1

CHAPTER ONE - Properties of Fluids

EXERCISE PROBLEM

1. If a certain gasoline weighs 7 KN/m3 , what are the values of its density,

specific volume, and specific gravity relative to water at 150C?

a.) =

b.) =

1

c.) s =

ws

=7 /3(1000)

9.81 /2 =

1

713.56 /3 =

7 /3

9.81 /3

= 713.56 Kg/m3 = 0.0014 m3/Kg s = 0.714

2. A certain gas weighs 16N/m3 at a certain temperature and pressure.

What are the values of its density, specific volume, and specific gravity

relative to air weighing 12N/m3?

a.) =

b.) =

1

c.) s =

ws

=16 /3

9.81 /2 =

1

1.63 /3 s =

16 /3

12 /3

= 1.63 Kg/m3 = 0.613 m3/Kg s = 1.33

3. If 5.30m3 of a certain oils weighs 43,860 N, calculate the specific weight,

density and specific gravity of this oil.

a.) w =

b.) =

c.) s =

ws

= 43.860

5.30 3 =

43860 ./2

(9.81

2)(5.30 3)

= 8.28 /3

9.81 /3

w = 8.28 KN/m3 = 843.58 kg/m3 s = 0.844

2

4. The density of alcohol is 790 Kg/m3 . Calculate its specific weight,

specific gravity and specific volume.

a.) w = g b.) s =

ws c.) =

1

= (790 kg/m3)(9.81 m/s2) = 7.75 /3

9.81 /3 =

1

790 /3

w = 7.75 KN/m3 s = 0.79 = 0.00127 m3/kg

5. A cubic meter of air at 101.3 KPa and 150C weighs 12 N. What is its

specific volume?

wa= 12 N/m3 s =

s

s = 12 /3

12.7 /3 a = (1.29 kg/m

3)(0.94) = 1

=

1

1.21 /3

s = 0.94 a = (1.21 kg/m3) = 0.82 m3/kg

6. At a depth of 8 km in the ocean the pressure is 82.26 MPa. Assume the

specific weight on the surface to be 10.10 KN/m3 and that the average

bulk modulus is 2344 MPa for that pressure range. (a) What will be the

change in specific volume between at the surface and at the depth? (b)

What will be the specific volume at that depth? (c) What will be the

specific weight at that depth?

a.) =

=

10.10(1000)

9.81 p = wh = 10.10(1000)(8000) b.) =

1

=

1

1043 /3

= 1029.6 kg/m3 p = 80.80 MPa = 9.5 x 10-4

m3/kg

= 3.3 x 10-5 m3/kg c.) w =

=

82.26 106

8000

w = 10282. 5 N/m3

3

7. To two significant figures what is the bulk modulus of water in KN/m2 at

500C under a pressure of 30 MN/m2?

W = 9.689 KN/m3 Ev = -v1

=

=

9.689

9.81 = -( 1 x 10-3)(

30,000,000

1 x 1031.012x103)

= 987.67 kg/ m3 = 2,500,000 Pa

= 1

=

1

987.67 Bv = 2.5 x 106 Pa

= 1.012 x 10-3 m3/kg

8. If the dynamic viscosity of water at 20 degree C is 1x10-3 N.s/m2, what is

the kinematic viscosity in the English units?

=

=

1103 . ./22

1000 /3

= 1x10-6 m2/s (3.28

1 )2

= 1.08 x 10-5 ft2/s

9. The kinematic viscosity of 1 ft2/sec is equivalent to how many stokes? (1

stoke= 1cm2/sec).

1 inch = 2.54 cm

1 ft2/s ( 12 2

1 2)(

2.54 2

1 2) = 929 stokes

10. A volume of 450 liters of a certain fluids weighs 3.50 KN. Compute the

mass density. (1 m3= 1000 liters).

450 liters (13

1000 ) = 0.45 m3

=

=

3.5(1000)

9.81(0.45) = 792.85 kg/m3

4

11. Compute the number of watts which equivalent to one horsepower. (1

HP = 550 ft-lb/sec; 1 W = 107 dyne-cm/sec; 1 lb = 444,8000 dynes).

1 Hp = 500

(

12

) (

2.54

1 ) (

444,800

1 )

1 Hp = 7456627200

100000000 /

1 Hp = 745.66 W

12. A city of 6000 population has an average total consumption per person

per day of 100 gallons. Compute the daily total consumption of the city in

cibic meter per second. (1 ft3 = 7.48 gallons).

100 Gallon (1 3

7.48 ) (

1 3

3.28 3) = 0.379 m3

P = 6000 (0.379 m3)

P = 2274 m3

D.C. = (

) = (

2274 3

606024)

D.C. = 0.026 m3/s

13. Compute the conversion factor for reducing pounds to newtons.

32.18

2(0.3048

1 ) (

1

2.205 ) (

1

/2)

= 4.448 N

5

CHAPTER TWO Principles of Hydrostatics

EXERCISE PROBLEM

1. If the pressure 3 m below the free surface of the liquid is 140 KPa,

calculate its specific weight and specific gravity.

Solution:

a.) P=wh b.)

W=p/n S=W/ws

=140kPa/3m =46.67/9.81

W=46.67KN/m3 S=4.76

2. If the pressure at the point in the ocean is 1400 KPa, what is the

pressure 30 m below this point? The specific gravity of salt water is 1.03.

Solution:

P=1400kPa+whs

=1400kPa+9.81(30)(1.03)

P=1,703kPa

3. An open vessel contains carbon tetrachloride (s = 1.50) to a depth of 2 m

and water above this liquid to a depth of 1.30 m. What is the pressure at

the bottom?

Solution:

Ht=1.50(2) P=wh

=3m =9.81(4.3)

P=42.18kPa

6

4. How many meters of water are equivalent to a pressure of 100 KPa?

How many cm. of mercury?

Solution:

a.) P=wh

b.)h=P/w=100kPa/9.81(13.6)

h=P/w=100kPa/9.81 h=0.75m

h=10.20m of water h=75cm of Hg

5. What is the equivalent pressure in KPa corresponding to one meter of air

at 15C under standard atmospheric condition?

Solution:

P=wh

=(12N/m3)(1m)

P=12Pa

6. At sea level a mercury barometer reads 750 mm and at the same time

on the top of the mountain another mercury barometer reads 745 mm.

The temperature of air is assumed constant at 15C and its specific

weight assumed uniform at 12 N/m3. Determine the height of the

mountain.

Solution:

P1=wsh1 ; P2=wsh2

wsh1+wh=wsh2

w(13.6)(0.745)+12h=w(13.6)(0.750)

h=(13.6)[0.75-0.745](9810)/12

h=55.60m

7

7. At ground level the atmospheric pressure is 101.3 KPa at 15C.

Calculate the pressure at point 6500 m above the ground, assuming (a)

no density variation, (b)an isothermal variation of density with pressure.

Solution:

a.)P2=P1+wh b.)P1=P2e-gh/RT

=101.3-12(6500) =(101.3)e-9.81(6500)(287/239)

P1=23.3kPa P1=47kPa

8. If the barometer reads 755 cm of mercury, what absolute pressure

corresponds to a gage pressure of 130 KPa?

Solution:

Patm=wsh

=9.81(13.6)(0.775)

Patm=100.72kPa

Pabs=Patm=Pgage

=100.72+130

Pabs=220.752kPa

9. Determine the absolute pressure corresponding to a vacuum of 30 cm of

mercury when the barometer reads 750 mm of mercury.

Solution:

Pv=-whs Patm=whs

=-9.81(0.30)(13.6) =9.81(0.75)(13.6)

Pv=-40.02kPa Patm=100.06kPa

Pabs=Patm-Pv

=100.06-40.02

Pabs=60kPa

8

10. Fig. shows two closed compartments filled with air. Gage (1) reads 210

KPa, gage (2) reds 25 cm of mercury. What is the reading of gage (3)?

Barometric pressure is 100 KPa.

(1) (2)

11. If the pressure in a gas tank is 2.50 atmospheres, find the pressure in

KPa and the pressure head in meter of water.

Solution:

a.)P=2.5(101.3kPa) b.)P=wh

P=253.25kPa h=P/w=253.25/9.81

H=25.81m

12. The gage at the sunction side of a pump shows a vacuum of 25 cm of

mercury. Compute (a) Pressure head in meter of water, (b) pressure in

KPa, (c) absolute pressure in KPa if the barometer read 755 cm of

mercury.

Solution:

a.)h=P/w=33.35/9.81 b.)Pv=-whs

h=3.40m =-0.25(9.81)(13.6)

Pv=-33.35kPa

c.)Pabs=Patm+Pv

=9.81(13.6)(0.775)-33.35

Pabs=67.38kPa

9

13. Oil of specific gravity 0.80 is being pumped. A pressure gage located

downstream of the pump reads 280 KPa. What is the pressure head in

meter of oil?

Solution:

H=P/ws

=280/9.81(0.80)

H=35.70m

14. The pressure of air inside a tank containing air and water is 20 KPa

absolute. Determine the gage pressure at point 1.5 m below the water

surface. Assume standard atmospheric pressure.

Solution:

Pabs=20+1.5(9.81)

=34.72kPa

Pabs=Patm+pg

34.72=101.3=pg

Pg=-66.60kPa

15. A piece of 3 m long and having a 30 cm by 30 cm is placed in a body of

water in a vertical position. If the timber weights 6.5 12 KN/m3what

vertical force is required to hold it to its upper end flush with the water

surface?

Solution:

W=wV F=Wa-Ww

=(9.81)(0.3x3x0.3) =2.65kN-1.756kN

W=2.65kN F=0.894kN

VWw=6.5(0.3x0.3x3)

Vw=1.755/9.81

Vw=0.179m3

Ww=wV

=0.179(9.81)

Ww=1.756kN

10

16. A glass tube 1.60 m long and having a diameter of 2.5 cm is inserted

vertically into a tank of oil (sg = 0.80) with the open end down and the

close end uppermost. If the open end is submerged 1.30 m from the oil

surface, determine the height from which the oil will rise from the tube.

Assume barometric pressure is 100 KPa and neglect vapor pressure.

17. A gas holder at sea level contains illuminating gas under a pressure

equivalent under a 5 cm of water. What pressure in cm of water is

expected in a distributing pipe at a point of 160 m above sea level?

Consider standard atmospheric pressure at sea level and assume the

unit weighs of air and gas to be constant at all elevations with values of

12 N/m3and 6 N/m3respectively.

18. If the barometric pressure is 758 mm of mercury, calculate the value h of

figure.

