980806Solutons Manual(1 2章) 修正搞

Solutons Manual 1.3 Classify the following substances (maintained at constant temperature): (a) du/dy, rad/s

0

3

4

6

5

4

τ,1b/ft2

2

4

6

8

6

4

du/dy, rad/s

0

0.5 1.1 1.8

τ,N/m

0

2

du/dy, rad/s

0

0.3 0.6 0.9 1.2

τ,1b/ft2

0

2

(b)

2

4

6

(c)

4

6

8

Solution (a) Non Newtonian. Hysteresis suggests thixotropy. (b) Nay Newtonian (c) Newtonian (linear relationship between τ and

du ) dy

The constant slope of the line gives the absolute viscosity μ：



 80 lb s     6.67 1.2  0 ft 2  du    dy

1.4 A Newtonian liquid flows down an inclined plane in a thin sheet of thickness t(Fig. 1.8). The upper surface is in contact with air, which offers almost no resistance to the flow. Using Newton’s law of viscosity, decide what the value of du/dy, y measured normal to the inclined plane, must be at the upper surface. Would a linear variation of u with y be expected? Solution Since air offers no resistance to the flow, τ=0.

At free surface : τ=0 

At the plane : τ≠0 

du dy

du dy

0 y t

0 y 0

Therefore, since we have change in

du (slope) the relationship dy

between μ and y is non-linear. 1.11 Does the weight of a 20 N bag of flour at sea level denote a force or the mass of the flour? What is the mass of the flour in kilograms? What are the mass and weight of the flour at a location where the gravitational acceleration is one-seventh that of the earth’s standard? Solution The weight always denotes force. W 20 N W  m g m  m  m 2.04kg g 9.806m / s 2 The mass is the same under any conditions. 1 1 20 N W '  m g '  m g  W W '  W ' 2.86 N 7 7 7 1.12 A Newtonian fluid is in the clearance between a shaft and a concentric sleeve. When a force of 600 N is applied to the sleeve parallel to the shaft, the sleeve attains a speed of l m/s. If a 1500 N force is applied, what speed will the sleeve attain? The temperature of the sleeve remains constant. Solution

 

du u F  ;   dy t A

Therefore : F  

A u  ￠．u ； ￠=constant t

F1  ￠． 1 and F2  ￠  2 and : F2 1500 m 1  (1) 2 2.5 F1 600 s

2 

1.16 Determine the viscosity of fluid between the shaft and sleeve in Fig. 19. Solution

  

du u  dy t

F A

A=Π．D．L Therefore: F t F t    A u  ( D L)

 0.003  lb S  12     0.02387 2 3 8 ft 3.14    12 12

(20)  

(0.4)  

1.17 A flywheel weighing 600 N has a radius of gyration of 300 mm. When it is rotating at 600 rpm, its speed reduces 1 rpm/min owing to fluid viscosity between the sleeve and shaft. The sleeve length is 50 mm; shaft diameter is 20 mm; and radial clearance is 0.05 mm. Determine the fluid viscosity. Solution I = mass moment of inertia K = radius of gyration T = torque and :

I = mk2 ; T  I

dw T ;  dt A r

A = 2ΠrL ; u  wr ;   

t u

Therefore : mk 2t dw   3 2  wr L dt 2

300 0.05  600           9.806  1000 1000  2  3.14 1          3 60 60 2  3.14  10    50  2  3.14   600      60   1000 1000     N S  0.0243 2 m    0.0243

N S m2

1.18 A 25-mm-diameter steel cylinder 300 mm long falls, because of its own weight, at a uniform rate of 0.1 m/s inside a tube of slightly larger diameter. A castor-oil film of constant thickness is between the cylinder and the tube. Determine the clearance between the tube and the cylinder. The temperature is 38℃. The specific gravity of steel = 7.85. Solution D 2 F  W      L 4

  

du u  dy t

F ; A =Π．D．L A

So : t

 A u   D L  u 4 u    D2 F  D  L 4

Therefore : N S   m 4  2.8  101 2  0.1  m   s  5.82 105 m t  25   7.85  9806     1000 so : t = 5.82 × 10-5m = 0.06mm Note : for T = 38℃ from figure C.1 we find that : μ= 2.8 × 10-1 N．S/m2 1.20 A 12 kg cube slides down an inclined plane making an angle of 30° with the horizontal. A fluid film 0.1 mm thick separates solid and surface. The fluid viscosity is 0.04 N．s/m2. Assuming the velocity distribution in the film is linear, find the terminal velocity of the block. The area of the cube in contact with the film is 0.25 m2. Solution Assume that the cube has reached its terminal velocity (u) and that the temperature (T) is constant.

 

du u F  ;  ; F = W．sin 30° dy t A

Therefore :

 w sin 30 F u    u  A t  A

o



 12  9.806  sin 30   o

so :

u

(0.04)(0.25)

t

0.1   1000  u  0.59m / s

1.24 A body weighing 120 lb with a flat surface area of 2 ft2 slides down a lubricated inclined plane making a 30° angle with the horizontal. For viscosity of 0.002 lb．s/ft2 and body speed of 3 ft/s, determine the lubricant film thickness.

