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 15°C?
Density:
ρ = 7 / 9.81
ρ = 713.56 kg/m3
Specific Volume:
v = 1 / 713.56
v = 1.401 x 10-3 m3/kg
Specific Gravity:
s = 7000 / 9798 , specific weight of water at 15°C is 9.789 N/m3
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 12 N/m3?
Specific Gravity:
s = w / wair
s = 16 / 12
s = 1.33
Density:
ρ = 16 / 9.81
ρ = 1.631 kg / m3
Specific Volume:
v = 1 / 1.631
v = 0.613 m3 / kg
3) If 5.30 m3 of a certain oil weighs 43,860 N, calculate the specific weight, density and specific gravity of this
oil.
Specific Weight:
w = weight / volume
w = 43860 / 5.30
w = 8275.47 N / m3
Density:
ρ = specific weight of oil / g
ρ = 8275.47 / 9.81
ρ = 843.58 kg / m3
Specific Gravity:
s = density of oil / density of water
s = 843.58 / 1000
s = 0.84
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.) ρ = 𝑤 𝑔 b.) ѵ = 1 ρ c.) s = 𝑤 ws = 7 𝐾𝑁/𝑚3(1000) 9.81 𝑚/𝑠2 = 1 713.56 𝐾𝑔/𝑚3 = 7 𝐾𝑁/𝑚3 9.81 𝐾𝑁/𝑚3 ρ =
713.56 Kg/m3 ѵ = 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.) ρ = 𝑤 𝑔 b.) ѵ = 1 ρ c.) s = 𝑤 ws = 16 𝑁/𝑚3 9.81 𝑚/𝑠2 ѵ= 1 1.63 𝐾𝑔/𝑚3 s = 16 𝑁/𝑚3 12 𝑁/𝑚3 ρ = 1.63
Kg/m3 ѵ = 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 = 𝑊 𝑔 b.) ρ = 𝑊 𝑔𝑉 c.) s = 𝑤 ws = 43.860 𝐾𝑁 5.30 𝑚3 = 43860 𝑘 .𝑚/𝑠2 ( 9.81𝑚 𝑠2 )(5.30 𝑚3) = 8.28
𝐾𝑁/𝑚3 9.81 𝐾𝑁/𝑚3 w = 8.28 KN/m3 ρ = 843.58 kg/m3 s = 0.844
4. The density of alcohol is 790 Kg/m3 . Calculate its specific weight, specific gravity and specific volume.
a.) w = ρg b.) s = 𝑤 ws c.) ѵ = 1 ρ = (790 kg/m3)(9.81 m/s2) = 7.75 𝐾𝑁/𝑚3 9.81 𝐾𝑁/𝑚3 = 1 790 𝑘𝑔/𝑚3 w =
7.75 KN/m3 s = 0.79 ѵ = 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/m 3 s = ρ ρs s = 12 𝑁/𝑚3 12.7 𝑁/𝑚3 ρa = (1.29 kg/m 3)(0.94) ѵ = 1 ρ = 1 1.21 𝑘𝑔/𝑚3 s = 0.94 ρa
= (1.21 kg/m 3) ѵ = 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.) ρ = 𝑤 𝑔 = 10.10(1000) 9.81 p = wh = 10.10(1000)(8000) b.) ѵ = 1 ρ = 1 1043 𝑘𝑔/𝑚3 ρ = 1029.6 kg/m3 p
= 80.80 MPa = 9.5 x 10-4 m3/kg Δѵ = 3.3 x 10-5 m3/kg c.) w = 𝑝 𝑕 = 82.26 𝑥 106 8000 w = 10282. 5 N/m3
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 𝛥𝑝 𝛥𝑣 ρ = 𝑤 𝑔 = 9.689 9.81 = -( 1 x 10-3)( 30,000,000 1 x 10−3−1.012x10−3 ) ρ
= 987.67 kg/ m3 = 2,500,000 Pa ѵ = 1 ρ = 1 987.67 Bv = 2.5 x 106 Pa ѵ = 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?
