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

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 – Problem Solutions 1.2.1 1.2.3 E1 = energy released in lowering steam temperature to 100C from 110°C E1 = energy needed to vaporize the water E1 = (500 L)(1000 g/L)(10C)(0.432 cal/g∙C) 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 100°C to 50C E3 = (500 L)(1000 g/L)(50C)(1 cal/g∙C) E3 = 2.50×10 cal; The energy remaining (E2) is: E2 = ETotal – E1 E2 = 2.00×108 cal – 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/g∙C)(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 95°C E1 = energy required to bring the water temperature to 95C from 15°C T = 70C, making the temperature T = 90C 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C)(0.465 cal/g∙C) + 10g(79.7 cal/g) E1 = 825 cal. This energy is taken from the water. E1 = (900 g)(95C – 15C)(1 cal/g∙C) The resulting temperature of the water will decrease to: E1 = 7.20×104 cal 825 cal = (0.165 L)(1000 g/L)(20C – T1)(1 cal/g∙C) E2 = energy required to vaporize the water T1 = 15.0C. Now we have a mixture of water at 0°C (formerly ice) and the original 165 liters that is now at 15.0 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C – T2)(1 cal/g∙C)] = Etotal = E1 + E2 = 6.09×10 cal 5 [(10 g)(T2 – 0C)( 1 cal/g∙C)] ; T2 = 14.1C 1 1.2.5 1.3.1 E1 = energy required to melt ice F = m∙a; Letting a = g yields: W = m∙g, (Eq’n 1.1) E1 = (5 slugs)(32.2 lbm/slug)(32F – 20F)* Then dividing both sides of the equation by volume, (0.46 BTU/lbm∙F) + (5 slugs)(32.2 lbm/slug)* (144 BTU/lbm) E1 = 2.41 x 104 BTU. Energy 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F – 4 T1)(1 BTU/lbm∙F) W/Vol = (m/Vol)∙g; γ = ρ∙g ___________________________________________ 1.3.2 SGoil = 0.976 = γoil/γ; where γ is for water at 4°C: γ = 9,810 N/m3 (Table 1.2). Substituting yields, 0.977 = γoil/9,810; γoil = (9810)(0.976) = 9,570 N/m3 T1 = 45.2F Also, γ = ρ∙g; or ρ = γoil/g The energy lost by the water (to lower its temp. to Substituting (noting that N ≡ kg·m/sec2) yields, 45.2F) is that required to melt the ice. Now you have 5 slugs of water at 32F and 10 slugs at 45.2F. ρoil = γoil/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F – T2)(1 BTU/lbm∙F)] By definition,  = W/Vol = 55.5 lb/ft3; thus, = [(5 slugs)(32.2 lbm/slug)(T2 – 32F)(1 BTU/lbm∙F)] W = ∙Vol = (55.5 lb/ft3)(20 ft3) = 1,110 lb (4,940 N) T2 = 40.8F _________________________________________  = /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C E1 = (7500 g)(100C – 20C)(1 cal/g∙C) E2 = 6.00×105 cal E2 = energy required to vaporize 2.5 kg of water E2 = (2500 g)(597 cal/g) SG = liquid/water at 4C = (55.5 lb/ft3)/(62.4 lb/ft3) = 0.889 ___________________________________________ 1.3.4 The mass of liquid can be found using ρ = γ/g and γ = weight/volume, thus γ = (47000 N – 1500 N)/(5 m3) = 9.10 x 103 N/m3 ρ = γ/g = (9.1 x 103 N/m3)/(9.81 m/sec2);  = 928 kg/m3 (Note: 1 N ≡ 1 kg·m/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) = γ/γwater at 4°C 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 (Eq’n 1.2). 