a) What is the flow speed in the pipe?
b) Is there a limit on how high the siphon top can rise above the surface of the water and still work? Explain. If so, what is the limit?
Question so kindly provided by Joe Kiskis
To solve this, simply apply Bernoulli's principle:
- P + ρgy + 1/2 ρv2 = C
- P - pressure
- ρ - Density
- g - acceleration due to gravity
- y - height from arbitrary reference point
- v - velocity of the fluid
- C - arbitrary constant
- 105N/m2 + 103 kg/m3*9.8m/s2*0m + 1/2*103 kg/m3*(0m/s)2 = 105N/m2
- We can arbitrarily set the origin for y as the surface of the water, but really any other point would do, just not as easily.
- We can assume the surface of the water isn't moving, since this is a lousy lower division physics class.
- C = 105N/m2
- 105N/m2 + 103 kg/m3*9.8m/s2*(-4m) + 1/2*103 kg/m3*v2 = 105N/m2
- y = -4m since the tube goes up 4m, then down 8m.
- Now just solve for v
- v = 8.85 m/s
There is an upper limit to how high the siphon can operate, since it's the surface air pressure that is pushing the water up the tube. Once the weight of the water exceeds the air pressure, the water can't raise any farther. Above this upper limit, the inside of the tube is a vacuum, so to find this limit, set P=0.
- 0 + ρgy + 0 = C
- Solve for y
- y = C / ρg
- y =
9.75 m10.2 m
Note: I was all kinds of confused on this problem. Thankfully, the professor was kind enough to drop into the classes chat room and point out all my failures in logic.
Now the real question is what happens between 9.75m and 10.2m? On the one hand, the siphon still works as long as the water is moving, but we've already calculated how fast the water has to exit the tube to conserve the lost potential energy of dropping 4m. How can the water slow down at the top of the tube?