5 Practice Problem Control Volume Approach and Continuity Principle

8/13/2019 5 Practice Problem Control Volume Approach and Continuity Principle

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Chapter 5

Control Volume Approach

and Continuity Principle

Problem 5.1

A 10-cm-diameter pipe contains sea water that ows with a mean velocity of 5m/s. Find the volumeow rate (discharge) and the mass ow rate.

Solution

The discharge is

=

where is the mean velocity. Thus

= 5

4012

= 00393m3/s

From Table A.4, the density of sea water is 1026 kg/m3

The mass ow rate is

= = 102600393 = 403kg/s

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30CHAPTER 5. CONTROL VOLUME APPROACH AND CONTINUITY PRINCIPLE

Problem 5.2

The velocity prole of a non-Newtonian uid in a circular conduit is given by

max =

1

212

wheremaxis the velocity at the centerline andis the radius of the conduit. Findthe discharge (volume ow rate) in terms ofmax and

Solution

The volume ow rate is

=

Z

For an axisymmetric duct, this integral can be written as

= 2

Z 0

Substituting in the equation for the velocity distribution

= 2max

Z 0

1

212

Recognizing that2 = 2 we can rewrite the integral as

= max

Z 0

1

2122

=max2

Z 0

1

212

2

or

= max2

Z 1

0

[1 ]12

= 2

3max

2 [1 ]32 |10

=23

max2


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Problem 5.3

A jet pump injects water at 120 ft/s through a 2-in. pipe into a secondary owin an 8-in. pipe where the velocity is 10 ft/s. Downstream the ows become fullymixed with a uniform velocity prole. What is the magnitude of the velocity wherethe ows are fully mixed?

Solution

Draw a control volume as shown in the sketch below.

Because the ow is steady X

V A= 0

Assuming the water is incompressible, the continuity equation becomesX

V A= 0

The volume ow rate across station isX

V A= 10

4

8

12

2

where the minus sign occurs because the velocity and area vectors have the oppositesense. The volume ow rate across station b isX

VA= 120

4

2

12

2


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The continuity equation is

Z

+X

V A= 0

The density inside the control volume is constant so

Z

+X

V A= 0

+X

V A= 0

The volume of the uid in the tank is = Mass crosses the control surface attwo locations. At the inlet

V A=

and at the outlet

V A=

Substituting into the continuity equation

+ = 0

or

=

=1 12

2= 01m/min


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Problem 5.5

Water ows steadily through a nozzle. The nozzle diameter at the inlet is 2 in.,and the diameter at the exit is 1.5 in. The average velocity at the inlet is 5 ft/s.What is the average velocity at the exit?

Solution

Because the ow is steady, the continuity equation isX

V A= 0

Also, because the uid is incompressible, the continuity equation reduces toX

V A= 0

Draw a control surface that includes the inlet and outlet sections of the nozzle asshown.

At the inlet, station 1,

(V A)1

= 5

4

2

122

At the exit, station 2,

(VA)2

= 2

4

15

12

2


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Substituting into the continuity equation

X

V A= 5

4

2

12

2+2

4

15

12

2= 0

or

2 = 5

20152

= 889ft/s

Problem 5.6

Air ows steadily through a 10 cm-diameter conduit. The velocity, pressure, andtemperature of the air at station 1 are 30 m/s, 100 kPa absolute, and 300 K. Atstation 2, the pressure has decreased to 95 kPa absolute, and the temperature re-mains constant between the two stations (isothermal ow). Find the mass owrate and the velocity at station 2.

Solution

The mass ow rate is=

The density is obtained from the equation of state for an ideal gas.