1 lb·sec/ft2 [(1 N)/(0.2248 lb)]·[(3.281 ft)2/(1 m)2] = F = W = mg = [ρ(Vol)] (g); ρ 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 (π ∙(1.25 ft)2 ∙ d)] (32.2 ft/sec2) Conversion: 1 lbsec/ft2 = 478.9 poise d = 3.00 ft (Note: 1 slug = 1 lb∙sec2/ft) ___________________________________________ (b) Note that: 1 stoke = 1 cm2/sec. Therefore, 1 ft2/sec [(12 in)2/(1 ft)2]· [(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 E’qn (1.1): m = W/g = (8,830 N)/(9.81 m/s2) 1.4.2 m = 900 kg [(air)/(H2O)]0C = (1.717×10-5)/(1.781×10-3) Note: mass on moon is the same as mass on earth [(air)/(H2O)]0C = 9.641×10-3 W (moon) = mg = (900 kg)[(9.81 m/s2)/(6)] [(air)/(H2O)]100C = (2.174×10-5)/(0.282×10-3) W(moon) = 1,470 N _________________________________________ [(air)/(H2O)]100C = 7.709×10-2 [(air)/(H2O)]0C = (1.329×10-5)/(1.785×10-6) 1.3.7 Density is expressed as ρ = m/Vol, and even though [(air)/(H2O)]0C = 7.445 volume changes with temperature, mass does not. [(air)/(H2O)]100C = (2.302×10-5)/(0.294×10-6) Thus, (ρ1)(Vol1) = (ρ2)(Vol2) = constant; or [(air)/(H2O)]100C = 78.30 Vol2 = (ρ1)(Vol1)/(ρ2) 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 () 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)] · 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 20C = 1.002×10-3 Nsec/m2; 20C = 1.003×10-6 m2/s (1.002×10-3 Nsec/m2)∙[(0.2248 lb)/(1 N)]∙ [(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 Newton’s law of viscosity (Eq’n 1.2): v = y2 – 3y, where y is in inches and v is in ft/s  = (dv/dy) = (Δv/Δy) v = 144y2 – 36y, where y is in ft and  is in ft/s  = (1.00 x 10-3 Nsec/m2)[{(4.8 – 2.4)m/sec}/(0.02 m)] Taking the first derivative: dv/dy = 288y – 36 sec-1  = 0.12 N/m2 _________________________________________  = (dv/dy) = (8.35 x 10-3 lb-sec/ft2)( 288y – 36 sec-1) 1.4.5 From Eq’n (1.2):  = (v/y) = Solutions: y = 0 ft,  = -0.301lb/ft2 y = 1/12 ft,  = -0.100 lb/ft2; y = 1/6 ft,  = 0.100 lb/ft2  = (0.0065 lbsec/ft )[(1.5 ft/s)/(0.25/12 ft)] y = 1/4 ft,  = 0.301 lb/ft2; y = 1/3 ft,  = 0.501 lb/ft2 _________________________________________  = 0.468 lb/ft2 1.4.9 F = ()(A) = (2 sides)(0.468 lb/ft2)[(0.5 ft)(1.5 ft)]  = (16)(1.00×10-3 Nsec/m2) = 1.60×10-2 Nsec/m2 F = 0.702 lb _________________________________________ Torque = (r )dF  r    dA  (r )( )( v )dA    2 R R R 0 0 0 y R 1.4.6 Summing forces parallel to the incline yields: Tshear force = W(sin15) = A = (v/y)A y = [()(v)(A)] / [(W)(sin15)] y = [(1.52 Nsec/m2)(0.025 m/sec)(0.80m)(0.90m)]/ [(100 N)(sin15)] y = 1.06 x 10-3 m = 1.06 mm ___________________________________________ 1.4.7 Torque = (r )( )( ( )(r )  0 )(2r )dr 0 y R Torque = (2 )( )( ) (r 3 )dr  y 0 2 2 4 Torque = (2 )(1.60 10 N  sec/ m )(0.65rad / sec)  (1m)  0.0005m  4    Torque = 32.7 Nm _________________________________________ 1.4.10  = /(dv/dy) = (F/A)/(v/y); Torque (T) = Force∙distance = F∙R where R = radius Using Newton’s law of viscosity (Eq’n 1.2):  = (dv/dy) = (v/y)  = (0.04 Nsec/m )[(15 cm/s)/[(25.015 – 25)cm/2] 2  = 80 N/m2 Fshear resistance = A = (80 N/m2)[()(0.25 m)(3 m)] Fshear resistance = 188 N Thus;  = (T/R)/[(A)(v/y)] = T/R T  y  (2 )( R)(h)(  R / y ) (2 )( R 3 )(h)( ) = (1.10lb  ft )[(0.008 / 12) ft ]  2rad / sec   (2 )((1 / 12) ft )3 ((1.6 / 12) ft )(2000 rpm )  60rpm   = 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)(1)(sin1)] / [()(D)] 1) The surface tension acts along the perimeter of the tube pulling the column of water upwards due to h1 = [(4)(1)(sin30)] / [()(0.8 mm)] adhesion between the water and the tube. Condition 2: h2 = [(4)(2)(sin2)] / [()(D)] 2) The surface tension is multiplied by the tube h2 = [(4)(0.881)(sin50)] / [()(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)()(sin )] / [()(h)] D = [4(0.57 N/m)(sin 50)]/[13.6(9790N/m3)(1.0×10-3 m)] D = 0.0131 m = 1.31 cm Note: 50° was used instead or 40° because it produces the largest D. A 40° angle produces a smaller error. _________________________________________ h2/h1 = [(0.88)(sin50°)] / (sin30) = 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)()(sin)] / [()(D)] where σ is from Table 1.4 and γ from Table 1.2. Thus,  = (6.90 x 10-2)(1.2) = 8.28 x 10-2 N/m and  = (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)()(sin θ)] / [()(D)] But sin 90˚ = 1,  = 62.3 lb/ft3 (at 20˚C), and h = 1.58×10-3 m = 0.158 cm _________________________________________ 1.5.6 σ = 4.89×10-3 lb/ft (from inside book cover) thus, D = [(4)()] / [()(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 – Pe (internal pressure minus external pressure) for h = 1.0 in., D = 3.77×10-3 ft = 0.0452 in. Fx = 0; P()(R2) – 2(R)() = 0 for h = 0.5 in., D = 7.54×10-3 ft = 0.0904 in. P = 2/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 – 30 bar = -27 bar = -2.7×105 N/m2 Vol/Vol = -(14.7 psi – 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= [()(1.50 m)2/(4)]∙(2000 m) = 3530 m3 / = -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  = 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 –14.7 psi)/(3.20×105 psi)](120 ft3) Vol = -0.546 ft3 new = (233 slug)/(120 ft3 – 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) γ/γ = -Vol/Vol = -0.727% (specific wt. increase) Specific weight at the surface: γs = 9,810 N/m3 Specific weight at the bottom: γb = (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 – 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 = γ·Vol = γ(A·h); where h = water depth 14,700 lbs = (62.3 lb/ft )[π·(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/[π·(5 ft)2] = 187 lbs/in.2 b) Pressure due to depth of water using Eq’n 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 = γ∙h = (62.3 lb/ft3)(3 ft) = 187 lb/ft2 _________________________________________ 2.2.5 2.2.2 P = γ∙h = (62.3 lb/ft3)(10 ft) = 623 lb/ft2 The absolute pressure includes atmospheric pressure. Fbottom = P∙A = (623 lb/ft2)(100 ft2) = 6.23 x 104 lbs Therefore, Pabs = Patm + (water)(h)  5(Patm); Pressure varies linearly with depth. So the average pressure on the sides of the tank occur at half the depth. Thus from Eq’n 2.4: h = 4(Patm)/water Pavg = γ∙h = (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 ∙A = (312 lb/ft2)(100 ft2) = 3.12 x 104 lbs ___________________________________________ 2.2.6 2.2.3 water at 30C = 9.77 kN/m3 (from Table 1.2) Pvapor at 30C = 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 – 85.0 kN/m2)/(89.2 kN/m2) Error = 0.0471 = 4.71% Pressure and force on the bottom of both containers is: P = (water)(h) = (9790 N/m3)(10 m) = 97.9 kN/m2 Also, F = P∙A = (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.) oil = (SG)(water) = (0.85)( 62.3 lb/ft3) = 53.0 lb/ft3 P10ft = Pair + (oil)(12 ft) or Pair = P10ft – (oil)(12 ft) W1 P1 ·A1 Pair = 25.7 psi (144 in2/ft2) – (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 = γh). This is opposed by an equal and opposite pressure on the water below. The resulting force is P1·A1 which equals γ·h·A1 or γ·Vol. 