Francis turbines • • • • Examples Losses in Francis turbines NPSH Main dimensions Traditional runner X blade runner SVARTISEN P = 350 MW H = 543 m Q* = 71,5 m3/S D0 = 4,86 m D1 = 4,31m D2 = 2,35 m B0 = 0,28 m n = 333 rpm La Grande, Canada P = 169 MW H = 72 m Q = 265 m3/s D0 = 6,68 m D1e = 5,71m D1i = 2,35 m B0 = 1,4 m n = 112,5 rpm Outlet draft tube Outlet runner Inlet runner Outlet guide vane Inlet guide vane Hydraulic efficiency ηh = c1u ⋅ u1 − c2u ⋅ u2 = g ⋅ Hn 1 c 32 c12 g ⋅ h1 + + z1 − g ⋅ h 3 + + z 3 − losses 2 2 c 32 c12 g ⋅ h1 + + z1 − g ⋅ h 3 + + z 3 2 2 Losses in Francis Turbines Hydraulic Efficiency [%] Draft tube Output Energy Head [m] Hydraulic Efficiency [%] Losses in Francis Turbines Output Energy Output [%] Friction losses between runner and covers Friction losses Gap losses Gap losses Friction losses Friction losses in the spiral casing and stay vanes Guide vane losses Gap losses Runner losses Draft tube losses Relative path Absolute path without the runner installed Absolute path with the runner installed Velocity triangles Net Positive Suction Head, NPSH c 32 c 32 c 22 h2 + + z 2 = hb + h3 + + z3 + ζ ⋅ 2⋅g 2⋅g 2⋅g c 32 − H s = h3 + + z3 − z 2 2⋅g c 22 + z2 h2 + 2⋅g c 22 h2 + 2⋅g c32 = h b − Hs + z 2 + ζ ⋅ 2⋅g J = h b − Hs + J 1 h2 = hb + h2' NPSH h2 c 22 = hb − H s − − J > hva 2⋅g NPSH NPSH = h b − h 2 − H s NB: HS has a negative value in this figure. NPSH required NPSH R < h b − h va − H s = NPSH A cm2 2 u22 NPSH R = a ⋅ +b⋅ 2⋅ g 2⋅ g Turbines a 1.05 < a < 1.15 Pumps 1.6 < a < 2.0 b 0.05 < b < 0.15 0.2 < b < 0.25 Main dimensions • Dimensions of the outlet • Speed • Dimensions of the inlet D1 D2 Dimensions of the outlet We assume cu2= 0 and choose β2 and u2 from NPSHR: cm2 2 u22 a ⋅ (u2 ⋅ tan β 2 ) + b ⋅ u22 NPSH R = a ⋅ +b⋅ = 2⋅ g 2⋅ g 2⋅ g 2 13o < 35 < 1,05 < 0,05 < β2 u2 a b < < < < 22o (Lowest value for highest head) 43 m/s (Highest value for highest head) 1,15 0,15 Diameter at the outlet c m 2 = u 2 tan β 2 Connection between cm2and choose D2 : D 22 ⋅ cm2 Q = π⋅ 4 ⇒ cm2 Q⋅4 = π ⋅ D 22 D1 4⋅Q ⇒ D2 = π ⋅ cm 2 D2 Speed Connection between n and choose u2 : π ⋅ D2 ⋅ n u2 = 60 u 2 ⋅ 60 ⇒ n= π ⋅ D2 Correction of the speed The speed of the generator is given from the number of poles and the net frequency 3000 n= zp for f = 50Hz Example Given data: Flow rate Head Q = 71.5 m3/s H = 543 m We choose: a = 1,10 b = 0,10 β2 = 22o u2 = 40 m/s ( ) D 2 1,1⋅ 40 ⋅ tan 22 + 0,1⋅ 40 2 NPSH R = = 22,8 m 2⋅g Find D2 from: 4⋅Q 4⋅Q D2 = = π ⋅ cm 2 π ⋅ u2 ⋅ tan β 2 4 ⋅ 71,5 D2 = = 2.37 m D π ⋅ 40 ⋅ tan 22 Find speed from: u2 ⋅ 60 40 ⋅ 60 n= = = 322 rpm π ⋅ D2 π ⋅ 2.37 Correct the speed with synchronic speed: 3000 = 9 .3 zp = n 3000 choose z p = 9 ⇒ n K = = 333 rpm 9 We keep the velocity triangle at the outlet: u2 u2K β2 c2 c2K 4⋅Q π ⋅ D22 w2 4⋅Q π ⋅ D22 K cm 2 cmK tan β 2 = = = = π π u2 u2 K ⋅ nK ⋅ D2 K ⋅ n ⋅ D2 60 60 ⇓ nK ⋅ D23K = n ⋅ D23 ⇓ D2 K n ⋅ D23 322 3 3 = = 2.373 ⋅ = 2.35m nK 333 Dimensions of the inlet u 1 ⋅ c u1 − u 2 ⋅ c u 2 = 2 ⋅ (u1 ⋅ c u1 − u 2 ⋅ c u 2 ) ηh = g⋅H At best efficiency point, cu2= 0 u 1 ⋅ c u1 ηh ≈ 0,96 = = 2 ⋅ u 1 ⋅ c u1 g⋅H ηh 0,96 c u1 = = 2 ⋅ u1 2 ⋅ u1 Diameter at the inlet We choose: 0,7 < u1 < 0,75 D1 n ⋅ 2 ⋅ π D1 = ⋅ u1 = ω ⋅ 2 60 2 ⇓ D1 u1 ⋅ 60 D1 = n⋅π D2 Height of the inlet Continuity gives: c m1 ⋅ A1 = c m 2 ⋅ A 2 We choose: cm2 = 1,1 · cm1 1.1 ⋅ π ⋅ D B1 ⋅ D1 ⋅ π = 4 2 2 B1 = B0 Inlet angle cu1 u1 c1 cm1 β1 w1 c m1 tan β 1 = u1 − c u1 Example continues Given data: Flow rate Head Q = 71.5 m3/s H = 543 m We choose: ηh = 0,96 u1 = 0,728 cu2 = 0 η h = 2 ⋅ u 1 ⋅ c u1 ηh 0.96 ⇒ c u1 = = = 0,66 2 ⋅ u1 2 ⋅ 0,728 Diameter at the inlet u1 = u1 ⋅ 2 ⋅ g ⋅ h = 0,728 ⋅ 2 ⋅ 9,81⋅ 543 = 75,15 m s u1 ⋅ 60 75,15 ⋅ 60 D1 = = = 4,31 m n ⋅π 333 ⋅ π D1 Height of the inlet 1,1⋅ π ⋅ D 1,1⋅ 2.35 B1 = = = 0,35 m 4 ⋅ D1 ⋅ π 4 ⋅ 4,31 2 2 B1 2 Inlet angle cu1 u1 c1 cm1 β1 w1 cm 2 sin β 2 ⋅ u2 sin 22D ⋅ 40,9 m s = = = 13,9 m s cm1 = 1,1 1,1 1,1 c m1 = c m1 = 2⋅g⋅h 13,9 = 0,135 2 ⋅ 9,81 ⋅ 543 c m1 0,135 = arctan = 62.9D 0,728 − 0,659 u 1 − c u1 β1 = arctan SVARTISEN P = 350 MW H = 543 m Q* = 71,5 m3/S D0 = 4,86 m D1 = 4,31m D2 = 2,35 m B0 = 0,28 m n = 333 rpm