Calculating Force from Pressure in Water Rocket Building

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To calculate the force exerted by air pressure in a water rocket, the thrust can be determined using the equation Thrust = mdot * v_exit, where mdot is the mass flow rate and v_exit is the fluid velocity at the nozzle exit. As water is expelled, the gas expands, leading to a decrease in pressure and flow rate, which results in diminishing thrust over time. The weight of the rocket also decreases as water is expelled, affecting acceleration. Unlike gases, liquids like water are generally considered incompressible, meaning temperature remains stable while pressure drops and velocity increases through the nozzle. Understanding these dynamics is crucial for effective water rocket design and performance.
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Hello I am currently building a water rocket. I would like some help on finding the equation for the force that is exerted when I add air pressure to an cylindrical tube filled with water. How can I figure out the force that the water would create when I release it. I hope that someone can help with this problem of mine. I appreciate the help that anyone can offer me.

Thank you,
Dru Jouhnson
 
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For a rocket, the force (thrust) generated is given by:

Thrust = mdot * v_exit

where,

mdot = mass flow rate
v_exit = velocity of fluid at exit of nozzle

For a water rocket pressurized with air, you could figure out the mass flow and velocity from fluid mechanics. As the water is expelled, the gas expands, pressure drops, flow rate drops, and the thrust drops. Thus, the thrust decreases with time. But the weight of the rocket decreases too, so the acceleration may increase or decrease depending on the variables.
 
edgepflow said:
As the water is expelled, the gas expands, pressure drops, flow rate drops, and the thrust drops.

Does the nozzle of the rocket act like the throttle valve in a refrigerator? The pressure of the refrigerant is sharply reduced by the throttle valve which also reduces the temperature of the refrigerant.

Jagella
 
Not exactly. Since the water is basically incompressible, the temperature will largely be unaffected by the water's travel through the nozzle. The pressure will drop dramatically with a corresponding increase in velocity though (as a rough estimate, you could assume that it is an ideal nozzle, and thus the water's exit velocity is equal to sqrt(2P/rho)).
 
cjl said:
Not exactly. Since the water is basically incompressible, the temperature will largely be unaffected by the water's travel through the nozzle. The pressure will drop dramatically with a corresponding increase in velocity though (as a rough estimate, you could assume that it is an ideal nozzle, and thus the water's exit velocity is equal to sqrt(2P/rho)).

Are liquids typically incompressible and gases typically compressible?

Jagella
 
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