# Homework Help: Pressure in a bent pipe

1. Apr 6, 2016

### foo9008

1. The problem statement, all variables and given/known data
the water flow in the pipe in upward direction . the pipe with internal diameter 60mm carries water with velocity = 3m/s under head of 30cm . Determine the resultant force on the bend of 70 degree and its direction.

2. Relevant equations

3. The attempt at a solution
my question is what does the head refer to ? from where to where ? red pen refer to from datum to the inlet , blue pen means from inlet to outlet. which is correct , blue or red?

second is is P1 = P2 ????? ( assuming no friction loss)

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2. Apr 6, 2016

### BvU

OK, for the head you can take the center of the flow lines, so where you wrote the 'circumcircled' F1 and F2

And the pressures are not identical: $\Delta p = \rho g \Delta h$ (Archimedes, Bernoulli)

 Oops, I suppose 'head of 30 cm' refers to the pressure at F1. So at F1 the pressure is $p = \rho g \times 0.3 {\rm \ m}$

3. Apr 6, 2016

### foo9008

if inlet diameter = outlet diameter , the pressure at both inlet and outlet also not the same ? assuming no friction loss

4. Apr 6, 2016

### foo9008

do you mean the red line is correct?

5. Apr 6, 2016

### Staff: Mentor

The head H is defined as $H=\frac{p}{\rho g}+z$, where z is the elevation above a fixed datum. In this case, the datum for z = 0 is to be taken as the elevation of the pipe inlet (red location). So, at this location, H = 0.3 m. At the outlet, the pressure may be different and, of course, the elevation is different. This datum applies to the entire pipe and bend.

The head H can be interpreted physically as the height to which fluid in a static column would rise above the datum elevation (if the column were attached to the pipe) at any given location along the pipe.

6. Apr 6, 2016

### foo9008

so , the 0.3 refers to the distance of inlet and outlet ? ( blue line ) ?

7. Apr 6, 2016

### Staff: Mentor

No.
The 0.3m refers to how high water would rise vertically within a tube that is attached to the pipe at its inlet cross section. There is no tube actually attached there, but, if there were, this is how high the fluid would rise.

8. Apr 6, 2016

### foo9008

ok, so the Fx = rho(g)(v2-v1) - P1A1 +P2A2cos(theta) ?
Fx= 1000(0.848)(3cos70 -3) - (0.3)(9.81 x1000)(A1) , but how to find P2 ?

9. Apr 6, 2016

### Staff: Mentor

I'm not checking your equation but, if it is a section of pipe in which there is just a bend in the pipe, the assumption is that the exit cross section area is equal to the inlet cross section area. What does the Bernoulli equation tell you about the relationship between P1 and P2? (The equation you are using to find the force is not the Bernoulli equation; it is the macroscopic momentum balance equation). What does the Bernoulli equation tell you about H1 and H2?

10. Apr 6, 2016

### foo9008

so P 1= P 2?

since bernoullis equation =
P1 / (rho)(g) + z1 + (V1)^2 / 2g = P2 / (rho)(g) + z2 + (V2)^2 / 2g

since Q=Av = constant , cancelling off z and v , P 1= P2 ?

11. Apr 6, 2016

### Staff: Mentor

No. In this problem, z1 and z2 are not equal. The inlet and outlet are at different elevations. If you rewrite Bernoulli's equation in terms of head H, then
$$H_1+\frac{v_1^2}{2g}=H_2+\frac{v_2^2}{2g}$$
What does this tell you about H1 and H2?

12. Apr 6, 2016

### foo9008

Like this? Sorry, I am not in front of computer now

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13. Apr 6, 2016

### Staff: Mentor

14. Apr 6, 2016

### foo9008

H1 - H2 =[ ( (V1 ^2 ) - (V2 ^2) ) / 2g ]
2g(H1 - H2) = (V1 ^2 ) - (V2 ^2)
P1- P2 = (V1 ^2 ) - (V2 ^2) ??
like this ?

15. Apr 6, 2016

### foo9008

i asked my senior , he told me that this is a closed pipe , so the Q =constant , as area for inlet and outlet same velocity = constant (v1=v2) , this act like the hydarulic jack , so the pressure transferred is constant throughout the pipe .... i doubt is the statement correct ? can someone explain ?

16. Apr 7, 2016

### Staff: Mentor

If that's what he said, it's not correct. Suppose that Q = 0. Certainly your analysis should be able to handle this specific case. If Q = 0, the system is in hydrostatic equilibirum.

Multiple Choice Question:
Which of the following answers are correct for a system in hydrostatic equilibrium (like a hydraulic jack)?

(a) the pressure is uniform throughout
(b) the pressure is constant horizontally
(c) the pressure varies vertically
(d) the head is uniform throughout

Chet

17. Apr 7, 2016

### foo9008

for MCQ , my ans is (a) the pressure is uniform throughout
so , the conclusion is P1 are not the same with P2 ?
relationship between P1 and P2 must b e
H1 - H2 =[ ( (V1 ^2 ) - (V2 ^2) ) / 2g ]
2g(H1 - H2) = (V1 ^2 ) - (V2 ^2)
P1- P2 = (V1 ^2 ) - (V2 ^2) ??

18. Apr 7, 2016

### Staff: Mentor

(b), (c), and (d) are all correct. (a) is incorrect.

19. Apr 7, 2016

### foo9008

in hydraulic jack , why the pressure isnt uniform throughout ? the hydarulic jack has the same pressure at both ends , enabling the small force applied to carry big force , right ?

20. Apr 7, 2016

### Staff: Mentor

In a system at hydrostatic equilibrium, does the pressure increase with depth or doesn't it? Why should the pressure within a hydraulic jack be uniform?

21. Apr 7, 2016

### foo9008

i mean the pressure is constant at both ends

22. Apr 7, 2016

### foo9008

maybe i heard it wrongly . i forgt he gt said 'hydraulic jacks' or not , but most importantly , he stressed that , the Q is constant (not equal to 0 ) , so v1 = v2 , leading to P1 = P2 (consequence of bernoulli's principle)

23. Apr 7, 2016

### Staff: Mentor

Only if the two ends are at the same elevation. Otherwise, there is a difference in the pressures. In analyzing a hydraulic jack, we usually neglect the hydrostatic pressure variations in the vertical direction because they are small compared to the applied pressure. In your problem, this is not the case. Otherwise, they wouldn't show the inlet and outlet at different elevations. Of course, at hydrostatic equilibrium, the hydraulic head is independent of vertical location, and is thus uniform.

24. Apr 7, 2016

### Staff: Mentor

Bernoulli's principle does not predict this. You left out the z terms.

25. Apr 7, 2016

### foo9008

ok , so
relationship between P1 and P2 is
H1 - H2 =[ ( (V1 ^2 ) - (V2 ^2) ) / 2g ]
2g(H1 - H2) = (V1 ^2 ) - (V2 ^2)
P1- P2 = (V1 ^2 ) - (V2 ^2) ??
am i correct ?