6-Practice Problems.pdf

Chapter 6

Momentum Principle

Problem 6.1

Water at 20 o C is discharged from a nozzle onto a plate as shown. The ß ow rate of

the water is 0.001 m 3 ! s, and the diameter of the nozzle outlet is 0.5 cm. Find the

force necessary to hold the plate in place.

Solution

This is a one-dimensional, steady ß ow. Since the system is not accelerating, the ve-

locities with respect to the nozzle and plate are inertial velocities. The momentum

equation in the " -direction (horizontal direction) is

# ! =

ú$ " % " ! !

ú$ # % # !

Draw a control volume with the associated force and momentum diagrams.

CHAPTER 6. MOMENTUM PRINCIPLE

From the force diagram

# ! = !&

From the continuity equation, the mass ß ow in is equal to the mass ß ow out so

ú$ " = ú$ # = ú$

The velocity at the inlet is ' . The component of velocity in the " -direction at the

outlet is zero, so the momentum ß ux is

ú$ " % " ! !

ú$ # % # ! = ! ú$'

Equating the forces and momentum ß ux

!& = ! ú$'

& = ú$'

The volume ß ow rate is 0.01 m 3 ! s, so the mass ß ow rate is ú$ = () = 1000×0*001 =

1 kg/s. The velocity is

' = )

+ =

4 (0*005) 2 = 50*9 m/s

0*001

The restraining force is

& = 50*9×1 = 50*9 N

Problem 6.2

A water jet with a velocity of 30 m/s impacts on a splitter plate so that 4 of

the water is de ß ected toward the bottom and 4 toward the top. The angle of the

plate is 45 " . Find the force required to hold the plate stationary. Neglect the

weight of the plate and water, and neglect viscous e ! ects.

Solution

The pressure is constant on the free surface of the water. Because frictional e ! ects

are neglected, the Bernoulli equation is applicable. Without gravitational e ! ects,

the Bernoulli equation becomes

2 (' 2 = constant

Since pressure is constant, the velocity will be constant. Therefore, each exit

velocity is equal to the inlet velocity.

Momentum and force diagrams for this problem are

The forces acting on the control surface are

F = # ! i+# % j

,+ 1

CHAPTER 6. MOMENTUM PRINCIPLE

The momentum ß ux from the momentum diagram is

ú$ " v " !

ú$ # v # = 3

4 ú$ # (30iCOS45+30jSIN45)

+ 1

4 ú$ # (!30iCOS45!30jSIN45)! ú$ # 30i

Equating the force and momentum ß ux

# ! i+# % j = ú$ # (!19*4i+10*6j)

The inlet mass ß ow rate is

ú$ # = (+' = 1000×0*1×30 = 3000 kg/s

The force vector evaluates to

# ! i+# % j = !5*82×10 4 i+3*18×10 4 j (N)

Thus

# ! = !5*82×10 4 N

# % = 3*18×10 4 N

Problem 6.3

A 12-in. horizontal pipe is connected to a reducer to a 6-in. pipe. Crude oil

ß ows through the pipe at a rate of 10 cfs. The pressure at the inlet to the reducer

is 60 psi. Find the force of the ß uid on the reducer. The speci Þ c gravity of crude

oil is 0.86. The Bernoulli equation can be used through the reducer.

Solution

Draw a force and momentum diagram as shown.

At the control surface, forces due to pressure are acting. Also, a force ( # ) is needed

to hold the reducer stationary.

# ! = , 1 + 1 !, 2 + 2 !#

(1)

The control volume is not accelerating and the ß ow is steady, so the momentum

change becomes

ú$ " % "(! !

ú$ # % #(! = ú$' 2 ! ú$' 1 = ú$(' 2 !' 1 )

(2)

The velocity at the inlet is

' 1 = )

+ 1

= 10

4 1 2 = 12*7 ft/s

The velocity at the exit is

' 2 = )

+ 2

4 (0*5) 2 = 50*9 ft/s

The pressure at the outlet can be found by applying the Bernoulli equation.

- + ' 2. = , 2

- + ' 2.

2 (' 1 !' 2 )

= 60×144+ 0*86×1*94

(12*7 2 !50*9 2 )

= 8640!2027 = 6613 psf = 45*9 psi

Equating the force and momentum ß ux by combining Eqs. (1) and (2) gives

60×

4 ×12 2 !45*9×

4 ×6 2 !# = 10×0*86×1*94×(50*9!12*7)

# = 4850 lbf

, 1

, 2 = , 1 + (

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