Gage reads 25 cm Hg

sunction

mercury h

Solution:

P = (13.6)(9.81)(7.08) p =wh

P = 1,011.29 kpa h = p/w

h = 1,011.29/9.81

h = 103.08 m

11

19. The manometer of figure is tapped to a pipeline carrying oil (sg = 0.85).

Determine the pressure at the center of the pipe.

mercury

75 cm

oil

150 cm

20. Determine the gage reading of the manometer system of figure.

air

water 20cm

Gage 3m

Mercury

Solution:

P = wsh + wsh

P = (9.81) (13.6) (0.75) + (9.81) (0.85) (1.5)

P = 112.6 kpa

12

Solution:

P = -wsh Pg = wsh - wsh

P = - (9.81) (0.2) (13.6) Pg = 9.81 (3) (9.81) (13.6) (0.2)

P = -26.68 kpa Pg = 2.75 kpa

21. In fig. calculate the pressure at point m.

Liquid (s= 1.60)

water

55 cm

m

30 cm

. Solution:

Pm = wsh wsh

Pm = (9.81) (1.60) (0.55) (9.81) (3)

Pm = 5.70 kpa

22. In fig. find the pressure and pressure at point m ; Fluid A is oil (s= 0.90),

Fluid B is carbon tetrachloride (s= 1.50) and fluid C is air.

B

C

60 cm

A 45 cm

m

13

Solution:

a) Pb = -wsh Pm = -8.829 + 0

Pb = - (9.81) (1.5) (0.6) Pm = -8.829 kpa

Pb = - 8.829 kpa

23. Compute the gage and absolute pressure at point m at the fig. ; Fluids A

and C is air, Fluid B is mercury.

C

A

m 2 cm

B 6 cm

Solution:

Pg = - wsh Pabs = Patm + Pg

Pg = - (9.81) (13.6) (0.06) Pabs = 101.3 10.67

Pg = - 10.67 kpa Pabs = 90.63 kpa

24. The pressure at point m is increased from 70 KPa to 105 KPa. This

causes the top level of mercury to move 20 cm in the sloping tube. What

is the inclination ?

Water mercury

b) h = p/w

h = -8.82/9.81

h = -1.0 m

14

. Solution:

P = wsh 10.5 26.68sin = 0

P = (9.81) (13.6) (0.20) 26.68sin = 10.5

P = 26.68 kpa = 22.6

25. In fig. determine the elevation of the liquid surface in each piezometer.

EL. 7 m

(s= 0.75)

EL. 4.5 m

(s= 1.00)

EL. 4.35 m

EL. 2.15 m

EL. 2 m

(s= 1.50)

26. In fig. fluid A is water, fluid B is oil(s= 0.85). Determine the pressure

difference between points m and n.

Solution:

1.02 = y x

68 x = z

170 y = 68 x

Pm/w y 0.68 (0.85) + x = Pn/w

Pm Pn = [ ( y x ) + ( 0.65 ) (0.85) ] 9.81

Pm Pn = ( 1.02 + 0.578) (9.81)

Pm Pn = 15.67 kpa

15

Solution:

Pm = wsh + wsh

Pm = (9.81) (0.4) (3) + (0.4) (9.81) (0.9)

Pm = 14.13 kpa

A

27. In fig. determine .

water

n

m

90 cm

52 cm

105 cm 65 cm 45 cm

Mercury

Solution:

Pm/w + 1.05 (13.6) (0.65) + 0.45 (13.6) (0.52) 0.38 = Pn/w

Pm Pn = [ (13.6) (0.65) (1.05) 1.05 - 0.45 + 0.52 (13.6) + 0.38] 9.81

Pm Pn = 149 kpa

28. In fig. Fluid A is has a specific gravity of 0.90 and fluid B has a specific

gravity of 3.00. Determine the pressure at point m.

B

12 mm. D

3 mm. D 36 cm

12 cm, D 40 cm

m

16

CHAPTER THREE Hydrostatic Force on Surfaces

EXERCISE PROBLEM

1. A rectangular plate 4m by 3m is emmersed vertically with one of the

longer sides along the water surface. How must a dividing line be drawn

parallel to the surface so as to divide the plate into two areas,the total

forces upon which shall be equal?

Solution:

F1 = F2

Awh1 = Awh2

(12.0)(1.50) = h(4.0)( h/2 )

2h2 =18.0

2h = 18

h = 4.24/2

h = 2.12 m below w.s

2. A triangle of height H and base B is vertically submerged in a liquid. The

base B coincides with the liquid surface.Derive the relation that will give

the location of the center of pressure.

3. The composite area shown in Fig. A is submerged in a liquid with

specific gravity 0.85. Determine the magnitude and location of the total

hydrostatic force on one face of the area.

Solution:

e = g

=

2

12

hp = + F1 = wA

e =

2

12

=

3.52

12

3.25 hp = 3.25 + 0.31 F1 = 9.81(3.5)(1.5)(3.25)(0.85)

e = 0.31 m hp = 3.56 m F1 = 142.28 KN

17

e = g

=

2

12

hp = + F1 = wA

e =

2

12

=

1.52

12

4.25 hp = 4.25 + 0.04 F1 = 9.81(1.5)(1.5)(4.25)(0.85)

e = 0.04 m hp = 4.29 m F1 =79.74 KN

Ft = F1 + F2 Pt = P1 + P2

Ft = 142. 28 + 79.74 Pt h = F1 h + F2h

Ft = 222.02 KN 222.02

222.02 =

142.28 3.56 + 79.76(4.29)

222.02

h = 3.83 m , below w.s

18

4. The gate in fig. B is subjrcted to water pressure on one side and to air

pressure on the other side. Determine the value of X for which the gate

will rotate counterclockwis if the gate is (a) rectangular, 1.5m by 1.0m (b)

triangular, 1.5m base and 1.0m high.

Solution:

F = PA a.) F = w A e =

F = 30(1.0)(1.5) F = (9.81)(x+0.5)(1.5)(1.0) e = 1

12+6

F = 45 KN F = 14.72x + 7.36

1 = 0

14.42x + 7.36(0.5 + 1

12+6) = 45(0.5)

86.5x2 168.16x 105.56 = 0

=(168.16) (168.16)24 86.5 (105.56)

2(86.5)

= 2.40

5. A vertical circular gate 1m in diameter is subjected to pressure of liquid

of specific gravity 1.40 on one side. Thefree surface of the liquid is 2.60m

above the uppermost part of the gate. Calculate the total force on the

gate and the location of the center of pressure.

Solution:F = w A e = g

=

(4)2

4

2

F =9.81(1.4)(3.1)()(0.52) e = (0.52)

4(3.1)

F = 33.44 KN e = 0.02m (below the center)

19

6. A horizontal tunnel having a diameter of 3m is closed by a vertical gate.

When the tunnel is (a) full (b) full of wter, determine the magnitude

and location of the total force.

Solution:

a.) full

= 4

3 F = w A

= 4(1.5)

3 F = (9.81)(

(1.52)

(2))(0.64)

= 0.64 m F = 22.15

b.) = 1.5+0.64

2 F = w A

= 1.08 m F = (9.81)(3(1.52)

(4))(1.08)

F = 56.25 KN

=g

hp = +

=0.1098(1.5)

4

3.53(0.64) hp = 0.64 + 0.25

e = 0.25 m hp = 0.89 m (below center)

7. In Fig. C is a parabolic segment submerged vertically in water.

Determine the magnitude and location of the total force on one face of

the area.

Solution:

F = w A

F = 9.81(1.8)(23)(3)(3)

F = 105.95 KN

=g

=

8 3 (3)2

1752 3 3

3(1.8)

= +

= 0.34 = 1.8 + 0.34

= 2.14

20

8. A sliding gate 3m wide by 1.60m high is in a vertical position. The

coefficient of friction between the gate and guides is 0.20. If the gate

weighs 18KN and its upper edge is 10m below the water surface, what

vertical force is required to lift it? Neglect the thickness of the gate.

Solution:

= =

= 9.81 1.6 (10.8) = 0.2(508.55)

= 508.55 = 101.71

F = 508.55 KN

=0

= +

= 18.0 + 101.71

= 119.71

9. The upper edge of a vertical rapezoidal gate is 1.60m long and flush with

the water surface. The two edges are vertical and measure 2m and 3m,

respectively. Calculate the force and location of the center of pressure on

one side of the gate.

10. How far below the water surface is it necessary to immerse a vertical

plane surface, 1m square, two edges of which are horizontal, so that the

center of pressure will be located 2.50cm below the center of gravity?

Solution:

=

2

12

0.025 =12

12

0.5

= 2.83 m

21

11. The gate shown in fig. D is hinged at B and rest on a smooth surface at

A. If the gate is 1.60m wide perpendicular to the paper, find BH and BH

Solution:

= 1(3

2) =

= 56.31 = 9.81 3.61 1.6 (2.8)

= 158.66

= =

= 158.6656.31 = 158.6656.31

= 132.01 = 80.70

22

12. In fig. E gate AB is 2m wide perpendicular to the paper. Determine FH to

hold the gate in equilibrium.

Solution:

= 3.2 2 = 6.4 =g

= =

2

12

= 9.81 1.21 (6.4) =3.20

12(1.26)

= 77.85 = 0.22

=3.20.48

2 0.53 = 0

= 1.38 3.20 1.38 77.85 = 0

= 42.50

23

16. A triangular gate having a horizontal base of 1.30m and an altitude of 2m

is inclined 45o from the vertical with the vertex pointing upward. The base

of the gate is 2.60m below the surface of oil (s=0.80). What normal force

must be applied at the vertex of the gate to open it?

17. What depth of water will cause the rectangular gate of Fig. I to fall?

Neglect weight of the gate.

Solution:

=0.5

60 . 1 =

g

=

2

12

=

60

12( 0.560)

= 0.19 . 3

= 0

60

0.5

60+

0.19

60 = 22.5(5.0)

5.953 = 112.5(5.0)

= 18.91

= 2.66

=

= 9.81

60 2.6 (

0.560

)

= 17.02 . 2

24

18. Determine the horizontal and vertical components of the total force on

the gate of Fig. J. The width of the gate normal to the paper is 2m.

Solution:

A1 = AAOBC A2 = ( 1

2)(6)(6)(c0s30)

1

60=

(6)2

360 A2 = 15.59 2

A1 = 18.85 2

=

A = A1 A2 Fh = 9.81(6)(3)(2) FV = 9.81(3.26)(2)

A = 18.85 15.59 Fh = 353.16 KN FV = 63.96 KN

A = 3.26 2

25

19. The corner of floating body has a quarter cylinder AB having a length

normal to the paper of 3m. Calculate the magnitude and location of each

of the components of the force on AB. Fig. K.

Solution:

= = 0

= 9.81 1.5 3 (1.03) = 147.48 30

= 147.78 = 128.56

20. The cylindrical gate of Fig. L is 3m long. Find the total force on the gate.

What is the minimum weight of the gate to maintain equilibrium of the

system?