ѵ = µ 𝑝 = 1𝑥10−3 𝑘 .𝑚 .𝑠/𝑚2𝑠2 1000 𝑘𝑔/𝑚3 ѵ = 1x10-6 m2/s ( 3.28 𝑓𝑡 1 𝑚 )2 ѵ = 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 𝑖𝑛 2 1 𝑓𝑡2 )( 2.54 𝑐𝑚 2 1 𝑖𝑛 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 ( 1𝑚3 1000 𝐿 ) = 0.45 m3 ρ = 𝑊 𝑔𝑉 = 3.5(1000) 9.81(0.45) = 792.85 kg/m3
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 𝑓𝑡−𝑙𝑏 𝑠𝑒𝑐 ( 12 𝑖𝑛 𝑓𝑡 ) (
2.54 𝑐𝑚 1 𝑖𝑛 ) ( 444,800 𝑑𝑦𝑛𝑒𝑠 1 𝑙𝑏 )
1 Hp = 7456627200 𝑑𝑦𝑛𝑒𝑠 −𝑐𝑚 𝑠𝑒𝑐 100000000 𝑑𝑦𝑛𝑒 −𝑐𝑚/𝑠𝑒𝑐 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 𝑓𝑡3 7.48 𝑔𝑎𝑙 ) ( 1 𝑚3 3.28 𝑓𝑡3 ) = 0.379 m3 P = 6000 (0.379 m3) P =
2274 m3 D.C. = ( 𝑝 𝑑 ) = ( 2274 𝑚3 60𝑥60𝑥24 ) D.C. = 0.026 m3/s
13. Compute the conversion factor for reducing pounds to newtons.
32.18 𝑓𝑡 𝑠2 ( 0.3048 𝑚 1 𝑓𝑡 ) ( 1 𝑘𝑔 2.205 𝑚 ) ( 1 𝑁 𝑘𝑔/𝑚2 ) = 4.448 N
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
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 15®C 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 15®C 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)[0.750.745](9810)/12 h=55.60m
7. At ground level the atmospheric pressure is 101.3 KPa at 15®C. 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
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
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=0.179m 3 Ww=wV =0.179(9.81) Ww=1.756kN
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
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 θ? Water mercury θ b) h = p/w h = -8.82/9.81 h = -1.0 m
14 . Solution: P = wsh 10.5 – 26.68sin𝜃 = 0 P = (9.81) (13.6) (0.20) 26.68sin𝜃 = 10.5 P = 26.68 kpa 𝜃 = 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
27. In fig. determine𝑝𝑚 − 𝑝 . 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 = 𝐼g 𝐴𝑦 = 𝑏𝑕2 12 𝑏𝑕𝑦 hp = 𝑕 + 𝑒 F1 = wA𝑕 e = 𝑕2 12 𝑦 = 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 = 𝐼g 𝐴𝑦 = 𝑏𝑕2 12 𝑏𝑕𝑦 hp = 𝑕 + 𝑒 F1 = wA𝑕 e = 𝑕2 12 𝑦 = 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 = 𝐼 𝐴𝑦 F = 30(1.0)(1.5) F = (9.81)(x+0.5)(1.5)(1.0) e = 1
12𝑥+6 F = 45 KN F = 14.72x + 7.36 ∑𝑀𝑃1 = 0 14.42x + 7.36(0.5 + 1 12𝑥+6 ) = 45(0.5) 86.5x2 – 168.16x – 105.56 =
0 𝑥 = −(−168.16)± (−168.16)2−4 86.5 (105.56) 2(86.5) 𝑥 = 2.40 𝑚 5. A vertical circular gate 1m in diameter is
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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 = 𝐼g 𝐴𝑦 = 𝜋(𝑟4)2 4 𝜋𝑟2𝑦 F =9.81(1.4)(3.1)(𝜋)(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) ½ full (b) ¾ full
of wter, determine the magnitude and location of the total force. Solution: a.) ½ full 𝑕 = 4𝑟 3𝜋 F = w𝑕 A 𝑕 = 4(1.5)
3𝜋 F = (9.81)( 𝜋(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)( 3𝜋(1.52)
(4) )(1.08) F = 56.25 KN 𝑒 = 𝐼g 𝐴𝑦 hp = 𝑕 + 𝑒 𝑒 = 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)( 2 3 )(3)(3) F = 105.95 KN 𝑒 = 𝐼g
𝐴𝑦 = 8 3 (3)2 175 2 3 3 3 (1.8) 𝑕𝑃 = 𝑕 + 𝑒 𝑒 = 0.34 𝑚 𝑕𝑃 = 1.8 + 0.34 𝑕𝑃 = 2.14 𝑚 𝑏𝑒𝑙𝑜𝑤 𝑐𝑒𝑛𝑡𝑒𝑟
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: 𝐹 = 𝐴𝑤𝑕 𝐹𝑓 = 𝜇𝑁 𝐹 = 9.81 1.6 (10.8) 𝐹𝑓 = 0.2(508.55)
𝐹 = 508.55 𝐾𝑁 𝐹𝑓 = 101.71 𝐾𝑁 F = 508.55 KN ∑𝐹𝑣=0 𝐹 = 𝑤 + 𝐹𝑓 𝐹 = 18.0 + 101.71 𝐹 = 119.71 𝐾𝑁 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: 𝑒 =
𝑕 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: 𝜃 = 𝑡𝑎𝑛−1( 3 2 ) 𝐹 = 𝐴𝑤𝑕 𝜃 = 56.31° 𝐹 = 9.81 3.61 1.6 (2.8) 𝐹
= 158.66 𝐾𝑁 𝐵𝑉 = 𝐹𝑠𝑖𝑛𝜃 𝐵𝑕 = 𝐹𝑠𝑖𝑛𝜃 𝐵𝑉 = 158.66𝑠𝑖𝑛56.31° 𝐵𝑕 = 158.66𝑠𝑖𝑛56.31° 𝐵𝑉 = 132.01 𝐾𝑁 𝐵𝑉 = 80.70 𝐾𝑁
22 12. In fig. E gate AB is 2m wide perpendicular to the paper. Determine FH to hold the gate in equilibrium.
