fluid_mechanics
678 Chapter 12 ■ Turbomachines E XAMPLE 12.6 Pelton Wheel Turbine Characteristics GIVEN Water to drive a Pelton wheel is supplied through a pipe from a lake as indicated in Fig. E12.6a. The head loss due to friction in the pipe is important, but minor losses can be neglected. FIND (a) Determine the nozzle diameter, D 1 , that will give the maximum power output. (b) Determine the maximum power and the angular velocity of the rotor at the conditions found in part (a). (0) z 0 = 200 ft = 1000 ft, f = 0.02 z 1 = 0 D = 8 in D 1 (1) β = 150 deg 2R = 3 ft F I G U R E E12.6a SOLUTION (a) As indicated by Eq. 12.51, the power output depends on the flowrate, Q m /, and the jet speed at the nozzle exit, V 1 , both of which depend on the diameter of the nozzle, D 1 , and the head loss associated with the supply pipe. That is The nozzle exit speed, V 1 , can be obtained by applying the energy equation (Eq. 5.85) between a point on the lake surface (where V 0 p 0 0) and the nozzle outlet (where z 1 p 1 0) to give where the head loss is given in terms of the friction factor, f, as (see Eq. 8.34) The speed, V, of the fluid in the pipe of diameter D is obtained from the continuity equation We have neglected minor losses associated with the pipe entrance and the nozzle. With the given data, Eq. 2 becomes or where D 1 is in feet. W shaft QU1U V 1 211 cos 2 z 0 V 2 1 2g h L h L f / D V 2 2g V A 1V 1 A aD 1 D b 2 V 1 z 0 c 1 f / D aD 1 D b 4 d V 2 1 2g 1/2 2gz 0 V 1 £ 1 f / D aD 1 § D b4 2132.2 ft/s 2 21200 ft2 C 1000 ft 1 0.02 a 8/12 ft b a D 4 1 8/12 b § 113.5 21 152 D 4 1 1/2 (4) (1) (2) (3) By combining Eqs. 1 and 4 and using Q D 2 1V 1 /4 we obtain the power as a function of D 1 and U as W shaft 323 UD2 1 CU 21 152 D 4 1 where U is in feet per second and W shaft is in ft lb/s. These results are plotted as a function of U for various values of D 1 in Fig. 12.6b. As shown by Eq. 12.52, the maximum power (in terms of its variation with U) occurs when U V 1 /2, which, when used with Eqs. 4 and 5, gives The maximum power possible occurs when which according to Eq. 6 can be found as dW shaft dD 1 35,000 30,000 25,000 20,000 −W shaft , ft • lb/s 15,000 10,000 5,000 F I G U R E E12.6b W 1.04 10 6 D 2 1 shaft 11 152 D 4 12 3/2 113.5 21 152 D 4 1 1.04 10 6 2 D 1 c 11 152 D 4 12 3/2 a 3 2 b 411522 D 5 1 11 152 D 4 12 5/2 d 0 D 1 = 0.300 ft D 1 = 0.239 ft § (5) (6) dW shaft/dD 1 0, 0 0 20 40 60 80 100 U, ft/s D 1 = 0.200 ft
12.8 Turbines 679 or Thus, the nozzle diameter for maximum power output is (Ans) (b) The corresponding maximum power can be determined from Eq. 6 as or W shaft 1.04 106 10.2392 2 31 15210.2392 4 4 3 2 3.25 104 ft lb/s W shaft 3.25 10 4 ft lb/s 59.0 hp 304 D 4 1 1 D 1 0.239 ft (Ans) The rotor speed at the maximum power condition can be obtained from U R V 1 2 where V 1 is given by Eq. 4. Thus, 113.5 V 1 2R 21 15210.2392 4 ft/s 2a 3 2 ftb 1 hp 550 ftlb/s COMMENT The reason that an optimum diameter nozzle exists can be explained as follows. A larger diameter nozzle will allow a larger flowrate, but will produce a smaller jet velocity because of the head loss within the supply side. A smaller diameter nozzle will reduce the flowrate but will produce a larger jet velocity. Since the power depends on a product combination of flowrate and jet velocity (see Eq. 1), there is an optimum-diameter nozzle that gives the maximum power. These results can be generalized (i.e., without regard to the specific parameter values of this problem) by considering Eqs. 1 and 3 and the condition that U V 1 /2 to obtain W shaft UV 12 p r 11 cos b2 16 12gz 0 2 3/2 D 2 1^a1 f / 3/2 D 5 D4 1b By setting dW shaft/dD 1 0, it can be shown (see Problem 12.67) that the maximum power occurs when D 1 D^ a2f / D b 1/4 which gives the same results obtained earlier for the specific parameters of the example problem. Note that the optimum condition depends only on the friction factor and the length-to-diameter ratio of the supply pipe. What happens if the supply pipe is frictionless or of essentially zero length? 