26

21. The gate if Fig. O is 3m long. Find the magnitude and location of the

horizontal and vertical components of the force on the gate AB.

Solution:

=

= 9.81 1.06 3 (2.12)

= 66.14

= 0.88 + 3 2.12

2

= 4.11 2

= 2

360=

(3)2(45)

360

= 3.58 2

22. A pyramidal object having a square base (2m on a side) and 1.50m high

weighs 18KN. The base covers a square hole (2m on a side) at the

bottom of a tank. If water stands 1.50m in the tank, what force is

necessary to lift the object off the bottom? Assume that atmospheric

pressure acts on the water surface and underneath the bottom of the

tank.

27

23. The hemesphirical dome of Fig. P surmounts a closed tank containing a

liquid of specific gravity 0.75. The gage indicates 60KPa. Determine the

tension holding the bolts in place.

Solution:

= =

60 = 9.81 0.75 = 9.81 39.23 (0.75)

= 8.15 = 288.63

= 2 43

6

= 1.5 2(8.15) 4(1.5)3

6

= 39.233

24. Fig. Q shows semi-conical buttress. Calculate the components of the

total force acting on the surface of this semi-conical buttress.

Solution:

=2

3 =

= 0.15 2(3)

3 = 9.81 1.463 (7.07)

= 7.07 2 = 101.47

= + ( 1.5 4

4) =

= 1.3 + 0.163 =1.3

3 1.3 1.5 +

1.5 2(3)

3

= 1.463 = 0.12 9.81

= 7.8

28

25. In Fig. R a circular opening is closed by a sphere. If the pressure at B is

350KPa absolute, what horizontal force is exerted by the sphere on the

opening?

Solution:

= 2 =

= (0.125)2 = 9.81 178.39 0.099 (0.71)

= 0.099 2 = 7.8

=

350 = 9.81 0.20

= 178.39 2

26. Calculate the force required to hold the cone of Fig. S in position.

Solution:

2 = 1 + = 2

2 = 3.5 9.81 0.8 1.5 = (0.375)2(0.8)(9.81)(2.5)

2 = 8.26 = 8.66

1 = 2 = 1

32

1 = 2 2 = 9.81(0.8)

1

3(0.375)2

1 = (0.375)2(8.26) 2 = 1.16

1 = 3.65

= 0

+ 2 + 1 =

= 8.66 3.65 1.16

= 3.85

29

27. A steel pipe having a diameter of 15cm and wall thickness of 9,50mm

has an allowable stress of 140,000KPa. What is the maximum allowable

internal pressure in the pipe?

Soln:

Sa = T/t FB = PiD

T = Sat = 14,000(0.0095) Pi = FB/D = 2T/D =

(2(1330)/1000)/0.15

T = 1330 kN/m Pi = 17.73 Mpa

28. A pipe carrying steam at a pressure of 7,000KPa has an inside diameter

of 20cm. If the pipe is made of steel with an allowable stress of

400,000Kpa, what is the factor safety if the wall thickness is 6.25mm?

Soln:

S = PD /2t fc =

=

0.714

0.2

D = 2

=

2(400,000)(0.00625 )

7000 fc = 3.60

D = 0.714 m

29. A 60 cm cast iron main leads from a reservoir whose water surface is at

EL. 1590m. In the heart of the city the main is at EL. 1415m. What is the

stress in the pipe wall if the the thickness of the wall is 12.5mm and the

external soil pressure is 520Kpa? Assume static condition.

Soln:

EL = 1 - 2 S =

2 =

1716.75320

2(0.0125 )

= 1590 1415 S = 28,709 kPa

= 175 m = 28.7 MPa

P = wh

= 9.81 (175m)

P = 1716.75 kPa

30

30. Compute the stress in a 90cm pipe with wall thickness of 9.50mm if

water fills under a head of 70m.

Soln:

FB = PiD T = FB/2 = 618.03/2 = 309.01 kN/m

= whD Sa = T/t = 309.01/0.0095

= 9.81(70)(0.9) Sa = 32,527 kPa

FB = 618.03 kN/m

31. A wood stave pipe, 120cm in inside diameter, is to resist a maximum

water pressure of 1,200KPa. If the staves are bound by steel flat bands

(10cm by 2.50cm), find the spacing of the bands if its allowable stress is

105MPa.

Soln:

FB = PiD T = FB/2

= 1200kPa(1.2) = 1440/2

= 1440 kN/m T = 720 kN/m

S = SaAH/T = (105(2.5))/0.72

S = 36.46 cm

32. A continuous wood stave pipe is 3m in diameter and is in service under a

pressure head of 30m of water. The staves are secured by metal hoops

2.50cm in diameter. How far apart should the hoops be spaced in order

that the allowable stress in the metal hoop of 105MPa be not exceeded?

Assume that there is an initial tension in the hoops of 4.50KN due to

cinching.

33. A vertical cylindrical container, 1.60m diameter and 4m high, is hel

together by means of hoops,one at the top and the other at the bottom. A

liquid of specific gravity 1.40 stands 3m in the container. Calculate the

tension in each hoop.

Soln:

F = wAh MCD = 0 Mab = 0

= 9081(1.4)(3)(1.6)(1.5) 4(2TU) = 1F 4(2TL) = 3F

F = 98.9 kN TU = 12.40 kN TL= 37.09 kN

e = h^2/12h = 3^2/(12(1.5)) = 0.5 h_p = 1.5 + 0.5 = 2m

31

34. A masonry dam has trapezoidal section: one face is vertical, width at the

top is 60cm and at the bottom is 3m. The dam is 7m high with the vertical

face subjected to water pressure. If the depth of water is 5m, where will

the resultant force intersect the base? Determine the distribution of

pressure along the base, (a) assuming there is no uplift pressure; (b)

assuming that the uplift pressure varies uniformly from full hydrostatic at

the heel to zero at the toe. Specific weight of masonry is 23.54KN/m3.

Soln:

a.)G_1 = wVs R.M = G_1 + G_2 = 266.95 + 316.38

= 23.54(0.6)(7(1) = 583.33 kN.m

= 98.87 kN O.M = F_1 = 204.38 kN.m

G_2 = 23.54(0.5)(7)(2.4)(1)

= 197.74 kN x = (R.M-EO.M)/RV = (583.33-204.38)/296.61

F1 = 1/2wh^2 x = 1.28 m (from toe)

= 0.5(9.81)(5^2) e = b/2 x = 3/2 1.28

= 122.63 kN e = 0.22

Moment Forces:

G_1 = 2.7(98.87) = 266.95 kN.m

Smax = Rv/b (1 + 6e/b) = 296.61/3(1 + (6(0.22))/3)

G_2 = 1.6(197.74) = 316.38 kN.m

Smax = 142.38 kPa

F_1 = 1/3 (5)(122.63) = 204.38 kN.m

Smin = 296.61/3(1 - (6(0.22))/3)

FV = G_1 + G_2 Smin = 55.38 kPa

= 98.87 + 197.74

RV = 296.61 kN

FH = F_1 = 122.63 kN

b.)U_1 = 1/2 whb x = (583.33-351.54)/223.03 = 1.04 m (from toe)

= (1/2)(9.81)(5)(3)(1) e = 3/2 1.04

U_1 = 73.58 kN e = 0.46

Moment forces: Smax = 223.03/3 (1 + 6x0.46/3)

U_1 = 2/3 (3)(73.58) = 147.16 kN Smax = 142.74 kPa

RV = G_1 + G_2 - U_1

= 98.87 + 197.74 73.58 Smin = 223.03/3 (1 - 6x0.46/3)

RV = 223.03 kN Smin = 5.95 kPa

R.M = 583.33 kN.m

O.M = 204.38 + 147.16

= 351.54 kN.m

32

35. The masonry dam of Problem 40 has its inclined face subjected to

pressure due to a depth of 5m of water. If there is no uplift pressure ,

where will the resultant intersect the base? Specific weight to concrete is

23.54KN/m3

Soln:

a/5 = 2.4/7 R.M = W_1 + W_2+ W_3

a = 1.71 m = 23.91 + 316.38 + 266.95

= 607.24 kN.m

W_1 = wV O.M = 204.38 kN.m

= 9.81(0.5)(5)(1.71)(1)

= 41.94 kN x = (607.24-204.38)/338.55

W_2 = 1/2 (2.4)(1)(7)(23.54) x = 1.19 m

=197.74 kN

W_3 = 0.6(7)(1)(23.54)

= 98.87

F = 1/2 (9.81)(5^2)

= 122.63 kN

Moment Forces:

W_1= 0.57(41.94) = 23.91 kN.m

W_2 = 1.6(197.74) = 316.38 kN.m

W_3 = 2.7(98.87) = 266.95 kN.m

F_1 = 1/3 (5)(122.63) = 204.38 kN.m

RV = W_1+W_2+W_3

= 41.94 + 197.74 + 98.87

RV = 338.55 kN

RH = F = 122.63 kN

33

36. A masonry dam of trapezoidal cross section, with one face vertical has

thickness pf 60cm at the top, 3.70m at the base, and has height of

7.40m. what is the depth of water on the vertical face if the resultant

intersect the base at the downstream edge of the middle third? Assume

that the uplift pressure varies uniformly from full hydrostatic at the heel to

zero at the toe.

Soln:

G_1 = 104.52 kN R.M = G_1+G_2-U

G_2 = 270 kN = 355.37 + 558.9 44.77h

F = 1/2 (9.81)h^2 O.M = 1.635h^3

= 4.905h^2

U = 1/2wh(3.7) x = (R.M- O.M)/Rv

= 1/2 (9.81)h(3.7) 1.23 = ((355.37+558.9-44.77h)-

U = 18.15h1.635h^3)/(374.52-18.15h)

h = 5.83 m

Moment Forces:

G_1 = 3.4 (104.52) = 355.37 kN.m

G_2 = 2.07(270) = 558.9 kN.m

F = 1/3 h(4.905h^2) = 1.635h^3 kN.m

U = 2/3 (3.7)(18.15h) = 44.77h kN.m

Fv = G_1+G_2-U

= 104.52 + 270 18.15h

Rv = 374.52 18.15h

34

37. A concrete dam is triangular in cross section and 30 m high from the

horizontal base. If water reaches a depth of 27 m on the vertical face,

what is the minimum length of the base of the dam such that the

resultant will intersect the base within the middle third? What minimum

coefficient of friction is required to prevent sliding? Determine the

pressure distribution along the base.

Soln:

a.)

G = wV RVx = R.M - O.M

= 23.54(1/2)(30)(1)B 353.1B(B/3) = 235.4B^2 32181.75

G = 353.1B (235.4 117.7)B^2 = 32181.75

B = 16.54m

F = 1/2 (9.81)(27^2)

F = 3575.75 kN

Moment Force:

G = 2/3 B(253.1B)

G = 235.4B^2

F = 1/3 (27)(3575.75)

= 32181.75 kN.m

b.)