Solution: 𝐴 = 3.2 2 = 6.4 𝑒 = 𝐼g 𝐴𝑦 𝐹 = 𝐴𝑤𝑕 𝑒 = 𝑕 2 12𝑦 𝐹 = 9.81 1.21 (6.4) 𝑒 = 3.20 12(1.26) 𝐹 = 77.85 𝐾𝑁 𝑒 = 0.22
𝑚 𝑥 = 3.2−0.48 2 − 0.53 ∑𝑀𝑣 = 0 𝑥 = 1.38 𝑚 3.20𝐹𝑕 − 1.38 77.85 = 0 𝐹𝑕 = 42.50 𝐾𝑁
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𝑕 𝑠𝑖𝑛60° 𝑒𝑞. 1 𝑒 = 𝐼g 𝐴𝑦 𝑒 = 𝑕 2 12𝑦 𝑒 = 𝑕 𝑠𝑖𝑛60° 12( 0.5𝑕𝑠𝑖𝑛60°)
𝑒 = 0.19𝑕 𝑒𝑞. 3 ∑𝑀𝑣 = 0 𝐹 𝑕 𝑠𝑖𝑛60 − 0.5𝑕 𝑠𝑖𝑛60 + 0.19𝑕 𝑠𝑖𝑛60 = 22.5(5.0) 5.95𝑕3 = 112.5(5.0) 𝑕 = 18.91 𝑕 = 2.66
𝑚 𝐹 = 𝐴𝑤𝑕 𝐹 = 9.81 𝑕 𝑠𝑖𝑛60° 2.6 ( 0.5𝑕 𝑠𝑖𝑛60° ) 𝐹 = 17.0𝑕2 𝑒𝑞. 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°) 𝐴1 60° = 𝐻(6)2 360° A2 = 15.59
𝑚2 A1 = 18.85 𝑚 2 𝐹 = 𝐴𝑤𝑕 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 𝑚2
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: 𝐹𝑕 = 𝐴𝑤𝑕 𝐹𝑣 = 𝐹𝑕𝑐0𝑠𝜃
𝐹𝑕 = 9.81 1.5 3 (1.03) 𝐹𝑣 = 147.48 𝑐𝑜𝑠30° 𝐹𝑕 = 147.78 𝐾𝑁 𝐹𝑣 = 128.56 𝐾𝑁 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: 𝐹𝑕 = 𝐴𝑤𝑕 𝐹𝑕 = 9.81 1.06 3 (2.12) 𝐹𝑕 = 66.14 𝐾𝑁 𝐴𝐴𝐵𝐶 = 0.88 + 3 2.12 2 𝐴𝐴𝐵𝐶
= 4.11 𝑚 2 𝐴𝑠𝑒𝑐𝑡𝑜𝑟 = 𝜋𝑟2𝜃 360° = (3)2(45°) 360° 𝐴𝑠𝑒𝑐𝑡𝑜𝑟 = 3.58 𝑚 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: 𝑃 = 𝑤𝑠𝑕 𝑇 = 𝑤𝑉𝑠 60 = 9.81 0.75 𝑕
𝑇 = 9.81 39.23 (0.75) 𝑕 = 8.15 𝑚 𝑇 = 288.63 𝐾𝑁 𝑉 = 𝜋𝑟2𝑕 − 4𝜋𝑟3 6 𝑉 = 𝜋 1.5 2(8.15) − 4(1.5)3 6 𝑉 = 39.23𝑚3 24.
