# Solution Manual for Fundamentals of Hydraulic Engineering Systems, 5th Edition

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SOLUTIONS MANUAL
Contents
1
FUNDAMENTAL PROPERTIES OF WATER
1
2
WATER PRESSURE AND PRESSURE FORCES
7
3
WATER FLOW IN PIPES
19
4
PIPELINES AND PIPE NETWORKS
31
5
WATER PUMPS
68
6
WATER FLOW IN OPEN CHANNELS
83
7
GROUNDWATER HYDRAULICS
97
8
HYDRAULIC STRUCTURES
109
9
WATER PRESSURE, VELOCITY, AND
DISCHARGE MEASUREMENTS
120
10 HYDRAULIC SIMILITUDE AND MODEL STUDIES
126
11 HYDROLOGY FOR HYDRAULIC DESIGN
135
12 STATISTICAL METHODS IN HYDROLOGY
162
Chapter 1 โ Problem Solutions
1.2.1
1.2.3
E1 = energy released in lowering steam temperature to
100๏ฐC from 110ยฐC
E1 = energy needed to vaporize the water
E1 = (500 L)(1000 g/L)(10๏ฐC)(0.432 cal/gโ๏ฐC)
E1 = 2.16×10 cal
E1 = (300 L)(1000 g/L)(597 cal/g)
E1 = 1.79×108 cal
6
E2 = energy released when the steam liquefies
E2 = (500 L)(1000 g/L)(597 cal/g)
E2 = 2.99×108 cal
E3 = energy released when the water temperature is
lowered from 100ยฐC to 50๏ฐC
E3 = (500 L)(1000 g/L)(50๏ฐC)(1 cal/gโ๏ฐC)
E3 = 2.50×10 cal;
The energy remaining (E2) is:
E2 = ETotal โ E1
E2 = 2.00×108 cal โ 1.79×108 cal
E2 = 2.10×107 cal
The temperature change possible with the remaining
energy is:
2.10×107 cal = (300 L)(1000 g/L)(1 cal/gโ๏ฐC)(๏T)
7
Thus, the total energy released is:
Etotal = E1 + E2 + E3 = 3.26×108 cal
_________________________________________
1.2.2
First, convert kPa pressure into atmospheres:
84.6 kPa(1 atm/101.4 kPa) = 0.834 atm
From Table 1.1, the boiling temperature is 95ยฐC
E1 = energy required to bring the water temperature to
95๏ฐC from 15ยฐC
๏T = 70๏ฐC, making the temperature
T = 90๏ฐC when it evaporates.
Therefore, based on Table 1.1,
๏P = 0.692 atm
_________________________________________
1.2.4
E1 = energy required to warm and then melt the ice
E1 = (10 g)(6๏ฐC)(0.465 cal/gโ๏ฐC) + 10g(79.7 cal/g)
E1 = 825 cal. This energy is taken from the water.
E1 = (900 g)(95๏ฐC – 15๏ฐC)(1 cal/gโ๏ฐC)
The resulting temperature of the water will decrease to:
E1 = 7.20×104 cal
825 cal = (0.165 L)(1000 g/L)(20๏ฐC – T1)(1 cal/gโ๏ฐC)
E2 = energy required to vaporize the water
T1 = 15.0๏ฐC. Now we have a mixture of water at 0ยฐC
(formerly ice) and the original 165 liters that is now at
15.0๏ฐ C. The temperature will come to equilibrium at:
E2 = (900 g)(597 cal/g)
E2 = 5.37×105 cal
[(0.165 L)(1000 g/L)(15.0๏ฐC – T2)(1 cal/gโ๏ฐC)] =
Etotal = E1 + E2 = 6.09×10 cal
5
[(10 g)(T2 – 0๏ฐC)( 1 cal/gโ๏ฐC)] ; T2 = 14.1๏ฐC
1
1.2.5
1.3.1
E1 = energy required to melt ice
F = mโa; Letting a = g yields: W = mโg, (Eqโn 1.1)
E1 = (5 slugs)(32.2 lbm/slug)(32๏ฐF – 20๏ฐF)*
Then dividing both sides of the equation by volume,
(0.46 BTU/lbmโ๏ฐF) + (5 slugs)(32.2 lbm/slug)*
(144 BTU/lbm)
E1 = 2.41 x 104 BTU. ๏จEnergy taken from the water.
The resulting temperature of the water will decrease to:
2.41 x 10 BTU = (10 slugs)(32.2 lbm/slug)(120๏ฐF โ
4
T1)(1 BTU/lbmโ๏ฐF)
W/Vol = (m/Vol)โg; ฮณ = ฯโg
___________________________________________
1.3.2
SGoil = 0.976 = ฮณoil/ฮณ; where ฮณ is for water at 4ยฐC:
ฮณ = 9,810 N/m3 (Table 1.2). Substituting yields,
0.977 = ฮณoil/9,810; ฮณoil = (9810)(0.976) = 9,570 N/m3
T1 = 45.2๏ฐF
Also, ฮณ = ฯโg; or ฯ = ฮณoil/g
The energy lost by the water (to lower its temp. to
Substituting (noting that N โก kgยทm/sec2) yields,
45.2๏ฐF) is that required to melt the ice. Now you have
5 slugs of water at 32๏ฐF and 10 slugs at 45.2๏ฐF.
ฯoil = ฮณoil/g = (9,570 N/m3) / (9.81 m/sec2) = 976 kg/m3
___________________________________________
Therefore, the final temperature of the water is:
1.3.3
[(10 slugs)(32.2 lbm/slug)(45.2๏ฐF โ T2)(1 BTU/lbmโ๏ฐF)]
By definition, ๏ง = W/Vol = 55.5 lb/ft3; thus,
= [(5 slugs)(32.2 lbm/slug)(T2 – 32๏ฐF)(1 BTU/lbmโ๏ฐF)]
W = ๏งโVol = (55.5 lb/ft3)(20 ft3) = 1,110 lb (4,940 N)
T2 = 40.8๏ฐF
_________________________________________
๏ฒ = ๏ง/g = (55.5 lb/ft3)/(32.2 ft/s2) = 1.72 slug/ft3 (887 kg/m3)
1.2.6
E1 = energy required to raise the temperature to 100๏ฐC
E1 = (7500 g)(100๏ฐC โ 20๏ฐC)(1 cal/gโ๏ฐC)
E2 = 6.00×105 cal
E2 = energy required to vaporize 2.5 kg of water
E2 = (2500 g)(597 cal/g)
SG = ๏งliquid/๏งwater at 4๏ฐC = (55.5 lb/ft3)/(62.4 lb/ft3) = 0.889
___________________________________________
1.3.4
The mass of liquid can be found using
ฯ = ฮณ/g and ฮณ = weight/volume, thus
ฮณ = (47000 N โ 1500 N)/(5 m3) = 9.10 x 103 N/m3
ฯ = ฮณ/g = (9.1 x 103 N/m3)/(9.81 m/sec2);
๏ฒ = 928 kg/m3 (Note: 1 N โก 1 kgยทm/sec2)
E2 = 1.49×106 cal
Etotal = E1 + E2 = 2.09×106 cal
Time required = (2.09×10 cal)/(500 cal/s)
6
Time required = 4180 sec = 69.7 min
Specific gravity (SG) = ฮณ/ฮณwater at 4ยฐC
SG = (9.10 x 103 N/m3)/9.81 x 103 N/m3)
SG = 0.928
2
1.3.5
1.4.1
The force exerted on the tank bottom is equal to the
(a) Note that: 1 poise = 0.1 N๏sec/m2. Therefore,
weight of the water body (Eqโn 1.2).
