# Homework Help: [Hyrdaulics] Bernoulli's Equation w/Pumps

1. Dec 26, 2011

### MarleyDH

1. The problem statement, all variables and given/known data

Water is flowing from left to right through the system shown below. If the reading on
the pressure gauge is 384 kPa, what is the power input from the pump? Ignore energy
losses.

2. Relevant equations

Conservation of Mass (Q = vA)
Manometry (Pa = Pb)

3. The attempt at a solution

I figured that you can use the Pressure from the pressure gauge to get the pressure in the venturi, but beyond that I am lost.

I'm guessing because the diameter [of the pipe] before the pump is the same as the diameter directly after it the velocities will be equal. (v1 = v2)

The pressure before the pump should in my mind be negative as the flow is from left to right. This can also be seen with the manometer on the left hand side, the mercury is higher on the left tube than on the right tube, indicating suction (vacuum).

This however is where I get stuck. I can't see past this point, as I need the velocities to calculate the differential in head so as to calculate the power of the pump.

2. Dec 26, 2011

### nvn

MarleyDH: Hint 1: Can you write a conservation of mass equation between the flow immediately after the pump, and the flow at the 150-mm-diameter venturi meter?

3. Dec 27, 2011

### MarleyDH

@nvn: That would be the standard equation, for example:

Q300 mm = Q150 mm
∴ v300 mmA300 mm = v150 mmA150 mm
∴ v300 mm = v150 mm(D150 mm/D300 mm)2

Edit:

Bah, right figured it out. Thank you.

I get a value of roughly 65.3 kW.

Last edited: Dec 27, 2011
4. Dec 27, 2011

### nvn

MarleyDH: Your final answer currently differs from mine. Would you be able to post your final, power equation, showing the numerical values you used to compute the final answer, so we can track down which quantity differs? There is no need to post the whole solution. Let's first just pinpoint exactly which term differs in only the final equation, thereby showing as little of the solution as possible.

5. Dec 27, 2011

### MarleyDH

Very well.

The flow rate I got was 0.158 m^3/s.
From this, the differential head I got was 42.16m.
The equation for pump power I used was:
Powerpump = ρgΔHQ

6. Dec 27, 2011

### nvn

MarleyDH: Both values currently differ from mine. Could you show how you computed either Q or the energy head gain? The equation for one of these two values might help us trace back toward a discrepancy.

By the way, always leave a space between a numeric value and its following unit symbol. E.g., 42.16 m, not 42.16m.

7. Dec 27, 2011

### MarleyDH

Q was computed by using Q = vA.

My method of approach was:

Find the pressure inside the smaller part of the venturi using manometry and the given information. I then used manometry to get the pressure at the left hand manometer. As the right hand tube is exposed to the atmosphere it eliminates the need for a second equation.

Using Bernoulli's equation I then got the velocity inside the 300 mm pipe to be 2.23 m/s.

Using Q = vA got my flow rate, going over this I think I may have made an error in the Bernoulli's Equation.

8. Dec 27, 2011

### nvn

MarleyDH: Your velocity currently looks close, but slightly incorrect. Verbal descriptions do not help too much. The actual numeric equation you used to compute an answer allows me to check the math, to pinpoint a discrepancy, if you wish. E.g., 2.230 m/s shows a slight discrepancy. Therefore, if you could now post just the numeric equation you used to compute 2.230 m/s, that might help us trace back toward a discrepancy source.

By the way, generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits.

9. Dec 27, 2011

### MarleyDH

Okay.

The subscript 1 pertains to the 300 mm pipe after the pump and the subscript 2 pertains to the 150 mm pipe in the venturi.

v1A1 = v2A2
=> v1 = v2(D2/D1)2

Bernoulli:

H1 = H2
=> z1 + P1/γ + v12/2g = z2 + P2/γ + v22
=> z1 = z2 therefore they cancel.
=> 384 000/9810 + (v2(D2/D1)2)2/2g = 346918.2/9810 + v22/2g
=> v2 = 8.90 m/s

Substituting back:

v1 = 2.23 m/s

10. Dec 27, 2011

### nvn

MarleyDH: 346 918.2 Pa shows a very small discrepancy. I do not know why yet. Generally always maintain (list) at least four significant digits throughout all your intermediate calculations. Please do not round these intermediate answers to three significant digits.