Fig. Q shows semi-conical buttress. Calculate the components of the total force acting on the surface of this semiconical buttress. Solution: 𝐴 = 𝜋𝑟2𝑕 3 𝐹𝑕 = 𝑤𝑕 𝐴 𝐴 = 𝜋 0.15 2(3) 3 𝐹𝑕 = 9.81 1.463 (7.07) 𝐴 = 7.07 𝑚2 𝐹𝑕 =
101.47 𝐾𝑁 𝑕 = 𝑦𝑝 + 𝑒( 𝜋 1.5 4 4 ) 𝐹𝑉 = 𝑤𝑉 𝑕 = 1.3 + 0.163 𝐹𝑉 = 1.3 3 1.3 1.5 + 𝜋 1.5 2(3) 3 𝑕 = 1.463 𝑚 𝐹𝑉 = 0.12
9.81 𝐹𝑉 = 7.8 𝐾𝑁
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: 𝐴 = 𝜋𝑟2 𝐹𝑕 = 𝑤𝑕 𝐴 𝐴 = (0.125)2 𝐹𝑕 = 9.81 178.39 0.099
(0.71) 𝐴 = 0.099 𝑚2 𝐹𝑕 = 7.8 𝐾𝑁 𝑃 = 𝑤𝑕 𝐴 350 = 9.81 0.20 𝐴 𝐴 = 178.39 𝑚2 26. Calculate the force required to
hold the cone of Fig. S in position. Solution: 𝑃2 = 𝑃1 + 𝑤𝑠𝑕 𝑊 = 𝜋𝑟 2𝑃 𝑃2 = 3.5 − 9.81 0.8 1.5 𝑊 = 𝜋(0.375)






2(0.8)(9.81)(2.5) 𝑃2 = −8.26 𝐾𝑝𝑎 𝑊 = 8.66 𝐾𝑁 𝐹1 = 𝐴𝑃 𝑇2 = 𝑤𝑕 1 3 𝜋𝑟2 𝐹1 = 𝜋𝑟 2𝑃 𝑇2 = 9.81(0.8) 1 3 𝜋(0.375)2 𝐹1
= 𝜋(0.375) 2(8.26) 𝑇2 = 1.16 𝐾𝑁 𝐹1 = 3.65 𝐾𝑁 ∑𝐹𝑉 = 0 𝐹 + 𝑃2 + 𝐹1 = 𝑊 𝐹 = 8.66 − 3.65 − 1.16 𝐹 = 3.85 𝐾𝑁
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? Sol’n: 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? Sol’n: S = PD′ /2t fc = 𝐷′ 𝐷 = 0.714 𝑚 0.2 D’ =
2𝑠𝑡 𝑃 = 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. Sol’n:
∆EL = 𝐸𝐿1 - 𝐸𝐿2 S = 𝑃𝑑 − 𝑃𝑠 2𝑡 = 1716.75−320 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. Sol’n: 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. Sol’n: 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. Sol’n: 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. Sol’n: 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)
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


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 µ = RH/RV = 235.4(〖16.54〗^2) = 3575.75/5840.27 G = 64398.76 kN.m µ = 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.) 𝐺1 = 5(8)(1)w ∑Fv = 𝐺1 + 𝐺2 + 𝐺3 + 𝐺4 + 𝐺5 + 𝐺6 − 𝑈1 − 𝑈2 = 40w =
(40+25+60+252+176.4+4.18- 73.67-56.67)w 𝐺2 = 1 2 (5)(10)(1)w Rv = 427.24w = 25w 𝐺3 = 1 2 (5)(10)(1)(2.4)w
∑Fh = 𝐹1 − 𝐹2 = 60w = (162-12.5)w 𝐺4 = 5(21)(1)(2.4)w Rh = 149.5w = 252w 𝐺5 = 1 2 (7)(21)(1)(2.4)w Moment
Forces: = 176.4w 𝐺1 = 14.5(40w) = 580w 𝐺6 = 1 2 (1.67)(5)(1)w 𝐺2 = 15.33(25w) = 383.25w = 4.18w 𝐺3 =
13.67(60w) = 820.2w 𝐹1 = 1 2 (182)w 𝐺4 = 9.5(252w) = 2394w = 162w 𝐺5 = 4.67(176.4w) = 823.79w 𝐹2 = 1 2
(52)w 𝐺6 = 0.56(4.18w) = 2.34w = 12.5w 𝑈1 = 11.33(73.67w) = 834.67w 𝑈1 = 1 2 (17)[ 2 3 (176.58 – 49.05)] 𝑈2 =
8.5(56.67w) = 481.70w = 73.67w 𝐹1 = 6(162w) = 972w 𝑈2 = 2 3 (17)(49.05) 𝐹2 = 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 = 𝐹1 + 𝑈1 + 𝑈2 =
(972+834.68+481.7)w = 2288.38w
36 x = ∑𝑅.𝑀− ∑𝑂.𝑀 𝑅𝑣 = 5024 .46−2288.38 𝑤 427.24𝑤 x = 6.40 m (from toe) b.) F.S. vs. Overturning = ∑𝑅.𝑀 ∑𝑂.𝑀
= 2024.46𝑤 2288.38𝑤 = 2.20 c.) F.S vs. Sliding = µ𝑅𝑣 𝑅𝑕 = 0.6(427.24) 149.5𝑤 = 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 : 𝐹𝑏=W d=25cm. (wV)displaced mercury = (wV)cone r=12.5cm=0.125 (9.81)(13.60)Vm =