1 lbยทsec/ft2 [(1 N)/(0.2248 lb)]ยท[(3.281 ft)2/(1 m)2] =
F = W = mg = [ฯ(Vol)] (g); ฯ found in Table 1.2
47.9 N๏sec/m2 [(1 poise)/(0.1 N๏sec/m2)] = 478.9 poise
920 lbs = [1.94 slugs/ft3 (ฯ โ(1.25 ft)2 โ d)] (32.2 ft/sec2)
Conversion: 1 lb๏sec/ft2 = 478.9 poise
d = 3.00 ft
(Note: 1 slug = 1 lbโsec2/ft)
___________________________________________
(b) Note that: 1 stoke = 1 cm2/sec. Therefore,
1 ft2/sec [(12 in)2/(1 ft)2]ยท [(1 cm)2/(0.3937 in)2] =
929.0 cm2/sec [(1 stoke)/(1 cm2/sec)] = 929.0 stokes
1.3.6
Weight of water on earth = 8.83 kN
Conversion: 1 ft2/sec = 929.0 stokes
_________________________________________
From Eโqn (1.1): m = W/g = (8,830 N)/(9.81 m/s2)
1.4.2
m = 900 kg
[๏ญ(air)/๏ญ(H2O)]0๏ฐC = (1.717×10-5)/(1.781×10-3)
Note: mass on moon is the same as mass on earth
[๏ญ(air)/๏ญ(H2O)]0๏ฐC = 9.641×10-3
W (moon) = mg = (900 kg)[(9.81 m/s2)/(6)]
[๏ญ(air)/๏ญ(H2O)]100๏ฐC = (2.174×10-5)/(0.282×10-3)
W(moon) = 1,470 N
_________________________________________
[๏ญ(air)/๏ญ(H2O)]100๏ฐC = 7.709×10-2
[๏ฎ(air)/๏ฎ(H2O)]0๏ฐC = (1.329×10-5)/(1.785×10-6)
1.3.7
Density is expressed as ฯ = m/Vol, and even though
[๏ฎ(air)/๏ฎ(H2O)]0๏ฐC = 7.445
volume changes with temperature, mass does not.
[๏ฎ(air)/๏ฎ(H2O)]100๏ฐC = (2.302×10-5)/(0.294×10-6)
Thus, (ฯ1)(Vol1) = (ฯ2)(Vol2) = constant; or
[๏ฎ(air)/๏ฎ(H2O)]100๏ฐC = 78.30
Vol2 = (ฯ1)(Vol1)/(ฯ2)
Note: The ratio of the viscosity of air to water increases
Vol2 = (999 kg/m3)(100 m3)/(996 kg/m3)
with temperature. Why? Because the viscosity of air
Vol2 = 100.3 m3 (or a 0.3% change in volume)
____________________________________________
increases with temperature and that of water decreases
with temperature magnifying the effect. Also, the
values of kinematic viscosity (๏ฎ) for air and water are
1.3.8
(1 N๏m)[(3.281 ft)/(1 m)][(0.2248 lb)/(1 N))]
= 7.376 x 10-1 ft๏lb
_________________________________________
1.3.9
(1 N/m ) [(1 m)/(3.281 ft)] [(1 ft)/(12 in)] ยท
2
2
[(1 lb)/(4.448 N)] = 1.450 x 10-4 psi
2
much closer than those of absolute viscosity. Why?
_________________________________________
1.4.3
๏ญ20๏ฐC = 1.002×10-3 N๏sec/m2; ๏ฎ20๏ฐC = 1.003×10-6 m2/s
(1.002×10-3 N๏sec/m2)โ[(0.2248 lb)/(1 N)]โ
[(1 m)2/(3.281 ft)2] = 2.092×10-5 lb๏sec/ft2
(1.003×10-6 m2/s)[(3.281 ft)2/(1 m)2] = 1.080×10-5 ft2/s
3
1.4.4
1.4.8
Using Newtonโs law of viscosity (Eqโn 1.2):
v = y2 โ 3y, where y is in inches and v is in ft/s
๏ด = ๏ญ(dv/dy) = ๏ญ(ฮv/ฮy)
v = 144y2 โ 36y, where y is in ft and ๏ฎ is in ft/s
๏ด = (1.00 x 10-3 N๏sec/m2)[{(4.8 โ 2.4)m/sec}/(0.02 m)]
Taking the first derivative: dv/dy = 288y โ 36 sec-1
๏ด = 0.12 N/m2
_________________________________________
๏ด = ๏ญ(dv/dy) = (8.35 x 10-3 lb-sec/ft2)( 288y โ 36 sec-1)
1.4.5
From Eqโn (1.2): ๏ด = ๏ญ(๏v/๏y) =
Solutions: y = 0 ft, ๏ด = -0.301lb/ft2
y = 1/12 ft, ๏ด = -0.100 lb/ft2; y = 1/6 ft, ๏ด = 0.100 lb/ft2
๏ด = (0.0065 lb๏sec/ft )[(1.5 ft/s)/(0.25/12 ft)]
y = 1/4 ft, ๏ด = 0.301 lb/ft2; y = 1/3 ft, ๏ด = 0.501 lb/ft2
_________________________________________
๏ด = 0.468 lb/ft2
1.4.9
F = (๏ด)(A) = (2 sides)(0.468 lb/ft2)[(0.5 ft)(1.5 ft)]
๏ญ = (16)(1.00×10-3 N๏sec/m2) = 1.60×10-2 N๏sec/m2
F = 0.702 lb
_________________________________________
Torque = (r )dF ๏ฝ r ๏ ๏ด ๏ dA ๏ฝ (r )(๏ญ )( ๏v )dA
๏ฒ
๏ฒ
๏ฒ
2
R
R
R
0
0
0
๏y
R
1.4.6
Summing forces parallel to the incline yields:
Tshear force = W(sin15๏ฐ) = ๏ด๏A = ๏ญ(๏v/๏y)A
๏y = [(๏ญ)(๏v)(A)] / [(W)(sin15๏ฐ)]
๏y = [(1.52 N๏sec/m2)(0.025 m/sec)(0.80m)(0.90m)]/
[(100 N)(sin15๏ฐ)]
๏y = 1.06 x 10-3 m = 1.06 mm
___________________________________________
1.4.7
Torque = (r )(๏ญ )( (๏ท )(r ) ๏ญ 0 )(2๏ฐr )dr
๏ฒ0
๏y
R
Torque = (2๏ฐ )(๏ญ )(๏ท ) (r 3 )dr
๏ฒ
๏y
0
๏ญ2
2
4
Torque = (2๏ฐ )(1.60 ๏10 N ๏ sec/ m )(0.65rad / sec) ๏ฉ (1m) ๏น
0.0005m
๏ช 4 ๏บ
๏ซ
๏ป
Torque = 32.7 N๏m
_________________________________________
1.4.10
๏ญ = ๏ด/(dv/dy) = (F/A)/(๏v/๏y);
Torque (T) = Forceโdistance = FโR where R = radius
Using Newtonโs law of viscosity (Eqโn 1.2):
๏ด = ๏ญ(dv/dy) = ๏ญ(๏v/๏y)
๏ด = (0.04 N๏sec/m )[(15 cm/s)/[(25.015 โ 25)cm/2]
2
๏ด = 80 N/m2
Fshear resistance = ๏ด๏A = (80 N/m2)[(๏ฐ)(0.25 m)(3 m)]
Fshear resistance = 188 N
Thus; ๏ญ = (T/R)/[(A)(๏v/๏y)]
๏ญ=
T/R
T ๏ ๏y
๏ฝ
(2๏ฐ )( R)(h)(๏ท ๏ R / ๏y ) (2๏ฐ )( R 3 )(h)(๏ท )
๏ญ=
(1.10lb ๏ ft )[(0.008 / 12) ft ]
๏ฆ 2๏ฐrad / sec ๏ถ
๏ท๏ท
(2๏ฐ )((1 / 12) ft )3 ((1.6 / 12) ft )(2000 rpm )๏ง๏ง
๏จ 60rpm ๏ธ
๏ญ = 7.22×10-3 lb๏sec/ft2
4
1.5.1
1.5.4
The concept of a line force is logical for two reasons:
Condition 1: h1 = [(4)(๏ณ1)(sin๏ฑ1)] / [(๏ง)(D)]
1) The surface tension acts along the perimeter of the
tube pulling the column of water upwards due to
h1 = [(4)(๏ณ1)(sin30๏ฐ)] / [(๏ง)(0.8 mm)]
adhesion between the water and the tube.
Condition 2: h2 = [(4)(๏ณ2)(sin๏ฑ2)] / [(๏ง)(D)]
2) The surface tension is multiplied by the tube
h2 = [(4)(0.88๏ณ1)(sin50๏ฐ)] / [(๏ง)(0.8 mm)]
perimeter, a length, to obtain the upward force used
in the force balance developed in Equation 1.3 for
capillary rise.
_________________________________________
1.5.2
To minimize the error (< 1 mm) due to capillary action,
apply Equation 1.3:
D = [(4)(๏ณ)(sin ๏ฑ)] / [(๏ง)(h)]
D = [4(0.57 N/m)(sin 50๏ฐ)]/[13.6(9790N/m3)(1.0×10-3 m)]
D = 0.0131 m = 1.31 cm
Note: 50ยฐ was used instead or 40ยฐ because it produces
the largest D. A 40ยฐ angle produces a smaller error.
