"Ans. 5.81As shown in the above diagram, there is a pump which is pumping the fluid from the lake tothe reservoir, so calculate the power of the pump required to reach the water in the reservoir,Apply the Bernoulli’s principle at the lake point and the reservoir point,So,? 2 P v P V 1 1 2 2 ?Z ? ?h ?h ? ?Z ? 12 pL r22 g r g whereP ? 0,Z ? 0,V ? 01 1 1 Z ? 20 ft 2 So put the values in the above equation,So,P 2 h ?h ? ?Z pL 2 r Calculate the rate of flow of water, and convert the unit,3 ?? 1000 gal 1ft 1 Q ? ?( ) ?( ) ?? 10min 7.48gal 60?? 3 Q ? 0.223 ft /s Now,W 3 ?550 p h ? ? ? 119 ft p rQ 62.4 ?0.223 If P ? 2atm ? 2 ?14.7 ?144 2 4230 h ?h ? ? 20 ?h ? 87.8 ft p L L 62.4 Thus,if h ?h ?87.8 ft ?119 ?87.8 ? 31.2 ft 1 pThe given pump will work for P=2atmIf lb P?? 3atm 6350 ,then2 2 ft 6350 h ?h ? ? 20 ?h ?122 p L L 62.4 Thus, If this pump is to work 119 ft?? h 122 ft L h ??3 ft L Since it is not possible to have h ? 0Thus pump will not work for P ? 3atm .L 2 Ans 3.61Apply Bernoulli’s principle at both the points,22 P V P V 1 1 2 2 ? ?ZZ ? ? ? 12 r22 g r g So,() PP ? 21 Vg ? 2 1 r Here the total height term applies for the equilibrium,So,P ?rl ?r h ? P ?r() l ?h 12 m P ?P ?() r ?r h 21 m Put the value in the equation (1) () rr ? m V ? 2g h 2 r 3 1.07?? 9.8 10 Vm ?? 2(9.81)( 1)0.02 2 r applythecantinuety principle Q ? AV 10.49 ?3 Q ? 3.99 ?10 ?1 r 3 here,/ riskN m Now there are three kinds of fluid given and according to that the illustration is given asbelowkN Fluid r( ) Q,(S) 3 m ?3 (a) Water 9.80 1.06 ?10 ?3 (b) gasoline 6.67 3.02 ?10 ?3 (c) air 12 ?10 0..118 Ans.3.67As shown in the figure, the (1) pipe delivers the fluid as the cross section of the other pipes,First of all calculate the velocity at the first section, apply the continuity principle,Q ?Q ?Q ?Q 1 2 3 4 HereQisthedisch argeof the fluid. So by the continuity principle,AV ? AV ? AV ? AV 1 1 2 2 3 3 4 4 10 ? 0.07 ?VQ ? 0.2 ? 20 ? 24 10 ? 0.07VQ ? 4 ? 24 QV?? 0.07 6 ......(1) 42 Apply the Bernoulli’s principle for the section first and the second pipe, 2 P V P V 1 1 2 2 ? ?zz ? ? ? 12 ?? g22 g g g 2 V 10 10 5 2 ? ? ? ?? g22 g g g 2 5 V 2 ?? 5 1.94 2 V ? 3.89 ft / sec 2 So the discharge through the second pipe,Q ? AV 2 2 2 Q?? 0.07 3.892 3 Q ? 0.27 ft / sec 2 Calculate the value of Q4 from the equation (1)Q ? 6 ? 0.07 ? 0.27 4 3 Q ? 5.98 ft / sec 4 Ans. 5.76From the energy equation,22 P V P V 1 1 2 2 ?Z ? ?h ?h ? ?Z ? 12 sL r22 g r g HereP ? 0,P ? 0,andV ? 0 1 2 1 Thus, 2 V 2 h ?() Z ?Z ?h ? sL 12 2g Note: Since this is a turbine, h ? 0Let hh ?? , where h ? 0s Ts T From the above,2 V 2 h ?() Z ?Z ?h ?TL 12 2g Also, the power is given by2 V 2 W ?rQh ?rQ((Z ?Z ) ?h ? )turb T 12 L 2g The maximum power output would occur if there were no losses (h ? 0) and negligibleL kinetic energy at the exit (V ? 0,;l argediameteroutlet) 2 Thus, 2 W ?rQ(Z ?Z ) ? 1000 ? (3.14 ? 0.5 ? 6)(100) turb 12 max W ? 471000 HP turb max Ans. 3.24(a) Apply Bernoulli’s equation,1 2 P ? ?v ? z ?cons. 2 So, Z?? Z Z 1 2 3 11 22 P ??? v ? P ? v,0 P ? 1 1 3 3 1 22 1 22 P?? ?() v v 3 1 3 2 Here, 88ft / sec V?? 200mph( ) 293ft / sec 1 60mph 88ft /s V?? 273mph( ) 400ft /s 3 60mph So ?3 3 2 2 2 2 P ? 0.5(2.05 ?10 slugs / ft )(293 ? 400 )ft /s 3 2 P ??76lb / ft 3 () b 2 ?3. 3 2 2 P ? 0.5 ?v ? 0.5(2.05 ?10 s lg/ ft )(293ft /s) ? 88lb / ft 2 1Ans.3.32Apply Bernoulli’s principle,22 P V P V 1 1 2 2 ?ZZ ? ? ? ? 12 r 22 g r g where 2 2 2 2 P?? 10lb /in (144in / ft ) 1440lb / ft 1 P ? 0 2 Z ? 15ft,Z ? 0, V ? 0 and V ? 40ft /s 1 2 1 2 So,22 1440lb / ft (40ft /s) ?? 15ft 2 r 2(32.2ft /s ) 3 r ? 146.3lb / ft Hence 3 r 146lb / ft SG ? ? ? 2.34 3 r 62.4lb / ft water Ans.3.48The collapse of the pipe is called as cavitations, so we have to avoid cavitations,At the water level in the surge tank the particulars are,P ? 0, V ? 0,Z ? 0 1 1 1 At the stagnation point of the pipe,P ? 1atm,V ? ?,Z ? 2m2 2 2 Apply Bernoulli’s principle,2 101kpa V 2 02 ? ? ? ? 2 2 v 2 0 ? ?10.3 ? ? 22 v ?4/ m s 2Now, the cross section of the siphone is same at each point so the velocity of the dischargewill be same, Apply Bernoulli’s theorem,V ? 2gh disc 2 v disc h ? 2g 16 h ? 2 ? 9.8 hm ? 0.81 Ans. 3.54At the entrance the particulars are,A eV?? 0.5m /s, 1.8e A ar Calculate the pressure difference,Apply Bernoulli’s principle,22 VV ? 21 PP ??12 2g Apply the continuity principle,AV ? AV 1 1 2 2 AV 12 ?? 1.8 AV21 VV ? 1.8 21 V ? 0.9m /s 2 Put the values in the above equation 22 VV ? 21 PP ?? 12 2g 22 0.9 ? 0.5 Pp?? 12 2 ? 9.8 2 P?? P 0.0285N /m 12 Ans. 3.20Some parts missing(figure) "