The Integral Forms of the Fundamental Laws

Transcription

1 CHAPTER The Integral Frms f the Fundamental Laws a) N net frce may act n the system: ΣF b) The energy transferred t r frm the system must be zer: Q - W c) If ˆ ˆ n nˆ i ( j) is the same fr all lume elements then ΣF D Dt dm, r Σ D F ( m ) Since mass is cnstant fr a system Dt Σ F m D Since D a, Σ F ma Dt Dt Etensie prperties: Mass, m; Mmentum, m ; kinetic energy, m ; () ptential energy, mgh; enthalpy, H Assciated intensie prperties (diide by the mass): unity, ; elcity, ; /; gh; H/m h (specific enthalpy) Intensie prperties: Temperature, T; time, t; pressure, p; density, ρ; iscsity, µ System ( t) c( t) System ( t + t) + c( t + t) 5 System ( t) c( t) + + System ( t + t) + c( t+ t) + pump 5

2 6 a) The energy equatin (the st law f Therm) b) The cnseratin f mass c) Newtn s nd law d) The energy equatin e) The energy equatin 7 ˆn ˆn ˆn ˆn ω ˆn 8 ˆn ˆn ˆn ˆn 9 ˆn ˆn ˆn ˆn ˆn ˆn $ $ n i $ j ( i $ $ 77 + j) n$ $ 5 $ 866i j n$ $ j ˆ ˆ ˆ n nˆ i [ 77( i + j)] 77 fps n n$ i$ ( 866 i$ 5 j$) 866 fps nˆ iˆ ( jˆ) n flu ηρ$n A flu ηρ[ 77( i $ + $ j)] ia $ / 77 ηρa flu ηρ( 866i$ 5 $ j) ia $ / 866 ηρa flu ηρ( $ j ) $ ia 5

3 ( n$) A 5( 5 i $ + 866$ j ) $ j ( ) cm lume 5 sin cm The cntrl lume must be independent f time Since all space crdinates are integrated ut n the left, nly time remains; thus, we use an rdinary deriatie t differentiate a functin f time ut, n the right, we nte that ρ and η may be functins f (, y, z, t); hence, the partial deriatie is used system ( t) is in lumes and c () c ( t) lume 5 system ( t) + + c ( t) + 6 system bundary at (t + t) 7 If fluid crsses the cntrl surface nly n areas A and A, ρn$ da ρn$ da + ρn$ da c s A Fr unifrm flw all quantities are cnstant er each area: ρ n$ da + ρ n$ da A A A Let A be the inlet s $n and A be the utlet s n$ r A ρ A ρ + ρ A ρ A Then 5

4 8 Use Eq with m representing the mass in the lume: dm dm + ρ$ n da + A A ρ ρ c s dm + Q m ρ & Finally, dm m& ρ Q 9 Use Eq with m S representing the mass in the spnge: dm S dms + ρ$ n da + ρa + ρa ρa dm S + m + A Q & ρ ρ Finally, dm S ρq m& ρa (D) p m& ρa A π 7 87 kg/s RT 87 9 A A π 5 6 π 5 5 ft/sec m& ρa 9 π slug/sec Q A π ft / sec A A π 5 (π 6 ) 76 m/s m& ρa π 5 96 kg/s Q Aπ 5 96 m / s &m in ρa + ρa π Q Q 59 m / s p ρ 655 slug/ft 7 ρ 96 slug/ft RT m& m& ρa 55 fps ρ A ( π / ) 655 m& 96 ( / ) 98 fps 5

5 p 5 kg 6 kg 5 ρ A ρ A ρ ρ 8 7 RT 87 9 m 87 5 m π π 5 98 m/s &m ρ A 89 kg/s Q A 7 m / s Q 5 m / s 6 ρ A ρ A p RT A p RT A π 5 π 9 T T 89 9 K r 8 C 7 a) A A ( ) π d b) ( ) π d d 78 m c) ( ) R πr 866R R 58 m d 76 m d 67 m csθ / θ 6 θ R 8 (A) Refer t the circle f Prblem 7: 757 Q A ( π sin755 ) 56 m /s 6 r r r r r r 9 a) r da rdr r dr r r π π π r r r r m& ρa π m/s 6 75kg / s Q A 675 m / s r r r r r b) r rdr r r π π π 5 m/s m& ρa π 5 5 kg/ s Q A 5 m / s r r r c) πr rdr + r r π π / 58 m/s r r / m& ρa π 58 9 kg / s Q 9 m / s 5

