Calculating PSI to KW Conversion | Dale's Questions

[daleb]

although this may appear to be a homework question – it is not.

I understand that in order to determine PSI at 100 feet of water, you would divide 100 x 2.31.

42.39 PSI

in order to create that much PSI in an air chamber that is 7.3333 cubic feet; how much power in KW would be required?

I was unable to find an online calculator to do this.

also if you would be so kind could you answer a second question?

How many KW is generated by a 400 KG weight falling 100 feet?

Regards,

Dale

 

[Forensics]

How will you be generating this pressure? A simple oiled air compressor or some other means?
How long will it remain pressurized?
Will it be gaining and losing pressure frequently?

When you say how much power, do you mean kw hours? Will you only need to know the power required to achieve that psi once? Multiple times? Maintain that psi for days?

I believe I can calculate the kw needed if you tell me more about the project or application.

As for the second question I cannot answer.

 

[daleb]

it will be part of a sophisticated pneumatic system.

I need to know how much kWh is required to pressurize it once.

Losing pressure or pressurizing it multiple times is not at issue in this case.

 

[Forensics]

Do you know how many amps this device will draw and at what voltage it operates at?

It's hard to answer the question without knowing roughly the size of the motor that will be pumping air into this chamber. Since it's only doing it once, you don't really need the kWh, just the kW that the motor will use while pressurizing the chamber.

The amount of watts needed to pressurize this air chamber with a small motor would naturally be different than the amount needed to do the job with a large motor.

This is assuming you are using a motor to pump air into the chamber which is what I have experience with.

 

[Baluncore]

Consider a diving bell flooded with water. It is suspended with it's base at a depth of 100ft (which is predicted to be 42.39 PSI). To eliminate pressure gradient complexities I will assume the height of the chamber in the bell is zero. Now you must pump in air to push out 7.3333 cuft of water.

The energy needed to compress the air into the bell is the same as the energy needed to lift 7.3333 cu ft of water 100 ft.

As a potential energy question PE = mass * gravity * height
Start by converting all units to SI. (metre, kilogram, second)
Mass = 7.3333 cu ft = 28.32(litre/cuft) * 7.3333(cuft) = 207.68 litre.
Because one litre of water weighs 1 kg, the mass is 207.68 kg.
Gravity = 9.8 m/sec-2. Height = 0.3048(m/ft) * 100 ft = 30.48 metre.
Therefore PE = 207.68 * 9.8 * 30.48 = 62034.85 joule.
Energy is measured in joules, power in watts.
A power of one watt will transfer or convert one joule per second.
The energy needed is 62034.85 joule which is the same as 62034.85 watt seconds.
62034.85 watt seconds = 62034.85 / 3600 = 17.23 watt hours of energy.

You can apply the same PE equation to your second question.

 

[Forensics]

I was thinking about this the wrong way. Thank you Baluncore.

 

[Baluncore]

Here is another simpler view of your first problem as an industrial hydraulics question;
Energy = pressure * volume / efficiency.

Firstly; the conversion factor between head of water and psi.
1 foot = 12 * 2.54 = 30.48 cm. 1 sq inch = 6.4516 square cm.
1 pound = 0.453592 kg = 453.592 cubic cm of water.
453.592 / 6.4516 = 70.307 cm of water per psi.
70.307 / 30.48 = 2.30665 ft of water per psi.
So 100 ft of water = 43.353 psi, not 42.39 psi.

Then, the energy calculation;
207.68 litres of fluid pumped through a pressure difference of 43.353 psi.
1 psi = 6.89476 kPa, therefore pressure = 6.89476(kPa/psi) * 43.353(psi) = 298.9 kPa
Assuming 100% efficiency, energy = pressure * volume.
Energy = 298900.(Pa) * 0.20768.(m3) = 62075. joule.
This compares well with the 62034.85 joule from the PE calculation.

 

[Borek]

Do you know the difference between power and energy?

Generally speaking question as posted doesn't make physical sense. Baluncore did some calculations (with an extravagant number of digits which mostly don't matter) and found amount of energy, but power required will depend on how fast you want the work done.