- #1
RemingtonSteele
- 2
- 0
Sorry for the noobie question, but I've got myself in a mental bind trying to figure out static hydraulics and why things happen the way they do.
Could somebody give me an equation that shows me why the fluid level on the outside of the straw in my drink is equal to the fluid level on the inside of my drink.
For ease of math and my understanding, let's assume:
1. The air pressure is 16 psi
2. The area of water exposed to air on the inside of the straw is 1 sq in (it's a big straw )
3. The area of water exposed to air on the outside of the straw is 10 sq in (it's a huge glass )
For some strange reason, I keep thinking that with 160 total pounds pushing on the water in the glass, any straw would produce a nice fountain affect at rest. How do I mathmatically show that the backpressure in the straw is 16 psi and is in equilibrium with the pressure of the water outside the straw?
Thanks in advance.
Could somebody give me an equation that shows me why the fluid level on the outside of the straw in my drink is equal to the fluid level on the inside of my drink.
For ease of math and my understanding, let's assume:
1. The air pressure is 16 psi
2. The area of water exposed to air on the inside of the straw is 1 sq in (it's a big straw )
3. The area of water exposed to air on the outside of the straw is 10 sq in (it's a huge glass )
For some strange reason, I keep thinking that with 160 total pounds pushing on the water in the glass, any straw would produce a nice fountain affect at rest. How do I mathmatically show that the backpressure in the straw is 16 psi and is in equilibrium with the pressure of the water outside the straw?
Thanks in advance.