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Problem6‐80

 ( + → ) ∑ Fx

=

⎞ d ⎛ ⎜ ∫ v x ρdV ⎟ + ∑ m ov ox − ∑ m iv ix dt ⎝ CV ⎠

p i A i − p o A o − D = 0 + ( m ov ox )1 + (m ov ox )2 − m iv ix

 100 × (0.5 × 0.5) − 90 × (0.5 × 0.5) − D = [1.2 × (0.25× 0.5)× 8]× 8 + [1.2× (0.25× 0.5)× 12]× 12 − [1.2 × (0.5 × 0.5) × 10 ] × 10 D = 1.3 ( N )



( + ↑) ∑ F

y

1

1

0

0

=

⎞ d ⎛ ⎜ ∫ v y ρdV ⎟ + ∑ m ov oy − ∑ m i v iy dt ⎝ CV ⎠

− ∫ pu dA + ∫ p l dA − L = 0 + 0 − 0

 1

1

0

0

− ∫[100 − 10x − 20x (1− x )] × 0.5× dx + L = 3.33 (N )

 

∫[100 − 10x + 20x (1− x )]× 0.5× dx − L = 0

Problem6‐10

vj vb

  WeselectaControlVolumewhichismovingwiththeblock.TheFDandMDareshownon theboard. 1stMethod:FixedCoordinate System

( + → ) ∑ Fx

=

⎞ d ⎛ ⎜ ∫ v x ρdV ⎟ + ∑ m ov ox − ∑ m iv ix dt ⎝ CV ⎠

 i v ix − Ff = 0 + m o vox − m − Ff = m ov b − m iv

j

⎞ d⎛ ⎜ ∫ ρ dV ⎟ + m o − m i = 0 ⇒ dt ⎝ CV ⎠

m o = m i

F f = m i (v j − v b ) m i = ρ A i (v j −v b ) Ff = ρ A i (v j − v b )

2



2ndMethod:MovingCoordinateSystem

( + → ) ∑ Fx

=

⎞ d ⎛ ⎜ ∫ v x ρdV ⎟ + ∑ m ov ox − ∑ m iv ix dt ⎝ CV ⎠

 i v ix − Ff = 0 + m o vox − m − Ff = 0 − m i (v j − v b ) m i = ρ A i (v j −v b ) Ff = ρ A i (v j − v b )

                

2



Problem6‐88           ThemassofthecartisM=100kg.Initiallythecartisatrest.Themassflowrateofthe waterjetis10(kg/s).Thereisnofrictionforceonthecart.Findthetimetfrequiredsothat thespeedofthecartvcbecomeshalfofthespeedofjetvj. 

Solution Weselectacontrolvolumewhichisattachedtothecartasshown.TheFDandMDareas shownontheboard.However,wehavetouseafixedcoordinatesystem.Applyingthe momentumequationinthex‐direction. 

( + → ) ∑ Fx 0=

⎞ d ⎛ ⎜ ∫ v x ρdV ⎟ + ∑ m ov ox − ∑ m iv ix dt ⎝ CV ⎠

d (v x mCV ) + m ov c − m i v j dt

dv c

M

=

dt

+ m ov c − m iv j = 0

⎞ d ⎛ ⎜ ∫ ρdV ⎟ + ∑m o − ∑ m i = 0 dt ⎝ CV ⎠ m o = m i dv c + m i (v c −v j ) = 0 dt

M

m i = ρ A i (v j −v c )

M

dv c − ρA i (v c −v dt

M

dv c = ρA i (v c − v dt

(v

dv c c

−v j )

0.5v j

∫ 0`

(v

2

=

ρAi M

tf

dv c c

−v j )

tf = 10 ( s)

dt

2

=∫ 0

)

j

j

2

)

=0

2

⇒

(v

dv c c

−v j

)

2

=

ρ A iv j Mv j

dt

10 dt 100v j

...