CHAPTER 4 The Integral Forms of the Fundamental Laws

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1 CHAPTER 4 The Integral Forms of the Fundamental Laws FE-type Exam Review Problems: Problems 4- to (B) 4 (D) 4 (A) 44 (D) p m ρa A π kg/s RT 87 9 Refer to the circle of Problem 47: 757 Q A ( π 4 4 sin 755 ) 56 m /s 6 W P Q g p p W P W P 4 kw and energy req'd 47 kw (A) p p + g p + p 7 Pa Manometer: ρ H + p g + p or g p ρ g g (C) 47 (B) 796 Energy: K K Combine the equations: 98 8 m/s Δp Q 4 hl K 796 m/s g A π K K Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

2 48 (C) 49 (D) 4 (A) 4 (A) 4 (C) W P Q g W P Δp + W P 6 QΔ p kw 8 kw η 89 pb pb 46 Pa In the above energy equation we used Q hl K with 44 m/s g A π Q A π m/ s Energy surface to entrance: p H K P g g H P m W QH / η 98 4 / 75 6 W P P P After the pressure is found, that pressure is multiplied by the area of the window The pressure is relatively constant over the area p p (65 ) p Pa g g 98 pa F ρq ( ) 85 π 5 F 7(65 ) F 7 5 N 4 (D) Fx m ( x x) 5(5 cos 6 5) 5 N 44 (A) 45 (A) F m ( ) π 6 (4cos 45 4) 884 N x rx rx Power F W x B Let the vehicle move to the right The scoop then diverts the water to the right Then, F m ( x x) 5 6 [6 ( 6)] 7 N 8 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

3 Basic Laws 46 b) The energy transferred to or from the system must be ero: Q W 4 b) The conservation of mass d) The energy equation System-to-Control-olume Transformation ˆ ˆ ˆ 77( ˆ ˆ) n i j i+ j ˆ ˆ ˆ n nˆ i 77( + ) i j 77 fps nˆ ˆi (866ˆi 5 ˆj ) 866 fps n nˆ ˆi ( ˆj ) n nˆ 866ˆi 5ˆj nˆ ˆj ( Bn ˆ ) A 5(5ˆi ˆj) ˆj ( ) cm olume 5 sin cm Conservation of Mass Use Eq 44 with m representing the mass in the volume: 4 dm dm + ρ ˆ da n + ρa ρa cs dm + ρq m dm Finally, m ρq 44 A A π 5 6 π m ρa 94 π slug/sec 44 5 ft/sec Q A π ft /sec 9 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

4 48 p 5 kg 46 kg ρ ρ ρ ρ RT 87 9 m 87 5 m A A π π 5 98 m/s m ρ A 89 kg/s Q A 47 m /s Q 5 m /s 44 b) ( ) π d 4 d 4478 m cosθ / θ 6 o θ R a) Since the area is rectangular, 5 m/s 44 m ρ A kg/s Q m ρ m /s c) m/s m ρ A kg/s Q m 48 m /s ρ 444 If dm /, then ρ A ρ A + ρ A In terms of m and Q becomes, letting ρ ρ ρ, π m + m 58 kg/s this 446 min m out + m ρ ρ (y y )dy+ ρ + m Note: We see that at y m the velocity u() m/s Thus, we integrate to y, and between y and the velocity u : 4 4ρ ρ+ ρ + m m 6667ρ 8 kg/s / ( ) m ρ da 545 y ( 6 y 9 y ) 5 dy 448 / u ( ) 6y 7y 9y + 9y dy 4 max fps (See Prob 44b) 458 slug/sec 4 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

5 ρ slug/ft 4 ρa slug/sec Thus, ρa m since ρ ρ(y) and (y) so that ρ ρ m of HO 4 m of air π (5h) m of air s h 565 m m in m + m m/s (see Prob 44c) π + π 4 m/s The control surface is close to the interface at the instant shown i interface velocity e ˆn i ˆn 454 ρ A ρ A e e e i i i 8 5 π 5 π i i 44 m/s For an incompressible flow (low speed air flow) 456 / 5 uda A y 8dy π 5 A 8 5 6/ 5 π 5 7 m/s Draw a control volume around the entire set-up: dm tissue ρ ρ + A A d d + m h ( h tan ) h tissue ρπ ρπ φ 4 or m d d h h h tissue ρπ + tan φ 4 4 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