6 a) Since the area is rectangular, 5 m/s m& ρa kg / s Q &m ρ m / s y y b) with y at the lwer wall h h h y y h hw wdy w 6667 m/s h h 6 m& ρa kg / s Q 67 m / s c) m/s m& ρa kg / s & & Q m 8 m / s ρ a) A da π r π π r 6 rdr r r ma ma With r, ma fps ( r) ( 576r ) fps b) A da With h c) A da d) ˆn With r m, h 6 y w h h wdy w h ma ma, 9 fps ( y) ma ( y ) fps r r r ma rdr ma r π π π m/s r () ( r ) m/s ma h y ma h h h w wdy maw With h m, m/s y ( ) ( y ) m/s ma If dm/, then ρa ρ A + ρ A In terms f &m and Q this becmes, letting ρ ρ ρ, π m & + & 5 8 kg / s m r r r da A ma rdr 5 π π r 5 r π ma π 5 ma m/s ( r) 5 m / s 55

7 m& & in m ut + m& ρ ρ ( y y ) dy + ρ + m& Nte: We see that at y m the elcity u() m/s Thus we integrate t y, and between y and the elcity u ρ ρ + ρ + m & m& 6667ρ 8 kg/s h h 5 h u( y) dy 5 ( y y ) dy h h 6667 h h This can be sled by trial-and-errr: h 6: 576? h 7: 75? h 8: 99? h 8: 997? h 8: 6? h 8: r 8 cm Nte: Fluid des nt crss a streamline s all the flw that enters n the left leaes n the right The streamline simply mes further frm the wall / 6 ( ) m& ρ da 55 y ( 6 y 9 y ) 5 dy / ( ) 6y 7y 9y + 9y dy 58 slug/sec uma fps (See Prb b) + 9 ρ 7 slug/ft ρ A slug/sec Thus, ρ A m& since ρ ρ(y) and (y) s that ρ ρ 7 A A π 8 ( π ) cs m f H O m f air 8 π m f air s 577 m/s 5 (5h) h 565 m 9 Use Eq : ρ + ρ t d n $ A n$ ρ ρ A t ( 7 + 7) tire ρ π 96 t 56

8 ρ t slug 5 ft sec m& m& + in m& m/s (see Prb c) π + π m/s d m m d & m m & m & m & c net c d m m m m & & & c π π 57 kg/s The cntrl surface is clse t the interface at the instant shwn i interface elcity ρ A ρ A e e e i i i 8 5 π 5 π i i m/s e ˆn i ˆn Assume an incmpressible flw: Q A 5/ 6 ( ) 5 fps Fr an incmpressible flw (lw speed air flw) / 5 uda A y 8dy π 5 A / 5 π 5 7 m / s 5 A + da A e e 5 r π( 5 ) + π π 5 rdr e e m / s 6 Draw a cntrl lume arund the entire set-up: dm tissue + A A ρ ρ d d & + & m h ( h tan ) h & tissue ρπ ρπ φ e 57

9 r m & d d h & h h & tissue ρπ + tan φ 7 The wih w f the channel is cnstant thrughut the flw Then dm d + ρa ρ A ( ρwhl) + ρa ρ A dh ρ + ρ 8 ρ w w w h & 8 m / s dm 8 + ρa ρ A 6 m& + ( π / 6) m& 99 kg / s 9 ρ A ρ A &m ρ A / 6 e 9 π 5 e 7 m / s dm 5 + ρ Q ρ A m& where m ρ Ah a) π 6 h & + 6/ 6 π h& m / s r mm / s b) π 6 h & + h& 88 m / s r 88 mm/ s c) π 6 h& + /6 π 5 h& 9 m/s r 9 mm/s 5 A A where A is an area just under the tp surface t/ dh a) π e π ( h tan 6 ) h t/ dh e h e t / + Finally, h( t) ( e t / / ) t/ b) e ( htan 6 ) h& t/ hdh 9e Finally, / / ht () 5( e t ) e h e t /