9.81(6.95)Vcone Vcone=4.9087x10−3 13.60Vm = 6.95𝜋𝑟2𝑕 3 Vm = 6.95𝜋 0.1252 (0.30) 3(13.60) Vm =
2.50x10−3𝑚3 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? Sol’n: a.) b.) FB = 9.81 (4/3 πr^3)W_hy = 11,120 – 9.81(13.6)(4/3)(π)(〖0.3〗



^2) = 9.81(4/3)(〖0.3〗^2) = -3976 N FB = 1.11 kN W = 11.12 – 1110 = 10.01 N 𝑉𝑐𝑜𝑛𝑒 𝑉𝑚 = ( 0.30 𝑥 )3 𝑥3 =
(0.30)3𝑉𝑚 𝑉𝑐𝑜𝑛𝑒 x = 0.30 3(2.50𝑥10−3) 4.91 𝑥 10−3 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=FbW 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: 𝐹𝑏=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: 𝑉𝑝 𝑉𝑑 = ( 𝑎 0.45 )3 𝑉𝑝 1.20𝑉𝑑 = 𝑎3 0.453 1 1.20 = = 𝑎3 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? Sol’n:
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 = ρ_agV - ρ_hgV [W = V(ρ_ag - ρ_hg)]1/(〖(ρ〗_a g - ρ_h g)) V = w/(g(ρ_a-ρ_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
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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 = 4/3π^3 Wa =
4/3π(0.53)^3 = 0.52m³ A.) W=fb =WsLVs/2 Ws=9.81Kn/m 3(0.80)(0.52m3)/2 Ws=2.05KN B.) Wa=Fba+Fbs–W
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=2.04kn4.08kn 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=WvƩfv=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=δhA X=1.28m
=9.81(2.5)(5) F=122.625 e=b/2-x Rx=122.625d S=Ry/b(1±6e/b) Ry=296.604d S=142.36992kPa RM=98.868(30.3)+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)/1.030.92 Vt=6554.55m 3
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=0.1575m 2 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=0.02m 3 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 𝑥 = −𝑏 ± 𝑏2 − 4𝑎𝑐 2𝑎 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 2.5cosθWt=(5-0.5x)cosθFb 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: 𝐹𝐵𝐻 = 9.81 4𝜋 (0.603) 3 𝐹𝐵𝐻 = (9.81)( 1 2 )( 4𝜋(0.60)3 3 ) 𝐹𝐵𝐻 = 8.8759 kN𝐹𝐵𝐻 = 4.4379 kN T = 𝑊𝑆𝐻 𝐹𝐵𝐻 T = Wss - 𝐹𝐵 T =12 kN - 8.8759 kN T = 4 kN - 4.4379 kN T = 3.12 kN T = 0.4379 kN ∑Fy=0] 𝐹𝐵𝑆 = 𝑊𝑆𝑆 - T
9.81Vss = 4 + 3.12 Vss = 0.72579𝑚3
49 𝑊𝐻 = T + 𝐹𝐵𝐻 𝑊𝐻 = 0.44 kN + 8.88 kN Vs = 𝜋 3 𝐷2(3𝑟 − 𝐷)𝑊𝑆𝐻 = 9.32 kN 0.72579 = 0.60π𝐷2 - 0.33π𝐷3 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: 𝐹𝐵 =
𝑊𝐵 (𝑤𝑉)𝑆𝐿 = (𝑤𝑉)𝑆𝐵 (9.81)(1.20)𝑉𝑆𝐿 = (9.81)(0.80)𝑉𝑇 (1.20)𝑉𝑆𝐿 = (0.80)𝑉𝑇 𝑉𝑆𝐿 = 2 3 𝑉𝑇 2 3 of 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
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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: 𝑉𝑃 = 0.04m (0.20m) (1m) 𝑉𝑃 = 8x10 −3𝑚3𝑉2 = 𝑉1= 4x10
−3𝑚3 Θ = 𝑡𝑎𝑛−1 ( 0.04 0.20 ) Θ = 11.31𝑜 𝑊𝑃 + T = 𝐹𝐵 𝑊𝑃= 𝐹𝐵- T 𝑊𝑃 = 9810(4x10 −3) – 15 𝑊𝑃 = 39.24 - 15 𝑊𝑃 =