Solution W = 120 lb A = 2ft2 μ= 0.002 lb．s/ft2 u = 3 ft/s

 

du u F  ;  ; F = W sin 30° dy t A

Therefore : F u u A u A   t    A t F w sin 30o so : t  (0.002)

(3)(2)  t  0.0002 ft 120 sin 30o

or t = 0.0024 in. 1.25 What is the viscosity of gasoline at 25℃ in poises? Use Fig. C.1, Appendix C. Solution From figure C.1 : ◎T = 25℃ μ= 6 × 10-6 lb．s/ft2 so :

   6 106    479  poise    2.874 10 3 poise 1.26 A liquid has a specific weight of 48 lb/ft3 and a dynamic viscosity of 3.05 lb．s/ft2. What is the kinematic viscosity? Solution

  48lb / ft 3 ;   3.05lb s / ft 2    g

;    

so :



 g  3.05   32.174      2.044 ft 2 / s  48

1.27 What is the specific volume in cubic feet per pound mass and cubic feet per slug

of a substance of specific gravity 0.75? Solution s 

1 1 1 ft 3    s  0.02136  s  (0.75)(62.43) lbm

and s 

1 1 1 ft 3    s  0.687  s  (0.75)(1.94) sluy

1.29 The density of a substance is 2.94 g/cm3. In SI units, what is its (a) specific gravity, (b) specific volume, and (c) specific weight? Solution

  2.94

g 103 kg kg  2.94    2.94  103 3 3 3 cm m  102  m3

 2.94 103   S  2.94 (a) S  w 103 (b) s 

1 1 m3     3.4014 s  2.94  103 kg

(c)    g  2.94 103   9.806   28.83

KN m3

1.42 Three wastewater samples were taken from the same site at the same time, and after analyzing them in the laboratory the following results were obtained Sample

Volume (mL)

Mass suspended solids (g)

1 2 3

75 83.2 80

23.0 35.6 Glass container broken

The three samples are of the same density (ρ), and the first one contains solids of specific gravity 1.93. Find the density ρ and the concentration of the suspended solids in the three samples. Take the mass fraction of solids in Sample 3 to be the average of the first two.

Solution Sample 1 :

  s  1  1.93 1000  1930kg / m3

1  75mL  75  106 m3 ;  ms  1  23 g  23  10 3 kg

Now  

m  s ; ws  s 1 ws   1  ws   1.93 1

Therefore :



s  0.93  ms  1  s  1.93  0.93ws 1.93 1.93 1



1930 0.93 23 103 kg     1147.8 3 6 1.93 1.93 75 10 m

All three samples have the same density ρ.

 Cs  1 

 Cs  2 

 ms  1 1

 ms  2 2

Now :  ws  1 

 ws  2 

 ms  2 m

23 103 kg kg    Cs  1  306.7 3 6 3 75  10 m m



35.6  103 kg kg   Cs  2  427.9 3 6 3 83.2 10 m m

 ms  1 m

23 103   0.26717  1147.8   75 106 

35.6  103   0.37279  1147.8  83.2 106 

Therefore :

 ws  3 

1 1  ws  1   ws  2    0.26717  0.37279  2 2

  ws  3  0.31998

and :

 ws  3 

 ws  3 3

  Cs  3 

 ws  3 3

   ws  3

Finally :

 Cs  3    ws  3  (1147.8)(0.31996)   Cs  3  367.27

kg m3

1.47 A gas with molecular weight 28 has a volume of 4.0 ft3 and a pressure and temperature of 2000 psfa (lb/ft2 abs) and 600°R, respectively. What are its specific volume and specific weight? Solution R

1545 1545 ft lb   R  55.179 o M 28 lbm R

s 

1 R T (55.179)(600)     P 2000

 s  532.60 ft 3 / slug  16.554 ft 3 / lbm 1 32.174 lb    g  g   0.06041 3 s 532.60 ft

1.53 An open tank contains water at 22℃ with 32.7-percent (by weight) suspended solids of specific gravity 2.32. If the volume of the mixture is 1.2 m3, calculate the concentration of the solids in 1bm/ft3 and kg/m3. Solution ws  0.327 ; m  1.2m3 ; T=22oC

From equation (1.5.8) for T=22℃ we find that :  w  997.783kg / m3 Therefore:  s  (2.32)(997.783)   s  2314.86

m 

s ws   1  ws 

  m  1225.86

s w



kg m3

2314.86 0.327  (1  0.327)(2.32)

kg m3

(mixture) mm   mm  (1225.86)(1.2)  mm  1471kg (solids) ms  ws mm (0.327)(1471) ms 481.02kg Therefore : Cs 

ms 481.02 kg   Cs  400.85 3 m 1.2 m

Now : 1lbm  0.453592kg ; 1m  3.28084 ft lbm 0.453592  16.02kg / m3 and : 1 ft 3  3  1/ 3.28084  Or 1kg / m3  0.0624lbm / ft 3 so: Cs  (400.85)(0.0624)  Cs  25.01

lbm ft 3

1.58 A steel container expands in volume 1-percent when the pressure within it is increased by 10,000 psi. At standard pressure, 14.7 psi absolute, it holds 0.5 slug of water; ρ= 1.94 slug/ft3. For K = 300,000 psi, when it is filled, how many slugs of water must be added to increase the pressure to 10,000 psi ? What is the weight of the water added? Solution o    p  14.7 psi  

mo 0.5   o  0.258 ft 3 0 1.94

1    p  10, 000 psi   1.01o  1  0.261 ft 3 Now :   o

p 10000 slug  (1.94)    0.065 3 k 300000 ft

and

1   o    1.94  0.065  1  2.005

slug ft 3

mo  o o  (1.94)(0.258)  mo  0.5slug m1  11  (2.005)(0.261)  m1  0.5233slug m  m1  mo  0.5233  0.5  m  0.0233slug Therefore 0.0233 slug of water must be added. △W = g．△m = (32.174) (0.0233) → △W = 0.75 lb 1.62 What is the isothermal bulk modulus for air at 0.4 MPa abs? Solution k 

dp mRT ; p d /  

so :

dp m RT dp m R T    p 2 d  d /  

Therefore : K = P = 0.4MPa abs. The relation K = P holds for any ideal gas, at any pressure. 1.66 A small circular jet of mercury 0.1 mm in diameter issues from an opening. What is the pressure difference between the inside and outside of the jet when at 20℃ ? Solution at T = 20℃ from table we find : σ= 0.51 N/m Now :