10 5 W& du Tω + pa + µ dy A belt 5 5 π / W 5 If the temperature is essentially cnstant, the internal energy f the c des nt change and the flu f internal energy int the pipe is the same as that leaing the pipe Hence, the tw integral terms are zer The lsses are equal t the heat transfer eiting the pipe 5 8% f the pwer is used t increase the pressure while % increases the internal energy ( Q & because f the insulatin) Hence, m& ~ u W& 8 T 5 T 86 C 55 (D) W& P Q g p p W& P + W& P kw and energy req'd 7 kw W& Q H P P η p W& T mg & Q a) W& T W b) W& T 89 ( 9 / 6) W c) & 6 W T 89 ( 8 / 6) W Q 656 m / s W& T ηt z m/s ρ Ag 6 98 p p z z g + + g + + ft h 6 h h Cntinuity: h h h 59

11 This can be sled by trial-and-errr h 8': 8? 8 h 79': 8? 8 Qh 7 9' h 8': 8? 8 h 75': 8? 8 Qh 76' 6 z z h g + g L h h / h + h Trial-and-errr prides the fllwing: h 5: 65? 6 h 5: 65? 59 h 7 m h 65: 65? 58 h 6: 65? 6 h 66 m 6 Manmeter: Psitin the datum at the tp f the right mercury leel z + p + ( 98 6) p p p Diide by 98: + z () g Energy: p p z z g + + g + + Subtract () frm (): With z m, () g 6 The manmeter equatin (see Prb 6) is p p + z g 6 99 m/s p p Energy: z z g + + g () g Subtract () frm (): With z m, and with (cntinuity) 8 g 6 7 m/s () 6 (A) p p + g p + p 7 Pa Q 8 π 5 fps 6

12 Cntinuity: π π 5 5 fps Energy: p p g g 7 g 5 5 p psf r 6 psi Q 6 - /6 π 7958 m/s y α da A w 667 A m/s A 6 Energy: h L 9 8 p p g + g + + h L m 98 wdy 8 66 Q / A 89 m/s 9 56 m/s π p p Energy: g g g 67 a) Acrss the nzzle: p π 6 Pa π p p Energy: + + g g 78 p Fr the cntractin: π 7 π 5 96 p p Energy: + + g g p p Manmeter: 5 + p 6 5+ p Subtract the abe eqns: g g g ( 96 ) 6 5 g 6 m/s p 9 Pa 6

13 Frm the reserir surface t sectin : p p z z g + + g H + m p p b) Manmeter: + p 6 + p 6 + p p Energy: + + Als, 96 g g g g 7 m/s The nzzle is the same as in part (a): p 5 7 Pa Frm the reserir surface t the nzzle eit: p p + + z + + z H 7 g g g m 68 a) Energy: p p z z g + + g + + gz m/s Q A m / s Fr the secnd gemetry the pressure n the surface is zer but it increases with depth The eleatin f the surface is 8 m z + h g( z h) m/s g Q m / s Nte: z is measured frm the channel bttm in the nd gemetry z H + h b) p p + + z + + z gz 6 + fps g g Q A ( ) 5 cfs Fr the secnd gemetry, the bttm is used as the datum: z + + h ( H + h) h g g gh fps Q 9 cfs 6

14 69 Frm the reserir surface t the eit: Cntinuity: p p z z K g g g g g 6 m/s Q 6 π 5 9 m / s The elcity in the pipe is 878 m/s Energy A: p + A Energy : p Energy C: 878 p C Energy D: p + D pa 65 5 Pa p 59 Pa pc 6 Pa pd 87 5 Pa 7 p p z z g + + g + + a) Q A π m / s b) Q A π m / s c) Q A π m / s m/s a) p z + g + 5 g m/s g g Q A π m / s b) A A 9 5 c) g g 8 m/s Q A 78 m / s m/s Q A g g 767 m / s 7 (C) Manmeter: Energy: ρ H + p g + p r g 796 K K p ρ g g

15 Cmbine the equatins: 98 8 m/s p 7 Manmeter: H + z + p 6H + z + p H + p 6 p p Energy: + g + g Cmbine energy and manmeter: 6H g Cntinuity: d d 6H g / d d Q d H g π 6 d d π / / d 5d d d H d / 7 Use the result f Prblem 7: a) Q b) Q / 65 / m / s 5 m / s H c) Using English units with g : Q 7d d d d Q 7 d) Q 7 / / / / 8 cfs 796 cfs / 75 () p Q hl K 796 m/s g A π 796 K K a) Energy frm surface t utlet: Energy frm cnstrictin t utlet: H gh g p + p g + g 6