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 TanΦ=0.24/4 C=Wx Φ=3.43˚
490(3)=39140x SinΦ=X/MG T = 𝐹 𝑙 15 𝑁 𝑚 = 𝐹 1𝑚 F = 15 X=MGSinΦ 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 tanθ = 1.21/4.575 = 1.53 m θ = 14.81° MG = MB_o - GB_o
MB_o = B^2/12D(1 + 〖tan〗^2θ/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(MGsinθ) =
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=(π(6^2))/(12(4)) ; V=(π(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 TanΦ=1.85/4.575 V=1/2(15.25m)(1.85m)(4.573)
Φ=222.02˚ V=64.54m3 Mbo=VL/V’SinΦ Gbo=2.30-1.375m =64.54m3(6.1)/383.73m3(Sin22.02˚) Gbo=0.925m
Mbo=2.74m MG=Mbo-Gbo RM=w(MGSinΦ) =2.74m-0.925m =(3764.39kn)(1.815)sinΦ22.02˚ 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 𝐹 = 𝑚𝑎 = 𝑊𝑎 𝑔 a = 2 dg L 𝐹 = 105.95 2.45 9.81 a = 2 0.75 9.81 6 𝐹 =
26.46 𝑘𝑁 𝑎 = 2.45 𝑚 𝑠2
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: 𝐹 = 𝑤𝑕𝐴 1 + 𝑎 𝑔
𝐹 = 22 1.20 𝜋. 602 4 (1 + 9.81 9.81 ) 𝐹 = 11.20 𝑘𝑁
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: θ tanθ= macos 20° mg −masin 20° = m(acos 20°) m(g−asin 20°) = acos 20° g−asin 20° Consider: tanα
= 0.9 1.5 a= 9.81 tan(30.96°) =30.96° a=5.885 m/s2 tanα = a g tan(30.96°) = 𝑎 𝑔 tan 𝜃 = 5.885 cos 20° 9.81− 5.885
sin 20° = 5.53 7.79 = 35.35° 3m θ 0.9 0.6 1.5 1.5 1.5 20° W REFн REFv 20° W REFн REFv 20° W REFн REFv a
20° a av aн W=mg θ REFн = maн =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:
tanθ = 0.15 4.575 = 1.878° tanθ = 𝑎 𝑔 tan(1.878°)= 𝑎 9.81 a = 9.81 tan(1.878°) a= 0.322 m/s2 a= 𝑉𝑓−𝑉𝑜 𝑡 t= 13.33
0.322 =41.44 s θ 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: tanθ = 𝑎 g = 1 4 𝑔 2 = 1 4
tanθ=_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: 𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 𝑤𝑎𝑕 𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 𝑤𝑕𝐴(1 + 𝑎 𝑔 ) 𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 9.81 1.3 1.5 (0.65)𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 =
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9.81 1.3 1.5 (0.65)(1 + 3 9.81 ) 𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 12.43 𝑘𝑁𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 16.23 𝐾𝑁 𝐹𝑟𝑖𝑔𝑡 𝑕 𝑒𝑛𝑑 = 𝑤𝑎𝑕 𝐹𝑟𝑖𝑔𝑡 𝑕 𝑒𝑛𝑑 =
𝑤𝑕𝐴(1 + 𝑎 𝑔 ) 𝐹𝑟𝑖𝑔𝑡 𝑕 𝑒𝑛𝑑 = 9.81 1.5 1.5 (0.65)𝐹𝑟𝑖𝑔𝑡 𝑕 𝑒𝑛𝑑 = 9.81 1.5 1.5 (0.65)(1 + 3 9.81 ) 𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 14.35
𝑘𝑁𝐹𝑟𝑖𝑔𝑡 𝑕 𝑒𝑛𝑑 = 18.74 𝑘𝑁 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 𝑤𝑎𝑕 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 𝑤𝑕𝐴(1 + 𝑎 𝑔 ) 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 9.81 1.5 1.3 1.3 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 9.81
1.5 1.3 (1.3) (1 + 3 9.81 ) 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 57.40 𝑘𝑁 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 74.95 𝑘𝑁