Fσ = 2σL

(1)

FP = P．D．L

(2)

H should be FP = Fσ

2 So : P D L 2 L p  D Therefore : p 

(3)

2(0.51)  p  10200 N / m 2 3 0.1 10

Or P = 10.2 KPa 1.69 Using the data given in Fig. 1.6, estimate the capillary rise of tap water between two parallel glass plates 0.20 in. apart. Solution For the two parallel plates (figure) : p A 2 ( L) cos a p (W L) 2 L cos a

(1)

P =  H

(2)

and

From (1) and (2) :

 cos a  2 H W

For a vertical tube : h 

 cos  2 cos    2 hr h D

or :

(3)

2 cos  r (4)

where D is the tube diameter. From (3) and (4) we find : cos a  D     W  2 cos 

H  h

(5)

Now letting a=θ and D=W, equation (5) gives : H 

h 2

From figure 1.6, for D  2 101 m we find that h = 0.09in (T = 68℉) So : H 

0.09  0.045in.  H  0.045in. 2

1.70 A method of determining the surface tension of a liquid is to find the force needed to pull a platinum wire ring from the surface (Fig. 1.10). estimate the force necessary to remove a 20-mm-diameter ring from the surface of water at 20℃.

Solution Let D be the ring diameter. The wire diameter is not important. Assume that the surface is vertical at the breaking point A. 2 From table C.1 :   7.36 10

N m



20  2   7.36 10   1000

F  2 D  2(3.14)   F  9.24 103 N

1.71 Calculate the capillary rise h in the tube of Fig. 1.11 in terms of θ,σ,γ, and r. Solution If F  2r cos  is the Component of surface tension down, then a force Balance gives : 0 2 r cos   W p2  p1  p  r 2 But Δp = γ．h So :  h 

2 cos  2 h cos  r r

1.73 What vertical force due to surface tension would be required to hold the tube of Fig. 1.11? Consider the tube wall thickness to be very small. Solution The total force required to hold the tube is : F  2 F  2  2r cos  

Therefore :

F  4 r cos 

1.75 Develop a formula for capillary rise h between two concentric glass tubes of radii R and r and contact angle θ. Soution The pressure force is : Fp  p A  p A Where : p   h A  ( R 2  r 2 ) 2 2 Therefore : Fp   h  R  r 

(1)

The total force due to surface tension down is : F   (2r  2 R ) cos 

(2)

Neglecting the weight W of the fluid between the two tubes, a force balance gives: Fp  F    h  R 2  r 2   2  ( R  r ) cos  or :  h  R  r   R  r   2  R  r  cos  h

2 cos    R r 

2.3 What is the pressure at a point 10 m below the free surface in a fluid that has a variable density in kilograms per cubic meter given by σ= 450+ah, in which a=12 kg/m4 and h is the distance in meters measured from te free surface ? Solution h  10m ;   b  ah ; a  12

kg kg ; b  450 3 4 m m

dp   dy   ( dh)  g  dh (b ah) g dh

(1)

From (1) by integration we obtain: p

 ah 2 dp  g ( b  ah ) dh  p  g bh    o o 2  h

(2)

Substitute the values for g, a, b and h, and find :  12  102 2 p  (9.806)  450 10    p  50 KN / m 2   2.10 Express a pressure of 50 kPa in (a) millimeters of mercury, (b) meters of water, (c) meters of acetylene tetrabromide, S = 2.94. Solution p   h S  w h ,

 w  62.4lb / ft 3  9806 N / m3

(a) S=13.57 hHg  50 /  13.57  9806   hHg  375.75mm So : p  50kpa  375.75mmHg 3 (b) hH 2O   50 10  /  9806   hH 2O  5.1m

So : p  50kpa  5.1mH 2O 3 (c) ha   50 10  /  294  (9806)  ha  1.734m

P = 50kPa = 1.734m acetylene tetrabromide 2.11 A Bourdon gage reads 2 psi suction, and the barometer is 29.5 in. Hg, Express the pressure in six other customary ways. Solution 2

lb lb 4.448 N  1in  2 2 2 2    13788.83 pa in in 1lb  0.0254m 2

lb 34 ft 0.3048m  2 psi  1.410mH 2O  0.104mHg 2 in 147 psi 1 ft

For absolute pressures : Pabs  Pbar  Pgage 

 24.5in  

2  12  0.0254m  0.646mHg abs  (0.435)(13.6) 1in

 0.646mHg abs   13.6   8.786mH 2O abs  8.786   9806   86155.5Pa abs

2.12 Express 4 atm in meters of water gage with a barometer reading of 750mm Hg. Solution Pbar  750mmHg 

750  13.6  10.2mH 2O 1000

psi 144 psf 1   atm 1 psi 62.4lb / ft 3

Pabs  4atm  14.7

So : Pabs = 135.63ft = 41.34m Pgage  41.34  10.2  Pgage  31.14mH 2O 2.15 The container of Fig. 2.41 holds water and air as shown. What is the pressure at A, B, C, and D in pounds per square foot and in pascals. PA   h  62.4(1  3)  249.6 psf N lb 2 m  0.3048 ft 2  

lb  249.6 2  ft 

4.448

 11950 Pa PB  PA    3  1  1  249.6  62.4  5  62.4 psf  2987.70 Pa Pc  PB ; PD  PC    1  1  3  62.4  62.4  5  374.4 psf  17926.27 Pa

