Có bài tập mẫu để thi cuối kì môn Máy thủy khí. Có bài tập mẫu để thi cuối kì môn Máy thủy khí. Có bài tập mẫu để thi cuối kì môn Máy thủy khí. Có bài tập mẫu để thi cuối kì môn Máy thủy khí. Có bài tập mẫu để thi cuối kì môn Máy thủy khí. Có bài tập mẫu để thi cuối kì môn Máy thủy khí. Có bài tập mẫu để thi cuối kì môn Máy thủy khí. Có bài tập mẫu để thi cuối kì môn Máy thủy khí. Có bài tập mẫu để thi cuối kì môn Máy thủy khí. Có bài tập mẫu để thi cuối kì môn Máy thủy khí. Có bài tập mẫu để thi cuối kì môn Máy thủy khí. Có bài tập mẫu để thi cuối kì môn Máy thủy khí.
Chapter 12 / Turbomachinery CHAPTER 12 Turbomachinery Elementary Theory ω = 800 × π / 30 = 83.8 rad/s u1 = ωr1 = 83.8 × 0.04 = 3.35 m/s u = ω r2 = 83.8 × 0.125 = 10.48 m/s Q = 2π r1b1Vn1 , but Vn1 = u1 since β1 = 45D 12.2 ∴Q = 2π × 0.04 × 0.05 × 3.35 = 0.0421 m3 / s Vn2 = Q 0.0421 = = 2.14 m/s 2π r2b2 2π × 0.125 × 0.025 Vt2 = u2 − Vt1 = Vn2 tan β = 10.48 − 2.14 = 6.77 m/s tan 30D (α1 = 90D under ideal conditions) ( ) ∴T = ρQ r2Vt2 − rV t1 = 1000 × 0.0421(0.125 × 6.77 − 0) = 35.6 N ⋅ m W P = ωT = 83.8 × 35.6 = 2980 W H t = ωT / γ Q = 2980 /(9810 × 0.0421) = 7.22 m Q = 7.6 ×10−3 m3/s 12.4 ω = 2000 × π 30 = 209 rad/s 175 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery Q 7.6 ×10−3 = = 1.94 m/s 2π r2b2 2π × 0.0625 × 0.01 Vn2 = u2 = ωr2 = 209 × 0.0625 = 13.06 m/s ∴ Ht = 13.06 u2 (13.06 − 1.94cot 60°) = 15.9 m (u2 − Vn2 cot β ) = 9.81 g = γ QH = 9810 × 0.8 × 7.6 ×10−3 ×15.9 = 948 W W t = 948/ 746 = 1.21 hp W Compute loss in suction pipe: ⎛ L ⎞ Q hL = ⎜ f + Σ K ⎟ ⎝ D ⎠ gA 11 ⎛ ⎞ = ⎜ 0.015 × + × 0.19 + 0.8⎟ ⎝ ⎠ 0.1 12.6 0.05 2 ⎛ π⎞ × 9.81 × ⎜ ⎟ × 0.1 ⎝ 4⎠ = 5.85 m Water at 20°C: γ = 9792 N/m3 , pv = 2340 Pa Substitute known data into NPSH relation, solving for Δz: ∴Δ z = patm − pv γ 101×103 − 2340 − 5.85 − = 1.23 m − hL − NPSH = 9792 Dimensional Analysis and Similitude Q1 = N ω1 970 Q = 1Q = × 0.8 = 0.65 m / s ω 2 N 2 1200 ∴from Fig 12.9, H1 ≅ 11.5 m and W ≅ 91 kW ⎛ω ⎞ ⎛ 1200 ⎞ ∴ H = ⎜ ⎟ H1 = ⎜ ⎟ ×11.5 = 17.6 m ⎝ 970 ⎠ ⎝ ω1 ⎠ 12.8 ⎛ 1200 ⎞ = ⎛ ω2 ⎞ W ∴W ⎜ ⎟ 1=⎜ ⎟ × 91 = 172 kW ⎝ 970 ⎠ ⎝ ω1 ⎠ 12.10 CQ = Q Q / 3600 = = 1.061× 10 −4 Q ω D 304 × 0.2053 (Q in