Question

A model of a Francis turbine is built to a scale of one fifth of full size and when tested it developed a power output of $3 \mathrm{~kW}$ under a head of $1.8 \mathrm{~m}$ of water, at a rotational speed of $360 \mathrm{rev} / \mathrm{min}$ and a flow rate of $0.215 \mathrm{~m}^3 / \mathrm{s}$.

Estimate the speed, flow rate, and power of the full-scale turbine when working under dynamically similar conditions with a head of $60 \mathrm{~m}$ of water. By making a suitable correction for scale effects, determine the efficiency and the power of the full-size turbine. Use Moody's formula and assume $n=0.25$.

Answer 1

Determine the efficiency and the power of the full-size turbine. Use Moody's formula and assume n=0.25.

From the group ψ=gH/(ND)2, we get

Np=Nm(Dm/Dp)(Hp/Hm)0.5=(360/5)(60/1.8)0.5=415.7 rev/min

From the group ϕ=Q/(ND3), we get

Qp=Qm(Np/Nm)(Dp/Dm)3=0.215×(415.7/360)×53=31.03 m3/s

Lastly, from the group P^=P/(ρN3D3), we get

Pp=Pm(Np/Nm)3(Dp/Dm)5=3×(415.7)3×55=14,430 kW=14.43 MW

This result has still to be corrected to allow for scale effects. First we must calculate the efficiency of the model turbine. The efficiency is found from

ηm=P/(ρQgH)=3×103/(103×0.215×9.81×1.8)=0.79

Using Moody's formula, the efficiency of the prototype is determined:

(1−ηp)=(1−ηm)×0.20.25=0.21×0.6687

hence,

ηp=0.8596

The corresponding power is found by an adjustment of the original power obtained under dynamically similar conditions, i.e.,

corrected Pp=14.43×0.8596/0.79=15.7 MW

There is the all true answer.