2.23 In Fig.2.11a S1 = 1.0, S2 = 0.95, S3 = 1.0, h1 = h2 = 280 mm, and h3 = 1 m. Compute PA – PB in millimeters of water. Solution hA  hB  h1S1'  h2 S 2'  h3 S3'  280 1  280  0.95  1000 1  454mmH 2O Therefore : P  PA  PB  454mmH 2O 2.24 In Prob. 2.23 find the gage difference h2 for pA – pB = -350 mm H2O. Solution From problem 2.23 h3  h1  h2  AB  1000  280  280  AB  AB  440mm Then h3  h1  h2  440 The manometer equation is : hA  hB  h1S1'  h2 S 2'  h3 S3'  350mm Thus :

 h1  h3   h2  0.95  350    h2  440   0.95h2  350  h2  1800mm 2.30 In Fig. 2.12 determine R, the gage difference, for a difference in gas pressure of 9 mm H 2O,  2  9.8kN / m3 , and a / A  0.01. Solution From equation 2.4.1

 a    a PC  P  R   3   2  1     1  A    A  Since  1   2 and  1   3 (that is,  1  0 ) We find that : R

P   3   2  1 

a  A



 2h  a  3   2 1    A



h  3  a  1    2  A

Therefore : R

9mm 10.5   1  0.01 9.8

 R  110.53mm

2.31 The inclined manometer of Fig. 2.13 reads zero when A and B are the same pressure. The reservoir diameter is 2.0 in. and that of the inclined tube is 1/4 in. For θ=30° and gage fluid S = 0.832, find pA – pB in pounds per square inch as a function of gage reading R in feet. Solution

A  y a R ;

 h R sin  ; PA   h  y   PB

Combination of the above equations yields: a  PA  PB  R    sin    A 

2

  1 a 4  4   1 Where :  A  2 2 64   4 Therefore : 1   1 p  R  62.4  0.832   0.5      64  144    p  0.1859 R ( ft ) in psi 2.35 The container shown in Fig. 2.48 has a circular cross section. Determine the upward force on the surface of the cone frustum ABCD. What is the downward force on the plane EF ? Is this force equal to the weight of the fluid? Explain. Solution

D1  2 ft ; D2  4 ft ; h1  2 ft ; h2  1 ft ; h3  8 ft D   D2  D1  / 2  1 ft

 3.14   4   D22 1   h1  h2   4 4  3.14   2   D12 2   h1   4 4 3 

2

2

 2  1  1  37.68 ft 3

 2   2  6.28 ft 3

2 2 1   D1  D2  D1D2   h2   3  7.327 ft 3 3 4

The force up on the come frustum ABCD is: FABCD    (62.4)(37.68 6.26 7.327) 

 FABCD  1502.16lb The weight of the water in the container is :   3.14   4  2  W  T   62.4   2  1  5        4     62.4   100.46   37.68  6.26  7.327     W  4767.80lb FEF

 3.14   4  D22  p A   h1 h2 h3   62.4   8  4 4

2



 FEF  6270lb . So : FEF  W  FABCD 2.38 A vertical right-angled triangular surface has a vertex in the free surface of a liquid (Fig. 2.50). Find the force on one side (a) by integration and (b) by formula. Solution

(a)

x y b   x  y ; dA  x dy  b h h h

h

o

o

 b  y dy   h 

F   pdA     y   xdy      y   A

b b  y 3  F    y 2 dy     h h  3 o h

Therefore : F 

 bh 2 3

h

 o

 bh 2 3

 2  1  h bh  3  2 

(b) F  pG A  h A   F



 bh 2 3

2.39 Determine the magnitude of the force acting on one side of the vertical triangle ABC of Fig. 2.51 (a) by integration and (b) by formula. (a) The low of cosines gives: 32  42  52  2  4   5  cos B   B  36.87o and H  4sin B  H  2.4 ft From similar triangles:

x 5  H Y  , so : 5 H

x  2.0833(7.4  y ) and : y2

7.4

y1

5

F    yxdy 

 (55 y)  2.0833(7.4  y)dy 

 F  1914lb 2.4  3  4  F   hA  (55)  5     (b) 3  2   F  1914lb 2.40 Find the moment about AB of the force acting on one side of the vertical surface ABC of Fig. 2.50.   9000 N / m3 . Solution

M AB   x p dA  xydA  I xy ; p  y A

A

Now from Appendix A : I x' y '

b2h2  24

∴ I x' y ' 

b2h2   24 

For figure 2 then : I xy 

b   3 

b2 h2   h  bh     72  3  2

b2h2 72

Then for our problem (figure 3) we have : b 2 h 2  2     b  bh I xy  I xy  x y A   h   72  3     3  2

b2 h 2  8

Finally :  b 2 h 2 9800 2 2 b h  M AB  1125b 2 h 2   8  8 

M AB   

Where MAB in N．m and b, h in m (y axes in Figures 2 and 3 point to opposite directions) 2.42 Find the moment about AB of the force acting on one side of the vertical surface ABC of Fig. 2.51. Solution From equation (2.5.10) yp 

IG y y A

IG 

bh3 36

A

1  4   3  6 ft 2 2

The law of cosines give :

32  42  52  2   4   5  cos B  B  36.8oF Then : H  4sin B  4sin(36.87 o)  H  2.4 ft y  5

