Energy needed to inflate approx 0.1m^3 under water

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To inflate a rubber cylinder with a volume of 0.1m³ at a depth of 5cm underwater, the energy required is similar to that needed to push an inflated balloon down to the same depth. The pressure exerted by the surrounding water must be calculated to determine the work done against it during inflation. Air compressor efficiency ranges from 65% to 85%, which impacts the total energy consumption for the task. The energy needed will also include considerations for stretching the rubber of the cylinder. Understanding these factors will help in calculating the total energy required for inflation.
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How much energy is required to inflate a volume of 102,102cm3 with ambient air at 5cm below the surface of a water tank. The surface of the water tank is 8 meters above sea level.

The object to inflate is a rubber cylinder which has a diameter of 50cm and 52cm in length with 0 resistance from the cylinders material. When fully inflated, I have calculated the surface area of the cylinder to be 12,106cm2.

The temperatures are fixed; the water is 12 degrees and ambient air temperature is 20 degrees Celsius.

The target is to use method of inflation e.g. an air compressor that uses the lowest amount of energy to inflate within 5 minutes.

Any help will be much appreciated.
 
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It will be much the same as that needed to push an already inflated balloon down that far.
What is the purpose here?
 
Thanks for your reply Simon - its a homework question.

My thoughts were the same i.e. an air compressor would require at least the same amount of energy as manually pushing down. I am wanting to understand how efficient air compressors are and try to calculate the amount of energy needed. Any idea how this can be solved?
 
You ask the manufacturer how efficient their air compressors are.
 
Thanks very much for your replies :-)
 
I am a bad engineer but let me give it a try :D
I couldn't understand the question properly but what i assume is that initially the rubber cylinder is flat and immersed in water, if it is immersed horizontally in water the total pressure is P=ro*g*H, now when you inflate the cylinder and keeping the upper surface 5cm below water level, assuming zero resistance from the material you are actually doing work against the pressure developed by the surrounding water, this work would be same as the energy needed for inflation.
this was my understanding of the problem, i am 50 % sure i would be wrong because i am a bad engineer but let's have a discussion lol
 
Hi Ali - your spot on. Btw what does the 'ro' part stand for in the P=ro*g*H?
Thanks for replying :-)
 
ro is the density
sorry i missed it lol
 
  • #10
I agree with Ali's method, if you can calculate the average pressure your balloon will be subject to underwater, and then treating it as stored energy in a pressure vessel could be a useful method. If the balloon is stretching, you will also have to take into account the energy required to stretch the rubber of the balloon as well.

Some information regarding isothermal pressure vessel energy storage can be found here: https://en.wikipedia.org/wiki/Compressed_air_energy_storage#Isothermal_Storage
 
  • #11
Since this is a homework question, let's let the OP do the work from here on... :smile:
andyg007 said:
Thanks for your reply Simon - its a homework question.
 
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