Solution Manual for Fluid Mechanics Russell C. Hibbeler 1st Edition

Click here to Purchase full Solution Manual at http://solutionmanuals.info 9–1.  Crude oil flows through the 2-mm gap between the two fixed parallel plates due to a drop in pressure from A to B of 4 kPa. If the plates are 800 mm wide, determine the flow.

2 mm

A

B

0.5 m

SOLUTION dh = 0 and U = 0. From Appendix A, m = 30.2 ( 10-3 ) N # s>m2 and dx r = 880 kg>m3.

Here,

Q = -

= -

a3b dp 12m dx

3 2 ( 10-3 ) m 4 3(0.8 m) 12 3 30.2 ( 10-3 ) N # s>m2 4

£

- 4 ( 103 ) N>m2 0.5 m

= 0.1413 ( 10-3 ) m3 >s = 0.141 ( 10-3 ) m3 >s

§ Ans.

To evaluate the maximum Reynolds number, we must first find the maximum velocity of the oil flow. u max = -

3 2 ( 10-3 ) m 4 2 - 4 ( 103 ) N>m2 a2 dp = £ § = 0.132 m>s 8m dx 0.5 m 8 3 30.2 ( 10-3 ) N # s>m2 4

Thus, the maximum Reynolds number is   Re =

ru max a = m

( 880 kg>m3 )( 0.132 m>s ) 3 2 ( 10-3 ) m 4 = 7.719 6 1400(OK) 30.2 ( 10-3 ) N # s>m2

930

Ans: 0.141 ( 10-3 ) m3 >s

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9–2.  Crude oil flows through the gap between the two fixed parallel plates due to a drop in pressure from A to B of 4 kPa. Determine the maximum velocity of the oil and the shear stress on each plate.

2 mm

A

B

0.5 m

SOLUTION dh = 0 and U = 0. From Appendix A, m = 30.2 ( 10-3 ) N # s>m2 and dx r = 880 kg>m3.

Here,

u max = -

3 2 ( 10-3 ) m 4 2 4 ( 103 ) N>m2 a2 dp = £ § 8m dx 0.5 m 8 3 30.2 ( 10-3 ) m 4

= 0.1324 m>s = 0.132 m>s

Ans.

Thus, the maximum Reynolds number is Re =

rUmax a = m

( 880 kg>m3 )( 0.1324 m>s ) 3 0.2 ( 10-3 ) m 4 = 7.719 6 1400 (OK) 30.2 ( 10-3 ) N # s>m2

At the top and bottom plate, y = a = 2 ( 10-3 ) m, and y = 0, respectively. tt =

4 ( 103 ) N>m2 2 ( 10-3 ) m dp a ay - b = £ § £ 2 ( 10-3 ) m § dx 2 0.5 m 2

= - 8 Pa tb =

4 ( 103 ) N>m2 2 ( 10-3 ) m dp a ay - b = £ § £0 § dx 2 0.5 m 2

= 8 Pa

Ans.

Ans.

Ans: u max = 0.132 m>s tt = - 8 Pa    tb = 8 Pa 931

Click here to Purchase full Solution Manual at http://solutionmanuals.info 9–3.  Air at T = 40° F flows with an average velocity of 15 ft>s past the charged plates of the air cleaner. The plates are each 10 in. wide, and the gap between them is 1>8 in. If fully developed laminar flow develops, determine the pressure difference pB - pA between the inlet A and the outlet B.

1 — in. 8 A

B 24 in.

SOLUTION

From Appendix A, at T = 40° F, ra = 0.00247 slug>ft 3 and ma = 0.363 ( 10-6 ) lb # s>ft 2. The Reynolds number of the flow is raVa Re = = ma

( 0.00247 slug>ft 3 )( 15 ft>s ) a

0.125 ft b 12

0.363 ( 10-6 ) lb # s>ft 2

= 1063 6 1400 (laminar flow) Here, the flow can be considered as horizontal flow caused by pressure difference between the inlet and outlet through two fixed parallel plates. Also the air is incompressible. Therefore,

Vavg = -

2

a

2 0.125 ft b 12

pB - pA a dp ;  15 ft>s = c d -6 2 # 12ma dx 12 3 0.363 ( 10 ) lb s>ft 4 ( 24>12 ft ) pB - pA = - 1.204 lb>ft 2 = - 1.20 lb>ft 2

Ans.