m /h) 176 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery CH = gH 9.81H = = 2.526 × 10−3 H (H in m) 2 2 ω D 304 × 0.205 Tabulate CQ and CH using selected values of Q and H from Fig 12.6: CQ×10 −3 0.53 1.06 1.59 2.12 2.65 3.18 Q (m3/h) 50 100 150 200 250 300 H (m) 54 53 52 50 47 41 33 CH ×10−1 1.36 1.34 1.31 1.26 1.19 1.04 0.83 The dimensionless curve shown in Fig 12.12 is for the 240-mm impeller Since the impellers are not the same (240 mm versus 205 mm), dynamic similitude does not exist, and thus the curves are not the same Compute the specific speed: Ω P = 12.12 ω Q ( gH P )3 / = 1800 × π × 0.15 30 = 1.30 , (9.81 × 22)3 / hence use a mixed flow pump As an alternate, since Ω P is close to unity, a radial flow pump could be employed Fig 12.13: At η = 0.75 (best eff ), CQ ≅ 0.048 , C H ≅ 0.018, C W ≅ 0.0011, C NPSH ≅ 0.023 ω = 750 × π 30 = 78.5 rad/s 1/3 (a) 12.14 ⎛ 1240 ×10−3 ⎞ D = ⎜⎜ ⎟⎟ ⎝ 0.049 × 78.5 ⎠ H= = 0.685 m 0.018 × 78.52 × 0.6852 = 5.3 m 32.2 H NPSH = 0.023 × 78.52 × 0.6852 = 6.78 m 9.81 = 0.0011×1000 × 78.53 × 0.6855 = 80,250 W = 80.25 kW W or 12.16 = 82.25/0.746 = 107.6 hp W ω = 600 × π 30 = 62.8 rad /s, Q = 22.7 / 60 = 0.378 m / s, 177 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery ΩP = ω Q ( gH P ) = 3/ 62.8 0.378 = 0.751 (9.81×19.5)3/ H ⎛ ω ⎞ ⎛ D2 ⎞ =⎜ ⎟ ⎜ ⎟ =2 H1 ⎝ ω1 ⎠ ⎝ D1 ⎠ Q2 ⎛ ω ⎞⎛ D2 ⎞ =⎜ ⎟⎜ ⎟ =2 Q1 ⎝ ω1 ⎠⎝ D1 ⎠ ⎛ ω ⎞ ⎛ D ⎞ ⎛ ω ⎞⎛ D ⎞ ∴ ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟⎜ ⎟ ⎝ ω1 ⎠ ⎝ D1 ⎠ ⎝ ω1 ⎠⎝ D1 ⎠ 12.18 ∴Use a radial flow pump or ω D2 = ω1 D1 ⎛ω ⎞ ∴ ⎜ ⎟ = 2, ω2 = ω1 = 1.19 ω1 and D2 = D1 = 1.19 D1 ⎝ ω1 ⎠ Assume a pump speed N = 2000 rpm or ω = 2000 × π 30 = 209 rad/s /(γ Q) = 200 ×103 /(8830 × 0.66) = 34.3 m HP = W f ∴Ω P = ω Q ( gH P ) 3/ = 209 0.66 = 2.16 (9.81× 34.3)3/ The specific speed suggests a mixed-flow pump However, if N = 1000 rpm, a radial-flow pump may be appropriate Consider both possibilities Mixed flow: from Fig 12.14, at best η : CW ≅ 0.0117, CQ ≅ 0.148, CH ≅ 0.067 Use 12.20 CQ = gH Q and CH = P2 ωD ωD Combining and solving for D and ω Q / CQ D= gHP / CH ρ= Q CQD3 0.66 / 0.148 = 0.251 m 9.81× 34.3/ 0.067 ∴D = ω= and ω = 0.66 = 282 rad/s or