G = 235.4B^2 = RH/RV

= 235.4(16.54^2) = 3575.75/5840.27

G = 64398.76 kN.m = 0.61

RV = 353.1B

= 353.1(16.54)

RV = 5840.27 kN

c.)

x = 1/3 (16.54) = 5.51 S = 5840.27/16.54 (1+6x2.76/16.54)

S = 706.20 kPa

e = 16.54/2-5.51

e = 2.76

35

44. The section of masonry dam is shown in Fig. U. If the uplift pressure varies

uniformly from full hydrostatic at the heel to full hydrostatic at the toe, but acts

only 2/3 of the area of the base, find: (a) the location of the resultant, (b) factor

safety against overturning, (c) factor of safety against sliding if the coefficient of

friction between base andfoundation is 0.60.

Soln:

a.)

1 = 5(8)(1)w Fv = 1 + 2 + 3 + 4 + 5 +

6 1 2

= 40w = (40+25+60+252+176.4+4.18-

73.67-56.67)w

2 = 1

2 (5)(10)(1)w Rv = 427.24w

= 25w

3 = 1

2 (5)(10)(1)(2.4)w Fh = 1 2

= 60w = (162-12.5)w

4 = 5(21)(1)(2.4)w Rh = 149.5w

= 252w

5 = 1

2 (7)(21)(1)(2.4)w Moment Forces:

= 176.4w 1 = 14.5(40w) = 580w

6 = 1

2 (1.67)(5)(1)w 2 = 15.33(25w) = 383.25w

= 4.18w 3 = 13.67(60w) = 820.2w

1 = 1

2 (182)w 4 = 9.5(252w) = 2394w

= 162w 5 = 4.67(176.4w) = 823.79w

2 = 1

2 (52)w 6 = 0.56(4.18w) = 2.34w

= 12.5w 1 = 11.33(73.67w) = 834.67w

1 = 1

2 (17)[

2

3 (176.58 49.05)] 2 = 8.5(56.67w) = 481.70w

= 73.67w 1 = 6(162w) = 972w

2 = 2

3 (17)(49.05) 2 = 1.67 (12.5w) = 20.88w

= 56.67w

R.M = (580+383.28+820.2+2394+823.79+2.34+20.88)w

= 5024.46w

O.M = 1 + 1 + 2

= (972+834.68+481.7)w

= 2288.38w

36

x = . .

=

5024 .462288.38

427.24

x = 6.40 m (from toe)

b.)

F.S. vs. Overturning = .

. =

2024.46

2288.38 = 2.20

c.)

F.S vs. Sliding =

=

0.6(427.24)

149.5 = 1.70

45. Shown in Fig. V is an overflow dam. If there is no uplift pressure, determine

the location of the resultant.

37

Soln:

G_1 = 2(3)(1)(9.81) Moment Force:

= 58.86kN G_1 = 6.5(58.86) = 382.59 kN.m

G_2 = 1/2 (3)(6)(1)(9.81) G_2 = 7(88.29) = 618.03 kN.m

= 88.29 kN G_3 = 6(211.86) = 1271.16

kN.m

G_3 = 1/2 (3)(6)(1)(23.54) G_4 = 4(38.24) = 156.96 kN.m

= 211.86 kN G_5 = 4(282.48) = 1129.92

kN.m

G_4 = 2(2)(1)(9.81) G_6 = 2(211.86) = 423.72 kN.m

= 39.24 kN G_7 = 0.67(39.24) = 26.29 kN.m

G_5 = 2(6)(1)(23.54) F_1 = 2.4(294.3) = 706.32 kN.m

= 282.48 kN F_2 = 1.33(78.48) = 104.38

kN.m

G_6 = 1/2 (3)(6)(23.54)

= 211.36 kN

G_7 = 1/2 (2)(4)(1)(9.81)

= 39.24 kN

R.M = 382.59+618.03+1271.16+156.96+1129.92

+423.72+26.29+104.38

F_1 = Awh = 4113.05 kN.m

= 6(1)(9.81)(5)

= 294.3 kN O.M = F_1 = 706.32 kN.m

F_2 = 1/2 (9.81)(4^2)

= 78.48 kN

Fv = G_1+G_2+G_3+G_4+G_5+G_6+G_7

= 58.86+88.29+211.86+39.24+282.48+211.86+39.24

Rv = 931.83 kN

Fh = F_1-F_2 x = (R.M- O.M)/Rv = (4113.05-

706.32)/931.83

= 294.3 78.48 x = 3.66 m (from the toe)

Rh = 215.82 kN

38

46. The base of a solid metal cone (Sp. Gr. 6.95) is 25 cm in diameter. The

altitude of the cone is 30 cm. If placed in a basin containing mercury (Sp.

Gr. 13.60) with the apex of the cone down, how deep will the cone float?

Fy=0] Given :

=W d=25cm.

(wV)displaced mercury = (wV)cone r=12.5cm=0.125

(9.81)(13.60)Vm = 9.81(6.95)Vcone Vcone=4.9087x103

13.60Vm = 6.952

3

Vm = 6.95 0.1252 (0.30)

3(13.60)

Vm = 2.50x1033

47. If a metal sphere 60 cm in diameter weighs 11,120 N in the air, what

would be its weight when submerged in (a) water? (b) mercury?

Soln:

a.) b.)

FB = 9.81 (4/3 r^3)W_hy = 11,120 9.81(13.6)(4/3)()(0.3^2)

= 9.81(4/3)(0.3^2) = -3976 N

FB = 1.11 kN

W = 11.12 1110

= 10.01 N

= (

0.30

)3

3 =(0.30)3

x = 0.30 3(2.50103)

4.91 103

3

x = 0.24m

x = 24cm

39

48. A rectangular solid piece of wood 30 cm square and 5 cm thick floats in

water to depth of 3.25 cm. How heavy an object must be placed on the

wood (Sp. Gr. 0.50) in such a way that it will just be submerged?

Given: dept=3.25cm

Req. F=?

30cm

5cm Fb=w

w.s. wv'=wsv

s=v/v

Fb s= (30) (30) (3.25)

w.s. 4500

S= 0.65// ans.

F=Fb-W

Fb F=wv-wsv

49. A hollow vessel in the shape of paraboloid of revolution floats in fresh

water with its axis vertical and vertex down. Find the depth to which it

must be filled with a liquid (Sp. Gr. 1.20) so that its vertex will be

submerged at 45 cm from the water surface.

Solution:

=W

9.81Vd = 9.81(1.20)Vp

Vd = 1.20Vp

W

W

F=wv(1-s)

F= (9.81)(4500)(1-0.65)

F=15.45 N

By Similar Solids:

= (

0.45)3

1.20=

3

0.453

1

1.20= =

3

0.453

a = 0.42m

a = 42cm

40

50. A barge is 16 m long by 7 m wide 120 cm deep, outside dimensions. The

sides and bottom of the barge are made of timber having thickness of 30

cm. The timber weighs 7860 N/cu.m. If there is to be freeboard of 20 cm

in fresh water how many cubic meters of sand weighing 15700 N/cu.m

may be loaded uniformly into the barge?

Soln:

V_t=V_o-V_i F_b-W_t-W_s = 0

= (16)(7)(1.2) (15.4)(16.4)(0.9)

9810(16)(7)(1) 7860(45.7) 15700

= 45.7 m^3

V_s = 0

V_s = 47.10 m^3

51. A brass sphere (Sp. Gr. 8.60) is placed in a body of mercury. If the

diameter of the sphere is 30 cm (a) what minimum force would be

required to hold it submerged in mercury? (b) what is the depth of

flotation of the sphere when it is floating freely?

hg.s. F F=Fb-W

F=wSmVs-wSsVs

F=wVs(Sm-Ss)

F=(9.81) (3/4)(3.14)(0.15)^3(13.60-8.6)

F=693.43N//ans.

y

Fb

W

Fb

41

V=4/3(3.14)(r)^3

V=4/3(3.14)(15)^3

V=14,137.17cm^3

Fb=w V=3.14/3 D^2(3r-D)

wSmV=wSsV 8939.68 =3.14/3(y^2)((3x15)-y)

V=8.60(14137.17)

13.60 y=17.10cm

V=8939.68cm^3

52. A spherical balloon weighs 3115 N. How many newton of helium have to

be put in the balloon to cause it to rise, (a) at sea level? (b) at an

elevation of 4570 m?

Soln:

W = Fb Fh

W = _agV - _hgV

[W = V(_ag - _hg)]1/((_a g - _h g))

V = w/(g(_a-_h))

= 3115/(9.81(1.29-0.179))

V = 286.1 m^3

53. The Sp. Gr. of rock used as concrete aggregate is often desirable to

know. If a rock weighed 6.15 N in the air and 3.80 N when submerged in

water, what would be the specific gravity of the rock?

Soln:

W = W_a - W_w S = 6.15/(9810(2.4 x 10^(-4)))

W = 6.15 3.8 S.g = 2.62

(9810)V = 2.35

V = 2.4 x 10^(-4) m^3

S.W = W_a/V = (6.15 N)/(2.4 x 10^(-4) )

S.W = 25625 N/m^3

42

54. A piece of wood weighs 17.80 N in air and piece of metal weighs 17.80

in water. Together the two weighs 13.35 N in water. What is the specific

gravity of the wood?

Solution:

Wwo=17.80N (air)

Wm=17.80N(water)

WT=Wwo+Wm ; WT=13.85N

Wwo=17.80(air)

55. A sphere 1.0 in diameter floats half submerged in tank of liquid (Sp. Gr.

0.80) (a) what is the weight of the sphere? (b) What is the minimum

weight of the anchor (Sp. Gr. 2.40) that will require to submerge the

sphere completely?

Given: Find: Sa = 7.40 Sl iquid = 0.80 Ws

Vs = 4/3^3 Wa

= 4/3(0.53)^3

= 0.52m

A.) W=fb

=WsLVs/2

Ws=9.81Kn/m3(0.80)(0.52m3)/2

Ws=2.05KN

B.) Wa=Fba+FbsW where: Va=Wa/Wsa

=WslVa+WslVs-WsVs SS=Ws/wVs

Wwo=17.80-FB

13.35N=17.80-FB+17.80

FB=22.25N (Displaced Water)

Gs=Wwo/FB

=17.80N/22.25N

Gs=0.80

43

Wa=w[0.80xwa/w2.40]+[0.80x0.32m3]-[2.04kn/9.81kn/m3]

Wa=0.33wa+4.08kn-2.04kn

Wa-0.33wa=2.04kn-4.08kn

0.67wa/0.67=2.04/0.67

Wa= 3.50kn

56. Fig. Z shows a hemispherical shell covering a circular hole 1.30 m in

diameter at the vertical side of a tank. If the shell weighs 12,450 N, what

vertical force is necessary to lift the shell considering a friction factor of

0.30 between the wall and the shell?