The negative sign indicates that the pressure drops from the inlet to outlet.

Ans: pB - pA = - 1.20 lb>ft 2 932

Click here to Purchase full Solution Manual at http://solutionmanuals.info *9–4.  Glue is applied to the surface of the plastic strip, which has a width of 200 mm, by pulling the strip through  the container. Determine the force F that must be applied to the tape if the tape moves at 10 mm>s. Take rg = 730 kg>m3 and mg = 0.860 N # s>m2.

10 mm/s

40 mm F 40 mm 300 mm

SOLUTION The Reynolds number of the flow is

Re =

rgVa mg

=

τA F

( 730 kg>m3 ) c 10 ( 10-3 ) m>s d (0.04 m)

τA

0.860 N # s>m2

(a)

= 0.3395 Since Re62300, steady laminar flow occurs. Also, the glue is incompressible. Here, the flow is horizontal which is caused by the top moving plastic strip. Therefore, the shear stress acting along the surface of the plastic strip can be determined using Eq. 9–14.

t =

Umg a

=

c 10 ( 10-3 ) m>s d ( 0.860 N # s>m2 ) 0.04 m

= 0.215 N>m2

Consider the horizontal equilibrium of the FBD of the plastic strip in Fig. a, + ΣFx = 0;  F - 2tA = 0 S

F = 2tA = 2 ( 0.215 N>m2 ) (0.3 m)(0.2 m) Ans.

= 0.0258 N

933

Click here to Purchase full Solution Manual at http://solutionmanuals.info 9–5.  The 20-kg uniform plate is released and slides down the inclined plane. If an oil film under its surface is 0.2 mm thick, determine the terminal velocity of the plate along the plane. The plate has a width of 0.5 m. Take ro = 880 kg>m3 and mo = 0.0670 N # s>m2.

0.75 m

15

SOLUTION Here, the terminal velocity is constant. Thus, the plate is in equilibrium. Referring to the FBD of the plate shown in Fig. a,

20(9.81)N

+ ΣFx = 0;  Fv - [20(9.81) N] sin 15° = 0  Fv = 50.78 N S

15°

Thus, the shear stress acting on the bottom plate’s surface is tp =

Fr

Fv 50.78 N = 135.41 N>m2 = Ap (0.75 m)(0.5 m)

We will assume that the steady laminar flow occurs and the oil is incompressible. dp dh = 0, = - sin u and U = u t. Then Here dx dx ut u0 a t = - g ay - b sin u a 2

N (a)

For this case u = 15° and a = 0.2 ( 10-3 ) m. At y = a = 0.2 ( 10-3 ) m, t = tp = 135.41 N>m2. Substitute this data into the above equation 135.41 N>m2 =

u t ( 0.0670 N # s>m2 ) 0.2 ( 10

-3

)m

- ( 880 kg>m3 )( 9.81 m>s2 ) c 0.2 ( 10-3 ) m -

0.2 ( 10-3 ) m

Ans.

u t = 0.4049 m>s = 0.405 m>s

2

d (sin 15°)

The Reynolds number is Re =

ru t a = m0

( 880 kg>m3 )( 0.4049 m>s ) 3 0.2 ( 10-3 ) m 4 0.0670 N # s>m2

= 1.06 6 1400 (laminar flow)

Ans: 0.405 m>s 934

Click here to Purchase full Solution Manual at http://solutionmanuals.info 9–6.  Using pins, the plug is attached to the cylinder such that there is a gap between the plug and the walls of 0.2 mm. If the pressure within the oil contained in the cylinder is 4  kPa, determine the initial volumetric flow of oil up the sides of the plug. Take ro = 880 kg>m3 and mo = 30.5 1 10-3 2 N # s>m2. Assume the flow is similar to that between parallel plates since the gap size is very much smaller than the radius of the plug.