N = 282 × 30 / π = 2674 rpm 0.148 × 0.2513 γ g = 8830 = 900 kg/m3 9.81 178 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery = C ρω 3D5 = 0.0117 × 900 × 2823 × 0.2515 ∴W P W = 2.35 × 105 W or 235 kW Radial flow: from Fig 12.12, at best η : CW ≅ 0.027, CQ ≅ 0.0165, C H ≅ 0.125 ∴D = ω= 0.66 / 0.0165 = 0.878 m 9.81× 34.3/ 0.125 0.66 59.1× 30 = 59.1 rad/s or N = = 564 rpm π 0.0165 × 0.878 = 0.0027 × 900 × 59.13 × 0.8785 = 2.62 × 105 W W P or 262 kW Hence, a mixed-flow pump is preferred Use of Turbopumps 12.22 The intersection of the system demand curve with the head-discharge curve yields Q = 2.75 m3 / min, H P = 12.6 m, W P = 7.2 kW N = 1350 rpm, Q = 2.75 m / min, H = 12.6 m, 12.24 W = 7.2 kW, N = 1200 rpm, D = D1 179 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery ⎛ N ⎞⎛ D ⎞ ⎛ 1200 ⎞ ∴Q2 = Q1 ⎜ ⎟⎜ ⎟ = 2.75 ⎜ ⎟ = 2.44 m /min 1350 N D ⎝ ⎠ ⎝ ⎠⎝ ⎠ 2 ⎛N ⎞ ⎛D ⎞ ⎛ 1200 ⎞ H = H1 ⎜ ⎟ ⎜ ⎟ = 12.6 ⎜ ⎟ = 9.96 m ⎝ 1350 ⎠ ⎝ N1 ⎠ ⎝ D1 ⎠ ⎛N ⎞ ⎛D ⎞ ⎛ 1200 ⎞ W = W ⎜ ⎟ ⎜ ⎟ = 7.2 ⎜ ⎟ = 5.06 kW ⎝ 1350 ⎠ ⎝ N1 ⎠ ⎝ D1 ⎠ Efficiency will remain approximately the same Compute system demand: 2 14 ⎛ L ⎞V ⎛ ⎞ = ⎜ 0.01 × + × 0.1⎟ = 0.40 m H P = ⎜ f + ΣK ⎟ ⎝ D ⎠ 2g ⎝ 0.3 ⎠ × 9.81 12.26 Q = VA = × ∴Ω P = π × 0.32 = 0.212 m2 ω Q ( gH P ) 3/ = 31.4 0.212 = 5.19 (9.81 × 0.4)3/ ω = 300 × π / 30 = 31.4 rad/s ∴Axial pump is appropriate (a) 12.28 The intersection of the pump curve with the demand curve yields H P ≅ 64 m and Q ≅ 280 m3 / h ∴ From Fig 12.6, W P ≅ 64 kW and NPSH ≅ 8.3 m 180 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery Use energy eqn to establish system demand: e⎤ ⎡ f ≅ 1.325 ⎢ln 0.27 ⎥ D⎦ ⎣ H P = Δz + f −2 0.000255 ⎤ ⎡ = 1.325 ⎢ln 0.27 × 0.45 ⎥⎦ ⎣ −2 = 0.017 0.017 × 5000 L Q2 = 192 + Q2 = 192 + 381Q2 D gA × 9.81× (π / 4) × (0.45) From Fig 12.6, at best ηP , Q ≅ 12.30 240 m3 = 0.067 m3/s, and HP ≅ 65 m 3600 s Assume three pumps in series, so that HP = 3× 65 = 195 m Then the demand head is H P = 192 + 381× (0.067) = 193.7 m Hence three pumps in series are appropriate The required power is WP = γ QHP/ηP = 9810× 0.86× 0.067 ×195/0.75 = 146,965 W W P = 146,965/746 = 197 hp or (a) For water at 80°C, pv = 46.4 × 103 Pa, and ρ = 9553 kg/m3 Write the energy equation from the inlet (section i) to the location of cavitation in the pump: pi γ 12.32 + Vi2 pv = + NPSH 2g γ ∴ NPSH = pi − pv γ Vi2 (83 − 46.4) ×103 62 + = + = 5.67 m 2g 9533 × 9.81 (b) NPSH1 = 5.67 m, N1 = 2400 rpm, N2 = 1000 rpm, D2/D1 = 2 ⎛N ⎞ ⎛D ⎞ ⎛ 1000 ⎞ ∴ NPSH = NPSH1 ⎜ ⎟ ⎜ ⎟ = 5.67 ⎜ ⎟ ⎝ 2400 ⎠ ⎝ N1 ⎠ ⎝ D1 ⎠ ( )2 = 15.8 m Given: 12.34 L = 200 m , D = 0.05 m , Δz = z − z1 = −3 m , V = m/s , ν = × 10 −7 m /s , ρ = 1593 kg/m , pv = 86.2 × 103 Pa , pa = 101 × 103 Pa Compute the pump head: 181 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery Re = × 0.05 × 10 −7 = 2.5 × 105 ( ) f = 1.325 ⎡ ln 5.74Re-0.9 ⎤ ⎣ ⎦ −2 = 1.325 ⎡ ln 5.74 × (2.5 × 105 ) −0.9 ⎤ ⎣ ⎦ −2 = 0.015 fL ⎞ V ⎛ ⎛ 0.015 × 200 ⎞ = −3 + ⎜ + = 25.0 m H P = Δz + ⎜ + ⎟ ⎟ 0.05 D ⎠ 2g ⎝ ⎠ × 9.81 ⎝ (a) Choose a radial-flow pump Use Fig P12.35 to select the size and speed: C H = 0.124, CQ = 0.0165, η = 0.75 Q = 3× ∴D = ∴ω = π 4 × 0.052 = 0.00589 m CHQ CQ2 gH P Q CQ D = = 0.124 × 0.00589 0.01652 × 9.81 × 25.0 0.00589 0.0165 × 0.090 = 0.090 m = 490 rad/s, or N = 490 × 30 π = 4680 rpm (b) Available net positive suction head: NPSH = pa − pv 101×103 − 86.2 ×103 − Δz = + = 3.95 m 1593 × 9.81 ρg (c) 182 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery Turbines ω = 120 × π / 30 = 12.6 rad/s α1 = cot −1 (2π r12b1ω / Q + cot β1 ) = cot −1 (2π × 4.52 × 0.85 × 12.6 /150 + cot 75D ) = 6.1D Vt1 = u1 + Vn1 cot β1 = ω r1 + = 12.6 × 4.5 + 12.36 Q cot β1 2π r1b1 150 cot 75D = 58.37 m/s 2π × 4.5 × 0.85 Vt2 = u2 + Vn2 cot β = ω r2 + = 12.6 × 2.5 + Q cot β 2π r2b2 150 cot100D = 29.52 m/s 2π × 2.5 × 0.85 ∴ T = ρQ(rV t1 − r2Vt2 ) = 1000 ×150(4.5 × 58.37 − 2.5 × 29.52) = 2.83 × 107 N ⋅ m = ωT = 12.6 × 2.83 × 107 = 3.57 × 108 W or 357 MW W T =W , hence Under ideal conditions ηT = 1, and W T f HT = W T / γ Q = 3.57 × 108 /(9810 × 150) = 243 m N = 240 rpm , W = 2200 kW , D = 0.9 m, W = 9kW , H1 = 7.5 m 2 W ⎛ N ⎞ ⎛ D2 ⎞ H ⎛ N ⎞ ⎛ D2 ⎞ From the similarity rules = ⎜ =⎜ ⎟ ⎜ ⎟ and ⎟ ⎜ ⎟ H1 ⎝ N1 ⎠ ⎝ D1 ⎠ W1 ⎝ N1 ⎠ ⎝ D1 ⎠ 12.38 Substitute second eqn into the first to eliminate ( N / N ), and solve for D : 1/ 3/ ⎞1/ ⎛ H ⎞3/ ⎛W ⎛ ⎞ ⎛ 45 ⎞ ∴ D1 = D2 ⎜ ⎟ ⎜ ⎟ = 0.9 ⎜ ⎟ ⎜ ⎟ = 0.221 m ⎝ 2200 ⎠ ⎝ 7.5 ⎠ ⎝ W2 ⎠ ⎝ H1 ⎠ 1/ ⎛H ⎞ N = N1 ⎜ ⎟ ⎝ H2 ⎠ ⎛ D2 ⎞ ⎛ 7.5 ⎞ ⎜ ⎟ = 240 ⎜ ⎟ D ⎝ 45 ⎠ ⎝ 1⎠ 1/ ⎛ 0.9 ⎞ ⎜ ⎟ = 399 rpm ⎝ 0.221 ⎠ 183 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery Write energy eqn from upper reservoir (loc 1) to surge tank (loc 2) and solve for Q: 1/ ⎡ ⎤ D Q = ⎢ gA2 ( z1 − z2 ) ⎥ fL ⎣ ⎦ 1/ ⎡ × 9.81 × (π / 4) × 0.855 ⎤ =⎢ (650 − 648.5) ⎥ 0.025 × 2000 ⎣ ⎦ = 0.401 m / s Apply energy eqn from loc to lower reservoir (loc 3) and determine H T : ⎛ L ⎞ Q HT = z2 − z3 − ⎜ f + K v ⎟ ⎝ D ⎠ gA2 12.40 ⎛ 0.02 × 100 ⎞ = 648.5 − 595 − ⎜ + 1⎟ ⎝ 0.85 ⎠ 0.4012 ⎛π ⎞ × 9.81× ⎜ ⎟ × 0.855 ⎝4⎠ = 53.4 m ∴ W T = γ QHTηT = 9810 × 0.401 × 53.4 × 0.9 = 1.89 × 105 W or 189 kW ∴ From Fig 12.32, use a Francis turbine A representative value of the specific speed is (Fig 12.20): ΩT ( gHT )5 / 2(9.81× 53.3)5 / ∴ω = = = 306 rad/s / ρ )1/ (W (2.67 × 105 /1000)1/ T or N = 306 × 30 / π = 2920 rpm ω = 200 × π / 30 = 20.94 rad/s and from Fig P12.42 at best efficiency (ηT = 0.8), φ ≈ 0.42, cv = 0.94 V1 = cv gHT = 0.94 × 9.81× 120 = 45.6 m/s ∴Q = 12.42 WT 4.5 × 10 = = 4.78 m / s γ H T η T 9810 × 120 × 0.8 This is the discharge from all of the jets Determine the wheel radius r : r = φ gHT 0.42 × 9.81× 120 = = 0.973 m 20.94 ω Hence, the diameter of the wheel is 2r = × 0.973 = 1.95 m Compute diameter of one jet: D j = 2r / = 1.95 / = 0.244 m, or 244 mm Let N j = no of jets Then each jet has a discharge of Q / N j and an area 184 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery Q / Nj V1 Nj = = π D2j ∴Solving for N j : Q 4.78 = = 2.24 V1 ⎛ π ⎞ ⎛π ⎞ D 45.6 0.244 × × ⎜ ⎜ ⎟ j ⎟ ⎝4 ⎠ ⎝4⎠ ∴Use three jets ΩT = ω (W T / ρ )1/ ( gHT )5 / = 20.94(4.5 × 106 /1000)1/ = 0.204 (9.81 × 120)5 / Q = 2082 / = 347 m3 / sec (one unit) HT = WT 427,300 × 746 = = 110.1 m γ QηT 9810 × 347 × 0.85 Write energy eqn from upper reservoir (loc 1) to lake (loc 2): ⎛ L ⎞ Q + HT + z2 z1 = ⎜ f + ΣK ⎟ ⎝ D ⎠ 2gA 12.44 ⎛ 0.01 × 390 ⎞ ∴ 309 = ⎜ + 0.5 ⎟ D ⎝ ⎠ which reduces to 2.4 − 347 ⎛π ⎞ × 9.81 × ⎜ ⎟ D ⎝4⎠ + 110.1 + 196.5 38,840 4980 − =0 D5 D ∴ Solving, D ≅ m ∴From Fig 12.32, a Francis or pump/turbine unit is indicated (a) Let H