Thrust Calculation: 1000 psi Tank, 1 in² Removed

In summary, the conversation discusses the concept of pressure and thrust in a tank with 1000 psi pushing in all directions. It also touches on the use of compressed air and water for generating thrust and the factors that affect the amount of thrust produced. The main question is whether removing one square inch from an end of the tank would result in a 1000 lb push in the opposite direction. The conversation also mentions a test fixture being constructed to monitor thrust and the potential difference between using compressed air and water.
  • #1
bluechipx
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Let's say we have a tank with 1000 psi pushing in all directions, if we remove one square inch from an end, now the tank has 1000 psi more at one end than the other. Does this mean that there is a 1000 lb push in the opposite direction of the removed square inch?
 
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  • #2
bluechipx said:
Let's say we have a tank with 1000 psi pushing in all directions, if we remove one square inch from an end, now the tank has 1000 psi more at one end than the other. Does this mean that there is a 1000 lb push in the opposite direction of the removed square inch?
If the force on that square inch jet of water is indeed 1000pounds force then Newton's Third law tells you that there is 1000pounds reaction force on the tank. The forces must be equal.
PS IS someone pumping water into the tank to keep the internal pressure constant?
 
  • #3
bluechipx said:
Let's say we have a tank with 1000 psi pushing in all directions, if we remove one square inch from an end, now the tank has 1000 psi more at one end than the other. Does this mean that there is a 1000 lb push in the opposite direction of the removed square inch?
Yes, and that's one way of explaining how a rocket generates thrust in a vacuum where "there's nothing to push against". The combustion chamber of a rocket is basically a high-pressure tank with an opening at the back.
 
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  • #4
sophiecentaur said:
If the force on that square inch jet of water is indeed 1000pounds force then Newton's Third law tells you that there is 1000pounds reaction force on the tank. The forces must be equal.
PS IS someone pumping water into the tank to keep the internal pressure constant?

I am making a test fixture that will monitor thrust which is still under construction. I will test both ways, just air and air pushing water. I'll try to post pics soon. The tank is a full size high pressure welding type cylinder rated for 4500 psi with a test pressure of 6750 psi. The ball valve is a two inch with a pressure rating of 5000 psi. With no measuring system, I started out with a mere 250 psi yesterday for the first test, then tried 600 next. I am focusing on snapping the valve open as quickly as possible for the best effect, which was very impressive for the low starting pressure.
 
  • #5
bluechipx said:
, just air and air pushing water
There will be a big practical difference between to two situations. You will need to shift a lot of air, fast, to maintain the sort of pressure compared with doing it with water. But there is a lot of Potential Energy stored in compressed air, compared with the Energy stored in a compressed liquid.
Perhaps you could go into some more detail to help us make the leap between your actual problem and the original question. Was that the 'right' question?
 
  • #6
Although the actual test will be what matters, I was just curious if there was an accurate way of knowing beforehand what to expect. I have a hydraulic fixture with a one square inch piston (1.128") that applies pressure to a 10,000 psi gauge. The gauge has a telltale pointer that records max pressure. The tank will be pressurized to 4500 psi and released instantly though a two inch dia ball valve. I will be testing both ways, air over water and air only. Anybody care to venture a guess as to the actual thrust both ways? The valve has an area of just over three square inches (3.14). Hopefully I may see 3.14X 4500 = 14,130 lbs, but I suspect in reality the figure will be much lower for reasons I am unaware of.
 
  • #7
Which direction of thrust is important for you? Is this for a 'projectile' or a ram of some kind?
There is an important factor which I hinted at earlier. If you actually want some Kinetic Energy out of this device then compressed air is probably the way to go because it 'stores' energy whereas a compressed liquid stores virtually none. Otoh, the lack of stored energy makes things more controllable, if you simply want a ram. It's no coincidence that hydraulics are used for working tools and vehicles.
You mentioned a combination of air and water. That could have its advantages in some specialised applications.
Give us a clue.
bluechipx said:
Anybody care to venture a guess as to the actual thrust both ways?
The initial thrust will be the same as the pressure times the area but things will change once the fluid starts to flow.
 

Related to Thrust Calculation: 1000 psi Tank, 1 in² Removed

What is thrust calculation and why is it important?

Thrust calculation is the process of calculating the amount of force generated by a propulsion system. It is important because it helps determine the performance and efficiency of the system.

How is thrust calculated?

Thrust is calculated by multiplying the mass flow rate of the propellant by its exhaust velocity. In the case of a 1000 psi tank with 1in² removed, additional factors such as pressure and area must also be taken into account.

What factors affect thrust calculation?

The main factors that affect thrust calculation include the mass flow rate of the propellant, the exhaust velocity, the pressure and temperature of the propellant, and the area of the nozzle or thruster.

What is the significance of a 1000 psi tank and 1in² removed in thrust calculation?

The 1000 psi tank and 1in² removed are specific parameters that affect the pressure and area of the system, which in turn impact the thrust calculation. These values must be taken into consideration in order to accurately determine the thrust generated by the system.

Are there any limitations to thrust calculation?

Thrust calculation is based on various assumptions and ideal conditions, and may not accurately reflect real-world performance. Factors such as atmospheric conditions, engine inefficiencies, and other external factors can also affect the actual thrust generated by a propulsion system.

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