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

[scottniblock]

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

 

[paisiello2]

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

 

[scottniblock]

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 ?

 

[scottniblock]

or do I need to use an Air properties table to find ρ at the temperature I am given?

 

[scottniblock]

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.

 

[paisiello2]

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.