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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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?
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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.
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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90
28.A pilot tube in a pipe in w/c air is