16 Cntinuity: With p p 5 Pa and p Pa, gh gh H 66 m b) With p psia, p 7 psia, g gh 6 g gh H ft 77 Cntinuity: Energy surface t eit: g Energy cnstrictin t eit: p p + 6 g Frm Table, T C p p + + g g H p 5H Pa 78 Energy surface t surface: z z + h L + g Cntinuity: 6 g g 6g ( 9 Energy surface t cnstrictin: + ) + z g 98 z m H + 6 m 79 Cntinuity: Energy: p p + + g g g g m/s Q π m / s 8 elcity at eit e elcity in cnstrictin elcity in pipe Energy surface t eit: g e H gh D Cntinuity acrss nzzle: d e Als, Energy surface t cnstrictin: H p + g e 65

17 D 97 a) g g 98 D m D ( 7) b) 5 6 g g ( / ) 6 8 Energy surface t eit: + 77 g g D 6 r Energy surface t A : + + ( H + ) H 857 m 8 m& A ρ 9 π 5 79 slug / sec & ft - lb W P /, r 5 Hp 6 sec 8 m& ρa π 5 7 kg / s p / p 88 6 Pa 8 (C) W& P p + Q g W& 6 W& 6 kw P P Q p 8 kw η & + W T We used Q / A 5 π m / s W& T 6 W fps 9 fps π π ( 8 ), η T 9 η T 66

18 87 a) Q& W& mg & p p c z z ( ) g g T T S The abe is Eq 57 with Eq 58 and Eq 7 pg N / m RT T T 6 T ( 5 ) ( T 9) T 57 K r 99 C J e careful f units! p 6 Pa, c 765 K kg b) same as abe T 56 K r 87 C pg 7 lb 6 lb RT 76 5 ft ft c mg Ag A ft - lb 96 & ρ π lb /sec slug - R Use Eq 57 with Eqs 58 and 7: Q& W& mg & p p c + ( ) g g T T z z c & W c ( 6) 76 ft - lb W& c 6 r 78 Hp sec 89 Energy surface t eit: W& mg & TηT + 5 g g 5 6 m / s mg & Q N / s π 6 & 6 W + & T WT 59 kw (D) p p 6 Pa In the abe energy equatin we used Q hl K with m/s g A π 67

19 9 Energy surface t C : W& + mg & P ( mg & ρag π N / s) W& P 5 7 W pa Energy surface t A : p 69 A Pa O p po Energy surface t : W& PηP mg & + + z zo + K g g p Pa p 9 Manmeter: + z + p 6 + z + p + ρ p p + z z + + g Energy: p p z HT z g g H T 56 fps g π 56 HT ' W& Q H T η T T ft - lb 6, 98 r 5 Hp sec 9 Energy acrss the nzzle: p p 65 g g + 58 m/s Energy surface t eit: H P H P W& QH / η 98 ( π ) 86 68/85 8 W P P P, A 76 m/s, 86 m/s 68 m 68

20 Energy surface t A : pa pa 9 Pa Energy surface t : 9 (A) p p 6 Pa Q A π 9 89 m / s Energy surface t entrance: p H z K P g g H P m W & QH / η 98 / 75 6 W P P P p 95 Energy surface t eit: + + z + g g 7 8 m / s Q 7 8 π d / d 57 m 96 Depth n raised sectin y Cntinuity: y Energy (see Eq 5): + + ( + y ) g g y , r y y g y Trial-and-errr: y :? y 8: + 5? y 85 m y :? y : +? y m The depth that actually ccurs depends n the dwnstream cnditins We cannt select a crrect answer between the tw m 97 Mass flu ccurs as shwn The elcity f all fluid elements leaing the tp and bttm is apprimately m/s The distance where u m /s is y ± m m m m 69

21 T find &m use cntinuity: m& m& + m& ρ ρ ( 8 + y ) dy + m& 8 m& + 6ρ ρ 8 5 ρ Rate f KE lss m& m& u ρ dy 8ρ 5 ρ ρ ( 8 + y ) dy ρ[ ] 5 W 98 The aerage elcity at sectin is als 8 m/s The kinetic-energycrrectin factr fr a parabla is (see Eample 9) The energy equatin is: p p + α + + h g g L h L 85 m h L 99 da y dy A ( ) 9 m / s α da + 9 ( 8 y ) dy A 5 7 [ / + 8 / 5 + / 7] 5 9 r a) da 5 m/s π rdr A π r α da π π 5 rdr A

