- #1
JulienB
- 408
- 12
Hi everybody! I'm training for an upcoming exam, and I'd like to know if I correctly use the Bernoulli equation.
The problem is kind of difficult to describe without a drawing, please check out the attached jpg to see the situation and the known data.
The questions are:
a) How much time t is needed for a volume V = 1L of fluid to cross the hopper?
b) What is the diameter of d2?
Bernoulli equation, equation of continuity, formulas to calculate the flux of a fluid
a) I use the Bernoulli equation at points (1) and (0):
P1 + ρgh1 + ½ρv12 = P0 + ρgh0 + ½ρv02
Now I think that P1 = 0 (if we take the gauge pressure), ½ρv12 ≈ 0 (because the diameter at point (1) is much larger than at point (0) (equation of continuity)) and ρgh0 = 0 (because I choose h0 = 0).
What I am not sure of is that I assume P0 = 0 because the fluid is not being stopped at that point, so it should have no pressure. Is that correct? If so, here is where that takes me:
ρgh1 = ½ρv02
⇔ v0 = √(2gh1)
We can then use the flux to determine the time it takes for a litre of fluid to cross the hopper:
R = V/t = A0 ⋅ v0
⇒ t = V/(¼ ⋅ π ⋅ d02 ⋅ √(2⋅g⋅h1)) = 23.5 s
b) I use the Bernoulli equation again at points (0) and (2).
P0 + ρgh0 + ½ρv02 = P2 + ρgh2 + ½ρv22
I assume P0 = 0 and P2 = 0 for the same reason as before (is that right?) and ρgh2 = 0 (because I now choose h2 = 0)
⇒ v2 = √(v02 + 2⋅g⋅h0)
The equation of continuity says A0 ⋅ v0 = A1 ⋅ v1
⇒ A2 = A0 ⋅ v0/v2
⇒ r = 1.5 mm
⇒ d2 = 3 mmMy results are plausible, but I am unsure about how to deal with the pressure when there is a flow. Is there no pressure on the right side of the equation when there is a flow? Is that always the case?Thank you very much in advance for your answers.
Homework Statement
The problem is kind of difficult to describe without a drawing, please check out the attached jpg to see the situation and the known data.
The questions are:
a) How much time t is needed for a volume V = 1L of fluid to cross the hopper?
b) What is the diameter of d2?
Homework Equations
Bernoulli equation, equation of continuity, formulas to calculate the flux of a fluid
The Attempt at a Solution
a) I use the Bernoulli equation at points (1) and (0):
P1 + ρgh1 + ½ρv12 = P0 + ρgh0 + ½ρv02
Now I think that P1 = 0 (if we take the gauge pressure), ½ρv12 ≈ 0 (because the diameter at point (1) is much larger than at point (0) (equation of continuity)) and ρgh0 = 0 (because I choose h0 = 0).
What I am not sure of is that I assume P0 = 0 because the fluid is not being stopped at that point, so it should have no pressure. Is that correct? If so, here is where that takes me:
ρgh1 = ½ρv02
⇔ v0 = √(2gh1)
We can then use the flux to determine the time it takes for a litre of fluid to cross the hopper:
R = V/t = A0 ⋅ v0
⇒ t = V/(¼ ⋅ π ⋅ d02 ⋅ √(2⋅g⋅h1)) = 23.5 s
b) I use the Bernoulli equation again at points (0) and (2).
P0 + ρgh0 + ½ρv02 = P2 + ρgh2 + ½ρv22
I assume P0 = 0 and P2 = 0 for the same reason as before (is that right?) and ρgh2 = 0 (because I now choose h2 = 0)
⇒ v2 = √(v02 + 2⋅g⋅h0)
The equation of continuity says A0 ⋅ v0 = A1 ⋅ v1
⇒ A2 = A0 ⋅ v0/v2
⇒ r = 1.5 mm
⇒ d2 = 3 mmMy results are plausible, but I am unsure about how to deal with the pressure when there is a flow. Is there no pressure on the right side of the equation when there is a flow? Is that always the case?Thank you very much in advance for your answers.