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 ∑Fx = 0; (Px )(Δy) – (Ps)(Δs)(sin θ) = 0 since (Δs∙sin θ) = Δy, Ps = Px ∑Fy = 0; and accounting for weight yields: (Py )(Δx) – (Ps)(Δs)( cos θ) – (Δx∙Δy/2)(Δz)(γ) = 0 since (Δs∙cos θ) = Δx and (Δy)(Δx)(Δz)  0; Ps = Py Hence, the pressure is the same in all seawater = (SG)() = (1.03)(62.3 lb/ft ) = 64.2 lb/ft 3 3 Ptank = (seawater)(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 + (liquid)(1.4 m); and liquid = (SG)(water) = (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)(liquid) = Pgage + (liquid)(1 m) h = (Pgage)/(liquid) + 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 Pascal’s 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)() = (5h)() + (h)(SGoil)() h = (hHg)(SGHg)/(5 + SGoil) h = (1.43 in.)(13.6)/(5 + 0.80) = 3.35 inches ____________________________________________ PA + (y)() = (h)(Hg) PA + [(1.34/12)ft]() = [(1.02/12)ft](Hg) 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)() = (hoil)(oil) = (hoil)()(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)() = (h)(Hg) + (0.6 m)(oil) (1 m)() = (h)(SGHg)() + (0.6 m)(SGoil)(); h = [1 m – (0.6 m)(SGoil)]/(SGHg) h = [1m – (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)(Hg) = P + (2 ft)() where Hg = (SGHg)() (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)() = (h2)(Hg) = (h2)(SGHg)() 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 “swim through” adding pressure when “swimming” down and subtracting when “swimming” up until you reach the pipe (Ppipe). Remember to jump across the manometer at surfaces of equal pressure (ES). Thus, 0 – (0.66 m)(ml) + [(0.66 + y + 0.58)m](air) – (0.58 m)(oil) = 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)() – (0.58 m)(SGoil)() P = 2,420 lb/ft2 = 16.8 psi Ppipe = 0 – (0.66 m)(0.8)(9790 N/m3) – (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/Hg = (2420 lb/ft2)/[(13.6)(62.3 lb/ft2)] height (head) of any liquid through P = ∙h. 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 – 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)() = (h2)(Hg) = (h2)(SGHg)() Volres = Voltube P + (0.5 ft)(62.3 lb/ft3) = (1.5 ft)(13.6)(62.3 lb/ft3) Δh = h (Atube/Ares) = h [(d2)2/(d1)2]; substituting yields P = 1,240 lb/ft2 = 8.61 psi P1 – P2 = h [(d2)2/(d1)2] (ρ1∙g) + (h)(ρ2∙g) – (h)(ρ1∙g) When the manometer reading rises or falls, volume balance is preserved for constant density. Therefore, P1 – P2 = h∙g [ρ2 – ρ1 + ρ1 {(d2)2/(d1)2}] Volres = Voltube or Ares∙h1 = Atube∙h2 or Ares∙(Δh) = Atube∙ h P1 – P2 = h∙g [ρ2 – ρ1 {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 “swim through” technique, start at the right tank where pressure is known and “swim through” the