Now that you point it out, the answer is pretty obvious: Boyle's Law to the rescue!
P1V1= P2V2
96 kPa * 3.79 L = 98.5 kPa * V
V = 96 kPa * 3.79 L / 98.5 kPa
V = 3.69 L
About 100 mL of water would enter the container. I was really overthinking the simplest thing. Thank you, Chestermiller...
I wanted to post this separately so that I don't confuse the question. Here's the short of my experiment. During the day the container warms up. As night passes, the container cools down and sucks up liquid. Using the Ideal Gas Law, one can determine a change in volume from the hottest point...
I'm trying to do the math for a backyard thermodynamic pump experiment, and getting stuck. Suppose an empty 3.79 liter container was airlocked by a vertical 25 cm long empty tube whose end sits at the top of an unlimited supply of water in a basin. The tube protrudes through the container such...
It is my understanding that the force between magnets relates directly to the shape of each magnet. To annul the shape of the magnet, I am seeking a formula that describes the force between two magnets in terms of magnetic flux density.
My Question.
If two permanent magnets faced each...
How's this for a late reply?
If your generator is what I think it is, then the potential of the circuit, E, is found as a function of the angular velocity of the disc, w, when the magnetic flux density is known, B, as well as the working radius of the disc, r by E = 1/2 B r^2 w. Current, I...
That doesn’t sound right either for this reason: the lift to drag ratio you have presented relates to a blade in water where the overall flow of the fluid does not change its direction. As such, the surface area of the blade would be relevant information. However, in nearly all impulse...
I'm not certain that it is: then again, I’m no engineer! Would you happen to know if force exerted upon a body, F, is equivalent to an impulse acting on the body, p, divided by the time in which the impulse has acted, t, or more simply F = p/t? An answer to this question would assist greatly.
Thank you Jeff for the link to Betz’ Law. It seems as though this law only applies to reaction turbines (like Kaplan, Francis, windmills, etc.) and not to impulse turbines (like Pelton and Turgo wheels). As an impulse turbine is not completely submerged like a reaction turbine, it does not...
Thank you Jeff. Your knowledge gave me a eureka moment:
In terms of impulse, the momentum of the water exhausted by a jet will change because its velocity changes after hitting the face of the paddle. This change in momentum is the impulse. If the paddle were traveling at the same...
I am interested in how impulse turbines work. Supposing a mass of water (in kilograms), m, is expelled by a jet at a given velocity (in meters per second), v, its momentum (in Newton-seconds), p, is the product of these quantities: p = mv. Also, a paddlewheel’s angular momentum (in...