1  2.4   5.8 ft  y  5.8 ft 3

IG 

1 3  5   2.4   I G  1.92 ft 4 36

yp 

1.92  5.8  y p  5.855 ft  5.8  6 

55  5.8   6 

F  pG A  y A

 F 1914  lb

Finally : M AB  F  arm  F   y p  5   1914   5.855  5  and : M AB  1636.47lb ft 2.43 Locate a horizontal line below AB of Fig. 2.51 such that the magnitude of pressure force on the vertical surface ABC is equal above and below the line. Solution Let F1 be the force on the area between AB and EF, and F2 on the area below EF. We need to find y* such that F1 = F2. Using the results from problem 2.34 we have : F

H = 2.4 ft;

y2

  2.0833   7.4 y  y  dy 2

(1)

y1

From equation (1) : y*

 7.4 y 2 y 3 F1  (2.0833 )   7.4 y  y  dy   2.0833     3  2 5

y*

2

5

 7.4 y 2 y 3 F2  (2.0833 )   7.4 y  y  dy   2.0833     3  2 y* 7.4

7.4

2

y*

Therefore for F1 = F2 we find that :  7.4 y 2 y 3  2  3  

y*

5

 7.4 y 2  y 3    2  3

7.4

 y*3  11.10 y*2  177.56  0 y*

Solving the latest equation by trial  ' error we find : y*  5.774 ft or that EF is 0.774 ft below AB 2.48 Locate the distance of the pressure center below the liquid surface in the triangular area ABC of Fig. 2.51 (a) by integration and (b) by formula. Solution The law of cosines gives : 32  42  52  2(4)(5) cos B  B  36.87o So : H = 4 sin B = 2.4 ft (a) From similar triangles x 5 H  y   x  2.0833(7.4  y ) 5 H Now 7.4

yp 

  y 2 dA   ydA



2  y x dy

 y x dy

  7.4  y  y dy 2



5 7.4

  7.4  y  ydy 5



97.805  5.8552  y p  5.8552 ft 16.704

(b) y p  y 

IG h 2.4 ; y  5  5  5.8 ft 3 3 y A



 2.4   I  1.92 ft 4 bh3 IG   5 G 36 36 3

A

1  2.4   5  A  6 ft 2 2

Therefore :

y p  5.8 

1.92  y p  5.8552 ft  5.8  6 

2.49 By integration locate the pressure center horizontally in the triangular area of ABC of Fig. 2.51. Solution

From problem 2.46 ; H = 2.4 ft ; B = 36.87° Now AD   AC 2  DC 2

1/2

  32  2.42 

1/2

 AD  1.8 ft

DB  AB  AD  5  1.8  DB  3.2 ft x1 

3.2  2.4  y  ; x12  1.778  5.76  4.8 y  y 2  2.4

x2 

1.8  2.4  y  ; x22  0.5625  5.76  4.8 y  y 2  2.4

Take moments about CD : dM   hx1dA   hx 2 dA    5  y  x1 Then :

x1 x dy    5  y  x2 2 dy 2 2

2.4

M CD 



 2  5  y  x

2 1

0

2.4



 0

 x22  dy

55  5  y   1.778  0.5625  5.76  4.8 y  y 2  dy 2

 862.4lb ft : F = 1914 lb, so : x 

From problem 2.39

M CD  0.4506 ft F

2.51 Determine by integration the pressure center for Fig. 2.50. Solution From similar triangles : x y b  x y b h h and  b  h

dA  xdy  

F y p M x 0

 y dy  and M   y   y  dA , so : A

h

 b   dy 3  h 0 A yp  h  h A   y  dA  y 2  b dy 4  h 0

 y   y  dA  y

F x p M y  0

3

 x and M       y  dA , so : 2 A h

 y  b   x A  2   y  dA 0  2  h  y xp   h b  y  dA   A 0 y y h

b y dy  h dy 

3  b 8

2.52 Locate the pressure enter for the annular area of Fig. 2.54. Solution Ro  1m

Ri  500mm 

500 m  0.5m 1000

yp 

IG y y A

IG 

 4 R0  Ri4   4

A    R02  Ri2  Therefore :

IG 1 2   R0  Ri2  A 4 1 2 R0  Ri2   yp  4 y y

Finally :

1 2 1  0.52   4 2 2  2.1564m or :

 y  2m 

y p  2.1564m

2.53 Locate the pressure center for the gate of Fig. 2.55. Solution yp 

IG y y A

IG 

1 bh3 12

A  b h where : b = 4ft y

;

h = 6ft

1 h  3 ft  from A  or y  3  4  7 ft 2

3  1    4  6 12 yp     7  y p  7.43 ft  7  4  6

2.56 Locate the pressure center for the vertical area of Fig. 2.56. Solution Ix y A

yp 

I xy

xp 

y A

1 h A  b h ; y  2 3 Ix 

1 3 1 bh ; I xy  b 2 h 2 12 24

Therefore :  1 2 2  bh b 24 xp     xp  4  h  1    bh  3  2 

yp  

 

 1 3   bh h 12    yp  h  1  2  bh 3  2 

2.59 Determine the pivot location y of the rectangular gate of Fig. 2.59 so that it will open when the liquid surface is as shown. Solution want y at pressure center (critical location) measured from the free surface. 1 3  1  1 IG yp  y   1.5  12 y A  1.5   11 = 1.5556 m Therefore : y = 2-1.556 → y = 0.444 m 2.66 The gate of Fig. 2.62 weighs 300 lb/ft normal to the paper. Its

center of gravity is 1.5 ft from the left face and 2.0 ft above the lower face. It is hinged at O. Determine the water-surface position for the gate just to start to come up. (The water surface is below the hinge.) Solution

