Find the speed of water out of a nozzle

[tomas123]

Homework Statement

Water is flowing through a nozzle with entrance data:
P₁ = 500 kPa
V₁ = 10 m/s
A₁ = π m²
ρ = 1000 kg/m³.

Exit data:

P₂ = atmospheric pressure, assume 100 kPa to make numbers easy
V₂ = ?
A₂ = π (0.5)²
ρ = 1000 kg/m³

With what speed does the water exit the nozzle?[/B]

Homework Equations

continuity equation: A₁V₁ = A₂V₂
Bernoulli's equation for horisontal flow:
V₁²/2 + P₁/ρ = V₂²/2 + P_atm/ρ

The Attempt at a Solution

I see that V₂ = V₁A₁/A₂ = 10 * π/(π*0.5²) = 40 m/s

But plugging this into Bernoulli's equation, you get:

10²/2 + 500k/1000 = 40²/2 +100k/1000
50 + 500 = 800 + 100
550 = 900

which isn't correct. My question is why doesn't it add up? If I were to use Bernoulli's equation to find the speed given the data above, I wouldn't need to know the area of the exit ... which is obviously fishy. Am I using Bernoulli's wrong? Why is it wrong?

[/B]

 

[drvrm]

which isn't correct. My question is why doesn't it add up? If I were to use Bernoulli's equation to find the speed given the data above, I wouldn't need to know the area of the exit ... which is obviously fishy. Am I using Bernoulli's wrong? Why is it wrong?

i think one should check again the data or give a copy of the data on the screen -as things are not looking correct?

 

[tomas123]

i think one should check again the data or give a copy of the data on the screen -as things are not looking correct?

I made up the numbers. With a pump, you can get pretty much w/e Pressure & speed situation inside the pipe you'd like.. no?

For the numbers I have, we can make up that upstream from where we are looking there is a 3 piston pump. The pistons have an area equal to A₁ = π m². To get P₁ = 500 kPa each piston would be delivering a konstant force of 159kN. You get the point.. To get 10 m/s I would make up some piston length and multiply it with the area and make up a constant number of rotations per second to get the correct flow rate and speed (10 m/s).

I'm obviously doing something wrong. My reasoning is incorrect somewhere.. But I don't know exactly what is wrong.

 

[haruspex]

which isn't correct. My question is why doesn't it add up? If I were to use Bernoulli's equation to find the speed given the data above, I wouldn't need to know the area of the exit ... which is obviously fishy.

Not at all. The problem is overspecified. You have two equations but only one unknown.
Imagine you set up a pipe with these diameters and inlet and outlet pressures. The flow rate would be a consequence. But you have also specified the flow rate, so if the numbers are made up you will almost surely have a contradiction.

 

[tomas123]

Not at all. The problem is overspecified. You have two equations but only one unknown.
Imagine you set up a pipe with these diameters and inlet and outlet pressures. The flow rate would be a consequence. But you have also specified the flow rate, so if the numbers are made up you will almost surely have a contradiction.

For a scenario w/o a pump, such as an elevated reservoir or something, the flow rate would be determined by the area at the outlet and the pressure difference. For a pump scenario on the other hand, the flow rate is determined by the pump, no?

When I do calculations for drill strings, their nozzles and mud pumps (triplex pumps) the pistons have a certain force, number rotations per minute and the flow rate is adjustable by choosing a piston diameter, i.e. area of piston. Larger area will give you a higher flow rate, but lower pressure. My point is that the pressure and flow rate inside the pipe are determined by pump settings.

 

[haruspex]

For a scenario w/o a pump, such as an elevated reservoir or something, the flow rate would be determined by the area at the outlet and the pressure difference. For a pump scenario on the other hand, the flow rate is determined by the pump, no?

Yes, but the same pump setting will determine both the flow rate and the upstream pressure. You cannot specify them independently.

 

[tomas123]

Yes, but the same pump setting will determine both the flow rate and the upstream pressure. You cannot specify them independently.

I don't think I did?

If you set up the area around the pump as the control volume, with the area of pipe being equal before and after, whatever energy/second is transferred to the flow, will be in the form of Pressure * flow rate. Kinetic energy, internal energy and potential energy will be the same before and after. With a lower pressure, I'll get a lower flow rate. They aren't independent.

So for a certain power setting on the pump, the following is one of many possible outcomes (numbers I used in OP):

P = 500 kPa
V = 10 m/s
A = π m2
ρ = 1000 kg/m3.

This is possible, no?

Then I'd find V2 using the continuity equation and then P2 using bernoulli's equation. But P2 is atmospheric pressure, so this can't be right. Hm..

Thank you for the help :)

 

[tomas123]

I think I understand now.. I don't have time to type what I'm thinking atm, but I'll be back later to see if you agree with my newfound reasoning.

Posting this because I don't want anyone to spend time explaining to me when I've figured it out :)