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 tanθ = 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 θ 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 2x1 2 2g =(20/3л)2(0.4)2 2(9.81) =3.577 m
y2== w 2x2 2 2g =(20/3л)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 л 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. 𝑦 = 𝑤2𝑥2 2𝑔 1.2 = 𝑤2 0.3 2 2 9.81 𝑤 = 16.17 𝑟𝑎𝑑 𝑠 𝑃 = 𝑤𝑕 𝑃 = 9810( 𝑤2𝑟2 2𝑔 ) 𝑃 =
9810( 16.172 × 0. 32 2 × 9.81 ) 𝑃 = 11.80 𝑘𝑃𝑎 𝑦 = 𝑤2𝑥2 2𝑔 0.4 = 𝑤2 0.3 2 2 9.81 𝑤 = 9.34 𝑟𝑎𝑑 𝑠 𝑑 = 𝐻 2 + 𝑦 2 𝑑 =
0.6 + 0.2 𝑑 = 0.8 𝑚
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 + 𝑑 = 𝜔2𝑥2 2𝑔
eq.1 𝑑 = 𝜔2 𝑥 2 2 2𝑔 eq.2 By equating eq 1 and eq 2 𝜔 = 18.68 𝑟𝑎𝑑 𝑠 𝑑 = 0.40 𝑚 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: 𝑦 𝑧 = 𝑎 𝑞 𝑦 0.30 = 1.22 9.81 𝑦 = 0.04𝑚 B A H G C D 30 cm 30 cm 30 cm y h1 h2 H3 𝑃𝐵 =
45.36 𝑘𝑃𝑎 𝑃𝑔 = 𝑤𝑕2 𝑃𝑔 = 40.02 𝑘𝑃𝑎 𝑃𝑑 = 𝑤𝑕3 𝑃𝑑 = 37.40 𝑘𝑃𝑎 𝑃𝐵 = 9.81 13.6 (. 34) 𝑃𝑔 = 9.81 13.6 (. 30) 𝑃𝑑 =
9.81 13.6 (. 26)
63 𝑃𝐵 = 𝑤𝑕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 = 𝑉2
2𝑔 = 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 = 𝑤1 2𝑥 1 2 2𝑔 ; in LL1 w= 𝑦 12𝑔 𝑥 2 = 0.30 2
(9.81) 0.302 w1=8.09 rad/s in LL1; h= Patm 𝑤 y2=0.30 +0.759 w2 = 𝑦 12𝑔 𝑥 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= 𝑤2𝑥 2 2𝑔 = 8 2 0.32 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 = w1 2x1 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 2𝑔 +0.3-z = 8.092𝑧 2 2 (9.81) +0.3 – z = 3.336z2 –z +0.3 P = wh ( 𝑑𝑃 𝑑𝑥 ) = 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 = 𝐹 𝑐 𝑤 = 𝑊 𝑔 𝑤2𝑥 𝑊 tan 45 = 𝑤2 𝑔 𝑥 𝑤 = 𝑔𝑡𝑎𝑛 45 𝑥 𝑤 = 9.81 tan 45 0.075 𝑤 = 11.44 𝑟𝑎𝑑
𝑠
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 = 5π rad s Solution: Y1 = 𝑤2𝑥 2 2𝑔 X = 𝑦 2𝑔 𝑤2 = 0.3 2 (9.81) (5𝜋 )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
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in it? Given : Find: Vr w = 60 rpm = 2π rad s Solution: y = 𝑤2𝑥 2 2𝑔 = (2𝜋 )2(0.3)2 2(9.81) = 0.1811 m Vr = Vcone –
Vpar = 1 3 𝜋 (𝑟 )3𝑕 − 1 2 𝜋𝑟 2𝑦 = 1 3 𝜋 0.3 3(1) − 1 2 𝜋 0.3 2(0.1811) Vr = 0.060 m 3 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 = 𝑔 𝑟 = 9.81 0.925 w = 3.26 𝑟𝑎𝑑 𝑠 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 = (π)2(1)2 2(9.81) = 0.503 m Vspill = 1 2 πr2y = 1 2 π 1 2(0.503) Vspill = 0.79 m 3 y 1.0 W = 30 RPM =
π 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 = (2π)2(0.65)2 2(9.81) = 0.850 m Y2 = w2x2 2 2g = (2π)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 = 𝑝 𝑤 = 140 9.81 = 14.271 m y = w2x2 2g =
(8π)2(0.3)2 2(9.81) = 2.897 m a.) Fside = Awh = 𝜋 2 0.6 9.81 14.271 + 2.897 + 1 = 671.90 KN b.) Ftop = Awh =𝜋
