Question

An axial flow hydraulic turbine operates with a head of 20m at turbine entry and develops 10MW when running at 250rev/min. The blade tip diameter is 3m, the hub diameter is 1.25m, and the runner design is based upon a "free vortex." Assuming the hydraulic efficiency is 94%, the overall efficiency is 92%, and the flow at exit is entirely axial, determine the absolute and relative flow angles upstream of the runner at the hub, mean, and tip radii.

Answer 1

Determine the absolute and relative flow angles upstream of the runner at the hub, mean, and tip radii.

Calculate volume flow rate.

Given: P=107W,ηo=0.92,ρ=1000kg/m3,g=9.81m/s2,HE=20m

Q=PηoρgHE=1070.92×1000×9.81×20=55.4m3/s

Calculate radial component of velocity at inlet.

Given: Q=55.4m3/s,dt=3m,dh=1.25m

cr1=Qπdt(dt−dh2)=55.4π×3×(3−1.252)=6.718m/s

Calculate axial velocity at exit.

Given: Q=55.4m3/s,dt=3m,dh=1.25m

cx2=Qπ4(dt2−dh2)=55.4π4×(32−1.252)=9.484m/s

Calculate blade tip velocity at exit.

Given: dt=3m,N=250rpm

U2t=πdtN60=π×3×25060=39.27m/s

Calculate tangential velocity at exit at the blade tip.

Given: ηh=0.94,g=9.81m/s2,HE=20m,U2r=39.27m/s

cθ2t=ηhgHEU2t=0.94×9.81×2039.27=4.696m/s

Calculate absolute flow angle at hub.

Given: cθ2t=4.696m/s,dt=3m,cx2=9.484m/s,dh=1.25m

α2=tan−1⁡(cθ21dtcx2dh)=tan−1⁡(4.696×39.484×1.25)=49.92∘

Calculate relative flow angle at hub.

Given: U2t=39.27m/s,dt=3m,cx2=9.484m/s,dh=1.25m,α2=49.92∘

β2=tan−1⁡(U21dhcx2dt−tan⁡α2)=tan−1⁡(39.27×1.259.484×3−tan⁡49.92)=28.22∘

Therefore, absolute flow angle at hub is 49.92∘, and relative flow angle at hub is 28.22∘.

Calculate absolute flow angle at mean radii.

Given: cθ2t=4.696m/s,dt=3m,cx2=9.484m/s,dh=1.25m

α2=tan−1⁡(cθ2tdtcx2(dt+dh2))=tan−1⁡(4.696×39.484×(3+1.252))=34.955∘

Calculate relative flow angle at hub.

Given: U2t=39.27m/s,dt=3m,cx2=9.484m/s,dh=1.25m,α2=34.955∘

β2=tan−1⁡(U2t(dt+dh2)cx2dt−tan⁡α2)=tan−1⁡(39.27×(3+1.252)9.484×3−tan⁡34.955)=65.885∘

Therefore, absolute flow angle at hub is 34.955∘ and relative flow angle at hub is 65.885∘.

Calculate absolute flow angle at tip.

Given: cθ2t=4.696m/s,cx2=9.484m/s

α2=tan−1⁡(cθ2tcx2)=tan−1⁡(4.6969.484)=26.34∘

Calculate relative flow angle at hub.

Given: U2t=39.27m/s,cx2=9.484m/s,α2=26.34∘

β2=tan−1⁡(U2tcx2−tan⁡α2)=tan−1⁡(39.279.484−tan⁡26.34)=74.66∘

Therefore, absolute flow angle at hub is 26.34∘, and relative flow angle at hub is 74.66∘.

Therefore, there is the answer we calcuate.