ACE Engineering Academy. Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru Chennai Vijayawada Vizag Tirupati Kukatpally Kolkata

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Eleanore Cook
5 years ago
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1 Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

: : Meanial Engg. _ ESE MAINS 0(a). Sol: 0(b). Sol: p F C.G. A F ga 9.8000 9 5 0.8 kn I M C G sin A p 5 6 5 5 m p F5 sin p sin 0.8 5 / 5 5 0.8 5 6.

2 : : Meanial Engg. _ ESE MAINS 0(a). Sol: 0(b). Sol: p F C.G. A F ga kn I M C G sin A p m p F5 sin p sin / kn-m Consider te ontrol volume as sown in te diagram. Te only external ore ating on te ontrol volume in orizontal diretion is te spring ore (F).v 60 k kn/m Applying linear momentum equation or te ontrol volume we get, F m out m in m.v () t F F ( F ats in ve ' x' diretion ) m out ( No mass is going out o C..) m a a 0 t in m 0 ( 0, ree surae veloity) equation () beomes F 0 A + 0 F A N F k Note: m m 000 Even toug mass is inreasing inside te C.. m is zero beause veloity o t tat mass (wi is equal to ree surae veloity) is zero. Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

3 : : Test Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

4 : : Meanial Engg. _ ESE MAINS 0(). Sol: Assumptions: (i) ressure is uniorm at every ross-setion. (ii) eloity over ea ross-setion is uniorm and is equal to average veloity. 0(d). Sol: Assumption: Flow over roug surae as turbulent boundary layer and low over smoot surae as laminar boundary layer. Smoot: aminar Flow Roug: Turbulent Flow et and be te pressures at setion () and () respetively. Te linear momentum equation or te ontrol volume is, F m out m in m t ( )A (A )( ) (A )( ) ( ) () Applying ernoulli s equation between () and () we get g g g g g i.e. g g g g (x) du less dy Smoot: aminar Flow (x) du more dy For streamlined objets like aerooils pressure drag is minimum (due to small wake) and skin rition drag ontributes mainly to total drag. Te turbulent boundary layer as more veloity gradient at surae. Hene it as more wall sear stress wi results in iger drag ore. For aerooils (F ) smoot < (F ) roug Roug: Turbulent Flow g g Note: Te ontrol volume starts just ater te joint/expansion. Hene pressure ore at inlet is A. 85 separation points 0 In ase o blunt objets like spere bot pressure drag and skin rition drag are present. Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

5 : 5 : Test Te turbulent boundary layer is more resistant to separation due to ig momentum lux. Hene omparatively it as delayed separation wi results in narrower wake. Even toug skin rition drag is more in ase o low over roug spere, overall drag is redued beause o omparatively mu less pressure drag due to narrower wake. 0(e). Sol: Given: For speres, (F ) smoot > (F ) roug r rsinos; rsin ontinuity equation in ylindrial oordinate system is, r r r r 0 r r r HS rr sin os Z r r + rsin r rsin os rsin os 0 RHS as ontinuity equation is satisied low is possible r r r r r r rsin rsin os r r r r r r r sin rsin r rsin r os. r sin [ sin ] 0(a). 0 low is not irrotational. Sol: et us onsider te oordinate system as sown in te igure. Te body ores ating on te luid are gravity (in downward diretion) and inertia ore (in ve x diretion) l Ater resolving g in x and y diretion we get g x a gsin g y gos rom pressure ield equations g x and y C x A x gsin a () g y gos () y Equations () and () sow tat pressure dereases in x as well as y diretion. 0 A +l(gos) a g sin a g os g Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

6 : 6 : Meanial Engg. _ ESE MAINS lgos () i.e. l(gsin+a) () C + l(gos) l(gsin+a) + l(gos) i.e. C l(gsin+gos+a) () As pressure distribution on ae A as linear variation. F 0 A +l(gsin+a) A A 0 gos F C similarly C gsin gos a gsin a gsin gos a (5) F C F A gsin a m(gsin+a) [ l m] F C F A mgsin ma i.e. F A gos () Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

7 : 7 : Test 0(b). Sol: Assumptions: (i) ressure at te needle opening is atmosperi pressure. (ii) Frition between piston and syringe surae is negleted. Applying ontinuity equation between () and () we get, F atm iston A A A A 0. m/s Applying ernoulli s equation between () and () we get, (or) (or) g g g g g atm g Case-I: Invisid luid, 0 atm () i.e. atm 80 a From ree body diagram o te piston F ( atm )A N Case-II: isous luid Te pressure drop () aross needle is given by (). 560 a g equation beomes atm a F ( atm )A N 0.0 g g Sol: Applying ernoulli s equation between () and () g g g g Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

8 : 8 : Meanial Engg. _ ESE MAINS g g (or) Q / a Q / a g Q (or) g a a a a Q g a a () Equation () gives atual disarge troug venture meter i.e. disarge wit loss. Te ideal or teoretial disarge is obtained wen loss is onsidered zero. Q t aa g () a a Te oeiient o disarge or enturi meter is deined as C d Q Q at t Applying ernoulli s equation between () and () we get g g g g Z Z i.e H Z g g i.e. H Z g g g Z [ Hydrostati law is valid between () and () as streamlines are parallel] 0 H H Z Z g g g Z gh g Z () Similarly applying ernoulli s equation between () and () we get g g g g Z Z 0(a). Sol: i.e g g () As water level in smaller tank remains same Q in Q out () d () () d () H Z d From () and () g g H d Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