30.9 rad/s 1 rev/2 rad 60 s/min 295 rpm (Ans) In previous chapters we mainly treated turbines 1and pumps2 as “black boxes” in the flow that removed 1or added2 energy to the fluid. We treated these devices as objects that removed a certain shaft work head from or added a certain shaft work head to the fluid. The relationship between the shaft work head and the power output as described by the moment-of-momentum considerations is illustrated in Example 12.7. E XAMPLE 12.7 Maximum Power Output for a Pelton Wheel Turbine GIVEN Water flows through the Pelton wheel turbine shown in Fig. 12.24. For simplicity we assume that the water is turned 180° by the blade. FIND Show, based on the energy equation 1Eq. 5.842, that the maximum power output occurs when the absolute velocity of the fluid exiting the turbine is zero. SOLUTION As indicated by Eq. 12.51, the shaft power of the turbine is given by W # shaft rQU1U V 1 211 cos b2 2rQ1U 2 V 1 U2 For this impulse turbine with b 180°, the velocity triangles simplify into the diagram types shown in Fig. E12.7. Three possibilities are indicated: (a) the exit absolute velocity, V 2 , is directed back toward the nozzle, (1) (b) the absolute velocity at the exit is zero, or (c) the exiting stream flows in the direction of the incoming stream. According to Eq. 12.52, the maximum power occurs when U V 12, which corresponds to the situation shown in Fig. E12.7b, that is, U V 12 W 1 . If viscous effects are negligible, then W 1 W 2 and we have U W 2 , which gives V 2 0 (Ans)
- Page 652 and 653: 628 Chapter 11 ■ Compressible Flo
- Page 654 and 655: 630 Chapter 11 ■ Compressible Flo
- Page 656 and 657: 632 Chapter 11 ■ Compressible Flo
- Page 658 and 659: 634 Chapter 11 ■ Compressible Flo
- Page 660 and 661: 636 Chapter 11 ■ Compressible Flo
- Page 662 and 663: 638 Chapter 11 ■ Compressible Flo
- Page 664 and 665: 640 Chapter 11 ■ Compressible Flo
- Page 666 and 667: 642 Chapter 11 ■ Compressible Flo
- Page 668 and 669: 644 Chapter 11 ■ Compressible Flo
- Page 670 and 671: 646 Chapter 12 ■ Turbomachines Ex
- Page 672 and 673: 648 Chapter 12 ■ Turbomachines Q
- Page 674 and 675: 650 Chapter 12 ■ Turbomachines E
- Page 676 and 677: 652 Chapter 12 ■ Turbomachines Th
- Page 678 and 679: 654 Chapter 12 ■ Turbomachines F
- Page 680 and 681: 656 Chapter 12 ■ Turbomachines Co
- Page 682 and 683: 658 Chapter 12 ■ Turbomachines Th
- Page 684 and 685: 660 Chapter 12 ■ Turbomachines 50
- Page 686 and 687: 662 Chapter 12 ■ Turbomachines Si
- Page 688 and 689: 664 Chapter 12 ■ Turbomachines (2
- Page 690 and 691: 666 Chapter 12 ■ Turbomachines cu
- Page 692 and 693: 668 Chapter 12 ■ Turbomachines SO
- Page 694 and 695: 670 Chapter 12 ■ Turbomachines h
- Page 696 and 697: 672 Chapter 12 ■ Turbomachines He
- Page 698 and 699: 674 Chapter 12 ■ Turbomachines Ro
- Page 700 and 701: V 1 W 1 = W 2 U 676 Chapter 12 ■
- Page 704 and 705: 680 Chapter 12 ■ Turbomachines U
- Page 706 and 707: 682 Chapter 12 ■ Turbomachines V1
- Page 708 and 709: 684 Chapter 12 ■ Turbomachines Fo
- Page 710 and 711: 686 Chapter 12 ■ Turbomachines Co
- Page 712 and 713: 688 Chapter 12 ■ Turbomachines Co
- Page 714 and 715: 690 Chapter 12 ■ Turbomachines us
- Page 716 and 717: 692 Chapter 12 ■ Turbomachines es
- Page 718 and 719: 694 Chapter 12 ■ Turbomachines 1
- Page 720 and 721: 696 Chapter 12 ■ Turbomachines 55
- Page 722 and 723: 698 Chapter 12 ■ Turbomachines Se
- Page 724 and 725: 700 Chapter 12 ■ Turbomachines Co
- Page 726 and 727: 702 Appendix A ■ Computational Fl