=

At station 1

1 = 100103 N/m2

287J/kgK300K= 116kg/m3

The ow rate is= 11630

4012 = 0273kg/s

Because the ow is steady, the continuity equation reduces toX

VA= 0


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which states that the rate of mass ow through station 1 will be equal to thatthrough station 2. The air density at station 2 is

2 = 95103 N/m2

287J/kgK300 K= 110kg/m3

The mass ow is the same at each station. Thus

( )1

= ( )2

So

2= 112

= 30116

110= 316m/s

Problem 5.7

Water ows steadily through a 4-cm diameter pipe that is 10-m long. The pipewall is porous, leading to a small ow through the pipe wall. The inlet velocity is

10 m/s, and the exit velocity is 9 m/s. Find the average velocity of the water thatis passing through the porous surface.

Solution

The ow rate is steady, and the uid is incompressible so the continuity equationreduces to X

V A= 0

Draw a control surface around the pipe. The entrance is station 1, the exit isstation 2, and the surface of the porous pipe is station 3.

For station 1(V A)

1= 10

40042


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For station 2(V A)

2= 9

40042

For the porous surface

(V A)3

= 300410

The continuity equation is

10

40042 + 9

40042 +300410 = 0

or3 = 0001m/s

Problem 5.8

Water is forced out of a 2-cm diameter nozzle by a 6-cm-diameter piston mov-ing at a speed of 5 m/s. Determine the force required to move the piston and thespeed of the uid jet (2). Neglect friction on the piston and assume irrotational

ow. The exit pressure (2)is atmospheric.

Solution

When ow is irrotational, the Bernoulli equation applies. Applying this equationalong the nozzle centerline between locations 1 and 2 gives

1

+ 21

2 = 2

+

22

2 (1)

The continuity principle is11 = 22

So

2 = (5m/s)0062

0022

= 45 m/s (2)

Letting2 = 0 kPa gage and combining Eqs. (1) and (2)

1 = 22

2

1

2= (1000)

452 52

2= 1MPa


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Since the piston is moving at a constant speed, the applied force is balanced bythe pressure force.

=11

= (1MPa)

00624 m2

= 283 kN

Problem 5.9

The sketch shows a fertilizer sprayer that uses a Venturi nozzle. Water movingthrough this nozzle reaches a low pressure at section 1. This low pressure drawsliquid fertilizer (assume fertilizer has the properties of water) up the suction tube,and the mixture is jetted to ambient at section 2. Nozzle dimensions are1 = 3mm,2 = 9 mm, and = 150 mm. Determine the minimum possible water speed(2) at the exit of the nozzle so that uid will be drawn up the suction tube.

Solution

Since we are looking for the lower limit of operation, assume inviscid ow so that theBernoulli equation applies. Also assume that the pressure at 1 is just low enough

to draw uid up the suction tube, meaning there is no ow in the suction tube.

Identify locations 1 to 3 as shown by the points in the sketch below.

Applying the hydrostatic principle (constant piezometric pressure in a uid of con-stant density) between 1 and 3 gives

3 = 1+(1 3)


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Let3 = 0kPa gage, and let (1 3) = (+12).

1 = (+12) (1)

Applying the Bernoulli equation between 1 and 2 gives

1

+

21

2 =

2

+

22

2 (2)

The continuity principle is11 = 22 (3)

Let2 = 0kPa gage, and combine Eqs. (2) and (3).

1 =2

2

2

1

42

41

(4)

Combine Eqs. (1) and (4).

(+12) =2

2

2 1

42

41

9800(015 + 00032) = 10002

2

2

1

9

3

4!

So2 = 0.193 m/s

Problem 5.10

Show that the velocity eld

V= 2i2

2j +

2

2

3

3

k

satises the continuity equation for an incompressible ow and nd the vorticity atthe point (1,1,1).

Solution

The continuity equation for the ow of an incompressible uid is

V=

+

+

= 0

Substituting in the velocity derivatives

2 + 2 0

so continuity equation is satised.


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The equation for vorticity is

= V

=i

+j

+ k

Substituting in the velocity derivatives

= i

2

2

3

3

+

2

2

+j (2 0) + k

0 2

Substituting values at point (1,1,1)

=2

3i+2j k

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5 Practice Problem Control Volume Approach and Continuity Principle