57. An iceberg has a specific gravity of 0.92 and floats in salt water (Sp. Gr.

1.03). If the volume of ice above the water surface is 700 cu.m, what is

the total volume of the iceberg?

given: find:

Si=0.92 Vt

Ssw=1.03

44

W=Wvfv=0] vt=v1+v2

W=WsiVt W=Fb v2=vt-v1

Fb=Wssw wsivt=wssw(vt-v1)

=Wsssv2 Sivt=wswvt-sswv1

sswv1=vt(ssw-si)

58. A concrete cube 60 cm on each edge (Sp. Gr. 2.40) rests on the bottom

of a tank in which sea water stands to a depth of 5 m. The bottom edges

of the block are sealed off so that no water is admitted under the block.

Find the vertical pull required to lift the block.

Solution:

W1=23.54(0.6x7xd) ;d=1

W1=98.868

W2=197.736[1/2(24)(7)d] X=(RM-OM)/RV

W2=197.736 =(583.3692-

204.375)/296.604

F=hA X=1.28m

=9.81(2.5)(5)

F=122.625 e=b/2-x

Rx=122.625d S=Ry/b(16e/b)

Ry=296.604d S=142.36992kPa

RM=98.868(3-0.3)+197.766[2/3(2.4)]

RM=583.3692kN.m

OM=122.625(5/3)

OM=204.375kN.m

vt=sswv1/ssw-si

=1.03(700)/1.03-0.92

Vt=6554.55m3

45

59. A 15 cm by 15 cm by 7 m long timber weighing 6280 N/cu.m is hinged at

one end and held in horizontal position by an anchor at the other end as

shown in Fig. AA. If the anchor weighs 23450 N/cu.m, determine the

minimum total weight it must have

Solution:

Vt=(0.15)2(7) Wa=WaVa

Vt=0.1575m2 Wa=23540Va Va=Wa/23540

Wt=WtVt

=(0.1575)(62.80)

Wt=989.1

Fbt=9810(0.1575) Fba=WVa

Fbt=1545.075 Fba=9810Va

Mh= 3.5Fbt+7Fba=3.5Wt+Wa

3.5(1545.075)+7(9810)

Va=3.5(989.1)+7(23540)Va

Va=0.02m3

Wa=WaVa

=23540N/m3(0.02m3)

Wa=470.8N

46

60. A cylinder weighing 445 N and having a diameter of 1.0 m floats in salt

water (Sp. Gr. 1.03) with its axis vertical as in Fig. BB. The anchor

consist 0f 0.0280 cu.m of concrete weighing 23450 N/cu.m. What rise in

the tide r will be required to lift the anchor off the bottom?

Solution:

Wa=23540(0.280) Fba=9810(1.05)(0.280)

Wa=659102N Fba=2829.204N

Wo=445N Fbc=9810(1.03)p(0.5)2(0.3+r)

Fbc=2380.769+7935.89866

Fv=0

Fba+Fbc=Wa+Wc

2829.204+2380.769+7935.898r=6591.2+445

r=0.23m ; 23cm

47

0.11x2-1.1x+1.375=0

X=1.46m

= 2 4

2

61. A timber 15 cm square and 5 m long has a specific gravity of 0.50. One

end is hinged to the wall and the other is left to float in water (Fig. CC).

For a=60 cm, what is the length of the timber submerged in water?

Solution:

Wt=9.81(0.5)(5)(0.15)2

Wt=0.5518125kN

Fb=9.81(0.15)2(x)

Fb=0.220725x

Mh=0

2.5cosWt=(5-0.5x)cosFb

2.5(0.5518125)=(5-0.5x)(0.220725x)

1.375=1.1x-0.11x2

62. A metal block 30 cm square and 25 cm deep is allowed to float on a

body of liquid which consist of 20 cm layer of water above a layer of

mercury. The block weighs 18,850 N/cu.m. What is the position of the

upper level of the block? If a downward vertical force of 1110 N is

applied to the centroid of the block, what is the new position of the upper

level of the block?

Solution:

a.) Fbm=wV Fbw=9.81(0.20)(009)

=9.81(13.6)(0.09)(0.05-x) Fbw=0.17658kN

Fbm=0.600372-12.00744x

W=18.85(0.09)(0.25)

W=0.424125kN

48

Fv=0

Fbw+Fbm=W

0.17658+0.600372-12.000744x=0.424125

X=0.0294m

X=2.94cm

b.) Fbm=9.81(13.6)(0.09)(0.25-x) W=0.4241225

Fbm=3.00186-12.00744x Wv=1.11kN

Fbw=9.81(0.09)(x)

Fbw=0.8829x

Fv=0

0.8829x+3.00186-12.00744x=0.4241225+1.11kN

X=0.132m

H=0.20m-0.132m

H=0.068m

H=68cm

63. Two spheres, each 1.2 m diameter, weigh 4 and 12 KN, respectively.

They are connected with a short rope and placed in water. What is the

tension in the rope and what portion of the lighter sphere produces from

the water? What should be the weight of the heavier sphere so that the

lighter sphere will float halfway out of the water?

Solution:

= 9.81 4 (0.603)

3 = (9.81)(

1

2)(

4(0.60)3

3)

= 8.8759 kN = 4.4379 kN

T = - T = Wss -

T =12 kN - 8.8759 kN T = 4 kN - 4.4379 kN

T = 3.12 kN T = 0.4379 kN

Fy=0]

= - T

9.81Vss = 4 + 3.12

Vss = 0.725793

49

= T +

= 0.44 kN + 8.88 kN

Vs =

32(3 ) = 9.32 kN

0.72579 = 0.602 - 0.333

D = 0.85m --- by trial & error

X = 1.20 D

X = 0.35m

68. If the specific gravity of a body is 0.80, what proportional part of its total

volume will be submerged below the surface of a liquid (Sp. Gr. 1.20)

upon which it floats?

Solution:

=

() = ()

(9.81)(1.20) = (9.81)(0.80)

(1.20) = (0.80)

= 2

3

2

3of the total Volume

69. A vertical cylinder tank, open at the top, contains 45.50 cu.m of water. It

has a horizontal sectional area of 7.40 sq.m and its sides are 12.20 m

high. Into its lowered another similar tank, having a sectional area of 5.60

sq.m and a height of 12.20 m. The second tank is inverted so that its

open end is down, and it is allowed to rest on the bottom of the first. Find

the maximum hoop tension in the outer tank. Neglect the thickness of the

inner tank.

50

70. A small metal pan of length of 1.0 m, width 20 cm and depth 4 cm floats

in water. When a uniform load of 15 N/m is applied as shown in Fig. DD,

the pan assumes the figure shown. Find the weight of the pan and the

magnitude of the righting moment developed.

Solution:

= 0.04m (0.20m) (1m)

= 8x10332 = 1= 4x10

33

= 1 (0.04

0.20)

= 11.31

+ T =

= - T

= 9810(4x103) 15

= 39.24 - 15

= 24.24 N

71. A ship of 39,140 KN displacement floats in sea water with the axis of

symmetry vertical when a weight of 490 KIN is mid ship. Moving a weight

3 m toward one side of the deck cause a plumb bob, suspended at the

end of a string 4 m long, to move 24 cm. Find the metacentric height.

Given:

W=39140kn

Tan=0.24/4 C=Wx

=3.43 490(3)=39140x

Sin=X/MG

T =

15

=

1

F = 15

X=MGSin

490(3)=39140(MGSin3.43)

MG=0.63m

51

72. A rectangular scow 9.15 m wide by 15.25 m long and 3.65 m high has a

draft of 2.44 m in sea water. Its center of gravity is 2.75 m above the

bottom of the scow. (a) Determine the initial metacentric height. (b) If the

scow tilts until one of the longitudinal sides is just at the point of

submergence, determine the righting couple or the overturning couple.

Soln:

a.) b.)

GB_o = 2.75 1.22 tan = 1.21/4.575

= 1.53 m = 14.81

MG = MB_o - GB_o MB_o = B^2/12D(1 + tan^2/2)

= 2.86 1.53 = 9.15^2/(12(2.46))(1 + tan

MG = 1.33 m^(2(14.81))/2)

MB_o = 2.96 m

Fv = 0 ; FB = W

FB = wV

= 9.15(15.25)(2.44)(9.81)(1.03)

FB = W

W = 3490.23 kN

RM = W(MGsin) = 3490.23(1.34sin14.81)

RM = 1257.7 kN.m

73. A cylindrical caisson has an outside diameter of 6 m and floats in fresh

water with its axis vertical. Its lower end is submerged to a depth of 6 m

below the water surface. Find: (a) the initial metacentric height; (b) the

righting couple when the caisson is tipped through an angle of 10

degrees.

Soln:

a.) b.)

MB_o = I/V ; I=((6^2))/(12(4)) ; V=((6)(6))/4

MG = MB_o+GB_o = 0.375 + 0.5

M_o = 0.375 m MG = 0.875 m

52

74. A rectangular scow 9.15 m wide by 15.25 m long has a draft of 2.44 m in

fresh water. Its center of gravity is 4.60 m above the bottom. Determine

the height of the scow if, with one side just at the point of submergence,

the scow is in unstable position.

75. A rectangular raft 3 m wide 6 m long has a thickness of 60 cm and is

made of solid timbers (Sp. Gr. 0.60). If a man weighing 890 N steps on

the edge of the raft at the middle of one side, how much will the original

water line on that side be depressed below the water surface?

Find: RM

V=(9.15m)(2.75m)(15.25) W=wV

=383.73m3=(9.81kn/m3)(383.73m3)

W=3764.39kn

Tan=1.85/4.575 V=1/2(15.25m)(1.85m)(4.573)

=222.02 V=64.54m3

Mbo=VL/VSin Gbo=2.30-1.375m

=64.54m3(6.1)/383.73m3(Sin22.02) Gbo=0.925m

Mbo=2.74m

MG=Mbo-Gbo RM=w(MGSin)

=2.74m-0.925m =(3764.39kn)(1.815)sin22.02

MG=1.815m RM=1242.60kn.m

53

CHAPTER FOUR Accelerated Liquids in Relative Equilibrium

EXERCISE PROBLEM

1. A car travelling on a horizontal road has a rectangular cross section, 6m

long by 2.40m wide by 1.50m high. If the car is half full of water, what is

the maximum acceleration it can undergo without spilling any water?

Neglecting the weight of the car, what force is required to produce

maximum acceleration?