6 46 dm + ρa ρ A 6 m + ( π / 6) m 4 99 kg/s dm + ρ Q ρ A m where m ρ Ah a) π 6 h + 6 / 6 π h m/s or mm/s Choose the control volume to consist of the air volume inside the tank The conservation of mass equation is d ρd + ρ da n C Since the volume of the tank is constant, and for no flow into the tank, the equation is CS d ρ + ρ A e e e p Assuming air behaves as an ideal gas, ρ At the instant of interest, RT dρ dp Substituting in the conservation of mass equation, we get RT ( )( ) π ( ) dp ρ 8 kg/m m/s 5 m ea e e kj ( RT ) K 5 m kg K dp 45 kpa/s Energy Equation 466 du W Tω+ pa + μ A dy belt 5 π / W % of the power is used to increase the pressure while % increases the 4 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

7 internal energy ( Q because of the insulation) Hence, mδ u W 4 8ΔT 5 Δ T 86 C W 47 T 4 89 b) W 4 89 (9 /6) W mg T p p g + + g + + ft h 64 4h 86 + h h Continuity: h This can be solved by trial-and-error: h 8': 84? 8 h 79' : 84? 8 h 79 ' h 8' : 84? 8 h 75' : 84? 8 h 76 ' Manometer: Position the datum at the top of the right mercury level p + ( 98 6) p p p Divide by 98: g Energy: h p p g + + g + + () () Subtract () from (): With m, g m/s Q 8 π 5 fps 476 Continuity: π π fps Energy: p p g + g + +7 g 4 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

8 5 544 p psf or 644 psi Q/ A π p p Energy: g + g + + g 89 m/s m/s p Pa 48 a) Energy: p p g + + g + + g m/s Q A m /s For the second geometry the pressure on the surface is ero but it increases with depth The elevation of the surface is 8 m: + h g( h) m/s g Q m /s Note: is measured from the channel bottom in the nd geometry H + h 48 p p g + + g + + b) Q A π m/s m /s p Manometer: H+ + p 6 H+ + p H + p Energy: p p + g + g Combine energy and manometer: 6H g Continuity: d d 4 d 6H g 4 d 44 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

9 / / d π 6H g H π d / d d d Q d d d a) Energy from surface to outlet: H gh g 486 p p Energy from constriction to outlet: + g + g Continuity: 4 With p p v 45 Pa and p Pa, gh gh H 66 m Energy: surface to surface: + h L + g 488 Continuity: 4 6 g g 6g ( 94 Energy: surface to constriction: + ) + g m H m elocity at exit e elocity in constriction elocity in pipe e Energy: surface to exit: g H gh e 49 Continuity across nole: D d e Also, 4 Energy: surface to constriction: H p v + g 4 D 97 a) g g 98 D m 49 m A ρ 94 π 5 79 slug / sec W P 44 ft-lb / 85,95 or 5 hp 64 sec 45 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

10 W T W T 4 6 W Q We used m/s A π 5 η T 94 a) p p c v Q WS mg ( T T) g g The above is Eq 457 with Eq 458 and Eq 7 pg N/m RT T T 6 T ( 5 ) ( T 9) T 57 K or 99 C Be careful of units p 6 Pa, c v 765 J/kg K Energy: surface to exit: W mg TηT g g 5 6 m / s mg Q N / s π 6 6 W + T WT 59 kw Choose a control volume that consists of the entire system and apply conservation of energy: H p H p h g g P T L 46 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

11 Carburetor Section () 5 m Pump Section () We recognie that H T, is negligible and h L /g where Q/A and Q/ A Rearranging we get: p p HP h g 6 ( ) π ( 5 m) L 6 m /s Q () m/s hl m A 98 6 Q 6 m /s 5 m/s π 4 ( 4 m) A Substituting the given values we get: H P ( 95 ) kn/m ( 5 m/s) + 5m + + m 885 m 666 kn/m 98 m/s The power input to the pump is: W ( ) 6 ( )( )( ) QH / η 666N/m 6 m /s 885m / 75 5 W P P P Energy: across the nole: p p 65 g g m/s m/s Energy: surface to exit: H P , A 76 m/s, 68 m H P 47 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