_________________________________________
h2/h1 = [(0.88)(sin50ยฐ)] / (sin30๏ฐ) = 1.35
alternatively,
h2 = 1.35(h1), about a 35% increase!
_________________________________________
1.5.5
Capillary rise (measurement error) is found using
Equation 1.3: h = [(4)(๏ณ)(sin๏ฑ)] / [(๏ง)(D)]
where ฯ is from Table 1.4 and ฮณ from Table 1.2. Thus,
๏ณ = (6.90 x 10-2)(1.2) = 8.28 x 10-2 N/m and
๏ง = (9750)(1.03) = 1.00 x 104 N/m3
h =[(4)( 8.28 x 10-2 N/m)(sin 35)] /
[(1.00 x 104 N/m3)(0.012m)]
1.5.3
For capillary rise, apply Equation 1.3:
h = [(4)(๏ณ)(sin ฮธ)] / [(๏ง)(D)]
But sin 90ห = 1, ๏ง = 62.3 lb/ft3 (at 20หC), and
h = 1.58×10-3 m = 0.158 cm
_________________________________________
1.5.6
ฯ = 4.89×10-3 lb/ft (from inside book cover)
thus, D = [(4)(๏ณ)] / [(๏ง)(h)]; for h = 1.5 in.
R
๏P
D = [(4)(4.89×10-3 lb/ft)] / [(62.3 lb/ft3)(1.5/12)ft]
D = 2.51 x 10-3 ft = 3.01 x 10-2 in.; thus,
for h = 1.5 in., D = 2.51×10-3 ft = 0.0301 in.
๏P = Pi โ Pe (internal pressure minus external pressure)
for h = 1.0 in., D = 3.77×10-3 ft = 0.0452 in.
๏ฅFx = 0; ๏P(๏ฐ)(R2) – 2๏ฐ(R)(๏ณ) = 0
for h = 0.5 in., D = 7.54×10-3 ft = 0.0904 in.
๏P = 2๏ณ/R
5
1.6.1
1.6.4
Pi = 1 atm = 14.7 psi. and Pf = 220 psi
Pi = 30 N/cm2 = 300,000 N/m2 = 3 bar
From Equation (1.4): ๏Vol/Vol = -๏P/Eb
๏P = 3 bar โ 30 bar = -27 bar = -2.7×105 N/m2
๏Vol/Vol = -(14.7 psi โ 220 psi)/(3.2×105 psi)
Amount of water that enters pipe = ๏Vol
๏Vol/Vol = 6.42 x 10-4 = 0.0642% (volume decrease)
Volpipe= [(๏ฐ)(1.50 m)2/(4)]โ(2000 m) = 3530 m3
๏๏ฒ/๏ฒ = -๏Vol/Vol = -0.0642% (density increase)
_________________________________________
๏Vol = (-๏P/Eb)(Vol)
1.6.2
m = W/g = (7,490 lb)/(32.2 ft/s ) = 233 slug
2
๏ฒ = m/Vol = (233 slug)/(120 ft3) = 1.94 slug/ft3
๏Vol = (-๏P/Eb)(Vol)
๏Vol = [-(-2.7×105 N/m2)/(2.2×109 N/m2)]*(3530 m3)
๏Vol = 0.433 m3
Water in the pipe is compressed by this amount. Thus,
the volume of H2O that enters the pipe is 0.433m3
๏Vol = [-(1470 psi โ14.7 psi)/(3.20×105 psi)](120 ft3)
๏Vol = -0.546 ft3
๏ฒnew = (233 slug)/(120 ft3 โ 0.546 ft3) = 1.95 slug/ft3
Note: The mass does not change.
_________________________________________
1.6.3
Surface pressure: Ps = 1 atm = 1.014 x 105 N/m2
Bottom pressure: Pb = 1.61 x 107 N/m2
From Equation (1.4): ๏Vol/Vol = -๏P/Eb
๏Vol/Vol = [-(1.014 x 105 – 1.61 x 107)N/m2]
(2.2×109 N/m2)
๏Vol/Vol = 7.27 x 10-3 = 0.727% (volume decrease)
๏ฮณ/ฮณ = -๏Vol/Vol = -0.727% (specific wt. increase)
Specific weight at the surface: ฮณs = 9,810 N/m3
Specific weight at the bottom:
ฮณb = (9,810 N/m3)(1.00727) = 9,880 N/m3
Note: These answers assumes that Eb holds constant for
this great change in pressure.
6
Chapter 2 โ Problem Solutions
2.2.1
2.2.4
a) The depth of water is determined using the
relationship between weight and specific weight.
Since mercury has a specific gravity of 13.6, the water
W = ฮณยทVol = ฮณ(Aยทh); where h = water depth
14,700 lbs = (62.3 lb/ft )[ฯยท(5 ft) ](h); h = 3.00 ft
3
2
b) Pressure on the tank bottom based on weight:
P = F/A = W/A = 14,700/[ฯยท(5 ft)2] = 187 lbs/in.2
b) Pressure due to depth of water using Eqโn 2.4:
height can be found from: hwater = (hHg)(SGHg)
hwater = (30 mm)(13.6) = 408 mm = 40.8 cm of water
Also, absolute pressure is: Pabs = Pgage + Patm
Pabs = 0.408 m + 10.3 m = 10.7 m of water
Pabs = (10.7 m)(9790 N/m3) = 1.05 x 105 N/m2
_________________________________________
P = ฮณโh = (62.3 lb/ft3)(3 ft) = 187 lb/ft2
_________________________________________
2.2.5
2.2.2
P = ฮณโh = (62.3 lb/ft3)(10 ft) = 623 lb/ft2
The absolute pressure includes atmospheric pressure.
Fbottom = PโA = (623 lb/ft2)(100 ft2) = 6.23 x 104 lbs
Therefore, Pabs = Patm + (๏งwater)(h) ๏ฃ 5(Patm);
Pressure varies linearly with depth. So the average
pressure on the sides of the tank occur at half the depth.
Thus from Eqโn 2.4: h = 4(Patm)/๏งwater
Pavg = ฮณโh = (62.3 lb/ft3)(5 ft) = 312 lb/ft2 Now,
h = 4(1.014 x 10 N/m )/(1.03)(9790 N/m )
5
2
Since force equals pressure on the bottom times area:
3
h = 40.2 m (132 ft)
___________________________________________
Fside = Pavg โA = (312 lb/ft2)(100 ft2) = 3.12 x 104 lbs
___________________________________________
2.2.6
2.2.3
๏งwater at 30๏ฐC = 9.77 kN/m3 (from Table 1.2)
Pvapor at 30๏ฐC = 4.24 kN/m2 (from Table 1.1)
Patm = Pcolumn + Pvapor
Patm = (8.7 m)(9.77 kN/m3) + (4.24 kN/m2)
Patm = 85.0 kN/m2 + 4.24 kN/m2 = 89.2 kN/m2
The percentage error if the direct reading is used and
the vapor pressure is ignored is:
Error = (Patm – Pcolumn)/(Patm)
Error = (89.2 kN/m2 โ 85.0 kN/m2)/(89.2 kN/m2)
Error = 0.0471 = 4.71%
Pressure and force on the bottom of both containers is:
P = (๏งwater)(h) = (9790 N/m3)(10 m) = 97.9 kN/m2
Also, F = PโA = (97.9 kN/m2)(2 m)(2 m) = 391 kN
This may be confusing since the water weights are
different. To clarify the situation, draw a free body
diagram of the lower portion of the L-shaped container.
(Solution explanation is continued on the next page.)
7
2.2.9
(Solution 2.2.6 cont.)
๏งoil = (SG)(๏งwater) = (0.85)( 62.3 lb/ft3) = 53.0 lb/ft3
P10ft = Pair + (๏งoil)(12 ft) or Pair = P10ft – (๏งoil)(12 ft)
W1 P1 ยทA1
Pair = 25.7 psi (144 in2/ft2) โ (53.0 lb/ft3)(12 ft)
W2
Pair = 3.06 x 103 lb/ft2 (21.3 psi); Gage pressure
Note the three vertical forces will be opposed at the
bottom. W1 is the weight of the water column above it.
W2 is the weight of water in the lower portion of the
container. But there is also an upward pressure force
from the water on the underside of the L indent (P =
ฮณh). This is opposed by an equal and opposite pressure
on the water below. The resulting force is P1ยทA1 which
equals ฮณยทhยทA1 or ฮณยทVol. In other words, the force equals
the weight of the imaginary column of displaced water
above it. Hence, the force on the bottom opposing these
force components is equal to the force on the bottom of
the water column without the indent.
