Turbine power - theory?

[cctdiag]

Hi,

Can anyone help and explain in 'laymans terms' why the power delivered to a turbine is proportional to: -

i) the density of a fluid, and
ii) the cube of the speed of a fluid.

My question application is related to hydro-electric power generation.

Many thanks

cctdiag

 

[willard3]

How heavy are the things are you dropping from how high (potential energy/density)

How fast are you dropping them (kinetic energy/speed)

total energy=potential energy + kinetic energy

 

[davefitz]

The kinetic energy developed ( and harnessed) is proportional to 0.5*rho*V^2 on the basis of KW-hr/kg ( or btu/lb), where rho is the density.

To get power, multiply this energy term by the mass flowrate of liquid thru the turbine, and get:

KW= c*rho*V^3 , where c is some proportional constant.

 

[kevindurette]

Blow air out of your mouth. Now blow water out of your mouth. It takes more work to blow the water because it's denser. The heavier something is, the more force it will take to accelerate it, and the more work you'll either achieve or give up to move it around.

I have a short discussion on work here:

http://kevinthenerd.googlepages.com/torque_vs_hp.html

Basically, without getting into nitpicking details, work is force times distance. Speed is distance over time. Power is work over time. Putting all that together (using a little algebra I guess), you get that power is force times speed.

Drag force is proportional to the square of the speed, and therefore, power is proportional to the square of the speed, times the force, times the speed, which lends to power being proportional to the cube of the speed.

I honestly don't remember the derivation of why drag is proportional the square of speed, but the units check out. Sorry.