Also, 384 000 Pa is wrong. Try again to obtain the pressure immediately after the pump.

Also, it is hard for me to tell so far, but it appears you might (?) also have the wrong density for mercury.

Is this given problem a school assignment, or a test question?

11. Dec 28, 2011

### MarleyDH

@nvn: The pressure after the pump is given as 384 kPa = 384x103 Pa. Unless I have missed the boat and assumed too much.

Its a tutorial question. I used a specific gravity of 13.6 for mercury.

I'll just scan in my written work, its going to be much easier than using clunky BBCode and typing it out.

12. Dec 28, 2011

### rcgldr

384 kPa is the pressure at the gauge. Wiki lists the specific gravity of mercury at 13.56 (and water at 1.00). I don't know if this is related to the discrepancy mentioned by nvn.

Last edited: Dec 28, 2011
13. Dec 28, 2011

### MarleyDH

Okay. So then if I understand correctly, the pressure in the pipe is (denoted here with the subscript 5):

P5 = Pgauge + 0.600*ρg
P5 = 389886 Pa

Wikipedia, not something I rely on too heavily. Got burn't back in my 1st year of varsity when citing wikipedia as a reference.

14. Dec 28, 2011

### rcgldr

In the original problem diagram, the distance from the center of the main pipe to the level of the mercury in the u-tube on the right (connected to main pipe and venturi) is not given, so it's not clear if your supposed to take the specific gravity of water into account.

I did a web search and I'm finding conflicting values for specific gravity of mercury, 13.546, 13.56, 13.6, ... .

Last edited: Dec 28, 2011
15. Dec 28, 2011

### MarleyDH

That can be eliminated when doing the manometry equations by adding a ρgx term to both equations which then cancel out when you equate the pressures.

For instance:

Pleft = P300 mm + ρgx + ρgh, where h = 0.300 m.
Pright = P150 mm + ρgx + SGmercuryρgh

Equating the two eliminates the ρgx term as stated above.

16. Dec 28, 2011

### rcgldr

Yeah, I didn't think that through before I replied.

As mentioned in my editted post above, I'm finding conflicting values for specific gravity of mercury, 13.546, 13.56, 13.593, 13.6, ... , so if you have a class related reference, such as a textbook, use that value.

17. Dec 28, 2011

### MarleyDH

I'm trying to log into my university's online class resources site, but it seems it down. I recall it saying 13.6 for the mercury.

18. Dec 28, 2011

### rcgldr

My ancient (1970's) physic's textbook (yes I'm that old) also has it at 13.6, so use that until you can get back online. My guess is the issue was the location of the pressure gauge, which you already figured out.

19. Dec 28, 2011

### MarleyDH

Attached is my solution in .pdf format. I get a pump power value of 65.95 kW if I remember correctly now.

#### Attached Files:

• ###### Hydraulics.pdf
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20. Dec 28, 2011

### nvn

MarleyDH: Mercury specific gravity of 13.54, 13.55, 13.6, or perhaps 13.53 (?) is fine.

Nice work. Your answer is now correct. Using SG_Hg = 13.60, the answer is 65.937 kW. Therefore, rounding the final answer to three significant digits gives P = 65.9 kW. (Rounding to four significant digits would be 65.94 kW.) Your answer rounds to 65.95 kW because you rounded some intermediate values; but your final answer would still round to 65.9 kW, which is correct, for SG_Hg = 13.60. Excellent work.

21. Dec 28, 2011

### MarleyDH

Awesome, thanks for the help. :)