Master Fluid Problems with Bernoulli's Equations | Expert Homework Help

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The discussion centers on applying Bernoulli's equation to fluid problems, particularly in determining the velocity of liquid exiting a straw. Initial confusion arises regarding the pressure variable (P) and the correct interpretation of heights in the equations. After analyzing the problem, it is clarified that P represents the pressure at specific points rather than the surface pressure. The final solution for the velocity of the liquid is determined to be 1.1 m/s, with the understanding that potential energy is based on a reference point. The participants emphasize the importance of correctly identifying reference points and pressure conditions in fluid dynamics.
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Homework Statement


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I often have much trouble with fluid problems in general. I know what equations are involved but am confused on how to apply them to these problems.

Homework Equations


bern's equations
P + pgh + 0.5pv^2 = constant
P - pressure
p = density
v = velocity
h= height from the surface of fluid?

The Attempt at a Solution


it is kinda unclear to me what velocity I am trying to find. I am thinking the problem wants to find the velocity of the liquid as it comes out of the straw out of the cup?
I'm thinking of comparing the constants at the bottom of the end of the straw where the velocity is unknown but that the pressure is just atm pressure with the end of the straw inside the water (where the pressure is unknown and velocity =0?)

So i get

(bottom of straw out of cup) 101300 Pa + pg(0.06m) + pv^2 = constant

(end of straw inside cup) P + pg(0.01m) + 0 = constant

set the constants equal to each other

101300 Pa + pg(0.06) + pv^2 = P + pg(0.01)

Kinda stuck from each as i have too many unknowns
 

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lc99 said:
(end of straw inside cup) P + pg(0.01m) + 0 = constant
What does P represent there?

Soapbox: In my view the question is flawed.
I believe your approach will yield the expected solution, but really it is the solution to a different problem, namely, what will the steady state speed be if the height of water in the cup is maintained by some inflow?
The Bernoulli equation is for steady state. In the problem as described, the water is still accelerating.
 
haruspex said:
What does P represent there?

Soapbox: In my view the question is flawed.
I believe your approach will yield the expected solution, but really it is the solution to a different problem, namely, what will the steady state speed be if the height of water in the cup is maintained by some inflow?
The Bernoulli equation is for steady state. In the problem as described, the water is still accelerating.
Im thinking P is just the pressure at the surface of the liquid?
 
lc99 said:
Im thinking P is just the pressure at the surface of the liquid?
Right.
 
haruspex said:
Right.
Then since the P is the same for both equations, then i have pg(0.06) + pv^2 = pg(0.01)

I understand now. I keep thinking P would be different for the 2 locations. I always forget it refers to the pressure at the top.
But i think the answer is 0.70 after i solve for v.
 
lc99 said:
i have pg(0.06) + pv^2 = pg(0.01)
Yes, if you are taking g as positive up (hence negative).

Edit: no, it was wrong - see below
 
Last edited:
haruspex said:
Yes, if you are taking g as positive up (hence negative).
0.7 is the incorrect answer though. What am i doing wrong?
 
haruspex said:
Yes, if you are taking g as positive up (hence negative).
I figured this out. The answer is 1.1 m/s.

I think i made a mistake cause i was thinking the P in both equations refer to the pressure on the surface. But it is just the pressure at the particular point. So for the equation (using Bern's equation) at the bottom of the straw inside the cup it should be = P + pgh +.5pv^2 = Patm + pg(0.01) + pg(0.05) + 0.

For the other end of the straw (where the liquid comes out) , i have the equation = P + pgh +0.5pv^2 = Patm +pg(0) + 0.5pv^2.

I was able to set those 2 equal to each other and found that v =1.1 m/s.
Also, i think i understood that pgh is just the PE of the particular point in fluid. the PE is based on a reference point, so i was able to set the h=0 for the end of the straw where the liquid comes out.
 
lc99 said:
I figured this out. The answer is 1.1 m/s.

I think i made a mistake cause i was thinking the P in both equations refer to the pressure on the surface. But it is just the pressure at the particular point. So for the equation (using Bern's equation) at the bottom of the straw inside the cup it should be = P + pgh +.5pv^2 = Patm + pg(0.01) + pg(0.05) + 0.

For the other end of the straw (where the liquid comes out) , i have the equation = P + pgh +0.5pv^2 = Patm +pg(0) + 0.5pv^2.

I was able to set those 2 equal to each other and found that v =1.1 m/s.
Also, i think i understood that pgh is just the PE of the particular point in fluid. the PE is based on a reference point, so i was able to set the h=0 for the end of the straw where the liquid comes out.
Yes, that all looks right.
Took me a while to figure out you were using the exit point of the tube as your height datum.
Note that the 1cm depth of the straw within the vessel was irrelevant.
 
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