___________________________________________
2.2.7
Pabs = Pgage + Patm = 21.3 psi + 14.7 psi
Pabs = 36.0; Absolute pressure
___________________________________________
2.2.10
โFx = 0; (Px )(ฮy) โ (Ps)(ฮs)(sin ฮธ) = 0
since (ฮsโsin ฮธ) = ฮy, Ps = Px
โFy = 0; and accounting for weight yields:
(Py )(ฮx) โ (Ps)(ฮs)( cos ฮธ) โ (ฮxโฮy/2)(ฮz)(ฮณ) = 0
since (ฮsโcos ฮธ) = ฮx and (ฮy)(ฮx)(ฮz) ๏จ 0;
Ps = Py Hence, the pressure is the same in all
๏งseawater = (SG)(๏ง) = (1.03)(62.3 lb/ft ) = 64.2 lb/ft
3
3
Ptank = (๏งseawater)(๏h) = (64.2 lb/ft3)(18 ft)(1 ft2/144 in2)
Ptank = 8.03 psi = 5.54 x 104 N/m2 (Pascals)
_________________________________________
2.2.8
Pbottom = Pgage + (๏งliquid)(1.4 m); and
๏งliquid = (SG)(๏งwater) = (0.85)(9790 N/m3) = 8,320 N/m3
๏Pbottom = 3.55×104 N/m2 + (8320 N/m3)(1.4 m)
Pbottom = 4.71 x 104 N/m2
The pressure at the bottom of the liquid column can be
determined two different ways which must be equal.
(h)(๏งliquid) = Pgage + (๏งliquid)(1 m)
h = (Pgage)/(๏งliquid) + 1m
h = (3.55×104 N/m2)/8320 N/m3 + 1 m = 5.27 m
directions at the same depth, thus omnidirectional.
___________________________________________
2.2.11
The mechanical advantage in the lever increases the
input force delivered to the hydraulic jack. Thus,
Finput = (9)(50 N) = 450 N;
The pressure developed in the system is therefore:
Psystem = F/A = (450 N)/(25 cm2) = 18 N/cm2
Psystem = 180 kN/m2 = 180 kPa
From Pascalโs law, the pressure at the input piston
should equal the pressure at the two output pistons.
๏ The force exerted on each output piston is:
Pinput = Poutput equates to: 18 N/cm2 = Foutput/250 cm2
Foutput=(18 N/cm2)(250 cm2)(1 kN/1000 N) = 4.50 kN
8
2.4.1
2.4.5
The gage pressure at the bottom of the tank is equal
Note the equal pressure surface: P1 = P2 Therefore,
to the pressure due to the liquid heights. Thus,
(hHg)(SGHg)(๏ง) = (5h)(๏ง) + (h)(SGoil)(๏ง)
h = (hHg)(SGHg)/(5 + SGoil)
h = (1.43 in.)(13.6)/(5 + 0.80) = 3.35 inches
____________________________________________
PA + (y)(๏ง) = (h)(๏งHg)
PA + [(1.34/12)ft](๏ง) = [(1.02/12)ft](๏งHg)
PA = [(1.02/12)ft](13.6)(62.3 lb/ft3)
– [(1.34/12)ft](62.3 lb/ft3)
2.4.2
PA = 65.1 lb/ft2 = 0.452 psi
_________________________________________
Note the equal pressure surface; P7 = P8 or,
2.4.6
(hwater)(๏ง) = (hoil)(๏งoil) = (hoil)(๏ง)(SGoil), thus
hwater = (hoil)/(SGoil) = (61.5 cm)(0.85) = 52.3 cm
_________________________________________
2.4.3
The mercury-water meniscus (left leg) will be lower
than the mercury-oil meniscus based on the relative
amounts of each poured in and their specific gravity.
Also, a surface of equal pressure can be drawn at the
mercury-water meniscus. Therefore, Pleft = Pright
(1 m)(๏ง) = (h)(๏งHg) + (0.6 m)(๏งoil)
(1 m)(๏ง) = (h)(SGHg)(๏ง) + (0.6 m)(SGoil)(๏ง);
h = [1 m โ (0.6 m)(SGoil)]/(SGHg)
h = [1m โ (0.6 m)(0.79)]/(13.6) = 0.0387m = 3.87 cm
_________________________________________
2.4.4
Equal pressure surface at Hg-H2O interface: Thus,
(3 ft)(๏งHg) = P + (2 ft)(๏ง) where ๏งHg = (SGHg)(๏ง)
(3 ft)(13.6)(62.3 lb/ft3) = P + (2 ft)(62.3 lb/ft3);
A surface of equal pressure surface can be drawn at the
mercury-water meniscus. Therefore,
Ppipe + (h1)(๏ง) = (h2)(๏งHg) = (h2)(SGHg)(๏ง)
Ppipe + (0.18 m)(9790 N/m3) =
(0.60 m)(13.6)(9790 N/m3)
Ppipe = 7.81 x 104 N/m2 (Pascals) = 78.1 KPa
___________________________________________
2.4.7
Use the swim-through-technique: Start at the end of
the manometer open to the atmosphere (Pgage = 0). Then
โswim throughโ adding pressure when โswimmingโ
down and subtracting when โswimmingโ up until you
reach the pipe (Ppipe). Remember to jump across the
manometer at surfaces of equal pressure (ES). Thus,
0 – (0.66 m)(๏งml) + [(0.66 + y + 0.58)m](๏งair)
โ (0.58 m)(๏งoil) = Ppipe
The specific weight of air is negligible when compared
to fluids, so that term in the equation can be dropped.
Ppipe = 0 – (0.66 m)(SGml)(๏ง) โ (0.58 m)(SGoil)(๏ง)
P = 2,420 lb/ft2 = 16.8 psi
Ppipe = 0 – (0.66 m)(0.8)(9790 N/m3)
โ (0.58 m)(0.82)( 9790 N/m3)
Pressure can be expressed as a fluid height:
Ppipe = – 9.83 kN/m2(kPa) Pressure can be converted to
h = P/๏งHg = (2420 lb/ft2)/[(13.6)(62.3 lb/ft2)]
height (head) of any liquid through P = ๏งโh. Thus,
h = 2.86 ft of Hg (or 34.3 inches)
hpipe = (-9,830 N/m2)/(9790 N/m3) = -1.00 m of water
9
2.4.8
2.4.10 โ cont.
A surface of equal pressure surface can be drawn at the
mercury-water meniscus. Therefore,
When the manometer reading (h) rises or falls, volume
balance must be preserve in the system. Therefore,
P + (h1)(๏ง) = (h2)(๏งHg) = (h2)(SGHg)(๏ง)
Volres = Voltube
P + (0.5 ft)(62.3 lb/ft3) = (1.5 ft)(13.6)(62.3 lb/ft3)
ฮh = h (Atube/Ares) = h [(d2)2/(d1)2]; substituting yields
P = 1,240 lb/ft2 = 8.61 psi
P1 โ P2 = h [(d2)2/(d1)2] (ฯ1โg) + (h)(ฯ2โg) – (h)(ฯ1โg)
When the manometer reading rises or falls, volume
balance is preserved for constant density. Therefore,
P1 โ P2 = hโg [ฯ2 – ฯ1 + ฯ1 {(d2)2/(d1)2}]
Volres = Voltube
or Aresโh1 = Atubeโh2
or Aresโ(ฮh) = Atubeโ h
P1 โ P2 = hโg [ฯ2 – ฯ1 {1 – (d2)2/(d1)2}]
_________________________________________
h1 = h2 (Atube/Ares) = h2 [(Dtube)2/(Dres)2]
2.4.11
h1 = (4 in.)[(1 in.)2/(5 in.)2] = 0.16 in.
_________________________________________
Using the โswim throughโ technique, start at the right
tank where pressure is known and โswim throughโ the
tanks and pipes, adding pressure when โswimmingโ
down and subtracting when โswimmingโ up until you
reach the known pressure in the left tank. Remember to
jump across the pipe at the surfaces of equal pressure
(ES). Solve for EA in the resulting equation.