W = 300 lb/ft  h F   h A     h 1  2  M o  W  arm  F  arm 1 h   300  2   h 2   5   2 3  and  M o  0

(for equilibrium)

Therefore :

1 h  9806   5   h2  600 2 3 

Which yields : h = 2.116 ft 2.70 For linear stress variation over the base of the dam of Fig. 2.64 (a) locate where the resultant crosses the base and (b) compute the maximum and minimum compressive stresses at the base. Neglect hydrostatic uplift. Solution (a) 1 F1   h 2  364.5 ; F2  7  3  21 2

1 F3  (20)(3 )  30 2 F4   4   27   2.5   270 F5 

3  20   2.5   75 2

F6 

11  20   2.5   275 2

Rx  F1  364.5 Ry  F2  F3  F4  F5  F6  671 Take moments about A :  27   364.5    1.5   21    1  30   3   2  3  270    3  1  75 

 M A  0   671  x  

11    3  4    275  3  Which yields : x = 11.588m (b) Taking moments about A :

x  11.588 

 min 18 

18 18 2  18    max   min    2 2 3 18   max   min   2

which gives :  max  13.563 min 1 Also :   max   min   18  Ry  671 2 Therefore :  max  5.12 ;  min  69.436 2.71 Solve Prob. 2.70 with the addition that the hydrostatic uplift varies linearly from 20 m at A to zero at the toe of the dam. Solution Refere to problem 270

(a) Ry  671 

1  20   18   491 2  27   3

 M A  0   491  x   364.5     21   1.5    30   1

  270   5    75   2 



11    275   7    0 3  x  13.636m

(b)

x  13.636 

 min 18 

18 18 2    max   min   18  2 2 3 18   max   min  2

The above equation yields :  max  4.6675 min  18    max   min   Ry  491  2

Also : 

Therefore :

 min  14.875 ;  max  69.431

2.74 To what height h will the water on the right have to rise to open the gate shown in Fig. 2.67? The gate is 5 ft wide and is constructed of material with specific gravity 2.5. Use the pressureprism method. Solution

w     2.5  62.4   3  4  5   w  9360lb

F1  F2 

1  3   3  5  F1  1404lb at 3ft below 0 2

  h  4   h   4  5   F2   20  62.4  h  2  62.4   20  2  F2  1248h  2496

The line of action of F2 is at

2  3h  4  below 0 3 h  2

For equilibrium it should be :  M o  0 So : F1  arm  F2  arm  w  arm  0   1404   3   1248h  2496 

2  3h  4   2    9360    4  0 3 h  2  3 

 h 2   5.021 h  2.71  0 Therefore : h1 = 0.615 ft If h  h1 h  h2

and

h2 = 4.406 ft

then the gate will open to the right

then the gate will open to the left

For h  4 ft we choose

h = 4.406 ft

2.78 (a) Determine the horizontal component of force acting on the radial gate (Fig. 2.70) and its line of action. (b) Determine the vertical component of force and its line of action. (c) What force F is required to open the gate, neglecting its weight? (d) What is the moment about an axis normal to the paper and through point O? Solution (a) h  3  1  4m A  2  2  4m 2

  9806 N / m3

Therefore : FH   hA  (9806)(4)(4)  156896 N 1 (2)(2)3 IG IG yp  y  h  4  12 (4)(4) y A h A  y p  4.0833m And : b  y p  3  b  1.0833m (b) 

 (2) 2  Fv    (9806)   2 3 2   2 4    179284.91N   2  2 4 2 a   18.283  1    2  3  2    3 4  a  0.948m 2

(d) Since the pressure force is normal to the surface, the resultant of FH and FV has to pass through O. Taking moments about O, we see that F = O (For equilibrium  M o  0 ). check：  Mo  0   2  F  156896  1.0833  179284.91 0.948  0 F 0 (d) Answered in question (c) Mo = 0 2.80 What is the force on the surface whose trace is OA of Fig. 2.53? The length normal to the paper is 3 m and   9kN / m3 Solution

  9000 N / m3 L  3m (width)

A

A

0

0

Fv    L ydx  L 

x2 dx 8



3 8 x3  x3   L A o    3 24 24



3

 8

Where x A  8 ; xo  0 FH   yA  1.5 The resultant fore is : F   Fv2  FH2 2    8  1.52  2  3.202  F  (3.202)(9000)  28814 N 1

1

Or : F = 28814 N = 28.814 KN 2.84 Calculate the force F required to hold the gate of Fig. 2.74 in a closed position when R = 2 ft. Solution Taking moments about A we see that Fx = 0 Hence F = horizontal Component of liquid force F  4 h 2  23  4   1 0.9  

 8(62.4)  (0.9)    F  549.12lb

2.85 Calculate the force F required to open or hold closed the gate of Fig. 2.74 when R = 1.5 ft. Solution From problem 2.84 :

 1 r   2

 p'  ' p   R  S   2  0.9  4 1 (62.4)  81.12 psf

F  Lr 

Therefore :  1  62.4  0.9   2   81.12  2   F  199.68lb F  (4)(2) 

2.89 A log holds back water and oil as shown in Fig. 2.76. Determine (a) the force per meter pushing it against the dam, (b) the weight of the cylinder per meter of length, and (c) its specific gravity. Solution (a) First observe that the horizontal forces due to water cancel out. Therefore, the horizontal force, Fx, is due to oil only. Consider unit length perpendicular to paper. Fx   hA  (9806  0.8)(1)(2 1)  Fx  15689.6

