## Velocity through a water turbine

## Velocity through a water turbine

(OP)

Hi all

I have a question about calculating power from a water turbine. If I have a rotor with an inlet area A1 and an inlet velocity V1, and an outlet area A2 and velocity V2, if the rotor tapers down, A2 is smaller than A1, therefore to satisfy the mass flow rate through the closed rotor, V2 (axial) must be greater than V1 (axial). But... for a water turbine I understand that V2 needs to be smaller than V1 as work is done and hence the dynamic pressure must decrease, not increase. I'm calculating the power at the inlet of the rotor and the outlet of the rotor using Power = the flow rate (Area x velocity) multiplied by the dynamic pressure (0.5 x area x density x velocity2 ). My issue is if I use axial velocity, which satisfies the mass flow rate, I get negative power as work is done on the liquid to increase the velocity. If I use the total velocity component (using CFD) in this equation, I get what seems a realistic power but the mass flow rate in is much greater than the mass flow out. Please help me! I am utterly confused!

Also, as I mentioned, the turbine I'm working on has a decreasing cross sectional area through the rotor... is it just me, or is this silly? Aren't you better with an increasing area through the rotor? I'd be interested in anyone's thoughts on this.

Thank you,

Kat

I have a question about calculating power from a water turbine. If I have a rotor with an inlet area A1 and an inlet velocity V1, and an outlet area A2 and velocity V2, if the rotor tapers down, A2 is smaller than A1, therefore to satisfy the mass flow rate through the closed rotor, V2 (axial) must be greater than V1 (axial). But... for a water turbine I understand that V2 needs to be smaller than V1 as work is done and hence the dynamic pressure must decrease, not increase. I'm calculating the power at the inlet of the rotor and the outlet of the rotor using Power = the flow rate (Area x velocity) multiplied by the dynamic pressure (0.5 x area x density x velocity2 ). My issue is if I use axial velocity, which satisfies the mass flow rate, I get negative power as work is done on the liquid to increase the velocity. If I use the total velocity component (using CFD) in this equation, I get what seems a realistic power but the mass flow rate in is much greater than the mass flow out. Please help me! I am utterly confused!

Also, as I mentioned, the turbine I'm working on has a decreasing cross sectional area through the rotor... is it just me, or is this silly? Aren't you better with an increasing area through the rotor? I'd be interested in anyone's thoughts on this.

Thank you,

Kat

## RE: Velocity through a water turbine

Check your units if you have:-

Area x velocity) multiplied by the dynamic pressure (0.5 x area x density x velocity2 ).

Power should be N*m/s from your units above unless I am making a mistake I can't get N*m/s.

formula I was looking at was :-

mass flowrate *(V-v)*v where V,v are the initial and

final velocities of the jet

respectively normal to the vane

thats the power output formula for jets on to a rotor with flat plates.

desertfox

## RE: Velocity through a water turbine

Kat

## RE: Velocity through a water turbine

Its difficult to say of the description you give any chance of uploading a file of what your looking at.

desertfox

## RE: Velocity through a water turbine

Have a look at this site it might help:-

htt

desertfox