Given:

d=0.75 m

L=6 m

Wide=2.4 m

H=1.5 m

Solution: W = Vw

W = . 75 6 2.4 9.81 = 105.95 kN

d =La

2g = =

a = 2dg

L =

105.95 2.45

9.81

a =2 0.75 9.81

6 = 26.46

= 2.45

2

54

1.20 m

0.90 m

Sg = 22 000 N/m3

0.60 m

a = 9.81 m/s2

F

2. A cylindrical bucket is accelerated upward with an acceleration of gravity.

If the bucket is 0.60m in diameter and 1.20m deep, what is the force on

the bottom of the bucket if it contains 0.90m depth of wet concrete whose

specific weight is 22,000 N/m3?

Solution:

= 1 +

= 22 1.20 . 602

4 (1 +

9.81

9.81)

= 11.20

55

3. A rectangular car is 3m long by 1.5m wide and 1.5m deep. If the friction

is neglected and the car rolls down a plane with an inclination of water

surface if the car contained 0.60m depth of water when the car was

horizontal?

Given:

Find:

tan= macos 20

mg masin 20

=m(acos 20)

m(gasin 20)

= acos 20

gasin 20

Consider:

tan =0.9

1.5 a= 9.81 tan(30.96)

=30.96 a=5.885 m/s2

tan =a

g

tan(30.96) =

tan =

5.885 cos 20

9.81 5.885 sin 20 =

5.53

7.79 = 35.35

3m

0.9

0.6

1.5

1.5 1.5

20

W

REF

REFv

20

W

REF

REFv

20

W

REF

REFv

a

20

a

av

a

W=mg

REF = ma

=masin20

REFv = mav

=macos20

F

F mg

masin20

macos20

56

4. An open tank, 9.15m long is supported on a car moving on a level track

and uniformly accelerated from rest to 48km/hr.When at rest the tank

was filled with water to within 15cm of its top. Find the shortest time in

which the acceleration may be accomplished without spilling over the

edge.

Given:

VF=48 Km/s= 13.33m/s

Find: t

Solution:

tan = 0.15

4.575

= 1.878

tan =

tan(1.878)=

9.81

a = 9.81 tan(1.878) a= 0.322 m/s2

a=

t= 13.33

0.322

=41.44 s

15 cm

a h

4.575 4.575

15 cm

57

5. A rectangular tank, 60cm long and containing 20 cm of water is given an

acceleration of a quarter of the acceleration of gravity along the length.

How deep will the water be at rear end? At the front end? What is the

pressure force at the rear end if it is 45 cm wide?

Given:

Find : hF, hr, F

Solution:

tan =

g =

1

4

2=

1

4

tan=_x_ 0.30 1 =_x_ 4 0.30 x= 0.075m or 7.5 cm hr=20+7.5 =27.5 cm hF=20-7.5 =12.5cm F=Awh =(0.45)(0.275)(9.81)(0.275/2) =0.16692 KN =166.92 N

F x

x

30 cm 30 cm

a= 1 g 4

20 cm

58

W.S.

600

3.0 m

1.3o m

6. Figure GG shows a container having a width of 1.50 m. Calculate the

total forces on the ends and bottom of the container when at rest and

when being accelerated vertically upward at 3m/s2?

Given:

a=3m/s2

w= 1.50m

Solution:

= = (1 +

)

= 9.81 1.3 1.5 (0.65) = 9.81 1.3 1.5 (0.65)(1 +3

9.81)

= 12.43 = 16.23

= = (1 +

)

= 9.81 1.5 1.5 (0.65) = 9.81 1.5 1.5 (0.65)(1 +3

9.81)

= 14.35 = 18.74

= = (1 +

)

= 9.81 1.5 1.3 1.3 = 9.81 1.5 1.3 (1.3) (1 +3

9.81)

= 57.40 = 74.95

59

7. A closed rectangular tank 1.20m high by 2.40 m long by 1.50 m wide is

filled with water and the pressure at the top is raised to 140 Kpa.

Calculate the pressures in the corners of this tank when it is accelerated

horizontally along its length at 4.60m/s2?

Given:

Find: P1 , P2

Solution:

h=p = 1.40 = 14.2712 m w 9.81 tan = a = y_ = 4.6_ g 2.4 9.81 y= 1.125 m P1 = wh1 = 9.81 (1.2 + 14.2712 + 1.125) = 162.81 Kpa P2 = wh2 = 9.81(1.2 +14.2712) = 151.77 Kpa

y

h

1.20 m

P1 P2

2.40 m

a=4.60m/s

60

8. A pipe 2.50 cm in diameter is 1.0 m long and filled with 0.60m water.,

what is the pressure at the other end of the pipe when it is rotating at 200

RPM?

Given:

Find: P Solution: y1 = w

2x12

2g =(20/3)2(0.4)2

2(9.81)

=3.577 m

y2== w2x2

2

2g

=(20/3)2(1)2

2(9.81)

=22.357 m

h2=22.357-3.577

=18.78 m

P=wh2 =9.81(18.78) = 184.23 Kpa

h2

0.6 X1=0.4

y1

y2

w= 200rpm =20/3 rad/s

x2= 1m

61

9. An open vertical cylindrical tank 0.60 m in diameter and 1.20 m high is

half full of water. If it is rotated about its vertical axis so that the water just

reach the top, find the speed of rotation. What will then be the maximum

pressure in the tank? If the water were 1.0 m deep, what speed will

cause the water to just reach the top? What is the depth of the water at

the center?

a.

=22

2

1.2 =2 0.3 2

2 9.81

= 16.17

=

= 9810(22

2)

= 9810(16.172 0. 32

2 9.81)

= 11.80

=22

2

0.4 =2 0.3 2

2 9.81

= 9.34

=

2+

2

= 0.6 + 0.2

= 0.8

62

10. If the tank of problem 9 is half full of oil (sp. Gr. 0.75) what speed of

rotation is necessary to expose one-half of the bottom diameter? How

much oil is lost in attaining this speed?

Given:

D=0.6 m

H=1.2 m

1.2 + =22

2 eq.1

=2

2

2

2 eq.2

By equating eq 1 and eq 2

= 18.68

= 0.40

11. The U-tube of figure HH is given a uniform acceleration of 1.22 m/ s2 to

the right. What is the depth in AB and the pressures at B, G and D?

Given: Find:

Solution:

=

0.30=

1.22

9.81

= 0.04

B

A H

G

C

D

30 cm

30 cm 30 cm

y

h1 h2

H3

= 45.36

= 2

= 40.02

= 3

= 37.40

= 9.81 13.6 (. 34)

= 9.81 13.6 (. 30)

= 9.81 13.6 (. 26)

63

= 1

12. The U-tube of figure HH is rotated about an axis through HG so that the

velocity at B is 3m/s. What are the pressures at B and G?

Given: Find: PR, PG

Solution :

PR= wh1 =9.81(3.6)(0.30) =40.02 Kpa

Y2 = 2

2

=3

2

2 (9.81)

=0.459 m PG= wh2 =9.81(13.6)(0.459-0.30) =21.18 Kpa it is below the point Therefore, PG = -21.18 Kpa

30

cm

30 cm

H

G

C

A

D

B

30 cm

Figure HH

30 cm 30 cm

H

G

A

D B

Figure HH

C

30 cm

64

13. The U-tube of figure HH is rotated about HG. At what angular velocity

does the pressure at G become zero gage? What angular velocity is

required to produce a cavity at G?

y1 = 1

2 12

2 ; in LL1

w= 12

2

= 0.30 2 (9.81)

0.302

w1=8.09 rad/s in LL1;

h=Patm

y2=0.30 +0.759 w2 =

12

2

=101.3

9.81 (13.6) =1.059 w2 =

1.059 2 (9.81)

(0.30)2

30 cm 30 cm

H

G

C A

D

B

Figure HH

0.30 m 0.30m Patm/w

w

0.30,0.30

0.30 m

0.30, y2

Y2

LL1

LL2

65

=0.759m w2 =15.19 rad/s 14. The tank of problem 9 is covered with a lid having a small hole at the

center and filled with water. If the tank is then rotated about its vertical

axis at 8rad/s, what is the pressure at any point of circumference of the

upper cover? Of the lower cover?

Given:

Solution:

y=2 2

2

=8

20.32

2(9.81)

=0.294 m

PU= wh1

= 9.81(0.294)

= 2.88 Kpa

PL = wh2

= 9.81 (1.20 +0.294)

= 14.65 Kpa

h2=1.20m

y

0.30 m 0.30 m

h2

66

15. The tank of problem 9 contains 0.60 m of water covered by 0.30m of oil

(sp. Gr. 0.75). What speed of rotation will cause the oil to reach the top?

What is then the pressure at any point on the circumference of the

bottom?

Given : Find: w, PB

Solution:

y1 = w12x1

2 ; in LL1 2g

w =y2g x2

= (0.6)(2)(9.81) ` (0.30)2 w = 11.44 rad/s

PB= woilh2 + whwh1

= 9.81(0.75)(0.3) + 9.81(0.9)

PB = 11.04 Kpa

oil

water

0.3 m 0.3 m

h2

h1 1.20 m

67

16. The tube of figure II is rotated about axis AB. What angular velocity is

required to make the pressures at B and C equal? At that speed where is

the location of the minimum pressure along BC?

Given: Y2 Find: w,z

Y1

Solution: w

Tan 45 = x

0.3 = 0.3m

Y = w2x2

2g

W= y2

2g

x2 =

2 (9.81)

0.3 = 8.08

rad

s

0.30 = y3 + z ; y3 = 0.3 z h = y1+ 0.3-z

= w2z2

2+0.3-z

= 8.092 2

2 (9.81) +0.3 z

= 3.336z2 z +0.3 P = wh

(

) = w(3.336z2 z + 0.3) = 0

= w (6.672z z) = 0 Z = 0.15 m

B

45

x

h

C

A

z z

Z

30c

m

68

17. A vessel 30 cm in diameter and filled with water is rotated about its

vertical axis with such a speed that the water surface at a distance of

7.50cm from its axis makes an angle of 45 degrees with the horizontal.

Determine the speed of rotation.

tan 45 =

=

2

tan 45 = 2

= 45

= 9.81 tan 45

0.075

= 11.44

69

18. A cylindrical vessel, 0.30 m deep, is half filled with water. When it is

rotated about its vertical axis with the speed of 150 RPM, the water just

rises to the rim of the vessel. Find the diameter of the vessel.

Given: Find : D

w = 150 rpm

= 5rad

s

Solution:

Y1 = 2 2

2

X = 2

2

= 0.3 2 (9.81)

(5 )2

= 0.154 m D = 2x = 2(0.154) = 0.3089 m = 30.89 cm

0.30

x

70

19. A conical vessel with vertical axis has an altitude of 1m and is filled with

water. Its base, 0.60m in diameter, is horizontal and uppermost. If the

vessel is rotated about its axis with a speed of 60RPM, how much water

will remain in it?