tanks and pipes, adding pressure when “swimming” down and subtracting when “swimming” 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 “swim through” technique, start at pipe A and “swim through” the manometer, adding pressure when “swimming” down and subtracting when “swimming” up until you reach pipe B. The computations are: PA + (5.33 ft)(oil) – (1.67ft)(ct) – (1.0 ft)() = PB PA – PB = (62.3 lb/ft3) [(1.0 ft) + (1.67 ft)(1.6) – (5.33)(0.82)] PA – PB = -43.5 lb/ft2 = -0.302 psi _________________________________________ 2.4.10 Using the “swim through” technique, start at P 2 and “swim through” the manometer, adding pressure when “swimming” down and subtracting when “swimming” up until you reach P1. The computations are: P2 + (Δh)(ρ1∙g) + (y)(ρ1∙g) + (h)(ρ2∙g) – (h)(ρ1∙g) (y)(ρ1∙g) = 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 – P2 = (Δh)(ρ1∙g) + (h)(ρ2∙g) – (h)(ρ1∙g) 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 “swim through” technique, start at both ends of the manometers which are open to the atmosphere and thus equal to zero gage pressure. Then “swim through” the manometer, adding pressure when “swimming” down and subtracting when “swimming” 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)() – (44)() = PA; PA = 269∙γ 0 + (46)(0.8)() + (20)(13.6)() – (40)(γ) = PB PB = 269∙γ; Therefore, PA = PB and h = 0 10 2.5.1 2.5.4 F    h  A = (62.3 lb/ft3)[4(2ft)/(3π)]∙[(1/2)π(2ft)2] The hydrostatic force and its locations are: F    h  A = (γ)(h)[π(D)2/4] F = 332 lb (Note: h  y in Table 2.1)     I0  ( D) 4 / 64 y h; Ay  ( D ) 2 / 4 ( h)   I 9 2  64 (2 ft ) 4 / 72 yP  0  y   [4(2ft)/(3 )] Ay [(1/2) (2ft) 2 ][4(2ft)/(3 )] yP  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  0.848 ft ) of the gate. Thus, summing moments: ∑ Mhinge = 0 ; _________________________________________ P(D/2) – F(yp – h) = 0 P(D/2) – {(γ)(h)[π(D)2/4][D2/(16h)]} = 0; 2.5.2 F    h  A = (9790 N/m3)[(9 m)/2]∙[(9 m)(1 m)] P = (1/32)(γ)(π)(D)3 _________________________________________ F = 3.96 x 105 N per meter of length 2.5.5 yP    The hydrostatic force and its locations are: I0 (1m)(9m) 3 / 12 y  4.5 m (9m)(1m)(4.5m) Ay yp = 6.0 m F    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 (∑MA), the weight acts to stabilize the dam (called a righting moment) and the hydrostatic force tends to tip it over (overturning moment). ∑MA = (Wt.)[(2/3)(4.5 m)] – (F)(3 m) = [1/2 (3 m)(9 m)∙(1 m)](2.78)(9790 N/m )∙(3 m) – 3 (3.96 x 105 N)∙(3 m) = -8.57 x 104 N-m ∑MA = 8.57 x 104 N-m (overturning; dam is unsafe) _________________________________________ 2.5.3 The hydrostatic force and its locations are: F  h  A F = (9790 N/m3)[(5 m)(sin 45˚)](π)(0.5 m)2 F = 2.72 x 104 N = 27.2 kN   I  (1m) 4 / 64 yP  0  y   5m  5.01 m Ay  (1m) 2 / 4 (5m)   F = 841 lb yP  y    I0 (3 ft )(3 ft )3 / 12 ; yp – ? ̿ = 0.500 ft  Ay 9 ft 2 (1.5 ft )   If the gate was submerged by 10 m (to top of the gate): F    h  A = (62.3 lb/ft3)(11.5 ft)(9 ft2) = 6,450 lb yP  y    I0 (3 ft )(3 ft )3 / 12 ; yp – ? ̿ = 0.0652 ft  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    I0 (2 ft )(3 ft )3 / 12 y  6.5  6.62 ft (2 ft )(3 ft )(6.5 ft ) Ay 11 2.5.7 2.5.9 Fleft    