- Page 728 and 729: 704 Appendix A ■ Computational Fl
- Page 730 and 731: 706 Appendix A ■ Computational Fl
- Page 732 and 733: 708 Appendix A ■ Computational Fl
- Page 734 and 735: 710 Appendix A ■ Computational Fl
- Page 736 and 737: 712 Appendix A ■ Computational Fl
- Page 738 and 739: A ppendix B Physical Properties of
- Page 740 and 741: 716 Appendix B ■ Physical Propert
- Page 742 and 743: 718 Appendix B ■ Physical Propert
- Page 744 and 745: 720 Appendix C ■ Properties of th
- Page 746 and 747: 722 Appendix D ■ Compressible Flo
- Page 748 and 749: 724 Appendix D ■ Compressible Flo
- Page 751 and 752: Answers to Selected Even-Numbered H
678 Chapter 12 ■ Turbomachines<br />
E XAMPLE 12.6<br />
Pelton Wheel Turbine Characteristics<br />
GIVEN Water to drive a Pelton wheel is supplied through a<br />
pipe from a lake as indicated in Fig. E12.6a. The head loss due to<br />
friction in the pipe is important, but minor losses can be<br />
neglected.<br />
FIND (a) Determine the nozzle diameter, D 1 , that will give<br />
the maximum power output.<br />
(b) Determine the maximum power and the angular velocity of<br />
the rotor at the conditions found in part (a).<br />
(0)<br />
z 0 = 200 ft<br />
= 1000 ft, f = 0.02<br />
z 1 = 0<br />
D = 8 in<br />
D 1<br />
(1)<br />
β = 150 deg<br />
2R = 3 ft<br />
F I G U R E E12.6a<br />
SOLUTION<br />
(a) As indicated by Eq. 12.51, the power output depends on the<br />
flowrate, Q m /, and the jet speed at the nozzle exit, V 1 , both of<br />
which depend on the diameter of the nozzle, D 1 , and the head loss<br />
associated with the supply pipe. That is<br />
The nozzle exit speed, V 1 , can be obtained by applying the<br />
energy equation (Eq. 5.85) between a point on the lake surface<br />
(where V 0 p 0 0) and the nozzle outlet (where<br />
z 1 p 1 0) to give<br />
where the head loss is given in terms of the friction factor, f, as<br />
(see Eq. 8.34)<br />
The speed, V, of the <strong>fluid</strong> in the pipe of diameter D is obtained<br />
from the continuity equation<br />
We have neglected minor losses associated with the pipe entrance<br />
and the nozzle. With the given data, Eq. 2 becomes<br />
or<br />
where D 1 is in feet.<br />
W shaft QU1U V 1 211 cos 2<br />
z 0 V 2 1<br />
2g h L<br />
h L f / D<br />
V 2<br />
2g<br />
V A 1V 1<br />
A aD 1<br />
D b 2<br />
V 1<br />
z 0 c 1 f / D aD 1<br />
D b 4<br />
d V 2 1<br />
2g<br />
1/2<br />
2gz 0<br />
V 1 £<br />
1 f / D aD 1 §<br />
D b4<br />
2132.2 ft/s 2 21200 ft2<br />
C 1000 ft<br />
1 0.02 a<br />
8/12 ft b a D 4<br />
1<br />
8/12 b §<br />
113.5<br />
<br />
21 152 D 4 1<br />
1/2<br />
(4)<br />
(1)<br />
(2)<br />
(3)<br />
By combining Eqs. 1 and 4 and using Q D 2 1V 1 /4 we obtain<br />
the power as a function of D 1 and U as<br />
W shaft 323 UD2 1<br />
CU <br />
21 152 D 4 1<br />
where U is in feet per second and W shaft is in ft lb/s. These results<br />
are plotted as a function of U for various values of D 1 in<br />
Fig. 12.6b.<br />
As shown by Eq. 12.52, the maximum power (in terms of its<br />
variation with U) occurs when U V 1 /2, which, when used with<br />
Eqs. 4 and 5, gives<br />
The maximum power possible occurs when<br />
which according to Eq. 6 can be found as<br />
dW shaft<br />
dD 1<br />
35,000<br />
30,000<br />
25,000<br />
20,000<br />
−W shaft , ft • lb/s<br />
15,000<br />
10,000<br />
5,000<br />
F I G U R E E12.6b<br />
W 1.04 10 6 D 2 1<br />
shaft <br />
11 152 D 4 12 3/2<br />
113.5<br />
21 152 D 4 1<br />
1.04 10 6 2 D 1<br />
c<br />
11 152 D 4 12 3/2<br />
a 3 2 b 411522 D 5 1<br />
11 152 D 4 12 5/2 d 0<br />
D 1 = 0.300 ft<br />
D 1 = 0.239 ft<br />
§<br />
(5)<br />
(6)<br />
dW shaft/dD 1 0,<br />
0<br />
0 20 40 60 80 100<br />
U, ft/s<br />
D 1 = 0.200 ft