Question

Water flows from one reservoir to another a height, 45 m below. A turbine (η=0.74) generates power from this flow. 1 m3/s passes through the turbine. If 11 m headloss occurs between the two reservoirs, determine the actual (i.e. useable) power generated by the turbine (in kW)

Answer 1

The power generated is 246.820 kW

Step-by-step explanation:

Imagine or draw the control volume: Its limits would be from the surface of the first reservoir, to the surface of the second one, which is above.

Consider the energy balance (its formulation can be studied in Cengel (Fluid mechanics, chapter five):

`Q+W=m*((P_(2)-P_(1))/(p)+g(z_(2)-z_(1))+(v_(2)^(2)-v_(1)^(2))/(2)+H)`

Where H is the energy head loss in squared velocity units. (i.e.
`(m^(2))/(s^(2))`. 2 is the final reservoir and 1 is the feeding reservoir. If the reservoirs are so big, it is possible to assume the velocity on both surfaces as zero (the mass flow is not enough to induce a appreciable velocity because of the decrease on the water level). Both reservoir surfaces should be at the same pressure, so the pressure difference will be zero (neglective). Additionally, the head loss may also consider the energy loss by heat (because there is not any considerable heat transfer), so the equation is finally:

`W=m*g*(z_(2)-z_(1)+H)`

Dividing by m*g, we find an equation for the turbine head:

`h_(turb)=z_(2)-z_(1)+h`

h is the head loss in length units.

So, calculate
`h_(turb)`:

`h_(turb)=-45m+11m=-34m` The turbine head is negative because the energy is going out of the fluid, but it is positive if we consider it as produced energy (the sign depends on what you are analyzing).

With
`h_(turb)` find the work:

`W=m*g*h_(turb)=1(m^(3))/(s)*(1000kg)/(m^(3))*9.81*-34m\\W=-333540W`

This work calculated before is the energy that the fluid delivery to the turbine, but the turbine is not 100% efficient, so it is not the energy that the turbine produces. The ratio between the delivered and the produced energy is the efficiency (I am going to write it as e):

e=(produced energy)/(delivered by the fluid energy)

So, the produced energy,
`W_(p)` can be calculated as:

`W_(p)=e*W_(d)` (Wd is the delivered energy).

Then,

`W_(p)=0.74*333540W=246820W=246.820kW`