Question

A small-scale Kaplan turbine has a power output of $8 \mathrm{MW}$, an available head at turbine entry of $13.4 \mathrm{~m}$, and a rotational speed of $200 \mathrm{rev} / \mathrm{min}$. The inlet guide vanes have a length of $1.6 \mathrm{~m}$ and the diameter at the trailing edge surface is $3.1 \mathrm{~m}$. The runner diameter is $2.9 \mathrm{~m}$ and the hub-tip ratio is 0.4. Assuming the hydraulic efficiency is $92 \%$ and the runner design is "free-vortex," determine

1. The radial and tangential components of velocity at exit from the guide vanes;
2. The component of axial velocity at the runner;
3. The absolute and relative flow angles upstream and downstream of the runner at the hub, mid-radius, and tip.

Answer 1

The radial and tangential components of velocity at exit from the guide vanes

As P=ηHρgQHE, then the volume flow rate is

Q=PηHρgHE=8×1060.92×9810×13.4=66.15m/s

Therefore,

cr1=Q2πr1L=66.152π×1.55×1.6=4.245m/scx2=4QπD2t2(1−v2)=4×66.15π×2.92×0.84=11.922m/s

Parameter	Ratio r/rt
	0.4	0.7	1.0
Cθ2(m/s)	9.955	5.687	3.982
tan⁡α2	0.835	0.4772	0.334
α2(deg)	39.86	25.51	18.47
U/Cx2	1.019	1.7832	2.547
β2(deg)	10.43	52.56	65.69
β3(deg)	45.54	60.72	68.57

Table 9.4 Calculated Values of Flow Angles

The component of axial velocity at the runner

As the specific work done is ΔW=U2cθ2 and ηH=ΔWgHE, then at the tip

cθ2=ηHgHEU2=0.92×9.81×13.430.37=3.892m/s

where the blade tip speed is U2=ΩD22=(200×π30)×2.92=30.37m/s,

cθ1=cθ2r2r1=3.892×1.451.55=3.725m/sα1=tan−1⁡(cθ1cr1)=tan−1⁡(3.7254.245)=41.26∘

The absolute and relative flow angles upstream and downstream of the runner at the hub, mid-radius, and tip.

Values α2,β2, and β3 given in Table 9.4 have been derived from the following relations:

α2=tan−1⁡(cθ2cx2)=tan−1⁡(cθ2tcx2rtr)β2=tan−1⁡(Ωrcx2−tan⁡α2)=tan−1⁡(U2tcx2rrt−tan⁡α2)β3=tan−1⁡(Ucx2)=tan−1⁡(U2tcx2rrt)

Finally, Figure 9.19 illustrates the variation of the flow angles, from which the large amount of blade twist mentioned earlier can be inferred.

There is the answer.