Turbine output power with gauge pressures and Bernoulli's equation

[Brabs23]

A turbine is installed in a vertical pipe line, 150mm diameter. A gauge is connected to the pipe at a point A above the turbine and another at point B below the turbine, the vertical distance between A and B being 1.2m. When the discharge is 0.06m^3/s, the gauages read 400kN/m^3 and 35kN/m^3 respectively.

Calculate the output power from the turbine, assuming it to be 85% efficient.



I've found the Cross sectional area = 0.01767m^2

Vol per sec = 0.06
therefore 0.06 = m/roh
0.06 = m/10^3
therefore mass flow rate = 60


Q = AV
0.06 = 0.01767V
therefore V = 3.395m/s

Then gone on to use Bernoulli's ... unsucessfully!

Any help? Am I on the right tracks or?

 

[Filip Larsen]

You may want to think about how pressure and flow rates relate to fluid power.

 

[Brabs23]

I know I need the pressure values to put into the bernoullis equation but that is what I am stuck on finding/working out

 

[Filip Larsen]

What quantity do the gauges measure? I assume its pressure (kN/m^2) and that you just wrote the unit as kN/m^3 by mistake.

If it is pressure they measure, you have the pressure and flow rate at two points in the flow so you can very easily calculate the fluid power those points without need of Bernoulli's equation. You also know how much pressure (and hence fluid power) increases over a 1.2 m drop so you can calculate how much power the fluid should have at point B if the turbine wasn't there. The difference between that value and the actual measured power at B is equal to the fluid power absorbed by the turbine (well, and pipe resistance, but that seems to be ignored in the problem text).

 

[Brabs23]

So I presume that the flow rate that I have is correct?

 

[Filip Larsen]

Yes, if the volume flow rate is 0.060 m^3/s and it is water we are talking about (density 1000 kg/m^3) then a mass flow rate of 60 kg/s is correct.

 

[Brabs23]

Thanks a lot!

 

[Brabs23]

I know how much the fluid power/pressure increases by 1.2m drop but how do I work it out if the turbine isn't there?

 

[Filip Larsen]

If there were no turbine (i.e. no pressure resistance between A and B), the pressure at B should be the pressure P_A at A plus the pressure increase P_h from the drop. If we now include the turbine and the measured pressure P_B at B, you now also have to include the pressure drop across the turbine P_T so that you get the relation P_A - P_T + P_h = P_B from which you can calculate P_T and the fluid power that it corresponds to (using the mass flow rate).

 

[Brabs23]

That makes more sense to me now, thank you :)