2.4.9
Using the โswim throughโ technique, start at pipe A and
โswim throughโ the manometer, adding pressure when
โswimmingโ down and subtracting when โswimmingโ
up until you reach pipe B. The computations are:
PA + (5.33 ft)(๏งoil) – (1.67ft)(๏งct) – (1.0 ft)(๏ง) = PB
PA – PB = (62.3 lb/ft3) [(1.0 ft) + (1.67 ft)(1.6) โ
(5.33)(0.82)]
PA – PB = -43.5 lb/ft2 = -0.302 psi
_________________________________________
2.4.10
Using the โswim throughโ technique, start at P 2 and
โswim throughโ the manometer, adding pressure when
โswimmingโ down and subtracting when โswimmingโ
up until you reach P1. The computations are:
P2 + (ฮh)(ฯ1โg) + (y)(ฯ1โg) + (h)(ฯ2โg) – (h)(ฯ1โg) (y)(ฯ1โg) = P1
where y is the vertical elevation difference between the
fluid surface in the left hand reservoir and the interface
between the two fluids on the right side of the U-tube.
Simplifying, P1 โ P2 = (ฮh)(ฯ1โg) + (h)(ฯ2โg) – (h)(ฯ1โg)
20 kN/m2 + (37 m – EA)(9.79 kN/m3) (35m – EA)(1.6)(9.79 kN/m3) – (5 m)(0.8)(9.79 kN/m3)
= -29.0 kN/m2
EA = 30 m
___________________________________________
2.4.12
Using the โswim throughโ technique, start at both ends
of the manometers which are open to the atmosphere
and thus equal to zero gage pressure. Then โswim
throughโ the manometer, adding pressure when
โswimmingโ down and subtracting when โswimmingโ
up until you reach the pipes in order to determine P A
and PB. With those pressures, you may solve for the
value of h. Remember to jump across the manometer at
surfaces of equal pressure (ES). The computations are:
0 + (23)(13.6)(๏ง) – (44)(๏ง) = PA; PA = 269โฮณ
0 + (46)(0.8)(๏ง) + (20)(13.6)(๏ง) โ (40)(ฮณ) = PB
PB = 269โฮณ; Therefore, PA = PB and h = 0
10
2.5.1
2.5.4
F ๏ฝ ๏ง ๏ h ๏ A = (62.3 lb/ft3)[4(2ft)/(3ฯ)]โ[(1/2)ฯ(2ft)2]
The hydrostatic force and its locations are:
F ๏ฝ ๏ง ๏ h ๏ A = (ฮณ)(h)[ฯ(D)2/4]
F = 332 lb (Note: h ๏ฝ y in Table 2.1)
๏
๏
๏
๏
I0
๏ฐ ( D) 4 / 64
๏ซy๏ฝ
๏ซh;
Ay
๏ฐ ( D ) 2 / 4 ( h)
๏
๏
I
9๏ฐ 2 ๏ญ 64 (2 ft ) 4 / 72๏ฐ
yP ๏ฝ 0 ๏ซ y ๏ฝ
๏ซ [4(2ft)/(3๏ฐ )]
Ay
[(1/2)๏ฐ (2ft) 2 ][4(2ft)/(3๏ฐ )]
yP ๏ฝ
yp = 0.853 ft (depth to center of pressure)
yp = D2/(16h) + h (depth to the center of pressure)
The center of pressure (0.853 ft) is deeper than the
centroid ( y ๏ฝ 0.848 ft ) of the gate.
Thus, summing moments: โ Mhinge = 0 ;
_________________________________________
P(D/2) โ F(yp โ h) = 0
P(D/2) โ {(ฮณ)(h)[ฯ(D)2/4][D2/(16h)]} = 0;
2.5.2
F ๏ฝ ๏ง ๏ h ๏ A = (9790 N/m3)[(9 m)/2]โ[(9 m)(1 m)]
P = (1/32)(ฮณ)(ฯ)(D)3
_________________________________________
F = 3.96 x 105 N per meter of length
2.5.5
yP ๏ฝ
๏
๏
The hydrostatic force and its locations are:
I0
(1m)(9m) 3 / 12
๏ซy๏ฝ
๏ซ 4.5 m
๏(9m)(1m)๏(4.5m)
Ay
yp = 6.0 m
F ๏ฝ ๏ง ๏ h ๏ A = (62.3 lb/ft3)(1.5 ft)(9 ft2)
(depth to the center of pressure)
In summing moments about the toe of the dam (โMA),
the weight acts to stabilize the dam (called a righting
moment) and the hydrostatic force tends to tip it over
(overturning moment).
โMA = (Wt.)[(2/3)(4.5 m)] โ (F)(3 m) =
[1/2 (3 m)(9 m)โ(1 m)](2.78)(9790 N/m )โ(3 m) โ
3
(3.96 x 105 N)โ(3 m) = -8.57 x 104 N-m
โMA = 8.57 x 104 N-m (overturning; dam is unsafe)
_________________________________________
2.5.3
The hydrostatic force and its locations are:
F ๏ฝ๏ง ๏h ๏ A
F = (9790 N/m3)[(5 m)(sin 45ห)](ฯ)(0.5 m)2
F = 2.72 x 104 N = 27.2 kN
๏
๏
I
๏ฐ (1m) 4 / 64
yP ๏ฝ 0 ๏ซ y ๏ฝ
๏ซ 5m ๏ฝ 5.01 m
Ay
๏ฐ (1m) 2 / 4 (5m)
๏
๏
F = 841 lb
yP ๏ญ y ๏ฝ
๏
๏
I0
(3 ft )(3 ft )3 / 12 ; yp โ ?
ฬฟ = 0.500 ft
๏ฝ
Ay
9 ft 2 (1.5 ft )
๏
๏
If the gate was submerged by 10 m (to top of the gate):
F ๏ฝ ๏ง ๏ h ๏ A = (62.3 lb/ft3)(11.5 ft)(9 ft2) = 6,450 lb
yP ๏ญ y ๏ฝ
๏
๏
I0
(3 ft )(3 ft )3 / 12 ; yp โ ?
ฬฟ = 0.0652 ft
๏ฝ
Ay
9 ft 2 (11.5 ft )
๏ ๏
The force increases tremendously as depth increases.
The distance between the centroid and the center of
pressure becomes negligible with increasing depth.
___________________________________________
2.5.6
The center of pressure represents the solution to both
parts of the question. Thus,
yP ๏ฝ
๏
๏
I0
(2 ft )(3 ft )3 / 12
๏ซy๏ฝ
๏ซ 6.5 ๏ฝ 6.62 ft
๏(2 ft )(3 ft )๏(6.5 ft )
Ay
11
2.5.7
2.5.9
Fleft ๏ฝ ๏ง ๏ h ๏ A = (9790 N/m3)(0.5 m)[(1.41m)(3m)]
F ๏ฝ ๏ง ๏ h ๏ A = (9790 N/m3)(2.5 m)[(ฯ){(1.5)2โ(0.5)2}m2]
Fleft = 20.7 kN (where A is โwetโ surface area)
๏
๏
F = 1.54 x 105 N = 154 kN
๏
๏
I0
๏ฐ / 64{(3m) 4 ๏ญ (1m) 4 }
๏ซy๏ฝ
๏ซ 2.5 m
Ay
๏ฐ {(1.5m) 2 ๏ญ (0.5m) 2 } (2.5m)
I
(3m)(1.41m) 3 / 12
yP ๏ฝ 0 ๏ซ y ๏ฝ
๏ซ 0.705m
๏(3m)(1.41m)๏(0.705m)
Ay
yP ๏ฝ
yp = 0.940 m (inclined distance to center of pressure)
yp = 2.75 m (below the water surface)
___________________________________________
Location of this force from the hinge (moment arm):
๏
๏
Yโ = 2 m โ 1.41 m + 0.940 m = 1.53 m
2.5.10
Fright ๏ฝ ๏ง ๏ h ๏ A = (9790 N/m3)(h/2 m)[(h/sin45ห)(3m)]
The total force from fluids A and B can be found as:
Fright = 20.8โh kN
2
๏
๏
FA ๏ฝ ๏ง ๏ h ๏ A = (ฮณA)(hA)โ[ฯ(d)2/4]
I
(3)(1.41 ๏ h) 3 / 12
yP ๏ฝ 0 ๏ซ y ๏ฝ
๏ซ 0.705 ๏ h
๏(3)(1.41๏ h)๏(0.705 ๏ h)
Ay
FB ๏ฝ ๏ง ๏ h ๏ A = (ฮณB)(hB)โ[ฯ(d)2/4]
yp = (0.940โh)m; Moment arm of force from hinge:
For equilibrium, forces are equal, opposite, & collinear.
