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

• scottniblock
In summary, the conversation discusses converting pressure drops into meters, using the formula Orifice pressure drop in meters = (Pa (N/m^2 ))/(ρg (N/m^3 )). The calculation is then demonstrated using the given values and it is noted that the result may seem too large due to the use of water instead of mercury in the formula.
scottniblock

## Homework Statement

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

Pressure drop across orifice = 470.72 Pa

## 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

Last edited:
Might help if you stated the problem clearly.

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.

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.

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,

## 1. What is the definition of pressure drop across an orifice?

The pressure drop across an orifice refers to the decrease in pressure that occurs when a fluid flows through a constriction, such as an orifice plate, in a pipe or channel.

## 2. What causes pressure drop across an orifice?

The pressure drop across an orifice is caused by the increase in fluid velocity as it flows through the constriction. This increase in velocity leads to a decrease in pressure, according to Bernoulli's principle.

## 3. How is pressure drop across an orifice calculated?

The pressure drop across an orifice can be calculated using the Bernoulli equation, which takes into account the fluid density, velocity, and the area of the orifice. Other factors such as the fluid viscosity and the shape of the orifice may also be considered in more complex calculations.

## 4. What factors affect the pressure drop across an orifice?

The pressure drop across an orifice is affected by several factors, including the size and shape of the orifice, the fluid density and viscosity, and the velocity of the fluid. Other factors such as the temperature and the roughness of the orifice surface may also have an impact.

## 5. How can pressure drop across an orifice be minimized?

The pressure drop across an orifice can be minimized by selecting an appropriate orifice size and shape, as well as optimizing the fluid velocity and viscosity. Additionally, reducing the roughness of the orifice surface and ensuring proper installation can also help minimize pressure drop.

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