100 mm 50 mm

SOLUTION The flow is assumed to be steady laminar, and the oil is incompressible. Positive x axis is directed in the direction of flow which is vertically upwards. r2 - r1 h2 - h1 0 (p + gh) = b + ga 0x L L =

0 - 4 ( 103 ) N>m2 0.05 m

= - 71.367 ( 103 )

+ ( 880 kg>m3 )( 9.81 m>s2 ) a

N>m2

0.05 m - 0 b 0.05 m

m

Substitute this value and U = 0 into u = 0 -

1

2 3 30.5 ( 10-3 ) N # s>m2 4

J - 71.367 ( 103 )

N>m2 m

u = 1.1700 ( 106 ) 3 0.2 ( 10-3 ) y - y2 4

R 3 0.2 ( 10-3 ) y - y2 4

Here b = 2p(0.05 m) = 0.1p m. Then the flow rate is Q =

=

LA L0

udA

0.2(10-3) m

1.1700 ( 106 ) 3 0.2 ( 10-3 ) y - y2 4 (0.1p)dy

= 0.36755 ( 106 ) J0.1 ( 10-3 ) y2 -

-3 y3 0.2(10 ) m R` 3 0

= 0.4901 ( 10-6 ) m3 >s = 0.490 ( 10-6 ) m3 >s

Ans.

The average velocity is V =

0.4901 ( 10-6 ) m3 >s Q = = 7.800 ( 10-3 ) m>s A (0.1p m) 3 0.2 ( 10-3 ) m 4

The Reynolds number is Re =

rVa = m0

( 880 kg>m2 ) 3 7.800 ( 10-3 ) m>s 4 3 0.2 ( 10-3 ) m 4 30.5 ( 10-3 ) N # s>m2

= 0.0450 6 1400 (laminar flow)

935

Ans: 0.490 ( 10-6 ) m3 >s

9–7.  The boy has a mass of 50 kg and attempts to slide down the inclined plane. If a 0.3-mm-thick oil surface develops between his shoes and the surface, determine his terminal velocity down the incline. Both of his shoes have a total contact area of 0.0165 m2. Take ro = 900 kg>m3 and mo = 0.0638 N # s>m2.

2

SOLUTION Since the boy is required to move with terminal velocity (constant), a = 0. Referring to the free-body diagram of the boy in Fig. a,

50(9.81) N a= 0

+ ΣFx = max;   350(9.81)N4 sin 2° - F = 0 d

2° x

F = 17.118 N

The shear stress on the oil layer in contact with the boy’s shoes is t =

Here, t =

F 17.118 N = = 1037.47 Pa A 0.0165 m2

F N

dp dh = 0 and = - sin 2°. dx dx

(a)

Um d a + c (p + gh) d ay - b a dx 2

1037.47

U ( 0.0638 N # s>m2 ) N = + 2 m 0.3 ( 10-3 ) m

U = 4.879 m>s = 4.88 m>s

30

+ ( 900 kg>m3 )( 9.81 m>s2 )( -sin 2° ) 4 c 0.3 ( 10-3 ) m -

0.3 ( 10-3 ) m

Ans.

2

d

Thus, the maximum Reynolds number is Re =

rUa = m

( 900 kg>m3 ) (4.879 m>s) 3 0.3 ( 10-3 ) m 4 0.0638 N # s>m2

= 20.65 6 1400 (OK)

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Ans: 4.88 m>s 936

*9–8.  The 2.5-lb plate, which is 8 in. wide, is placed on the 3° inclined plane and released. If its terminal velocity is 0.2  ft>s, determine the approximate thickness of oil underneath the plate. Take ro = 1.71 slug>ft 3 and mo = 0.632 1 10-3 2 lb # s>ft 2.

12 in. 3

SOLUTION + ΣF = max;  (2.5 lb) sin 30° - F = 0 S F = 0.1308 lb

2.5 lb a=0

Thus, the shear stress on the oil layer in contact with the bottom surface of the plate (y = a) is t =

P = A

dp dh = 0 and Here, = - sin 3°. dx dx

0.1963

lb = ft 2

0.1308 lb lb = 0.1963 2 12 8 ft a ft ba ft b 12 12

F N (a)

Um d a + c (p + gh) d ay - b a dx 2

t =

( 0.2 ft>s )( 0.632 ( 10-3 ) lb # s>ft 2 ) a

+

30

+ ( 1.71 slug>ft 3 )( 32.2 ft>s2 ) ( - sin 3°) 4 aa -

1.4409a2 + 0.1963a - 0.1264 ( 10-3 ) = 0 Solving for the positive root,

a = 0.6410 ( 10-3 ) ft = 7.69 ( 10-3 ) in.