12 QHP 98 ( ) W π P 8 W η 85 Energy: surface to A : P pa pa 9 4 Pa Energy: surface to B : pb pb 46 Pa Depth on raised section y Continuity: y Energy (see Eq 45): + + ( 4+y) g g y , or y y gy Trial-and-error: y y :? 8: + 5? y 85 m y :? y m y : +? The depth that actually occurs depends on the downstream conditions We cannot select a correct answer between the two The average velocity at section is also 8 m/s The kinetic-energy-correction factor for a parabola is (see Example 49) The energy equation is: p p + α + + h L g g h L 85 m 4 a) r da rdr 5 m/s A π π 4 r da π 5 α πrdr A h L 48 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

13 Engine power FD + m + u u mq f f F D + m + cv( T T) p p ν L α g + + g + gd m/s and Q 77 m /s Energy from surface to surface: p p H K P g g g 44 Q a) H P Q π 4 98 Try Q 5 : H 4 (energy) H 58(curve) P Try Q : H 446 (energy) H 48(curve) P P P Solution: Q m /s Momentum Equation p p + + g g d 4 ( d /) 46 b) m/s 4 π F π 7 ( ) F 679 N d) fps Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

14 π 5 F 94 π (5/) 74 (4 ) F 7 lb p p g + g + π 6 5 m/s e e Σ F m ( ) x x x 48 a) pa F m ( ) 56 m/s 4 π 5 F π 5 56(56 56) F 69 N c) 65 4 g g 4585 m/s 4 π 5 F π ( ) F 75 N p p 5 p g g g 4 98 b) m/s, 9 m/s 5 98 pa F m ( ) x x x F x 4 π 4 + π N F m ( ) π 4 7 ( 9 ) 7 N y y y A A m/s π 5 4 π(5 ) p p + g + g 4 44 p Pa 98 p A F m ( ) F 5 7π 5 π 5 4( 4) 496 N Continuity: 6 Energy (along bottom streamline): p A F 5 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

15 p p g + + g + + / , 6 m/s Momentum: F F F m( ) 98 (6 4) 98 ( 4) F ( o 4) 67(67 6) F 68 N (F acts to the right on the gate) Continuity: y y 4 y y 4 y F F F 46 Use the result of Example 4: a) y 4 8 m 8 y y y y g + + / / m/s Refer to Example 4: 48 4 y 6 yw 6 w ρ 6 w ( y 6 ) y y 6 ( y 6) 6ρ y or ( y+ 6) y 77 y 8 ft, 58 fps π 5 A A 5 m/s π 5 5 p Pa 98 Σ F m p A F m ( ) F p A + m ( ) x x x 7 5π 5 + π N p p + + g g 5 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

16 ( ), a) Fx m x x F m 4 m 8 m/s ρ A π 5 F 8 46 N F b) F m ( )(cos α ) r B 8 F ( 8 ) 65 N b) F m ( )(cos α ) 7 π 4 ( 8) ( 866 ) r 44 m/s B m ρa π kg / s B Rω 5 5 m / s R m ( )(cos α ) π 5 4 5(5 ) R x 98 N x B W R W x B 48 F m x ( B)(cos ) 4π ( 4 8) ( 5 ) R x 65 N B 5 8 m/ s W W The y-component force does no work b) sin r sinα α r r r α cos 4 cos 47, 685 m/s 44 sin 6 685sinα cos 6 685cosα 4 89 m/s, α 95 R m ( ) π 5 ( 89cos6 cos ) x x x R 75 N x 6 W R W B x 44 To find F, sum forces normal to the plate: F m ( ) Σ n out n n 5 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