22 y b) da wdy 667 m/s A w y α da A w / n R A da r u rdr u R ma π π R / n ma wdy + 5 n n n + n + K E R r da uma rdr uma R ρ ρ π ρπ R 5 5 a) uma 758 u 6 ma 5 5 K E ρπr uma ρπr uma 8 K E ρπr u ma α ρa ρπr 758 uma n n + n + n b) uma 87 u ma K E ρπu ma R 88 ρπr u 7 α ma K E 88ρπR uma ρa 87 u ρπr 87uma ma c) uma 85 u ma K E ρπr u ma ρπr u ma K E ρπr uma α ρa ρπr 85 u ma 7

23 Engine pwer F m D + & + u ~ u ~ m& g F m& f f D + + c ( T T) W& η F 5 m kg kj km 9 q 5 km m f kg 6 s 6 q f 8 kj / kg p p νl α + + z z + g g gd m / s and Q m / s p p 5 Energy frm surface t surface: H z z K P g + + g + g Q a) H + P Q π 9 8 Try Q 5: H (energy) H 58 (cure) P Try Q : H 6 (energy) H 8 (cure) Slutin: Q m / s P Q b) H + P + Q π 9 8 Try Q 5: H 5 7 (energy) H 58 (cure) P Slutin: Q 7 m / s Nte: The cure des nt allw fr significant accuracy P P P D 6 Cntinuity: A A + A π 6 5 π + π m / s Energy: energy in + pump energy energy ut m& p W& m& p m& p + P P η ρ ρ ρ 7

24 π + π + + & W P W& P 6 7 W 5 + π + 7 (A) After the pressure is fund, that pressure is multiplied by the area f the windw The pressure is relatiely cnstant er the area 8 p p + + g g d ( d /) a) p A F m& ( ) 56 m / s π F π 5 6( ) F 9 N b) c) d) e) f) m / s π F π 7 ( 7 7 ) F 679 N m/s π 6 F π 6 56( 56 56) F 56 N fps 6 π 5 F 9 π (5/) 7 ( ) F 7 lb fps 6 6 π 5 F 9 π (5/) 8 ( ) F 5 lb fps 6 π F 9 π (/) 7 ( ) F 59 lb 9 p p + + g g

25 p A F m&( ) m& 8 π 5 F π F 8 N p p + + π e 6 5 e m/s g g ΣF m&( ) a) 56 8 p A F m&( ) π 5 F π 5 56(56 56) F 69 N b) g 98 g 56 m / s 9 m / s π 5 F π 5 9( 778 9) F 79 N c) g 98 g 585 m / s π 5 F π 5 585( ) F 75 N d) g 98 g m / s π 5 F π 5 ( ) F 9 N (C) p p + + g g (65 ) 7 p Pa 98 pa F ρq ( ) 85 π 5 F 7(65 ) F 75 N fps p 6 g, 8 psf 5 5 F pa m & ( ),8 π 9 π ( ) 7 lb p p 5 p + + g g g 9 8 a) m / s, 7 m / s 5 98 p A F m&( ) F π + π N 7

26 Fy m&( y y ) F y π 5 6( 7) 57 N 9 8 b) 5 7 m / s, 9 m / s 5 98 p A F m&( ) F π + π 7 8 N Fy m&( y y ) π 7 ( 9 ) 7 N 9 8 c) m / s, m / s 5 98 F pa + ρa 8 π + π 56 N Fy m&( y ) π ( ) N 8 m / s p p + + g g p Pa p A F m& ( ) F 9 π π 5( 8 5) 5 N F p A 5 A A π 5 π( 5 ) m / s p + p g + g p Pa 98 pa F m & ( ) F 5 7 π 5 π 5 ( ) 96 N p A F 6 Cntinuity: 7 7 Energy: p p z z g + + g + + F R F , 67 m / s 9 8 Mmentum: F F R m&( ) 98 5( 7 5) 98 5( 5) R ( 5) 67( 67 95) R 986 N R acts t the left n the water, and t the right n the bstructin 75

27 7 Cntinuity: 6 Energy (alng bttm streamline): p p z z g + + g + + / , 6 m/s Mmentum: F F F m&( ) 98 ( 6 ) 98 ( ) F ( ) 67( 67 6) F 68 N (F acts t the right n the gate) F F F 8 a) 8 6 y F F m& ( ) y 6 ρ w y w w 8 y ρ y 8 8 ( y ) ( 6 + y ) y y 9 8 y + 6y 7 89 y 5 m (See Eample ) 8 b) y y y g y m 8 c) y y y g y ft 8 d) y y y g y ft 9 Cntinuity: y y y y y 8 Use the result f Eample : y y y g y + + a) y 8 m / b) y 8 ft / 8 86 m / s / fps 76

