homework solutions.week 10

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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2/7


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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3/7


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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4/7


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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5/7

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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6/7


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



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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7/7


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


Flow rate

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


X F = m (vo ) xT = mV cos60o

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

= 656 lbf

T = 656 lbf

22

homework solutions.week 10

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