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Fundamental similarity considerations • • • • • 1 Similarity Considerations Reduced parameters Dimensionless terms Classification of turbines Performance characteristics Similarity Considerations Similarity considerations on hydrodynamic machines are an attempt to describe the performance of a given machine by comparison with the experimentally known performance of another machine under modified operating conditions, such as a change of speed. 2 Similarity Considerations • Valid when: – Geometric similarity – All velocity components are equally scaled – Same velocity directions – Velocity triangles are kept the same – Similar force distributions – Incompressible flow 3 These three dynamic relations together are the basis of all fundamental similarity relations for the flow in turbo machinery. c = Const. u 1 F = p = r c 2 Const. A 2g H = Const. 2 c 4 2 3 2 g H = Const. 2 u Velocity triangles u = r c c = Const . u 5 w 1 Under the assumption that the only forces acting on the fluid are the inertia forces, it is possible to establish a definite relation between the forces and the velocity under similar flow conditions dc F = m dt In connection with turbo machinery, Newton’s 2. law is used in the form of the impulse or momentum law: c F = m t m = rQ t 6 F = r Q c For similar flow conditions the velocity change c is proportional to the velocity c of the flow through a cross section A. It follows that all mass or inertia forces in a fluid are proportional to the square of the fluid velocities. F 2 = p = Const r c A 2 p c = h = Const rg g 7 2 By applying the total head H under which the machine is operating, it is possible to obtain the following relations between the head and either a characteristic fluid velocity c in the machine, or the peripheral velocity of the runner. (Because of the kinematic relation in equation 1) gH = Const . 2 c 8 gH = Const. 2 u 3 For pumps and turbines, the capacity Q is a significant operating characteristic. c = Const . u Q D 2 = Q = Const. n D n D3 c is proportional to Q/D2 and u is proportional to n·D. gH = Const. 2 c H D Q 2 H D 4 Const. = = = Const. 2 Q g 2 gH = Const. 2 u 9 H n D 2 = H Const. = = Const. 2 2 n D g Affinity Laws Q = Const. 3 nD Q1 n1 D = Q2 n2 D 3 1 3 2 Q1 n1 = Q2 n2 10 This relation assumes that there are no change of the diameter D. Affinity Laws H = Const. 2 2 n D H1 n D = H2 n D 2 1 2 2 2 1 2 2 2 1 2 2 H1 n = H2 n 11 This relation assumes that there are no change of the diameter D. Affinity Laws Q = Const. 3 nD H = Const. 2 2 n D P = r g H Q 3 2 2 P1 r g H1 Q1 H1 Q1 n1 D1 n1 D1 n13 D15 = = = = 3 5 3 2 2 P2 r g H 2 Q2 H 2 Q2 n2 D2 n2 D2 n2 D2 P1 n13 = 3 P2 n2 12 This relation assumes that there are no change of the diameter D. Affinity Laws Q1 n1 = Q2 n 2 H1 n = H2 n P1 n = P2 n 13 2 1 2 2 3 1 3 2 This relations assumes that there are no change of the diameter D. Affinity Laws Example Change of speed n1 = 600 rpm n2 = 650 rpm Q1 = 1,0 m3/s Q2 = ? Q1 n1 = Q2 n2 3 n2 650 m Q2 = Q1 = 1,0 = 1,08 s 600 n1 14 Reduced parameters used for turbines The reduced parameters are values relative to the highest velocity that can be obtained if all energy is converted to kinetic energy 15 Reference line Bernoulli from 1 to 2 without friction gives: c12 c22 h1 z1 = h2 z2 2 g 2 g c22 = g z1 z2 = g H 2 c2 = 2 g H 16 Reduced values used for turbines c c= 2g H u u= 2g H w w= 2g H h = 2 cu1 u1 cu 2 u 2 = 2g H Q Q= 2g H h h= H 17 Dimensionless terms • Speed – – – – Speed number Specific speed Speed factor Specific speed W NQE nED, n11 n q, n s • Flow – Flow factor QED, Q11 • Torque – Torque factor TED, T11 • Power – Power factor 18 PED, P11 Fluid machinery that is geometric similar to each other, will at same relative flow rate have the same velocity triangle. For the reduced peripheral velocity: u ~ D = Const. For the reduced absolute meridonial velocity: cm ~ Q D 2 = Const. We multiply these expressions with each other: D 19 Q D 2 = Q = Const. 20 Speed number * W = * * Q D 21 Geometric similar, but different sized turbines have the same speed number Speed number cm = Q D 2 4 u =D Q D 2 = Const1 1 D = Const2 From equation 1: u = r cu 1 = D Const2 2 cm c w Inserted in equation 2: 2 Q = Const1 Const2 22 D Q = Const1 Const22 = W cm cm Speed Factor unit speed, n11 n 2 u = D = Const = D 60 2 g H nD = Const = n11 H u = r cu cm c If we have a turbine with the following characteristics: • Head H=1m • Diameter D = 1 m we have what we call a unit turbine. 23 nD n11 = H w Speed Factor nED n 2 u = D = Const = D 60 2 g H nD = Const = nED gH u = r cu cm c If we have a turbine with the following characteristics: • Energy E = 1 J/kg • Diameter D = 1 m nED 24 n D = E w Energy ctw2 c12 g h1abs g z1 = g hatm g ztw g Hn g g c12 ctw2 E = g H n = g h1abs g hatm g z1 g ztw g c1 h1 abs ztw z1 25 Reference line Specific speed that is used to classify turbines nq = n 26 Q H 0, 75 Specific speed that is used to classify pumps nq = n Q H 34 nq is the specific speed for a unit machine that is geometric similar to a machine with the head Hq = 1 m and flow rate Q = 1 m3/s n s = 333 n Q P 34 ns is the specific speed for a unit machine 27 that is geometric similar to a machine with the head Hq = 1 m and uses the power P = 1 hp 28 29 Flow Factor unit flow, Q11 cm = Q 4 D2 Q D 2 Q = Const = Q11 2 D H 30 = Const u = r cu cm c If we have a turbine with the following characteristics: • Head H=1m • Diameter D = 1 m we have what we call a unit turbine. Q Q11 = 2 D H w Flow Factor QED cm = Q 4 D2 Q D 2 = Const u = r cu Q = Const = QED 2 D gH cm c If we have a turbine with the following characteristics: • Energy E = 1 J/kg • Diameter D = 1 m QED 31 Q = 2 D E w Exercise • Find the speed number and specific speed for the Francis turbine at Svartisen Powerplant • Given data: 32 Speed number: 2 g H = 2 9,82 543 = 103m s n 2 333 2 = = = 34,9 rad s 60 60 34,9 rad 1 s * = = = 0,33m 2g H 103m s 3 m 71,5 Q s = 0,69 m 2 = 2gH 103 m s *Q = * 33 W = * * Q = 0,33 0,69 = 0,27 Specific speed: nq = n Q H 34 71,5 n q = 333 = 25,03 34 543 34 Performance characteristics 1.00 NB: H=constant Virkningsgrad [-] Efficiency 0.90 0.80 = = = 0.70 = = 0.60 0.50 200.00 400.00 600.00 Turtall [rpm] Speed [rpm] 35 800.00 Kaplan 36 Q g H D 2 = QED = = 1.0 1.3 0.6 0.7 0.8 0.9 0.7 = 0.3 0 nED 37 n D = gH