28 9 m / s + + y y : 5? Trial-and-errr: 7 9 m / s : 5? 6 y 7 m / 8 y m m / s y 6 Refer t Eample : y w 6 w ρ 6 w ( y 6 ) y y 6 ( y 6) 6ρ ( y + 6) y 7 7 y 8 ft, 5 8 fps y Cntinuity: 5 π π 5 m / s Mmentum: p A p A m&( ) p A p A 6 π p π π 5 ( 5 ) 5 kpa p π 5 A A 5 m/s π 5 p p 5 p 98 + g + g 9 8 ( ) Σ F m & pa F m& ( ) 7 5 Pa F p A + m & 7 5 π 5 + π 5 5 N m& π 9 kg / s & π 8 5 kg / s m m& m& m& 88 π 8 6 m / s R R y p A p A p A Energy frm : p p p 5 g g Pa 98 77

29 Energy frm : p p + + g g 8 p Pa 9 8 p A p A R m& + m& m& R 5 π 56 π N p A R m& + m& m& y y y y R y 5 π 5 ( 8) 759 N ΣF m& F m & 5 a) ( ) m& ρa 8 m/s π 5 F 8 6 N F b) F m& ( )(cs α ) r 8 F ( 8 ) 65 N 8 c) F m& ( )(cs α ) r 8 F ( 8 ( )) 8 5 N a) F m&( ) 9 π 55 fps 5 b) F m& r( )(cs α ) 9 π ( ) 85 fps 5 c) F m& r( )(cs α ) 9 π ( + ) 5 fps 7 a) F m&( ) 7 π ( cs ) m& ρa π 6 kg / s m / s b) F m& ( )(cs α ) 7 π ( 8) ( 866 ) m / s r m& ρa π kg/ s c) F m& ( )(cs α ) 7 π ( + 8) ( 866 ) m / s r m& ρa π 8 kg/ s 8 (D) F m & ( ) 5(5cs6 5) 5 N 78

30 9 a) R m&( ) 9π ( cs 6 ) R 5 lb Ry m y y & ( ) 9π ( 866) R y 58 lb b) R m& r( )(cs α ) 9 π 6 6(5 ) R 76 lb Ry mr & ( )sin α 9 π 6 ( 6 866) R y lb c) R m& r( )(cs α ) 9π 8 8(5 ) R 686 lb Ry mr & ( )sin α 9 π 8 ( 8 866) R y 88 lb Rω 5 5 m / s R m&( )(cs α ) π 5 5(5 ) R 98 N W & R W a) R m&( ) π ( cs 6 ) R m& ( ) π ( sin 6 ) y y y R 6 N R y 696 N b) R m& ( )(cs ) π ( 5 ) R 679 N r R m& ( )sin α π 866 R y 9 N y r c) R m& ( )(cs ) π 5 ( 5 ) R 885 N r R m& ( )sin α π R y 88 N y r F m& ( )(cs ) π ( 8) ( 5 ) R 65 N 5 8 m / s W & W The y-cmpnent frce des n wrk (A) F m & ( ) π 6 (cs5 ) 88 N r r Pwer F W β r r r 75sin sin5 57 fps a) Refer t Fig 6: 75csβ cs5 r Nte: csα + cs α (csα + cs α ) r r r + 5 R m & r(cs α cs α ) 5π 75 57(cs + cs 5 ) 8 9 lb 79

31 W & 5R 5 8 9, ft - lb sec r Hp 75sin β r sin6 b) r 55 fps r 75csβ r cs6 + 5 R m & r(cs α cs α ) 5π 75 55(cs + cs 6 ) 6 lb W & R, ft - lb r 8 Hp sec 75sin β r sin 9 c) r 687 fps r 75csβ r cs9 + 5 R m & r(cs α cs α ) 5π (cs + ) 65 lb W & R, ft - lb r 99 Hp sec sin r sin α 5 a) Refer t Fig 6: 6 9, 8 α r m / s cs r csα sin6 8sinα 75, α 8 cs6 8csα R m&( ) π 5 ( 75 cs6 cs ) R 865 N W & R W sin r sin α b) 7, 68 5 α r r m / s cs r csα sin6 685sinα 89 m/s, α 95 cs6 685csα R m&( ) π 5 ( 8 9 cs 6 cs ) R 75 N W & R W sin r sin α c) 5 8, 696 α r r m / s cs 5 r csα sin6 676sinα 9 m/s, α 566 cs6 696csα 5 8