Yโ = 2 m โ (h/sin 45)m + (0.940โh)m = 2m โ (0.474โh)m
FA = FB;
The force due to the gate weight: W = 20.0 kN
Moment arm of this force from hinge: X = 0.707 m
Summing moments about the hinge yields: โMhinge= 0
(20.8โh2)[2m โ (0.474โh)] โ 20.7(1.53) โ 32.3(0.707) = 0
h = 1.40 m (gate opens when depth exceeds 1.40 m)
_________________________________________
(ฮณA)(hA)โ[ฯ(d)2/4] = (ฮณB)(hB)โ[ฯ(d)2/4]
hA = [(ฮณB)/(ฮณA)](hB)
___________________________________________
2.5.11
F ๏ฝ ๏ง ๏ h ๏ A = (62.3 lb/ft3)[(20 ft)]โ[(10 ft)(6 ft)]
F = 7.48 x 104 lbs (Horizontal force on gate)
2.5.8
Logic dictates that the center of pressure should be at
the pivot point (i.e., the force at the bottom check block
is zero). As water rises above h = 2 feet, the center of
pressure will rise above the pivot point and open the
gate. Below h = 2 feet, the center of pressure will be
lower than the pivot point and the gate will remain
closed. For a unit width of gate, the center of pressure is
๏
๏
I
(1 ft )(8 ft ) 3 / 12
yP ๏ฝ 0 ๏ซ y ๏ฝ
๏ซ 6 ft
๏(1 ft )(8 ft )๏(6 ft )
Ay
yp = 6.89 ft (vertical distance from water surface to the
center of pressure)
Thus, the horizontal axis of rotation (0-0โ) should be
10 ft โ 6.89 ft = 3.11 ft above the bottom of the gate.
Summing vertical forces where T is the lifting force:
โ Fy = 0; T โ W โ F(Cfriction) = 0
T = 6040 + 74,800(0.2) = 21,000 lbs (lifting force)
___________________________________________
2.5.12
The hydrostatic force on the cover and its locations are:
F ๏ฝ ๏ง ๏ h ๏ A =(9790 N/m3)(1.5m)[(10m)(5m)] = 734 kN
yP ๏ฝ
๏
๏
I0
(10m)(5m)3 / 12
๏ซy๏ฝ
๏ซ 2.5m = 3.33 m
๏(10m)(5m)๏(2.5m)
Ay
โ Mhinge = 0; (734 kN)(3.33 m) โ W(2 m) = 0;
W = 1,220 kN
12
2.5.13
2.6.1
F ๏ฝ ๏ง ๏ h ๏ A = (62.3 lb/ft3)[(d/2)ft]โ[{(d /cos30)ft}(8ft)]
The resultant force in the horizontal direction is zero
(FH = 0) since equal pressures surround the viewing
F = 288โd2 lbs
I0
๏ซy
Ay
yP ๏ฝ
yP ๏ฝ
window in all directions. For the vertical direction, the
force equals the weight of the water above the window.
๏(8)(d / cos30๏ฐ) /12๏
3
๏(8)(d / cos30๏ฐ)๏(d / 2cos30๏ฐ)
FV ๏ฝ ๏ง ๏Vol = (1.03)(62.3 lb/ft3)[ฯ(1 ft)2(6 ft) โ
๏ซ (d / 2cos30)
(1/2)(4/3)ฯ(1 ft)3] = 1,080 lbs
___________________________________________
yp = [(0.192โd) + 0.577โd)] ft = 0.769โd
2.6.2
Thus, summing moments: โ Mhinge = 0
Obtain the horizontal component of the total hydrostatic
(288โd2)[(d/cos30)-0.769d] โ (5,000)(15) = 0
pressure force by determining the total pressure on the
vertical projection of the curved gate.
d = 8.77 ft
A depth greater than this will make the gate open.
FH ๏ฝ ๏ง ๏ h ๏ A = (0.82)(9790 N/m3)(5m)[(8 m)(2 m)]
Any depth less than this will make it close.
FH = 6.42 x 105 N = 642 kN โ
___________________________________________
The vertical component of the hydrostatic force equals
2.5.14
the weight of the water column above the curved gate.
The force and location on the vertical side of the gate:
F ๏ฝ ๏ง ๏ h ๏ A = (9790 N/m3)[(h/2)m][(h m)(1 m)]
F = (4.90 x 103)h2 N (per meter of gate width)
yP ๏ฝ
๏
3
๏
I0
(1)(h) /12
๏ซy๏ฝ
๏ซ h / 2 = (2/3)h m
๏(1)(h)๏(h / 2)
Ay
The force upward on the bottom of the gate::
F = pโA = (9790 N/m )(h)[(1m)(1m)]
3
F = (9.79 x 103)h N (per meter of gate width)
This force is located 0.50 m from the hinge.
Summing moments about the hinge; โ Mh = 0
[(4.90 x 10 )h ][h – (2/3)h] โ [(9.79 x 10 )h](0.5) = 0
3
h = 1.73 m
2
3
FV ๏ฝ ๏ง ๏ Vol = (0.82)(9790N/m3)[(4m)(2m)+ฯ/4(2m)2](8m)
FV = 7.16 x 105 N = 716 kN โ
The total force is
F = [(642 kN)2 + (716 kN)2]1/2 = 962 kN
ฮธ = tan-1 (FV/FH) = 48.1ห
Since all hydrostatic pressures pass through point A
(i.e., they are all normal to the surface upon which they
act), then the resultant must also pass through point A.
__________________________________________
2.6.3
Obtain the horizontal component of the total hydrostatic
pressure force by determining the total pressure on the
vertical projection of the viewing port.
FH = ๏ง ๏ h ๏ A = (62.3 lb/ft3)(4 ft)[ฯ(1 ft)2] = 783 lbs โ
F = [(783 lbs)2 + (261 lbs)2]1/2 = 825 lbs
ฮธ = tan-1 (FV/FH) = 18.4ห
13
2.6.4
2.6.7
The vertical component of the total hydrostatic pressure
First determine the oil column height due to pressure:
force is found by determining the weight of the water
column above the top of the viewing port (which
produces a downward force) and subtracting the
upward force on the bottom of the viewing port
(equivalent to the virtual weight of the water above it).
Note that the difference in the two columns of water
(vector addition) equals the weight of water in a hemispherical volume (the viewing port) acting upwards.
+
h = P/ ฮณ = [2.75 x 105 N/m2]/[(0.9)(9790 N/m3)]
h = 31.2 m of oil. The total upward force on the dome
is the weight of an oil column 31.2 m high minus the
volume of oil that is resident above the gage already.
FV = ๏ง ๏ Vol = (0.9)(9790 N/m3)[(31.2 m โ 0.75 m)
– (1/2)ฯ(1.0 m)2] (2.0 m)(8.0 m)
FV = 4.07 x 106 N โ; which is also the total tension
force in the bolts holding the tank top on.
___________________________________________
2.6.8
=
First find the height of the vertical projection of area:
(R)(sin 45ห) = (40 ft)(sin 45ห) = 28.3 ft. Now,
FV= ๏ง ๏ Vol = (9790 N/m3)[(1/2)(4/3)ฯ(1m)3] = 6530 N
__________________________________________
FH = ๏ง ๏ h ๏ A = (62.3 lb/ft3)(28.3ft/2)[(33 ft)(28.3 ft)]
2.6.5
FH = 8.23 x 105 lb
The vertical component of the total pressure force is the
FH = ๏ง ๏ h ๏ A = (1.03)(9790 N/m3)(3.875 m)ยท
[(1.75 m)(1m)] = 68.4 kNโ (per unit length)
FV = ๏ง ๏ Vol = (1.03)(9790 N/m )[(3 m)(1.75 m)(1 m)
3
+ ฯ/4(1.75 m) ] (1 m) = 77.2 kNโ; larger
__________________________________________
weight of the water column above the curved gate. The
volume of water above the gate is:
Vol = (Arectangle – Atriangle – Aarc)(length)
2
Vol = [(40ft)(28.3ft) – (1/2)(28.3ft)(28.3ft)
– (ฯ/8)(40ft)2](33 ft) = 3,410 ft3
2.6.6
FH is the total pressure on the vertical projection of the
FV = ๏ง ๏ Vol = (62.3 lb/ft3)(3410 ft3) = 2.12 x 105 lb;
curved surface ABC.
The total force is
FH ๏ฝ ๏ง ๏ h ๏ A = (62.3lb/ft3)(2ft)[(4ft)(1ft)] = 498 lbsโ
F = [(8.23 x 105 lb)2 + (2.12 x 105 lb)2]1/2
FV is the weight of the water above the curved surface
F = 8.50 x 105 lb;
ABC. The volume of water above this surface is:
ฮธ = tan-1 (FV/FH) = 14.4ห
Vol = (Aquaarter circle + Arectangle โ Aquarter circle )(unit length)
Vol = (Arectangle)(unit length) = (4ft)(2ft)(1ft) = 8 ft
FV ๏ฝ ๏ง ๏ Vol = (62.3 lb/ft3)[8 ft3] = 498 lbsโ
3
Since all hydrostatic pressures pass through point O
(i.e., they are all normal to the surface upon which they
act), then the resultant must also pass through point O.