9 : 9 : Test 0(b). H H H d d d Sol: Consider te ree body diagram o te blok wen it is travelling wit terminal veloity along te inline wen te terminal veloity is reaed te blok travels wit onstant veloity and ene aeleration is zero. Applying Newton s law along te inline A Wsin A Wsin d, dy A W Wsin A ma 0 Wsin A A assu min g linear proile Wos W Wsin mgsin () A sin m/s Consider te ree body diagram o te blok wen it is travelling along te roug surae. Again applying Newton s nd law to te orizontal motion o te blok we get. W k W k W ma k mg ma a k g () Tus blok retards on te roug surae by deeteration o magnitude g k. Applying tird kinemati equation to te blok. u +as 0 (7.88) + ( 0.9.8) S S 5.7 m X stopping S 5.7 m 0() (i). Sol: Given: m.600 kg/m 050 kg/m x 0.06 m Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

10 : 0 : Meanial Engg. _ ESE MAINS m 600 x m g m/s.5 km/r 0() (ii). Sol: M G H Given: s 850 kg/m 000 kg/m m H 0.8m or vertial equilibrium F W g s H g s 850 H m 000 0(a) m T K ggm I / A ggm / seonds / Sol: aminar oundary ayer: x (0.5m) 5.8x 5.8x Re x x i.e. (0.5m) we ave u x, y y x y x x () GM I min G Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

11 u 0.5m,.5mm () Solving above equation by trial and error we get Now w.5 m/s x Re x x 5.8 Re x mm C x x. a 0.7 Re x Turbulent oundary ayer: As ree stream veloity remains same.5 m/s Re x x x x x Re mm : : Test u x, y Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata y x 7.50 u(0.5m,.5mm) w 0(b). C x x Re x 0.. m/s w a.750 aminar Turbulent.5 m/s.5 m/s u (0.5m,.5 mm). m/s. m/s w (0.5 m). a 5 a (0.5 m). mm 7.9 mm Sol: Given: u We know w () (einition o ) Te ead loss by ary-weisba equation is given by () (einition o ) g Consider te equilibrium o luid mass lowing troug te pipe 7

12 : : Meanial Engg. _ ESE MAINS Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata w w g g w () equating equation () and () g g w w () Now, rom equation () and () we get u * u 8.75 u 8.75 u. 0(). Sol: Given, u u ( + y + y ). at y 0, 0 0 u ( ) 0. at y, u 0 0 u (0 + + ) (). at y, u u 0 u u Or [ rom () ] w w y

13 : : Test i.e., now, u u u y y y y u Q uda u Average veloity, i.e, 05(a). Sol: Given, 50 ka, H 5 m, 0.0, 0 m, 0 0 m, i 5 m, k entry 0, H s 0, s 0 y y 0 y y u u u dy u Q A u 0 Case I : Water is going out o tank. Assumption: ressure at pipe exit is atmosperi. Applying ernoulli s equation between water surae in tank and pipe exit we get. Z Z k g g g g Z g g g Now, 5m A o i () m Substituting in equation () we get m/s Q A 5 g o i i Q 0. m /s Case II: Water is going into te tank Assumption: 50 ka (Gauge) o 0 m, i 5 m 0 m entry g Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

14 : : Meanial Engg. _ ESE MAINS. Te pump takes water rom atmosperi ondition.. Sution ead o te pump is negligible. Applying ernoulli s equation between & we get 50 ka 05(b). Sol: Assumptions: (i) ine o ation o drag ore passes troug te entre o emispere. (ii) Torque generated by te drag ores on te two ups aligned to te low is negligible. F l ump H l l l + 6/ 5 m H s 0 F Z H Z K p s d exit g g g g i.e. H Z d K exit () g g [Note: Even toug pipe is well rounded at joint its exit loss oeiient will be beause all K.E. in pipe is lost due to mixing inside te tank] Sine te disarge is same, 0.9 m/s 500 H m gqh 0 p kw 0.85 g F C A N Similarly C F.7 A N Te torque generated on te sat o te anemometer is given by T F F N-m. Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

15 : 5 : Test Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

16 : 6 : Meanial Engg. _ ESE MAINS 05(). Sol: (i) Kinemati similarity: Te model and te prototype are said to be kinematially similar i te ratio o magnitudes o veloities at similar points is onstant and diretion o veloities at tose points is same. Te kinemati similarity is obtained wen streamline pattern or te model and prototype is same. p m m p onst (ii) istorted Model: I lengt sale ratio or model and prototype is not same in dierent diretion ten te model is alled distorted model i.e. Hm Hp vm vp In many situations it is impratial to maintain same sale ratios in dierent diretion tereore distorted models ave to be reated. For example i same sale ratio is maintained or river model in orizontal as well as vertial diretion ten vertial dimension o model river will be very small and ene unneessary surae tension eets will beome dominant wi are not present in atual river. (iii) Inomplete similarity: For omplete similarity all independent terms are needed to be mated or model and prototype. However it is always not possible to mat all te independent terms. I one or more independent term are not mated between model and prototype ten su a similarity is alled inomplete similarity. For example, te total drag o sip (rition drag + wave drag) depends on bot Reynolds number (Re) and Froude number (Fr). However it is not easy to mat Re & Fr simultaneously ( r / ondition as to be satisied). r Tereore, only Fr is mated or model and prototype to predit wave drag. Te ritional drag is alulated analytially using boundary layer teory. Hyderabad eli opal une ubaneswar uknow atna engaluru Cennai ijayawada izag Tirupati Kukatpally Kolkata

Department of Mechanical Engineering

Department of Mechanical Engineering Department o Mehanial Engineering AMEE41 / ATO4 Aerodynamis Instrutor: Marios M. Fyrillas Email: eng.m@it.a.y Homework Assignment #4 QESTION 1 Consider the boundary layer low on a lat plate o width b (shown