Bernoulli equation with losses dilemma

  • #1
williamcarter
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4

Homework Statement


Hello,Could you please lend me a hand with this problem?I would really appreciate it.
Question:[/B]
Capture.JPG


Their Answer:
Capture2.JPG


KL=2*0.75(2 elbows) D=5*10^-2 m ;f=0.001(fanning friction factor)
Assumptions I made:
Point 1 which is at the top of the liquid in the tank:
h1=23 m; u1=0m/s(neglected due to Area differences) P1=0 gauge(atmos press)

Point 2 which is at the mouthpiece of fountain:
h2=0m(datum); u2=? P2=0 gauge(atmos press)

Homework Equations


We know bernoulli for losses with fanning friction factor
P1/ro*g+h1+u1^2/2*g=P2/ro*g+h2+u2^2/2*g +delta hloss

where delta hloss=K*u^2/2g
where K=f*4*L/D+Ki
where Ki=sum of K in our case 2*0.75(2 elbows)
where f is fanning friction factor.

The Attempt at a Solution


P1/ro*g+h1+u1^2/2*g=P2/ro*g+h2+u2^2/2*g +delta hloss
As we made the assumption in first step.This can be rewritten as:
h1=delta h loss
=>h1=K*u^2/2*g
h1=(f*4*L/D+Ki)*u^2/2g
=>u=sqrt(h1*2g)/(f*4*L/D+Ki)
and substituting values in this formula gives u=10.05 m/s
However their answer is 9.0 m/s

I do not understand
1) why they did Ki=0.5+2*0.75, from where comes that "0.5"?
2)From where comes that "1+" in their sqrt formula
2)Why they picked L=20+10+2

Thank you in advance , much appreciated.
 
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  • #2
In my judgment, all three items that you cited are errors that they made.
 
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  • #3
Chestermiller said:
In my judgment, all three items that you cited are errors that they made.
Thank you very much for your quick response.
I would be more than glad if you could let me know how was the correct approach.
 
  • #4
The 0.5 should be a 1, and the 1 shouldn't be there. I was wrong about the L. The total length of tubing is 20 + 10 (horizontal) + 2 (vertical).
 
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  • #5
Chestermiller said:
The 0.5 should be a 1, and the 1 shouldn't be there. I was wrong about the L. The total length of tubing is 20 + 10 (horizontal) + 2 (vertical).

Thank you, could you please tell me why we need to add 20+10+2.
Why is it needed to add 20+10+2? And why we can't take in consideration just the height?
Why is it needed to do like that?
I mean why we cannot put just the height ,20 m?or 25m?
 
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  • #6
williamcarter said:
Thank you, could you please tell me why we need to add 20+10+2.
Why is it needed to add 20+10+2? And why we can't take in consideration just the height?
Why is it needed to do like that?
I mean why we cannot put just the height ,20 m?or 25m?
The total length of tubing is 32 meters. This is the length that experiences a frictional pressure drop.
 
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  • #7
Chestermiller said:
The total length of tubing is 32 meters. This is the length that experiences a frictional pressure drop.
Ok I understood this, thank you.But how do you know that?
 
  • #8
williamcarter said:
Ok I understood this, thank you.But how do you know that?
Waddya mean? How long is the tubing?
 
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  • #9
Chestermiller said:
Waddya mean? How long is the tubing?
The total length of tubing is 32 meters. This is the length that experiences a frictional pressure drop.

How do you know that?How did you identified that?What was the criteria to get to that value?What was the mechanism behind it?
 
  • #10
williamcarter said:
How do you know that?How did you identified that?What was the criteria to get to that value?What was the mechanism behind it?
If fluid flows in one end of a pipe and comes out the other end, doesn't it have to travel through the entire length of pipe to get from one end to the other? Isn't there viscous wall friction over the entire length?
 
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  • #11
Chestermiller said:
If fluid flows in one end of a pipe and comes out the other end, doesn't it have to travel through the entire length of pipe to get from one end to the other? Isn't there viscous wall friction over the entire length?
Yes,it is, thank you, now is clear, much appreciated.
 
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