14
2.6.9
2.6.11
The horizontal component of the hydrostatic force is:
The horizontal component of the hydrostatic force is:
FH ๏ฝ ๏ง ๏ h ๏ A = (62.3 lb/ft3)(7.0 ft)[(8.0 ft)(1.0 ft)]
FH ๏ฝ ๏ง ๏ h ๏ A = (62.3 lb/ft3)(7.8 ft)[(12.4 ft)(1.0 ft)]
FH = 3.49 x 103 lb = 3,490 lb (per unit length of gate)
FH = 6,030 lb โ (per unit length of surface)
The vertical component is the virtual (displaced) weight
of the water column above the curved gate.
Vol = (Arectangle + Aarc – Atriangle)(length) = [(4 ft)(3 ft)
+ (53.1ห/360ห)ฯ(10 ft)2 – (1/2)(8 ft)(6 ft)](1 ft) = 34.3 ft3
FV ๏ฝ ๏ง ๏ Vol = (62.3 lb/ft )(34.3 ft ) = 2,140 lb; Thus,
3
3
yP ๏ฝ
๏
๏
I0
(1 ft )(12.4 ft )3 / 12
๏ซy๏ฝ
๏ซ 7.8 ft = 9.44 ft
๏(1 ft )(12.4 ft )๏(7.8 ft )
Ay
The vertical component of the hydrostatic force is equal
to the weight of the water displaced by the quadrant and
the triangle. Solve in parts using Table 2.1 we have
FVTriangle ๏ฝ ๏ง ๏ Vol = (62.3 lb/ft3)[(1/2)(8 ft)(4.4 ft)](1 ft)
F = [(3,490 lb)2 + (2,140 lb)2]1/2 = 4,090 lb
FVTriangle = 1,100 lb upwards 1.47 ft from wall
ฮธ = tan-1 (FV/FH) = 31.5ห
Since all pressure vectors pass through the center of the
gate radius, the resultant force will do the same.
___________________________________________
2.6.10
FVquadrant ๏ฝ ๏ง ๏ Vol = (62.3 lb/ft3)[(ฯ/4)(4.4 ft)2](1 ft)
FVquadrant = 947 lb upwards 1.87 ft from wall
__________________________________________
2.6.12
The force on the end (E) of the cylinder is:
There are four vertical forces at work on the cone plug.
FHE = ๏ง ๏ h ๏ A = (9.79 kN/m )(4 m)[ฯ(1 m) ] = 123 kN
The cone unplugs when โFy = 0. The 4 forces are:
The horizontal force on the side (S) of the cylinder is:
The pressure force of fluid A on top of the plug (down):
FHS = ๏ง ๏ h ๏ A = (9.79 kN/m3)(4m)(8m2) = 313 kN
FATop ๏ฝ ๏ง ๏ Vol = (9790 N/m3)[ฯ(0.15m)2(0.3m)] = 208 N
The vertical force on the side of the cylinder is
The pressure force of fluid A on the cone sides (up):
3
2
downward and equal to the weight of the water in half
of the tank. (See Example 2.6.) Thus,
FASides ๏ฝ ๏ง ๏ Vol = [ฯ(0.15m)2(0.3m) + (ฯ/3) (0.15m)2(0.3m)
– (ฯ/3) (0.05m)2(0.1m)](9790 N/m3) = 274 N
FVS = ๏ง ๏ Vol = (9.79 kN/m3)[ฯ(1m)2/2](4m) = 61.5 kNโ
The pressure force of fluid B on the cone bottom (up):
Thus, F = [(313 kN)2 + (61.5 kN)2]1/2 = 319 kN
FBbottom ๏ฝ ๏ง ๏ Vol = (0.8)(9790 N/m3) [ฯ(0.05m)2(1.5m) +
ฮธ = tan (FV/FH) = 11.1ห
-1
The resultant force will pass through the center of the
tank since all pressures are normal to the tank wall and
thus pass through this point.
(ฯ/3) (0.05m)2(0.1m)] = 94.3 N
The weight of the cone is unknown. Solving yields ฮณ:
โFy = 208 โ 274 โ 94.3 + (ฮณcone)(ฯ/3) (0.15m)2(0.3m) = 0;
ฮณcone = 22,700 N/m3; S.G. = 2.32
15
2.6.13
2.8.1
Everything is the same as in problem 2.6.10 except:
Compare the weight of the pipe to the buoyant force.
The equivalent depth of oil based on the air pressure is:
W = mg = (25.3 kg)(9.81 m/sec2) = 248 N
h = P/ฮณ = (8,500 N/m2)/[(0.8)(9790 N/m3)] = 1.09 m
B = ฮณโVol = (1.26)(9790 N/m3)[ฯ(0.079 m)2(1 m)]
The pressure force of fluid B on the cone bottom (up):
B = 242 N < 248 N, thus it will sink.
___________________________________________
FBbottom ๏ฝ ๏ง ๏ Vol = (0.8)(9790 N/m3) [ฯ(0.05m)2(1.09m)
+ (ฯ/3) (0.05m)2(0.1m)] = 69.1 N
2.8.2
โFy = 208 โ 274 โ 69.1 + (ฮณcone)(ฯ/3) (0.15m)2(0.3m) = 0;
Neutral buoyancy๏จ Weight (W) = Buoyant force (B)
ฮณcone = 19,100 N/m3; S.G. = 1.95
__________________________________________
Also, B = wt. of water displaced = ฮณโVol ; therefore,
W = B = (62.4 lb/ft3)(4/3)ฯ[(4.3/12)ft]3 = 12.0 lb
2.6.14
m = W/g = 12.0 lb/32.2 ft/sec2 = 0.373 slugs
The horizontal force component is:
Also, the specific weight of the bowling ball is:
FHALeft ๏ฝ ๏ง ๏ h ๏ A = (0.9)(9790 N/m3)(6 m)[(1 m)(1.41 m)]
ฮณb = wt/Vol = 12lb/{(4/3)ฯ[(4.3/12)ft]3} = 62.3 lb/ft3
FHALeft = 74.5 kN
Finally, S.G. = ฮณb/ฮณ = 62.3 lb/ft3/62.4 lb/ft3 = 1.0
__________________________________________
FHARight ๏ฝ ๏ง ๏ h ๏ A = (0.9)(9790 N/m3)(5.65 m)[(1m)(0.707m)]
2.8.3
FHARight = 35.2 kN
Theoretically, the lake level will fall. When the anchor
is in the boat, it is displacing a volume of water equal to
its weight. When the anchor is thrown in the water, it is
only displacing its volume. Since it has a specific
gravity greater than 1.0, it will displace more water by
weight than by volume.
____________________________________________
FHBRight ๏ฝ ๏ง ๏ h ๏ A = (1.5)(9790 N/m3)(5.35 m)[(1 m)(0.707 m)]
FHBRight = 55.5 kN ; Thus, FH = 16.2 kN
The vertical force component is:
FVATop = (0.9)(9790N/m3)[(1.41m)(1m)(6m) – ฯ/2(0.707m)2(1.0m)]
FVATop = 67.6 kN
2.8.4
When the sphere is lifted off the bottom, equilibrium in
FVABottom = (0.9)(9790N/m )[(0.707m)(1m)(6m)+ฯ/4(0.707m) (1.0m)]
the y-direction occurs with W = B. Therefore,
FVABottom = 40.8 kN
W = ฮณsphere [(4/3)ฯ(0.14m)3] + ฮณbuoy[ฯ(0.25m)2(1.5m)]
FVBBottom = (1.5)(9790N/m3)[(0.707m)(1m)(5m)+ฯ/4(0.707m)2(1.0m)]
W = (13.5ฮณ)[0.0115m3] + (S.G.)ฮณ[0.295m3]
3
2
FVBBottom = 57.7 kN
WCylinder
= (2.0)(9790N/m3)[ฯ(0.707m)2(1.0m)] = 30.7 kN
Thus, FV = 0.2 kN
W = 0.155ฮณ + 0.295(S.G.)ฮณ
B = ฮณ[(4/3)ฯ(0.14m)3] + ฮณ[ฯ(0.25m)2(1.33m)] = 0.273ฮณ
Equating yields; (S.G.)buoy = 0.400
16
2.8.5
2.8.8
For floating bodies, weight equals the buoyant force.