Given : Find: Vr

w = 60 rpm

= 2rad

s

Solution:

y = 2 2

2

= (2 )2(0.3)2

2(9.81)

= 0.1811 m Vr = Vcone Vpar

= 1

3 ( )3

1

2 2

= 1

3 0.3 3(1)

1

2 0.3 2(0.1811)

Vr = 0.060 m3

1m

y

0.3

0.3

71

20. A cylindrical bucket, 35 cm deep and 30 cm in diameter, contains water

to a depth of 30 cm. A man swings this bucket describing a circle having

a diameter of 2.15 m. what is the minimum speed of rotation that the

bucket can have without permitting water to spill?

Given: Find : w

Solution:

r= R 0.30

2

= 1.075 0.15 = 0.925 m w = Fc mg = mwr

w =

= 9.81

0.925

w = 3.26

R = 1.075

S

w

72

21. If the water which just fills a hemispherical bowl of 1.0m radius be made

to rotate uniformly about the vertical axis of the bowl at the rate of 30

RPM, determine the amount of water that will spill out?

Given: Find : Vspill

Solution:

y = w2x2

2g

= ()2(1)2

2(9.81)

= 0.503 m

Vspill = 1

2r2y

= 1

2 1 2(0.503)

Vspill = 0.79 m3

y

1.0

W = 30 RPM

=

rad

s

73

22. The open cylindrical tank of figure JJ is rotated about its vertical axis at

the rate of 60 RPM. If the initially filled with water, how high above the

top of the tank will water rise in the attached piezometer?

Given:

Solution:

y1 = w2x1

2

2g

= (2)2(0.65)2

2(9.81)

= 0.850 m

Y2 = w2x2

2

2g

= (2)2(1)2

2(9.81)

= 2.012 m h= y2 y1 = 2.012 0.850 = 1.16 m

1.30

m

1

m

1.30

cm

Figure JJ

74

23. A closed cylindrical tank with axis vertical, 2m high and 0.60m in

diameter is filled with water, the intensity of pressure at the top being 140

Kpa. The metal making up the side is 0.25 cm thick. If the vessel is

rotated about its vertical axis at 240 RPM, compute (a) total pressure on

the side wall, (b) total pressure against the top, (c) maximum intensity of

hoop tension in pascals.

Given:

Solution:

h =

=

140

9.81= 14.271 m

y = w2x2

2g =

(8)2(0.3)2

2(9.81) = 2.897 m

a.) Fside = Awh

= 2 0.6 9.81 14.271 + 2.897 + 1 = 671.90 KN

b.) Ftop = Awh

= 0.3 2 9.81 14.27 + 2.879 = 47.62 KN

c.) P = wh

= 9.81(2.9 +14.27 + 2)

= 188.044 Kpa

FB = PD = 2T

T =188.04(0.6)

2

= 56.41 Kpa

S =

= 56.41

2.5 103

=22565.34 Kpa

y

0.6

w

2m Ftop Fside

= 240 rpm

= 8rad

s

75

24. A small pipe, 0.60m long, is filled with water and capped at both ends. If

placed in a horizontal position, how fast must it be rotated about a

vertical axis, 0.30m from one end, to produce maximum pressure of

6,900 Kpa?

Given:

Solution:

h =

= 6900

9.81

= 703.36 m h = y2 y1

= w2(0.9)2

2(9.81) -

w2(0.3)2

2(9.81)

703.3639 = 92

218 -

2

218

703.3639 = 42

109

w= 138.44 rad

s

0.3

0.6

0.3 + 0.6 = 0.9

Y2

w

h= p

w

76

25. A vertical cylindrical tank 2m high and 1.30m in diameter, two thirds full

of water, is rotated uniformly about its axis until it is on the point of

overflowing. Compute the linear velocity at the circumference. How fast

will it have to rotate in order that 0.170 m3 of water will spill out?

Given:

Solution:

y = w2x2

2g

w = y2g

x2

= 1.33 2 (9.81)

(0.65)2

= 7.869 rad

s

V = wx = 7.869(0.65)

=5.11 m

s

Vspill = 1

2r2

0.170 = 1

2(0.65)2h

h = 0.256 m

w = y2g

x2

= (1.33+0.256) 2 (9.81)

(0.65)2

= 8.59 rad

s

y

0.67

0.67

w

2m

1.3

h

With spillage

1.33

w

1.3

Vspill = 0.170m3

77

26. A steel cylinder, closed at the top, is 3m high and 2m in diameter. It is

filled with water and rotated about its vertical axis until the water

pressure is about to burst the sides of the cylinder by hoop tension. The

metal is 0.625 cm thick and its ultimate strength is 345 Mpa. How fast

must the vessel be rotated?

2 =

= 2

2(1)

=

=

= 345 1000 (6.25 103)

= 2156.25

=

= 9.81 + wh

2156.25 = 9.81 + 9.81 (3)

= 216.80

=2 2

2

216.80 =2 12

2(9.81)

= 65.22

78

27. A conical vessel with axis vertical and sides sloping at 30 degrees with

the same is rotated about another axis 0.60 m from it. What must be the

speed of rotation so that water poured into it will be entirely discharged

by the rotative effect?

Given:

Solution:

tan = w2x2

2g

w = gtan

x

= 9.81tan(60)

0.6

w = 5.32 rad

s

30 30

60

w

0.6 m

79

CHAPTER Five Principles of Hydrodynamics

EXERCISE PROBLEM

1. A fluid flowing in a pipe 30cm in diameter has a uniform velocity of 4m/s. the

pressure at the center of the pipe is 40kpa and the elevation of pipes centerline

above an assumed datum is 4.5m. compute the total energy per unit weight of

the flowing fluid if

(a)oil (sp.gr. 0.80) (b)gas(w=8.50N/m3)

GIVEN: a) E = V2 +P + Z b) E = V2 + P + Z

oil (sp.gr. 0.80) 2g w 2g w

gas (w=8.50N/m3) = (4)2 + 40__ + 4.5 = (4)2 + 40__ + 4.5

Z = 4.5m 2(9.81) (9.81).8 2(9.81)

V = 4m/s E= 10.41 J/N E = 4.7 J/N

2. A liquid of specific gravity 1.75 flows in a 6cm pipe. The total energy at

appoint in the flowing liquid is 80 J/N. the elevation of the pipe above a fixed

datum is 2.60m and the pressure in the pipe is 75kpa. Determine the velocity of

flow and the power available at the point.

GIVEN: E = V2 +P + Z P = QwE

Sp.gr = 1.7 2g w P = AVwES

P= 75kpa V2 = E P +Z P= (0.06)2 (37.85)(80)(1.75)

Z= 2.6m 2g w 4

V2 = 80 75__ +26 P= 147 kW

2g (9.81)(1.75)

V = 37.85m/s

4. A city requires a flow of 1.5m3/s for its water supply. Determine the diameter

of the pipe if the velocity of flow is to be 1.80m/s.

GIVEN: Q = AV : A= Q/V

Q = 1.5m3/s d2 = Q

V = 1.80m/s 4 V

d2 = 4Q

V

d2 = 4(1.5)

80

5. A pipe consists of three length 50cm, 40cm, and 30cm with a continuous

discharge of 300liters of oil (sp.gr. 0.75) compute the mean velocity in each

pipe.

Given

Q=300 L/s V1=Q/A1 = (0.3)(4) / (3.14)(0.5)2 = 1.53 m/s

S=0.75 V2=Q/A2 =(0.3)(0.4) / (3.14)(0.4)2 =

2.39m/s

V3=Q/A3 = (0.3)(4) / (3.14)(0.3)2 = 4.84m/s

6. A 30cm pipe is connected by a reducer to a 10 pipe points 1 and 2 are along

the same elevation. The pressure at 1 is 200KPa. The flows is 30 liters and the

energy lost between 1 and 2 is equivalent to 20KPa. Compute the pressure at 2

if the liquid flowing is water.

Given: V2+P1 + Z1 = V22 + P2 + Z2 + HL

2g w 2g w

(0.4246)2 + 200 =3.82 + P2 + 20

2(9.81) 9.81 2(9.81)

P2 =173 kPa

7. Compute the velocity head of the jet if the larger diameter is 10cm and the

smaller diameter is 30mm. The pressure head at point 1 is 30m of the flowing

water and the head lost between points 1 and 2 is 5% of the velocity head in jet.

Given

V2+P1 + Z1 = V22 + P2 + Z2 + HL

2g w 2g w

P1 =V22-V2 + 0.5V2

2 __________ 1

W 2g

Q=AV=A2 V2

V1= 0.333V2= 0.09V2 ______________2

Substitute 2 in 1

V2 =28.85 J/n

2g

81

9. In fig. a 5cm pipeline leads downhill from a reservoir and discharges into air,

If the loss of head between A and B is 44 J/N, determine the discharge.

V2+Pa + Za = V2b + Pb + Zb + HL

2g w 2g w

Vb2=Za-HL Q=AV

2g

=3.14(0.05)2(6.26)

V2=2(9.81)(46-44)

V=6.26m/s

Q=12.30L/s

11. In fig. shown a siphon discharging oil (sp.gr. 0.90). The siphon is composed

of 8cm pipe from A to B followed by 10cm pipe from B to C. The head losses

are: 1 to 2: 0.0 J/N; 2to3: 0.20J/N and 3 to 4: 1.00 J/N.

Compute the discharge and determine the pressures at points 2 and 3.

\

V2+Pa + Za = V2c + Pc + Zc + HL

2g w2g w

VC2 = ZA- HL= 1.5

2g

V2+Pa + Za = V2b + Pb + Zb + HL

Q2= 1.5(2)(9.81)[3.14(0.1)(.25)]2 2g w 2g w

Q= 0.042m3/s But V2= 8.36m/s

P2 = 3-(8.36)2 -5-0.3 P2=

W 2(9.81)

13. The 60cm pipe conducts water from reservoir A to a pressure turbine which

is discharging through another 60cm pipe into tailrace B. the head losses are: A

to 1:5V2/2g; 2 to B: 0.20V2/2g. if the discharge is .70m3/s what input power is

being given by the water to the turbine?

Solution:

Q=AV ; V=Q/A

V=0.7/(.5)2

V=2.4758m/s

V2+Pa + Za = V2b + Pb + Zb + 5V

2 + .20V2

2g w 2g w 2g 2g

Pi = QwHt

Pi =0.7(9.81)(68.3755)

Pi =469.53 kW

82

70 =5(2.4757)2 +.20(2.4758)2

19.62 19.62

Ht = 68.37551985m

14. A fire pump delivers water through 15cm main pipe to a hydrant to which is

connected an 8cm host, terminating in a nozzle 2cm in diameter. The nozzle

trained vertically up, is 1.60m above the hydrant and 12m above the pump. the

head losses are pump to hydrant is 3J/N; Nozzele;6% velocity head inthe

nozzle. If the gage pressure at the pump is 550Kpa to what vertical height can

the jet bo thrown? Neglect air friction.