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 = 𝑇 𝑡 = 56.41 2.5𝑥 10−3 =22565.34 Kpa y 0.6 w 2m Ftop Fside = 240 rpm = 8π rad 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 = 𝑝 𝑤 = 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 = 9𝑤2
218 - 𝑤2 218 703.3639 = 4𝑤2 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 2 πr2𝑕 0.170 = 1 2 π(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 = 0.170m 3
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? 2𝑇 = 𝑃𝐷 𝑇 = 𝑃2 2 (1) 𝑇 = 𝑃 𝑇 = 𝑆
𝑡 𝐴 𝑡 𝑇 = 345 1000 (6.25 × 10−3) 𝑇 = 2156.25 𝑘𝑁 𝑇 = 𝑃 𝑃 = 𝑦 9.81 + wh 2156.25 = 𝑦 9.81 + 9.81 (3) 𝑦 = 216.80 𝑚 𝑦
= 𝑤2𝑟 2 2𝑔 216.80 = 𝑤2𝑥 12 2(9.81) 𝑤 = 65.22 𝑟𝑎𝑑 𝑠
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 θ = w2x2 2g w = gtanθ 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 𝜋d2 = Q V = 1.80m/s 4 V d2 = 4Q
𝜋V 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 = V 2 2 + 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
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2 is 5% of the velocity head in jet. Given V2+P1 + Z1 = V 2 2 + P2 + Z2 + HL 2g w 2g w P1 =V2 2-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 = V 2 b + Pb + Zb + HL 2g w 2g w Vb 2=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 = V 2 c + Pc + Zc + HL 2g w2g w VC 2 = ZA- HL= 1.5 2g V2+Pa + Za = V 2 b + 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/𝜋(.5)2 V=2.4758m/s V2+Pa + Za =
V 2 b + 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,= 𝜋(0.01)v2 = 4V2 𝜋(0.075)2 225 V2+P1 + Z1 = V 2 2 +
P2 + Z2 + HL+ HL+ HL 2g w 2g w 4V2/225 + 550 = V2 +12 + 3 + 2 +12 + 0.06V2 2 19.62 9.81 19.62 19.62 V2 2=
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 V 2 +Pa + Za + ha = V 2 b + Pb + Zb V = 1.4/ 𝜋(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 = V 2 2 + P2 + Z2 + HL 2g w 2g w 0+ 480/9.81 + 46 – 50[8Q2/𝜋2g(0.2)2]-3[8Q2/𝜋2g(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 = V 2 2 + P2 + Z2 V = Q/A =4Q/𝜋D 4 2g w 2g
w V2 = 16Q2/ 𝜋2D4 16Q2/ 𝜋2D4 + P1 =16Q 2/ 𝜋2D4+ P2 2g w 2g w P1 – P2 = 1.38Q 2 ______________(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 = V 2 2 + P2 + Z2 + HL 2g w 2g w
V2 2-V1 2 =P1-P2 + Z1-Z2 2g w V2 = 6.29m/s V1= 0.3 2/(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 = V 2 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 = 0.06 2/.0252v
V2 = 28.73m/s Q2 = A2V2 Q2 =28.73𝜋(.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/ 𝜋(0.15) 2 = 14.147Q` BEE from 1 to 2 VN = Q/ 𝜋(0.05) 2 =127.324Q V2+P1 + Z1 = V 2 2 +
P2 + Z2 H2 = P2/w =0.2(13.6) =-2.72 2g w 2g w V, = Q/A = .218/𝜋(.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 = V 2 n + 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 = V 2
2 + P2 + Z2 Q = AV 2g w 2g w Q = 𝜋(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