Three forces act on the rod; the weight, buoyant force,
W = B; and using w & L for width & length of blocks
ฮณA(H)(w)(L) + ฮณB(1.5โH)(w)(L) = ฮณ (2โH)(w)(L)
ฮณA + (2ฮณA)(1.5) = ฮณ(2); ฮณA (1 + 3) = ฮณ (2);
ฮณA = 0.5ฮณ; and since ฮณB = 2โฮณA = ฮณ
__________________________________________
2.8.6
and the hinge force. The buoyant force is
B = (62.3 lb/ft3)(0.5 ft)(0.5 ft)(7 ft/sin ฮธ) = 109 lb/sin ฮธ;
B = 109 lb/sin ฮธ;
The buoyant force acts at the
center of the submerged portion. W = 165 lb
โMhinge = 0, and assuming the rod is homogeneous,
(109 lb/sin ฮธ)(3.5 ft/tan ฮธ) โ (165 lb)[(6 ft)(cos ฮธ)] = 0
When the anchor is lifted off the bottom, equilibrium in
the y-direction occurs (โFy = 0). Therefore,
T (sin60ห) + B = W;
where T = anchor line tension
Noting that tan ฮธ = (sin ฮธ/cos ฮธ) and dividing by cos ฮธ
(sin ฮธ)2 = 0.385; sin ฮธ = 0.621; ฮธ = 38.4ห
__________________________________________
B = (62.3 lb/ft3)ฯ(0.75 ft)2(1.2 ft) = 132 lb
2.8.9
W = (SG)(62.3 lb/ft3)ฯ(0.75 ft)2(1.2 ft) = 132(SG)
The center of gravity (G) is given as 6 ft up from the
Substituting: 244(sin60ยฐ) + 132 = 132(SG)
Solving yields: S.G.(concrete) = 2.60
__________________________________________
bottom of the caisson. The center of buoyancy (B) is
14 feet from the bottom since 28 feet is submerged.
Therefore, GB = 8.0 ft, and GM is found using:
I0
๏ฑ GB ; where Io is the waterline
Vol
2.8.7
GM ๏ฝ MB ๏ฑ GB ๏ฝ
The hydrostatic force on the gate: F ๏ฝ ๏ง ๏ h ๏ A
moment of inertia about the tilting axis. Chopping off
F = (9790 N/m3)(1.5m)[(1m/sin45ห)2] = 29.4 kN
a circle with a 30 ft diameter. Thus,
I0
๏ซy
Inclined location from water surface: y P ๏ฝ
Ay
yP ๏ฝ
๏(1m / sin 45๏ฐ)(1m / sin 45๏ฐ) /12๏
3
๏(1m / sin 45๏ฐ)(1m / sin 45๏ฐ)๏(1.5 / sin 45๏ฐ)
๏ซ (1.5 / sin 45๏ฐ)
yp = 2.20 m. The perpendicular distance from hinge is:
yโ = (2m/sin45ห) โ 2.20 m = 0.628 m
The buoyant force (on half the sphere) is
the caisson at the waterline and looking down we have
GM ๏ฝ
๏
๏
๏ฐ ๏จ30 ft ๏ฉ / 64
I0
๏ฑ GB ๏ฝ
๏ซ 8.0 ft = 10.0 ft;
Vol
๏จ๏ฐ ๏ฉ๏จ15 ft ๏ฉ2 ๏จ28 ft ๏ฉ
4
Note: Vol is the submerged volume and a positive sign
is used since G is located below the center of buoyancy.
M ๏ฝ W ๏ GM ๏ sin ๏ฑ
M =[(1.03)(62.3 lb/ft3)(ฯ)(15ft)2(28ft)](10ft)(sin ฮธ)
B = ฮณโVol = (9790 N/m3)(1/2)(4/3)ฯ(R)3 = 20.5(R)3 kN
M = 1.11 x 106 ft-lb (for a heel angle of 5ห)
โMhinge = 0, ignoring the buoy weight
M = 2.21 x 106 ft-lb (for a heel angle of 10ห)
29.4kN(0.628 m) + 6kN(0.5m) โ [20.5(R)3kN](1m) = 0
M = 3.29 x 106 ft-lb (for a heel angle of 15ห)
R = 1.02 m
17
2.8.10
2.8.12
First determine how much the wooden pole is in the
water. Summing forces in the y-direction, W = B
The center of gravity (G) is estimated as 17 ft up from
the bottom of the tube based on the depth of the water
inside it. The center of buoyancy (B) is 21 feet from the
h
hg
bottom since 42 feet is in the water.
hb
B
G
Therefore GB = 4.0 ft, and GM is found using
GM ๏ฝ MB ๏ฑ GB ๏ฝ
I0
๏ฑ GB ; where Io is the waterline
Vol
moment of inertial about the tilting axis. Chopping off
(Vols)(SGs)(ฮณ) + (Volp)(SGp)(ฮณ) = (Vols)(ฮณ) + (Volpโ)(ฮณ)
the tube at the waterline and looking down we have a
[(4/3)ฯ(0.25m)3](1.4)+[ฯ(0.125m)2(2m)](0.62) =
circle with a 36 ft diameter. Thus,
[(4/3)ฯ(0.25m)3]+[ฯ(0.125m)2(h)];
GM ๏ฝ
0.0916m3 + 0.0609m3 = 0.0654m3 + (0.0491m2)h
h = 1.77 m; Find โBโ using the principle of moments.
[(Vols)(ฮณ)+(Volpโ)(ฮณ)](hb)=(Vols)(ฮณ)(h+0.25m)+(Volpโ)(ฮณ)(h/2)
[(0.0654m3 + (0.0491m3)(1.77m)](hb) =
(0.0654m3)(1.77m + 0.25m) + [(0.0491m3)(1.77m)](1.77m/2)
hb = 1.37 m; Find โGโ using the principle of moments.
(W)(hg+0.23m) = (Ws)(2.0 m + 0.25 m) + (Wp)(2.0m/2)
[(0.0916m3 + 0.0609m3)](hg +0.23m) =
(0.0916m3)(2.25m) + (0.0609m3)(1.00m)
hg = 1.52 m; GB = hg – hb = 0.15 m
๏
๏
I0
๏ฐ ๏จ36 ft ๏ฉ4 / 64
๏ฑ GB ๏ฝ
๏ซ 4.0 ft = 5.93 ft;
Vol
๏จ๏ฐ / 4๏ฉ๏จ36 ft ๏ฉ2 ๏จ42 ft ๏ฉ
Note: Vol is the submerged volume and a positive sign
is used since G is located below the center of buoyancy.
M ๏ฝ W ๏ GM ๏ sin ๏ฑ
M = [(1.02)(62.3 lb/ft3)(ฯ/4)(36 ft)2(42 ft)](5.93 ft)(sin 4ห)
M = 1.12 x 106 ftโlb (for a heel angle of 4ห)
__________________________________________
2.8.13
The center of gravity (G) is roughly 1.7 m up from the
bottom if the load is equally distributed. The center of
buoyancy (B) is 1.4 m from the bottom since the draft is
MB = Io/Vol
MB = [(1/64)ฯ(0.25m)4] / [(0.0654m3 + (0.0491m3)(1.77m)]
MB = 1.26 x 10 m
-3
2.0 m. Therefore GB = 0.3 m, and GM is found using
GM ๏ฝ
๏
๏
I0
๏จ12m ๏ฉ๏จ4.8m ๏ฉ3 / 12 ๏ญ 0.3m = 0.660 m;
๏ฑ GB ๏ฝ
๏จ12m ๏ฉ๏จ4.8m ๏ฉ๏จ2m ๏ฉ
Vol
GM = MB + GB = 1.26 x 10-3 m + 0.15 m = 0.151 m
__________________________________________
Note: Vol is the submerged volume and a negative sign
2.8.11
M ๏ฝ W ๏ GM ๏ sin ๏ฑ
If the metacenter is at the same position as the center of
M = [(1.03)(9790 N/m3)(12m)(4.8m)(2m)](0.660m)(sin 15ห)
gravity, then GM = 0 and the righting moment is
M ๏ฝ W ๏ GM ๏ sin ๏ฑ = 0. With no righting moment, the
block will not be stable.
is used since G is located above the center of buoyancy.
M = 198 kNโm; The distance G can be moved is
d = GM (sin ฮธ) = (0.66 m)(sin 15ห) = 0.171 m
18

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