32 R m&( ) π 5 ( 9 cs 6 cs ) R 68 N W & R W 5sin r sinα 6 a) Refer t Fig 6: r + 5cs r csα sin6 r sinα r r 9+ + cs6 rcsα Cmbine the abe: 7 m / s Then, α 59, α R m&( ) π 5( cs 6 5 cs ) R 96 N W & 5 R W 5sin r sinα b) 5cs r csα sin7 r sinα cs7 rcsα r r m / s α & π,α 8 R m ( ) 5( cs7 5cs ) R 8 N W & 5 R W 5sin r sinα c) r + 5cs r csα 6 9 m / s sin8 rsinα r 9+ + α, α 5 7 cs8 rcsα R m&( ) π 5( cs 8 5 cs ) R 76 N W & 5 R W Σ n & ut n n 7 T find F, sum frces nrmal t the plate: F m ( ) a) [ ] F ( sin 6 ) 8 N (We hae neglected frictin) Σ Ft m & + m& ( ) m& sin ernulli: m& kg / s m& 5 m& m& m& Cntinuity: m& & + & & 8 kg/ s m m m b) F 9 ( sin6 ) 6 lb (We hae neglected frictin) ΣFt m& + m& ( ) m& sin ernulli: 8

33 & & & m& 75 m& slug/sec and m& 97 slug/sec m m 5m Cntinuity: m& m& m& 8 F m& ( ) ( + ) sin 6 9 N r r n F 9 cs 6 N W& 6 W 9 F m& r ( r ) n ( ) sin 6 F 8( )sin 6 & W F 8 ( ) 75 6( ) dw& 6( ) m / s d (A) Let the ehicle me t the right The scp then dierts the water t the right Then F m & ( ) 5 6 [6 ( 6)] 7 N F m& r( )(csα ) 6 ( )( ) At t : F N 6 a m/s 6 67 t F d d m t 6 67 t 6 6 sec F F m& r( )(csα ) ( 89)( ) 7 N 5 89 m/s W& W r 67 Hp 6 See the figure in Prblem F m& r( )(cs α ) 6 ( )( ) F m d d 5 d d 7 78 d d ln7 78 l n5 58 m

34 5 F m& r ( )(cs α ) 9 π ( ) ( ) d F ( ) d At t, Then With d 6, fps F Fr t >, d ( ) fps 5 Fr this steady-state flw, we fi the bat and me the upstream air This prides us with the steady-state flw f Fig 7 This is the same as bsering the flw while standing n the bat & 5 W F F F N ( 89 m / s) F m& ( ) π ( 89) 6 m / s Q A π 69 9 m / s 89 η p 65 r 65% 6 Fi the reference frame t the aircraft s that 5556 m / s m / s m& π 95 kg / s 6 F 95( ) 98 N pπ p 89 Pa W& F W r 88 Hp 88 7 Fi the reference frame t the bat s that 9 fps F m fps &( ) π ( ) 56 lb & ft - lb W F , r 9 Hp sec 8

35 m & π 86 slug/ sec 8 Fi the reference frame t the bat: m / s, m / s Thrust m&( ) ( ) N W & F W r 68 Hp 9 A m / s m / s ( y) ( y ) ma flu in ρ dy ( y) dy 8 67 N The slpe at sectin is ( y) y + A Cntinuity: A A m / s A / A 5 ( y) 5 y ( ) A A / ( 5) A 5 ( y 5) 8 flu ut ( 5 y) dy 8 [ 5] 8 N change 8 67 N ( y) dy da 5 a) β A b) See Prblem 9: ( y ) ( 5 y ), 5 y m / s 5 ( y 5) dy da β A 5 ( y 5) 5 5 Frm the c shwn: ( p p ) πr τ π r L p r du du 75 / τ w µ L dr w dr w 6 5 ft / sec 9 ft w τ w πr L p A p A r 5 Write the equatin f the parabla: ( r) ma r 8

