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  • 8/13/2019 Homework Solutions.week 10

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    5.80: PROBLEM DEFINITION

    Situation:A tank is draining through an ori ce.h1 = 3 m , h = 0.5 m.D

    T = 0 .6 m, D 2 = 3 cm

    Find:Time required for the water surface to drop the speci ed distance (3 to 0.5 m).

    SOLUTION

    From Example 5-6 the time to decrease the elevation from h1 to h is

    t = 2AT 2gA2 (h1 / 21 h1 / 2 )=

    2

    / 4 (0.6 m)2

    3

    0.5

    m1 / 2

    p 2 9.81 m/ s2 (/ 4) (0.03m)

    2

    t = 185 s

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    5.108: PROBLEM DEFINITION

    Situation:A sphere moves below the surface in water.D = 1 ft , h = 12 ft.

    Find:Speed at which cavitation occurs.

    Properties:Water ( 50 F), Table A.5: = 1.94 slug/ ft3 .

    PLAN

    Apply the Bernoulli equation between the free stream and the maximum width.

    SOLUTION

    Let po be the pressure on the streamline upstream of the sphere. The minimumpressure will occur at the maximum width of the sphere where the velocity is 1.5times the free stream velocity.Bernoulli equation

    po + 12

    V 2o + h o = p + 12

    (1.5V o )2 + (h o + 0 .5)

    Solving for the pressure p gives

    p = po 0.625V 2o 0.5

    The pressure at a depth of 12 ft is 749 lbf/ft 2 . The density of water is 1.94 slugs/ft 3

    and the speci c weight is 62.4 lbf/ft 3 . At a temperature of 50 o F, the vapor pressureis 0.178 psia or 25.6 psfa. Substituting into the above equation

    25.6 psfa = 749 psfa (0.625)(1.94) V 2o (0.5)(62.4)692.2 = 1.21V 2o

    Solving for V o gives

    V o = 23.9 ft/s

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    5.111: PROBLEM DEFINITION

    Situation:A hydrofoil is tested in water.h = 4 ft , V = 25ft / s.

    Find:Speed that cavitation begins.Properties:

    p0 = 2 .5 psi vacuum.Water ( 50 F) Table A.5, pv = 0.178 psia.

    PLAN

    Consider a point ahead of the foil (at same depth as the foil) and the point of minimumpressure on the foil, and apply the pressure coe ffi cient de nition between these twopoints.

    SOLUTION

    pmin = 2.5 144 = 360 psf gage p0 = 4 = 4 62.4 = 249.6 psf

    Then

    C p = ( pmin p0 )

    V 20 / 2 =

    ( 360 249.6)(1.94 slug/ ft3 / 2) (25ft/ s)2

    C p = 1.005

    Now let pmin = pvapor

    = 0 .178 psia = 14.52 psia = 2, 091 psfgThen

    1.005 = 249.6 psf + 2 , 091(1.94 slug/ ft3 / 2)V 20

    V 0 = 49.0 ft/s

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  • 8/13/2019 Homework Solutions.week 10

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    6.9: PROBLEM DEFINITION

    Situation:A water jet is lling a tank.m = 25kg, V = 25 L.

    d = 30 mm, v = 25 m/s.

    Find:Force on the bottom of the tank (N).Force acting on the stop block (N).

    Assumptions:Steady ow.

    Properties:Water (15 C), Table A.5: = 999kg/ m3 , = 9800N/ m3 .

    PLANApply the momentum equation in the x-direction and in the y-direction.

    SOLUTION

    Force and momentum diagrams

    Momentum equation ( x-direction)

    XF x = Xcs movox Xcs m i vixF = ( mv cos70o)

    = Av2 cos70o

    Calculations

    Av 2 = 999kg/ m3

    (0.03m)2

    4 !(25m/ s)2= 441.3 N

    F = (441.3 N)(cos70o)= 150.9 N

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    F = 150.9 N pushing to the left on the stop block

    y-direction

    XF y =

    Xcs

    movoy

    Xcs

    m i viy

    N W = ( mv sin70o)N = W + Av 2 sin70o

    Calculations:

    W = W tank + W water= (25 kg) 9.81 m/ s

    2

    + (0 .025m3 )(9800N/ m3 )

    = 490.3 N

    N = W + Av2

    sin70o

    = (490 .3 N) + (441 .3 N)sin70o

    N = 905 N acting upward

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    6.12: PROBLEM DEFINITION

    Situation:Horizontal round jet strikes a plate.Q = 2 cfs, F x = 200 lbf.

    Find:Speed of water jet (ft/s).Sketch:

    Properties:Water (70 o F) , Table A.5: = 1.94 slug/ft 3 .

    PLAN

    Apply the momentum equation to a control volume surrounding the plate.

    SOLUTION

    Force and momentum diagrams

    Momentum equation ( x-direction)

    X F x = mv1 xF x = ( mv1 ) = Qv1

    v1 = F x

    Q

    = 200lbf

    1.94 slug/ ft3 2 ft3 / s

    v1 = 51.5 ft/ s

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  • 8/13/2019 Homework Solutions.week 10

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    6.15: PROBLEM DEFINITION

    Situation:Water jet from a re hose on a boat.d = 4 in, V = 60 mph = 88.0 ft/s.

    Find:Tension in cable (lbf).

    Sketch:

    Properties:Water ( 50 F), Table A.5: = 1.94 slug/ ft3 .

    PLAN

    Apply the momentum equation.

    SOLUTION

    Force and momentum diagrams

    Flow rate

    m = AV = 1.94 slug/ ft3 (2/ 12ft)2 (88.0 ft/ s)= 14 .90 slug/ s

    Momentum equation ( x-direction)

    X F = m (vo ) xT = mV cos60o

    T = (14 .90 slug/ s)(88.0 ft/ s)cos60o

    = 656 lbf

    T = 656 lbf

    22