be the total head and Q the discharge delivered to the turbine; then HT = 0.95H = 0.95 × 305 = 289.8 m and 12.46 Q= W 10.4 × 10 T = = 4.30 m / s γHT ηT 9810 × 289.8 × 0.85 Write energy eqn from reservoir to turbine outlet: ⎛ L ⎞ Q H = HT +⎜ f +∑K⎟ ⎝ D ⎠ 2gA 4.32 ⎛ 0.02 × 3000 ⎞ ∴ 305 = 289.8 + ⎜ + 2⎟ D ⎝ ⎠ × 9.81× (π / 4)2 D 185 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery which reduces to 15.2 − 91.67 3.06 − =0 D5 D ∴Solving D = 1.45 m (b) Compute jet velocity: V1 = cv gHT = 0.98 × 9.81× 289.8 = 73.9 m/s The flow through one nozzle is Q/4 and the jet area is πD 2j / Hence π D 2j = Q / 4.3/ = = 0.0146 m V1 73.9 × 0.0146 ∴ Dj = π = 0.136 m 12.48 ⎡ 4.15 × (9.81 × 3.7)5 / ⎤ WT = 1000 ⎢ ⎥ = 4.99 × 10 W (one unit) × 50 / 30 π ⎣ ⎦ Total power developed is 9.21 × 10 W Hence, required number of units is 9.21/ 4.99 = 1.8 ∴ Use two turbines (a) H T = z1 − z − fL Q D gA 0.015 × 350 × 0.252 = 915 − 892 − = 11.8 m 0.3 × × 9.81× (0.7854 × 0.32 ) = γ QH η = 9810 × 0.25 ×11.8 × 0.85 = 2.46 ×104 W or 24.6 kW W T T T (b) Compute the specific speed of the turbine: 12.50 ω=N π 30 ∴ΩT = = 1200 × /ρ ω W T ( gHT ) 5/4 π 30 = = 126 rad/s 126 2.46 × 104 /1000 ( 9.81×11.8) 5/4 = 1.65 Hence, from Fig 12.20, a Francis turbine is appropriate (c) From Fig 12.24, the turbine with ΩT = 1.063 is chosen: CH = 0.23, CQ = 0.13, and ηT = 0.91 186 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery D= ω= CHQ2 = CQ2 gHT 0.23 × 0.252 = 0.293 m or approximately 0.30 m 0.132 × 9.81 × 11.8 Q 0.25 30 = = 71.2 rad/s or N = 71.2 × = 680 rpm 3 π CQ D 0.13 × 0.30 W T = γ QHTηT = 9810 × 0.25 × 11.8 × 0.91 = 2.63 × 10 W, or 26.3 kW Calculate a new specific speed based on the final design data: ΩT = 71.2 × 2.63 ×104 /1000 (9.81×11.8)5/4 = 0.96 An acceptable value according to Fig 12.20 187 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery 188 © 2012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part [...]... W or 24.6 kW W T T T (b) Compute the specific speed of the turbine: 12. 50 ω=N π 30 ∴ΩT = = 120 0 × /ρ ω W T ( gHT ) 5/4 π 30 = = 126 rad/s 126 2.46 × 104 /1000 ( 9.81×11.8) 5/4 = 1.65 Hence, from Fig 12. 20, a Francis turbine is appropriate (c) From Fig 12. 