Q = A,V, = A2V2

V,= (0.01)v2 = 4V2

(0.075)2 225

V2+P1 + Z1 = V2

2 + P2 + Z2 + HL+ HL+ HL

2g w 2g w

4V2/225 + 550 = V2 +12 + 3 + 2 +12 + 0.06V22

19.62 9.81 19.62 19.62

V22= 22. 3855m/s

H= V2 = 22.38552 = 25.54m

2g

15. The water from reservoir is pumped over a hill through a pipe 90cm in

diameter, and a pressure of 200kpa is maintained at the summit where the pipe

is 90m above the reservoir. The quantity pumped is 1.40m3/s and by reason of

friction there is a hed loss of 90% efficient, determine the input power furnished

to the water.

Q = Ab Vb V2+Pa + Za + ha = V

2b + Pb + Zb

V = 1.4/ (0.45)2 = 2.201m/s 2g w 2g w

Po = QwHp Hp = 113.63

Po= 1.4(9,81)(113)

Po = 1560.652

Pi = Po/ n

Pi = 1734.06 kN

83

16. The turbine shown in fig. extracts 50 J/N of water from the given pipe

system. At the summit S 480kpa is maintained. Determine the flow and the

pressure at the discharge side of the turbine considering the following losses:

summit to turbine : 4times the velocity head in the 20cm pipe; turbine to

reservoir 3times the velocity head in the 30cm pipe.

BEE from 1 to 2

V2+P1 -Z1 = V22 + P2 + Z2 + HL

2g w 2g w

0+ 480/9.81 + 46 50- [8Q2/2g(0.2)2]-3[8Q2/2g(0.2)2]

48.93 + 46 +50 -206.57 Q2 = 30.60Q2 = 16

Q=0.350m3/s or 350L/s

17. A horizontal Venturi meter 45cm by 60cm is used to measure the flow of air

through a 60cm pipeline. A differential gage connected to the inlet and throat

contains water which is deflected 10cm. considering the specific weight of air as

12.60 N/m3, find the flow of air. Neglect head losses.

V2+P1 + Z1 = V2

2 + P2 + Z2 V = Q/A =4Q/D4

2g w 2g w V2 = 16Q2/ 2D4

16Q2/ 2D4 + P1 =16Q2/ 2D4+ P2

2g w 2g w

P1 P2 = 1.38Q2 ______________(1)

w

sum-up pressure head from (1) (2) in meters

P1- P2 = 81.65 _____________(2)

W

Substitute (2) to(1)

81.65 = 1.38Q

Q = 7.5 m3/s

84

18. A venturi meter 60cm by 30cm has its axis inclined downward 30deg from

the horizontal. The distance, along the axis, from the inlet to the throat is 1.20m.

the differential manometer showns a deflection of 15cm of mercury. If the

flowing is water, find the discharge if C=0.98.

P1/w + v + x 0.15(13.6)-y + 0.15=P2/w C =Qa/Qt P1 - P2 = 1.89y Qa=CQt

W Qa = 0.98(444.46)

BEE for 1to 2 Qa =435.72 L/s

V2+P1 + Z1 = V2

2 + P2 + Z2 + HL

2g w 2g w

V22-V1

2 =P1-P2 + Z1-Z2

2g w

V2= 6.29m/s

V1= 0.32/(0.6)2V2

V1 = 0.25V2

Qt = V2A2

Qt = 6.9(0.3)2/4

Qt = 444.46L/s

19. A 6cm fire host water discharges through a nozzle having a diameter of

2.5cm. the head lost in the nozzle is 4% of the velocity head in the jet. If the

gage pressure at the base of the nozzle is 400kpa, find the flow and the

maximum horizontal range to which the stream can be thrown.

BEE from 1 to 2

V2+P1 + Z1 = V2

2 + P2 + Z2 + HL

2g w 2g w

0.051v2 + 40.77 = 0.051 v2 + 2.04x10-3v2

V= 4.987 m/s

Q,=Q2 ; A,v= A2v2

V 2 = 0.062/.0252v

V2 = 28.73m/s

Q2 = A2V2

Q2 =28.73(.025)2/4

Q2= 14.10L/s

85

20. Water is flowing through the pipe system of Fig. calculate the power of the

turbine, neglecting losses.

Solution:

V2 = Q/A = Q/ (0.15)2 = 14.147Q` BEE from 1 to 2

VN = Q/ (0.05)2 =127.324Q V2+P1 + Z1 = V

22 + P2 + Z2

H2 = P2/w =0.2(13.6) =-2.72 2g w 2g w

V, = Q/A = .218/(.1)2 =6.941m/s (6.941)2 + 345 HA = (3.084)2 2.72

BEE from to to nozzle 2(981) 9.81 2(9.81)

V2+P2 + Z2 = V2n + Pn + Zn HA = 39.733m

2g w 2g w

P = QwHa

14.147Q - 2.72 + 45 = 127.322 P = (.211)(9.81)(39.733)

2(9.81) 2(9.81) P= 82.24 kW

Q= 0.211 m3/s

21. Calculate the minimum power of the pump which will send the jet over the

wall shown in fig. neglect losses.

V2+P1 + Z1 = V2

2 + P2 + Z2 Q = AV

2g w 2g w Q = (0.075)2(31.63)

V2+ 55 = V22 + 60 76 -39 4

2g 2g Q = 0.14m3/s

V2 = 31.63m/s

HE = 101.98 -72.5

= 29.48

86

HL = (31.63)2

2(9.81)

P = Qwe HL = 101.98m

P = 0.14(9810)(29.48)

P = 40.49 kN

22. In fig. K h1 = 20cm and h2 = 30cm. If water is flowing, calculate the power

of the pump.

P1 + 0.6 + 0.2 (13.6) = -3.32m

W

BEE for 1 to 2 BEE for 3 to 4

V2+P1 + Z1 + Ha = V22 + P2 + Z2 V

2+ P1 = V

22 + P2

2g w 2g w 2g w 2g w

Ha =V22- V1

2 + P2 P1 V42- V3

2 = P3 P4

2g w w 2g w w

Ha = 415 - (-3.32) V42 - .25V4

2 = 29 P3_ P4

9.81 w

Ha = 45.62 V4 = [2(9.81)(3.78)/1-0.25]

V4= 8.89m/s

V3 = (A4/A3)V4

V3=(0.52/0.32)V4 P=QWHa

V3= 0.25V4 P=[8.89()(.152)/4](9.81)(45.62)

P= 71. 51 Kw

87

V1 = Q/A ; (0.220)(4)/ (0.3)2 Pout = QwHE

V1 = 3.11m/s Pout =23.14(9.81)(.22)(.8)

V2 = 0.70(4)/ (0.6)2 Pout = 39.95 kW

23. A 20cm pipe contains a short section in which the diameter is gradually

reduced to 7.5cm and then gradually enlarged to full size. The pressure of the

water at a point where the reduction starts is 520kpa. If the rate of flow is 35L/s,

determine the pressure at 7.5cm section. Neglect losses.

GIVEN: BEE from 1 to 2

Q1 = 20cm Q2 = A2V2 V2+P1 + Z1 = V

22 + P2 + Z2

Q2 = 7.5cm V2 = 0.35 2g w 2g w

P = 520kpa (0.075)2 520 +(71.78)2 = P2

Q = 35L/s V2 = 71.78m/s 9.81 2(9.81) (9.81)

P2 = 490.21kPa

24. The inlet in the end of a pipe is 2.50m above the discharge end. To maintain

a flow 5L/s through the 15cm pipe a pressure of 250kPa at the inlet end is kept.

Compute the head loss while passing through the pipe and determine the

energy per second it represents. Consider water flowing.

E1 = E2

V2+P1 + Z1 = V2

2 + P2 + Z2 + HE P = QwE

2g w 2g w P = 0.035(9.81)(27.98)

Therefore Q1 = Q2 ,V1 = V2 P = 9.61kW

V2+250 + 2.5 = V21 + P2 + HE

2g 9.81 2g 9.81

HE = 27.98m

25. A water motor is supplied from a horizontal 30cm pipe and uses 220L/s.

Discharge takes place through a 60cm vertical pipe. A differential gage tapped

into pipe close to the motor shows a deflection of 1.80m of mercury. The two

points where the gage was taped are separated by vertical distance of 1m. if

the motor is 80% efficient, determine its power output.

BEE from 1 to 2

V2+P1 + Z1 = V22 + P2 + Z2 + HE

2g w 2g w

HE = V12- 23

2 + P1 P2 + Z1- Z2

2g w

HE = (3.11)2- (0.78)2 + 21.68+1

2(9.81)

HE = 23

88

26 .A pump draws water from a pit through and vertical 30cm pipes w/c extends below the water surface. It discharges into a 15cm horizontal pipe 4.0m above the water surface. While pumping the y of flow in L/s, a pressure page on the discharge pipe reads 165kp2 and a page on the suction pipe shows a vacuum of 35kg2. Both pages are close to the pump and are separated by a vertical dist. Of 90cm. Compute the head lost in the sanction pipe. Compute the charge in energy per second between the pages. What is the power output of the pump?

Solution:

v2 = = 0.06(4)

03 2 = 0.85

v3 =

3=

0.06()

(0.15) = 3.40

B fr. 1-2

12

2+

12

+ 1 =

22

2+

22

+ 2+

H.L=22

2

2

- 2

=(0.85).2

2(9.81) -

(35)

9.81 -3.1

H.L=0.43

B fr. 2-3

22

2+

2

+ 2+ =

32

2+

3

+ 3

H.A= 3222

2+

32

+ 3 2

= (3.4)2(0.85)2

2(9.81)+

(16.5) (35)

(9.81)+4 3.1

H.A = 21.84

89

27. A free jet of water 5cm in diameter is discharge from a nozzle at an angle of

60 from the horizontal. If the pressure at the 10cm base, 30cm from the tip, is the maintained at 465kpa and cy = 0.97, what is the maximum distance that the nozzle can be placed from a building and still jet water into a window w/c is 20m above the nozzle?

Solution:

B fr. 1-2

12

2+

1

+ 1 =

22

2+

2

+ 2+

22

2=

465

9.81 0.360

1

0.97 2 0.25 2

2 =

465

9.81 0.360

1

0.97 2 0.25 2

2 9.81

2 = 30.41

2 = 0

2 2

= 0 +

1 =17.34

8.81= 1.77

2 =

=

30.4160

9.81= 2.68

T=1 + 2

T=1.8 +2.68 = 4.5s

X=2

= 30.4 cos60(4.5)

X= 68.42m

90

28.A pilot tube in a pipe in w/c air is