# What is the pressure in Kilopascals on a vertical plate

• ian richards
In summary, we have discussed the pressure and force exerted by a 42m/s wind on a 1400 x 700 MM glass plate at a height of 6 metres above the ground. The pressure is estimated to be around 0.11 kilopascals, taking into consideration factors such as air density and wind velocity. We have also mentioned the Betz limit, which could provide a more accurate calculation of the pressure. The wind speed mentioned is equivalent to a Category 1 hurricane or typhoon on the Saffir-Simpson wind scale.
ian richards
The plate (glass) is 1400 x 700 MM at right angles to the wind. The wind velocity is 42 Metres per second.

The height above the ground is 6 metres.

Your help to an 85yo would be enormously appreciated.

Ian Richards.

Hi Ian...
I was thinking that :
Pressure=Force /area
If we assume that the air particles collide elastically with the plate then for a small air particle of mass dm the momentum change will be:
$$2 (dm)v$$
Where ##v## is the velocity of air.
We know that force =change in momentum /change in time.
Then we get $$F=2 v (dm/dt)$$
And mass =volume×density=##V×\rho##
And volume here=area of plate × small length traveled by the particle=A×dl
So $$F=2v\rho A (dl/dt)$$
And we can take ##dl/dt=velocity=v##
So $$F=2\rho Av^2$$
And thus $$P=2\rho v^2$$
[Thats what I think, an assumption can also be taken that the air particle stops after collision ]

Interesting derivation, but because of your elastic collision assumption, you ended up at 4x the real world worst case. If the fluid flow stops entirely, all of its kinetic energy will be used to increase the pressure, which results in a pressure increase (compared to ambient) of ##1/2\rho v^2##. This is called the stagnation pressure, and is the worst case scenario.

Out of curiosity, where on Earth are you that has a 42 m/s wind speed?

billy_joule
cjl said:
Interesting derivation, but because of your elastic collision assumption, you ended up at 4x the real world worst case. If the fluid flow stops entirely, all of its kinetic energy will be used to increase the pressure, which results in a pressure increase (compared to ambient) of ##1/2\rho v^2##. This is called the stagnation pressure, and is the worst case scenario.

I realize what you meant here but I figured I would clarify for everyone else: the stagnation pressure is the pressure that occurs if that entire ##\rho v^2/2## is converted into pressure energy. The ##\rho v^2/2## term itself is commonly referred to as dynamic pressure.

cjl said:
Out of curiosity, where on Earth are you that has a 42 m/s wind speed?

I suppose he never explicitly said Earth... hmm.

True enough boneh3ad, and thanks for the clarification. I was ignoring the ambient pressure, but that is a good point to make.

42m/s ~= 151km/hr or about 93 mi/hr - hurricane force winds. I know there are geographically based building codes for areas subject to hurricane force winds. Don't know if they are applicable to large glass sheets. I would guess shuttering would be the only solution, if there is one.

cjl said:
Interesting derivation, but because of your elastic collision assumption, you ended up at 4x the real world worst case. If the fluid flow stops entirely, all of its kinetic energy will be used to increase the pressure, which results in a pressure increase (compared to ambient) of ##1/2\rho v^2##. This is called the stagnation pressure, and is the worst case scenario.

Out of curiosity, where on Earth are you that has a 42 m/s wind speed?
Sorry... but I didn't understand how will I get the ##1/2## part. If I assume that the particle will collide and stop then it will cause stagnation pressure. Therefore the change in momentum will be ##(dm)v## for the small particle. Which will give a ##1##. How will I get that ##1/2##?

jim mcnamara said:
42m/s ~= 151km/hr or about 93 mi/hr - hurricane force winds. I know there are geographically based building codes for areas subject to hurricane force winds. Don't know if they are applicable to large glass sheets. I would guess shuttering would be the only solution, if there is one.

To CJL - Where am I? I am 6 metres above the level ground at Sydney Australia's Mascot Airport on two separate days with the strongest winds recorded there since 1929 - 152 Kilometres per hr

As I understand it the Pressure in Kilopascals = p (which is ? - the figure and the terminology?) ) x v (velocity in m/s = 42)squared all divided by 2 == ? kps

Can you pse fill in the figures to give a reading for the Pressure in Kilopascals.

Is this figure a maximum? -an average ? - a?

I cannot begin to express my gratitude for your interest and help. Ian Richards.

It's ##\rho## the density of air.

So pressure = 1. 274×(42)2/2=1123. 668 Pascal =0. 112 kpa...
that's the value I am getting... And I ain't sure about it.

For identical Newtonian fluid the force to the plate must be 0. I think it is wrong to use Newtonian lows for this calculations because air viscosity. For example, there is not only the front superpressure but and the back subpressure. Without calculations, I think that is safe to take as energy density the double amount $$\Delta{p} = \frac{F}{A} = \rho v_0^2 \Leftrightarrow F=A\rho v_0^2$$

mooncrater

Air density ρ = 1.225kg/m^3 at ground level on average
Area of window A = 1.4m x .7m = .98m^2
Wind velocity v = 42m/s
We need something like a Betz limit factor,
... but I do not know enough to guess, so how about 1?
This number is usually less than one AFAIK.

So Force in Newtons = (1/2) * 1 * 1 (Betz #) * 1.225 * .98 * (42)^2

F ~= (1.225 * .98 * 1764) / 2 == 1058.8 rounded > 1059 Newtons ( 1 Newton - 1 Pa )
This is not too far from mooncrater's estimate. .11 Kilopascals is my sort-of answer.

Once somebody can provide a limit factor you can get a really decent answer.
Then multiply .11 times the Betz limit (or whatever it is called in Engineering) to get the true value note:
42m/s wind velocity is the upper limit for a Saffir-Simpson 1 - Hurricane (typhoon)

See: http://en.wikipedia.org/wiki/Saffir–Simpson_hurricane_wind_scale
This is the scale for estimating destruction of a storm in the North Atlantic. Scale that applies to Australia is RMSC Tokyo - and would be "Typhoon". Correct?

Last edited:

## 1. What is pressure?

Pressure is defined as the amount of force applied per unit area. In the case of a vertical plate, it is the force exerted on the plate divided by its surface area.

## 2. How is pressure measured?

The standard unit for pressure is the Pascal (Pa), which is equivalent to one Newton per square meter. In the case of a vertical plate, pressure can also be measured in Kilopascals (kPa), where 1 kPa equals 1000 Pa.

## 3. What is the formula for calculating pressure on a vertical plate?

The formula for pressure on a vertical plate is P = F/A, where P is pressure, F is the force exerted on the plate, and A is the surface area of the plate. This formula is based on the definition of pressure as force per unit area.

## 4. Is the pressure the same at all points on a vertical plate?

No, the pressure on a vertical plate may vary at different points depending on the distribution of force and the surface area at each point. For example, the pressure at the top of the plate may be different from the pressure at the bottom.

## 5. How does the pressure on a vertical plate change with depth?

The pressure on a vertical plate increases with depth due to the added weight of the fluid above it. This is known as hydrostatic pressure and follows the equation P = ρgh, where P is pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth of the fluid.

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