24, the turbine with ΩT = 1.063 is chosen: CH = 0.23, CQ = 0.13, and ηT = 0.91 186 © 2 012 Cengage Learning All Rights Reserved May not be scanned,... z2 z1 = ⎜ f + ΣK ⎟ 2 ⎝ D ⎠ 2gA 12. 44 ⎛ 0.01 × 390 ⎞ ∴ 309 = ⎜ + 0.5 ⎟ D ⎝ ⎠ which reduces to 2.4 − 347 2 ⎛π ⎞ 2 × 9.81 × ⎜ ⎟ D 4 ⎝4⎠ 2 + 110.1 + 196.5 38,840 4980 − 4 =0 D5 D ∴ Solving, D ≅ 8 m ∴From Fig 12. 32, a Francis or pump /turbine unit is indicated (a) Let H be the total head and Q the discharge delivered to the turbine; then HT = 0.95H = 0.95 × 305 = 289.8 m and 12. 46 Q= W 10.4 × 10 6 T = =... final design data: ΩT = 71.2 × 2.63 ×104 /1000 (9.81×11.8)5/4 = 0.96 An acceptable value according to Fig 12. 20 187 © 2 012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery 188 © 2 012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly... 9810 × 289.8 × 0.85 Write energy eqn from reservoir to turbine outlet: 2 ⎛ L ⎞ Q H = HT +⎜ f +∑K⎟ 2 ⎝ D ⎠ 2gA 4.32 ⎛ 0.02 × 3000 ⎞ ∴ 305 = 289.8 + ⎜ + 2⎟ D ⎝ ⎠ 2 × 9.81× (π / 4)2 D 4 185 © 2 012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery which reduces to 15.2 − 91.67 3.06...Chapter 12 / Turbomachinery Q / Nj V1 Nj = = π 4 D2j ∴Solving for N j : Q 1 4.78 = = 2.24 V1 ⎛ π 2 ⎞ ⎛π ⎞ 2 D 45.6 0.244 × × ⎜ ⎜ ⎟ j ⎟ ⎝4 ⎠ ⎝4⎠ ∴Use three jets ΩT = ω (W T / ρ )1/ 2 ( gHT )5 / 4 = 20.94(4.5 × 106 /1000)1/ 2 = 0.204 (9.81 × 120 )5 / 4 Q = 2082 / 6 = 347 m3 / sec (one unit) HT = WT 427,300 × 746 = = 110.1... chosen: CH = 0.23, CQ = 0.13, and ηT = 0.91 186 © 2 012 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 12 / Turbomachinery D= ω= 4 CHQ2 = CQ2 gHT 4 0.23 × 0.252 = 0.293 m or approximately 0.30 m 0.132 × 9.81 × 11.8 Q 0.25 30 = = 71.2 rad/s or N = 71.2 × = 680 rpm 3 3 π CQ D 0.13 × 0.30 W T = γ QHTηT... area is πD 2j / 4 Hence π D 2j 4 = Q / 4 4.3/ 4 = = 0.0146 m 2 V1 73.9 4 × 0.0146 ∴ Dj = π = 0.136 m 2 12. 48 ⎡ 4.15 × (9.81 × 3.7)5 / 4 ⎤ 6 WT = 1000 ⎢ ⎥ = 4.99 × 10 W (one unit) × 50 / 30 π ⎣ ⎦ Total power developed is 9.21 × 10 6 W Hence, required number of units is 9.21/ 4.99 = 1.8 ∴ Use two turbines (a) H T = z1 − z 2 − fL Q 2 D 2 gA 2 0.015 × 350 × 0.252 = 915 − 892 − = 11.8 m 0.3 × 2 × 9.81×