles, specifically pressure and flow dynamics.
  • Familiarity with the concept of pressure head and its conversion from pressure units.
  • Knowledge of the properties of air, including density at various temperatures.
  • Basic mathematical skills for performing unit conversions and calculations.
NEXT STEPS
  • Study the principles of compressible fluid flow and how they differ from incompressible flow.
  • Learn about the ideal gas law and its application in calculating properties of gases at different temperatures.
  • Research the use of air properties tables for various temperatures and pressures.
  • Explore advanced fluid dynamics topics, such as Bernoulli's equation and its applications in orifice flow calculations.
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scottniblock
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Homework Statement



ΔP = 1000 x 9.81 x (Orifice pressure drop in m)

Pressure drop across orifice = 470.72 Pa


2. Homework Equations



3. The Attempt at a Solution

I am not sure how this works. How can pressure be converted to meters? It does not make sense to me.

Any help would be much appreciated

Thanks
Scott
 
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To convert a pressure (Pa) to a pressure head (m) divide by ρg (N/m^3).
 
Hi,

Thank you for the reply. Still slightly confused.

It is ΔP that I need to find in order to calculate the ideal mass flowrate of air.

Do I already have the answer to the question, ie ΔP = 470.72 Pa ?
 
or do I need to use an Air properties table to find ρ at the temperature I am given?
 
Question

Orifice pressure drop in meters = (Pa (N/m^2 ))/(ρg (N/m^3 )) This gives answer in meters


Question 1

Given:
T = 312 K
ρ = 1.1333 kg/m^3 (From Air properties table)
g = 9.81 m/s^2

Orifice pressure drop = 470.72 Pa

Calculation:

Orifice pressure drop in meters = 470.72 / (9.81x1.1333) = 42.34 meters

This answer does not seem right, looks way too large.
 
Sorry, I thought you were only looking for the conversion and I assumed you were talking about an incompressible fluid not a compressible gas. So I can't help you there.
 

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