N m

(b) The vertical force up, Fy is : Fx  

 r2   S 2r 2 2

oil



on lower half due to water + oil head added 

 r 2  S  r   4   2

  2  (9806)  (0.8  9806)(2  2 2 ) 2   (2) 2  (9806  0.8)  (2) 2  4   2

 Fy  117637.28 N



on upper tourth (left) due to

(c) weight perfoot unit volume 117637.28   9366.025 (2) 2 (1)

 cyl .  S  

Therefore : S 

9366.025  S  0.955 9806

2.91 The hemispherical dome in Fig. 2.78 is filled with water. The dome arrangement as shown weighs 28 kN and is fastened to the floor by bolts equally spaced around the circum-ference at the base. Find the total force required to hold down the dome. Solution Let Fv be the vertical force (due to water) on the dome. Fv      (3  1.5) (1.5) 2      4     1         2 

 2  / 4 100  3  4    1.5    3  2

  9806 N / m3 So : Fv = 242.47 kN (up) The net force pulling up is : FNet  Fv  w  242.47  28  FNet  214.47kN Therefore the total force required to hold down the dome is : 214.47 kN 2.92 A cable and semicircular ring suspend a spherical container by the small piezometer tube, Fig. 2.79. The top of the tube is open to the atmosphere. Calculate (a) the force on the bottom half of the sphere, (b) the force on the top half of the sphere, and (c) the

total tension in the cable. Neglect the weight of the container. Solution (a) FB  1  S2  (9806)   (0.6) 2 (0.6  0.4)  1  4 3    (0.6)   2  3 

 (2.9)(9806) 

 FB  23.943kN  (b) FT     (0.6) 2 (0.6  0.4)  1  4        (0.6)3  2  3   FT  6.651kN  (assume that the diameter of the tube is very small) (c) T  FB  FT  23.943  6.651  17.3kN So : T = 17.3 kN 2.99 The cylinder gate of Fig. 2.80 is made from a circular cylinder and a plate hinged at the dam. The gate position is controlled by pumping water into or out of the cylinder. The center of gravity of the empty gate is on the line of symmetry 4 ft from the hinge. It is at equilibrium when it empty in the position shown. How many cubic feet of water must be added per foot of cylinder to hold the gate in its position when the water surface is raised 3 ft? Solution The shaded area has volume, 1 :   1   1   r 2 4  (per foot of length)

r 2  and : a     2.33 ft 3  4   Consider the moments about H “added” when the liquid level in the reservoir is raised 3ft : +7 M (water in cylinder) = 5 M (add, head on projection AA’) = 3  8  4  96  2  M (shaded area, vertical)   1   3   5  2.33   14.157 4  M (shaded area, horizontal)   1.5  3 1  4.5 Now :  MH  0   5  4.5  14.157  96  0    17.27 ft 3 2.100 A sphere 250 mm in diameter, S = 1.4, is submerged in a liquid having a density varying with the depth y below the surface given by ρ = 1000+ 0.03y kg/m3. Determine the equilibrium position of the sphere in the liquid. Solution In general, any homogeneous body completely submerged in a fluid with a linear density gradient will float such that the fluid density at the depth of the centroid is equal to the body density. Let the subscripts B and L indicate body and liquid. Given that  L   o   y ;

o  1000 and β = 0.03 we have, wB    B d     L d    o d     yd  or :  B    o    yd  . Let 









d   A( y )dy . Centroid is such that : yc 

y2

 A( y) y

dy , hence

y1

 B    o    yc  so : yc    B  o  /  and it is independent of the body shape. Complete submergence must be Confirmed before using this approach. Then : yc  (1.4 1000  1000) / 0.03  yc  13.3m 2.102 A cube 2 ft on an edge has its lower half of S = 1.4 and upper half of S = 0.6. It is submerged into a two-layered fluid, the lower S = 1.2 and the upper S = 0.9. determine the height of the top of the cube above the interface. Solution

a 3 ' a 3 '' w  xa S1  a  a  x  S 2   S   S 2 2 2

2

drop  and solve for x :  s ' s ''   s2 /  s1  s2   2   x  1.334 ft x  a

 0.6  1.4   1.2 2     0.9  1.2 2

2.103 Determine the density, specific volume, and volume of an object that weights 3 N in water and 4 N in oil, S = 0.83. Solution

Subscripts O, A, W refer to oil, air and water respectively. WA = actual weight (in air) ww  wA   w   

So :



  A   w  ;

wo  wA   o  ;

wo  ww w o

4 N  3N    5.9987 104 m3 N 9806 3  1  0.83 m

wA  wo   o   ww   w  3N   9806   5.9987  104   wA  8.882 N



wA 8.882 kg   1.51103 3 4 g (5.9987  10 )(9.806) m

3 1 4 m vs   6.623  10  kg

2.104 Two cubes of the same size, 1 m3, one of S = 0.80 and the other of S = 1.1, are connected by a short wire and placed in water. What portion of the lighter cube is above the water surface, and what is the tension in the wire? Soution From static equilibrium, the weight of the cubes (w) must equal the buoyant force FB FB   2a 3  a 2 x   w  a 3 s1  a 3 s2 , therefore a 2  2a  x    a 3  s1  s2    x  a  2  s1  s2  3 3 Since 1  2  a  1m , then a  1m