Ans.

Thus, the maximum Reynolds number is Re =

rUa = m

3°

( 1.71 slug>ft 3 )( 0.2 ft>s ) 3 0.6410 ( 10-3 ) ft 4 = 0.3469 6 1400 (OK) 0.632 ( 10-3 ) lb # s>ft

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937

a b 2

x

9–9.  The water tank has a rectangular crack on its side having a width of 100 mm and an average opening of 0.1 mm. If laminar flow occurs through the crack, determine the volumetric flow of water through the crack. The water is at a temperature of T = 20° C.

100 mm

2m 100 mm

SOLUTION Assuming that steady laminar flow occurs, and the water is incompressible. The flow can be considered as horizontal flow through two stationary parallel plates driven by a pressure gradient. The pressure of water at the inlet of the crack is pin = rwgh = ( 998.3 kg>m3 )( 9.81 m>s2 ) (2 m) = 19.587 ( 103 ) N>m2 At the outlet, pout = patm = 0. Thus, the pressure gradient is N>m2 0 - 19.587 ( 103 ) N>m2 pout - pin dp = = = - 195.87 ( 103 ) dx L 0.1 m m Q = -

3 0.1 ( 10-3 ) m 4 3(0.1 m) N>m2 a3b dp -3 ( ) J -195.87 10 R = 12m dx m 12 3 1.00 ( 10-3 ) N # s>m2 4

The average velocity is V =

= 1.632 ( 10-6 ) m3 >s = 1.63 ( 10-6 ) m3 >s Q = A

The Reynolds number is Re =

rVa = m

1.632 ( 10-6 ) m3 >s

3 0.1 ( 10-3 ) m 4 (0.1 m)

Ans.

= 0.1632 m>s

( 998.3 kg>m3 )( 0.1632 m>s )( 0.1 ( 10-3 ) m ) 1.00 ( 10-3 ) N # s>m2

= 16.29 6 1400 (laminar flow)

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938

Ans: 1.63 ( 10-6 ) m3 >s

9–10.  A solar water heater consists of two flat plates that rest on the roof. Water enters at A and exits at B. If the pressure drop from A to B is 60 Pa, determine the largest gap a between the plates so that the flow remains laminar. For the calculation, assume the water has an average temperature of 40° C.

A 3m

7.5

B

a

SOLUTION

From Appendix A at T = 40° C, rw = 992.3 kg>m3 and mw = 0.659 ( 10-3 ) N # s>m2. It is required that steady laminar flow occurs. Also, the water is incompressible. Here 60 N>m2 0 - (3 m) sin 7.5° 0 (p + gh) = + ( 992.3 kg>m3 )( 9.81 m>s2 ) c d 0x 3m 3m = - 1290.60

N>m2 m

u = u =

1

1 0 (p + gh) d ( ay - y2 ) c 2mw 0x

2 3 0.659 ( 10-3 ) N # s>m2 4

u = 0.9792 ( 106 )( ay - y2 )

a - 1290.60

N>m2 m

b ( ay - y2 )

The flow rate can be determined from Q =

L A

udA = 0.9792 ( 106 )

L0

a

( ay - y2 ) b dy = 0.1632 ( 106 ) a3b

The average velocity is V =

0.1632 ( 106 ) a3b Q = = 0.1632 ( 106 ) a2 A ab

To satisfy the condition of laminar flow, we require Re = 1400 rwVa = 1400 mw

( 992.3 kg>m3 ) 3 0.1632 ( 106 ) a2 4 a = 1400 0.659 ( 10-3 ) N # s>m2 a = 1.786 ( 10-3 ) m = 1.79 mm

Ans.

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Ans: 1.79 mm 939