Pressure drop across an orifice (orific drop in meters?)

[scottniblock]

Homework Statement

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

Pressure drop across orifice = 470.72 Pa

Homework Equations

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

 

[rude man]

Might help if you stated the problem clearly.

 

[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.

 

[SteamKing]

Pressure drops can be expressed as a head of a fluid, often water, but others like mercury can be used as well. The head would be measured in meters of fluid, or some other unit of length. That's what the reading on a barometer is, after all. The reading of 760 mm is the height of a column of mercury supported by the difference in pressure between a vacuum and atmospheric pressure. In working with modest pressure drops, water is used in place of mercury. A pressure drop of 1 meter of water is equivalent to 1000 kg/m^3 * 9.81 m/s^2 = 9810 pascals.

 

[HalfLostAndSearching]

Dear Scott,

Thank you for your question. The equation you have provided is the Bernoulli's equation, which relates the pressure drop across an orifice to the change in height of the fluid. The term "meters" in this equation refers to the unit of measurement for the pressure drop, which is usually expressed in units of meters of fluid column (mwc) or meters of water (mH2O). This is a common unit of measurement in fluid mechanics, and it represents the height of a column of fluid that would produce the same pressure as the pressure drop across the orifice.

To clarify, the equation should be read as follows:

ΔP = 1000 x 9.81 x (Orifice pressure drop in meters of fluid column)

Therefore, the pressure drop across the orifice in meters of fluid column would be:

ΔP = 1000 x 9.81 x (470.72 mwc) = 4,620,913.92 Pa

I hope this helps to clarify the concept. If you have any further questions, please don't hesitate to ask.

Best regards,