36 6 r Cntinuity: π 6 8 ma 6 Mmentum: pa pa FDrag ρ da m & 6 πrdr ma 6 m / s r π 6 FDrag 6 π π rdr 6 5 F Drag F Drag N 5 m& tp ρa ρ ( y) da ( 8 + y ) dy 65 6 kg / s F ρ da + m& m& tp ( 8 + y ) dy F 78 N 5 a) m& & & tp m m ρa ρ u( y) da 8 ( y y ) 8 dy 656 kg/ s (Nte: y fr u( y ) 8) Mmentum: FDrag ρ 6( y y ) dy ρ 8 b) T find h: 8h 8 ( y y ) dy h F N 667 m Mmentum: FDrag 6( y y ) dy Drag F Drag N 55 a) Energy: + z + z + h g g L See Prblem 8(a) h L h L 66 m lsses A h L 98 ( 6 ) W / m f wih b) See Prblem : + z + z + h g g L 85

37 h L h L 5 m lsses A h L W 5 7 c) See Prblem : h L h L 66 m lsses A h L W / m f wih 56 See Prblem : m / s, 5 m / s, p 6 kpa, p 5 kpa Then, p p hl g g hl hl 7 m K K K 56 g 9 8 d 57 Cntinuity: D d D Energy: + H( t) gh( t) g g d d s Mmentum: ΣF ( FI ) d + m a ρ & ( ) ut, c t d d a m( t) ρ π ( ) m ( t ) m ρ π ( t ) dh dh d dh d gd / gh g H D H D D t + H / t ρπd gd a g + + D t H d gd g D t H m ρ π 58 This is a ery difficult design prblem There is an ptimum initial mass f water fr a maimum height attained by the rcket It will take a team f students many hurs t wrk this prblem It inles cntinuity, energy, and mmentum 59 e m& 9 89 m / s elcity in arm ρa π e M r ( Ω ) ρd ri$ ( Ωk$ i$ ) ρ Adr I c 86

38 8ρAΩk$ rdr 6ρAΩk$ Σ d M and r d r n da i e j+ ek e A e ( ) ρ ( $) ρ $ ( 77 $ 77 $ ) ρ c c s The z-cmpnent f r ( n$) ρda 77 e A e ρ c s Finally, ( MI ) z 6ρAΩ 77 e Ae ρ Using A Aee, 6Ω Ω 6 9 rad / s y r 6 A mment M resists the mtin thereby prducing pwer One f the arms is shwn Ω e 5 / MI ri$ ( Ωk$ i$ ) ρadr 8 ρaωk$ rdr 778 ρ AΩk$ Σ d M Mk $, r d r n da e A e k ( ), ( $) and ρ ρ $ c c s Thus, M + 75 / π 9 π 9 M 9 ft - lb W& MΩ 9 97 ft - lb /sec 6 m& ρa π 8 m / s Cntinuity: π π + 6( r 5) π e 6 5 e m/s 9 ( r 5) e r 5 MI ri$ ( + Ωk$ i$ Adr + ri + k ) ρ $ [ Ω $ ( r) i $ ] ρ Adr 5 + Ω ρak$ rdr ΩρAk$ ( r r ) dr Ω 8 π k $ + Ω π 5 5 ( ) ( ) $ k e 87

39 ( 5Ω + Ω ) k $ 5Ωk $ ri$ ( j$ e ) e dr rdr k$ k$ ρ Ω 86 Ω 9 6 rad /s 6 MΩ M N m 5 MI ri $ r ( Ωk$ ( r) i $ r) ρπ r dr R 8 πω r ( r ) drk $ Cntinuity: ( r) πr cs πr ( r) 866R / r r r ( n$) ρda R( RΩ + r sin ) r cs ρπr k$ r( 5+ 5 r) k $ c s 5 6 r dr ( 5+ 5 ) 5 r r r r r r ( 5 ± 5 + ) 7 9 m / s The flw rate is Q A cs π m / s e r 6 See Prblem e 8 m / s m / s d MI ri k i + $ k ri Adr A Ae ( $ $ Ω ) $ $, Ω ρ π π 8 ρa Ω k $ rdr ρa d Ω 6 k$ r dr A Ω k$ 6Ad Ω k$ ( ) ( $) $ r z n ρda e A ek c s Thus, 6A 6A d Ω dω Ω + e Ae r + 8Ω 7 8t The slutin is Ω Ce + 7 The initial cnditin is Ω( ) C 7 Finally, Ω 7 8t ( e ) rad /s 6 This design prblem wuld be gd fr a team f students t d as a prject Hw large a hrsepwer blwer culd be handled by an aerage persn? r 88