So : x   1  2  1.1  0.8   x  0.1m  10cm The tension in wire is : F   s1  1 a 3  (0.1)a 3  980.6 N 2.113 A wooden cylinder 600 mm in diameter, S = 0.50, has a concrete cylinder 600 mm long of the same diameter, S = 2.50, attached to one end. Determine the length of the wooden cylinder for the system to float in stable equilibrium with axis vertical. Solution To float it should be : AL S1  Al S 2  A( L  l ) So : L  l L

s2  1 1  s1

(600)(2.5  1)  1  0.5

 L  1800mm



L l  l   AL S1  Al S 2 2 2 Taking moments : ZG   AL S1  Al S 2 

L2  1200 L  1.8  106  in mm  2 L  6000

Weigth     2 Z B      A    LS1 lS 2   A    2Z B  LS1  lS 2 hence : Z B  (0.5 L  600  2.5) / 2  ( L  3000) / 4 Then : GB  Z G  Z B 

L2  3600 L  5.4 106 4( L  3000)

I  4 / 4 45000 MB    2   r  2 Z B  L  3000  L2  3600 L  5.58  106 MG  MB  GB  4( L  3000) Set MG  0 to obtain the max. value of L : L2  3600 L  5.58 106  0 Therefore : L = 4769.85 mm (maximum) 2 2.120 In Fig. 2.86, ax  12.88 ft / s and ay = 0. Find the imaginary free

liquid surface and the pressure at B, C, D, and E. Solution The plane of zero pressure has slope : tan  

ax 12.88   0.4 g 32.174

  21.8o BM 

1  2.5 ft  tan 

NC  0.2 ft p A  0 ; pB  0.8  62.4 /144  0.347 psi Pc  0.2  0.8  62.4 /144  0.069 psi pD 

(0.2  3)  0.8  62.4  1.109 psi 144

PE 

2  0.8  62.4  0.693 psi 144

2.121 In Fig. 2.86, ax = 0 and ay = -8.05 ft/s2. Find the pressure at B, C, D, and E. Solution

a  p    1  y y  g  8.05   p  0.8  62.4   1   y 32.174  P = -37.43y Where p(psf) and y(ft) PB = -37.43 × (-1) = 37.43 psf = 0.26 psi PC = -37.43 × (-1) = 37.43 psf = 0.26 psi PD = -37.43 × (2) = -74.86 psf = -0.52 psi PE = -37.43 × (2) = -74.86 psf = -0.52 psi 2.123 In Fig. 2.87, ax = 9.806 m/s2 and ay = 0. Find the pressure at A, B, and C Solution tan  

ax ay  g

ax = 9.806 m/s2 = g ay = 0 tan  

ax  1    45o g

Consider unit length normal to paper air  const  0.3m 1.3m 1m  0.39m3 1 2 x  0.39  x  0.8832m  AN  1.3  0.8832  0.4168m 2 p A  0.4168m  9806 N / m3  4087.14 N / m 2

(or Pc)

PB  (1.3  0.4168) m  9806 N / m3  16834.94 N / m 2 pc  0.4168m  9806 N / m3  4087.14 N / m 2

2 2 2.124 In Fig. 2.87, ax  4.903m / s and a y  9.806m / s . Find the

pressure at A, B, and C. Solution ax  4.903m / s 2  g / 2 a y  9.806m / s 2  g tan  

ax g /2 1   g  ay g  g 4

  14.04o 1 1.3 air  const.  0.39m3  1.3 y  1.3   0.39 2 4 So y = 0.1375 m a  p    1  y y  9806(1  1) y  19612 y g  PA = 0 PB = (1.3-0.1375) × 19612 = 22,798.95 Pa 1.3  pc   1.3  0.1375   19612  16, 425.05 pa 4  2.127 The tube of Fig. 2.88 is filled with liquid, S = 2.40. When it is accelerated to the right 8.05 ft/s2, draw the imaginary free surface and determine the pressure at A. For PA = 8 psi vacuum determine ax. Solution a) tan  

ax 8.05 1   g 32.2 4

  14.036o

a  1  2 tan   0.5 ft p A  a S  0.5  62.4  2.4  74.88 psf  0.52 psi b) Given PA = 8 psi b 

vacuum = b s

8 144  7692 ft 62.4  2.4

tan  

1  b 8.692   4.346 2 2

  77.042o and ax  g tan   139.95 ft / s 2 2.133 The U tube of Fig. 2.88 is rotated about a vertical axis 6 in. to the right of A at such a speed that the pressure at A is zero gage. What is the rotational speed? Solution r1 = 0.5 ft r2 = 2.5 ft h = 1 ft Method is the same as for problem 2.132 w

2 gh 2  32.174 1   3.275 2   1 2.52  0.52 2 2

rad/sec

or w = 31.27 rpm. 2.134 Locate the vertical axis of rotation and the speed of rotation of the U tube of Fig. 2.88 so that the pressure of liquid at the midpoint of the U tube and at A are both zero. Solution

Due to symmetry dist. from A to axis equals dist. from mid-point to axis thus r1 = 1.5 ft and r2 = 0.5 ft Now equate PB and PO , Pc and Po w2 r12 2g

pB  po  H  h  w2 r22 pc  po  h  2g hence H 

w

and

w2 r12 w2 r22 w2 (r12  r22 )   2g 2g 2g 2 gH 2  32.174 1   5.672 rad/s 2 2 r1  r2 1.52  0.52

or w = 54.16 rpm 2.144 The U tbue of Fig. 2.88 is rotated about a vertical axis through A at such a speed that the water in the tube begins to vaporize at the closed end above A, which is at 70℉. What is the angular velocity? What would happen if the angular velocity were increased? Solution P = 14.7 psi = 2116.8 psf p   h  pr 

 w2 r 2 h 2g

then :

 p  pr  2 g 

w 

Thus

 r2





1

2

 (2116.8  52.33)  2  32.174   62.3  22 

1

2

w = 23.09 rad/sec = 220.5 rpm

If w increases then a vapor pocket will form.