4-Practice Problems.pdf

Chapter 4

Flowing Fluids and Pressure

Variation

Problem 4.1

A ß ow moves in the ! -direction with a velocity of 10 m/s from 0 to 0.1 second

and then reverses direction with the same speed from 0.1 to 0.2 second. Sketch

the pathline starting from ! = 0 and the streakline with dye introduced at ! = 0"

Show the streamlines for the Þ rst time interval and the second time interval.

Solution

The pathline is the line traced out by a ß uid particle released from the origin. The

ß uid particle Þ rst goes to ! = 1"0 and then returns to the origin so the pathline is

The streakline is the con Þ guration of the dye at the end of 0.2 second. During

the Þ rst period, the dye forms a streak extending from the origin to ! = 1 m.

During the second period, the whole Þ eld moves to the left while dye continues to

be injected. The Þ nal con Þ guration is a line extending from the origin to ! = !1

The streamlines are represented by

CHAPTER 4. FLOWING FLUIDS AND PRESSURE VARIATION

Problem 4.2

A piston is accelerating upward at a rate of 10 m/s 2 . A 50-cm-long water col-

umn is above the piston. Determine the pressure at a distance of 20 cm below the

water surface. Neglect viscous e ! ects.

Solution

Since the water column is accelerating, Euler’s equation applies. Let the # -direction

be coincident with elevation, that is, the $ -direction. Euler’s equation becomes

%$ (&+'$) = () !

(1)

Since pressure varies with $ only, the left side of Euler’s equation becomes

%$ (&+'$) = !

*$ +'

(2)

Combining Eqs. (1) and (2) gives

*$ = !(() ! +')

= !(1000 kg/m 3 ×10 m/s 2 +9810 N/m 3 )

(3)

= !19+810 N/m 3

Integrating Eq. (3) from the water surface ( $ = 0 m ) to a depth of 20 cm ( $ = -0.2

m ) gives

"(!=!0#2)

!=!0#2

*& =

!19"8 *$

"(!=0)

!=0

&($ = !0"2)!&($ = 0) =

!19"8 kN/m 3

(!0"2!0) m

Since pressure at the water surface ( $ = 0) is 0,

& !=!0#2 m =

!19"8 kN/m 3

(!0"2 m )

= 3.96 kPa-gage

CHAPTER 4. FLOWING FLUIDS AND PRESSURE VARIATION

Problem 4.3

A rectangular tank, initially at rest, is Þ lled with kerosene ( ( = 1"58 slug/ft 3 )

to a depth of 4 ft. The space above the kerosene contains air that is at a pressure

of 0.8 atm. Later, the tank is set in motion with a constant acceleration of 1.2 g to

the right. Determine the maximum pressure in the tank after the onset of motion.

Solution

After initial sloshing is damped out, the con Þ guration of the kerosene is shown in

Fig. 1.

Figure 1 Con Þ guration of kerosene during acceleration

In Fig. 1, , is an unknown length, and the angle - is

TAN- = ) $

. = 1"2

- = 50"2 %

To Þ nd the length , , note that the volume of the air space before and after motion

remains constant.

(10)(1)( width ) = (1/2)(1"2,)(,)( width )

, =

20/1"2 = 4"08 ft

The maximum pressure will occur at point 0 in Fig. 1. Before Þ nding this pressure,

Þ nd the pressure at 1 by integrating Euler’s equation from point ) to point 1 .

*! = () $

*& = !

() $ *!

& & = & ' +() $ (10!,)

= (0"8)(2116"2) lbf/ft 2 +(1.58)(1.2 ×32"2)(10!4"08) lbf/ft 2

= 2054 lbf/ft 2 absolute

To Þ nd the pressure at 0 , Euler’s equation may be integrated from 1 to 0 .

*(&+'$)

= 0

*$ = !'

(

*& =

!'*$

& ( !& & = !'($ ( !$ & )

& ( = & & +'(5 ft)

= 2054 lbf/ft 2 +(1"58×32"2 lbf/ft 3 )(5 ft)

= 2310 lbf/ft 2 -absolute

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