17 a) [ ] F 4 4 ( 4sin 6 ) 8 N (We have neglected friction) Σ F m + m ( ) m 4sin Bernoulli: t m m 5m Continuity: m m m m m 75m 75 4 kg/s + 8 kg/s r r n B x B F m ( ) 4(4 ) sin 6 F 8(4 ) sin 6 W F 8 (4 ) 75 6(6 8 + ) dw d B B x B B B B B B B B 6(6 6 + ) m/s F m ( )(cosα ) ( 89)( ) 4 7 N r B 5 89 m/s B W W or 647 hp 6 To solve this problem, choose a control volume attached to the reverse thruster vanes, as shown below The momentum equation is applied to a free body diagram: Momentum: [ ] ( ) R 5 m ( ) + ( ) m x r x r x r x Assume the pressure in the gases equals the atmospheric pressure and that r r r Hence, ( ) 8 m/s, and r x r ( ) ( ) sinα r x r x r where α Then, momentum is Rx 5m [ r sinα ] m r m sinα + r ( ) Momentum: R 5m sinα m m sinα + [ ] ( ) x r r r The mass flow rate of the exhaust gases is ( ) m m + m m kg/s air fuel air r r r R x 5 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

18 Substituting the given values we calculate the reverse thrust: ( )( )( ) R 5 kg/s 8 m/s + sin kn x Note that the thrust acting on the engine is in the opposite direction to Rx, and hence it is referred to as a reverse thrust; its purpose is to decelerate the airplane For this steady-state flow, we fix the boat and move the upstream air This provides us with the steady-state flow of Fig 47 This is the same as observing the flow while standing on the boat 5 W F F F 44 N ( 89 m/s) F m ( ) 44 π ( 89) 6 m/s Q A π 699 m /s 89 η p 65 or 65% 4 88 Fix the reference frame to the boat so that 9 fps fps F m ( ) 94 π (5867 9) 546 lb ft-lb W F , or 9 hp sec m 94 π 86 slug/sec A m/s m/s ( y) ( y) flux in max ρ dy ( y) dy 8 67 N The slope at section is ( y) y+ A Continuity: A A m/s 54 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

19 () A A / (5) A A A 5 ( y) 5 y 5 flux out ( y 5) 8 (5 y) dy 8 [5] 48 N change N From the cv shown: π τ w π ( p p ) r r L Δpro du τw μ L dr w du / 5 dr 6 w 9 ft/sec/ft τ w π r o L p A p A m ρ ρ ( ) top A y da 458 (8 + y )dy 656 kg/s top (8 ) 656 F ρ da m m y dy F 78 N Momentum and Energy a) Energy: h L See Problem 45(a) g g h L h L 66 m losses A h 98 (6 ) W/m of wih L 55 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

20 Moment of Momentum e m m/s elocity in arm ρ 4 π 4 A e M r ( Ω )ρd I cv 4 r ˆ ( Ω ˆ ˆ)ρ Adr i k i 464 8ρAΩ kˆ rdr 6ρA Ωkˆ d M and ( r )ρd cv cs ( ) ( ) ˆρ ˆ 77 ˆ 77 ˆ r n da i j + k ρa e e e e The -component of r ( n ˆ) ρda 77e A e cs ρ Finally, I ρ e eρ e e ( M ) 6 AΩ 4 77 A Using A A, 6Ω Ω 46 9 rad / s m ρa π 8 m/s Continuity: π π + 6( r 5) e π 6 5 m/s e 9( r 5) 44 r e e M ˆ ( ˆ ˆ) ρ ˆ [ ˆ (44 ) ˆ I r i + Ω k i Adr+ r i + Ω k r i ] ρadr 5 5 ˆ ˆ 4Ω ρak rdr+ 4 ΩρAk (44r r ) dr 5 44 ˆ ( 5 ) ( 5 ) k (5Ω+ Ω ) kˆ 5 Ωk ˆ 56 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part

21 e eρ 5 5 rˆi ( ˆj ) 6dr 6 rdr kˆ 86kˆ 5 Ω 86 Ω 96 rad/s 468 See Problem e 989 m/s m/s ˆ ˆ ˆ dω 4 ( ) ˆ ˆ MI ri Ω k i + k ri ρadr A π, A π 4 dω 8ρAΩkˆ 4ρ ˆ rdr A k r dr ˆ dω 6AΩk 6A kˆ ( r ) ( n ˆ) ρ da e A e k cs d Thus, 6 6 Ω dω AΩ+ A e Ae or + 8Ω 7 8 t ˆ The solution is Ω Ce + 7 The initial condition is Ω( ) C 7 Finally, 8t Ω 7( e ) rad/s e 57 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part