h  A = (9790 N/m3)(0.5 m)[(1.41m)(3m)] F    h  A = (9790 N/m3)(2.5 m)[(π){(1.5)2–(0.5)2}m2] Fleft = 20.7 kN (where A is “wet” surface area)   F = 1.54 x 105 N = 154 kN   I0  / 64{(3m) 4  (1m) 4 } y  2.5 m Ay  {(1.5m) 2  (0.5m) 2 } (2.5m) I (3m)(1.41m) 3 / 12 yP  0  y   0.705m (3m)(1.41m)(0.705m) Ay yP  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):   Y’ = 2 m – 1.41 m + 0.940 m = 1.53 m 2.5.10 Fright    h  A = (9790 N/m3)(h/2 m)[(h/sin45˚)(3m)] The total force from fluids A and B can be found as: Fright = 20.8∙h kN 2   FA    h  A = (γA)(hA)∙[π(d)2/4] I (3)(1.41  h) 3 / 12 yP  0  y   0.705  h (3)(1.41 h)(0.705  h) Ay FB    h  A = (γB)(hB)∙[π(d)2/4] yp = (0.940∙h)m; Moment arm of force from hinge: For equilibrium, forces are equal, opposite, & collinear. Y” = 2 m – (h/sin 45)m + (0.940∙h)m = 2m – (0.474∙h)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: ∑Mhinge= 0 (20.8∙h2)[2m – (0.474∙h)] – 20.7(1.53) – 32.3(0.707) = 0 h = 1.40 m (gate opens when depth exceeds 1.40 m) _________________________________________ (γA)(hA)∙[π(d)2/4] = (γB)(hB)∙[π(d)2/4] hA = [(γB)/(γA)](hB) ___________________________________________ 2.5.11 F    h  A = (62.3 lb/ft3)[(20 ft)]∙[(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  0  y   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 (0-0’) should be 10 ft – 6.89 ft = 3.11 ft above the bottom of the gate. Summing vertical forces where T is the lifting force: ∑ Fy = 0; T – W – 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    h  A =(9790 N/m3)(1.5m)[(10m)(5m)] = 734 kN yP    I0 (10m)(5m)3 / 12 y  2.5m = 3.33 m (10m)(5m)(2.5m) Ay ∑ Mhinge = 0; (734 kN)(3.33 m) – W(2 m) = 0; W = 1,220 kN 12 2.5.13 2.6.1 F    h  A = (62.3 lb/ft3)[(d/2)ft]∙[{(d /cos30)ft}(8ft)] The resultant force in the horizontal direction is zero (FH = 0) since equal pressures surround the viewing F = 288∙d2 lbs I0 y Ay yP  yP  window in all directions. For the vertical direction, the force equals the weight of the water above the window. (8)(d / cos30) /12 3 (8)(d / cos30)(d / 2cos30) FV   Vol = (1.03)(62.3 lb/ft3)[π(1 ft)2(6 ft) –  (d / 2cos30) (1/2)(4/3)π(1 ft)3] = 1,080 lbs ___________________________________________ yp = [(0.192∙d) + 0.577∙d)] ft = 0.769∙d 2.6.2 Thus, summing moments: ∑ Mhinge = 0 Obtain the horizontal component of the total hydrostatic (288∙d2)[(d/cos30)-0.769d] – (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    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 → ___________________________________________ 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    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   3  I0 (1)(h) /12 y  h / 2 = (2/3)h m (1)(h)(h / 2) Ay The force upward on the bottom of the gate:: F = p∙A = (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; ∑ Mh = 0 [(4.90 x 10 )h ][h – (2/3)h] – [(9.79 x 10 )h](0.5) = 0 3 h = 1.73 m 2 3 FV    Vol = (0.82)(9790N/m3)[(4m)(2m)+π/4(2m)2](8m) FV = 7.16 x 105 N = 716 kN ↓ The total force is F = [(642 kN)2 + (716 kN)2]1/2 = 962 kN θ = tan-1 (FV/FH) = 48.1˚ 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 =   h  A = (62.3 lb/ft3)(4 ft)[π(1 ft)2] = 783 lbs ← F = [(783 lbs)2 + (261 lbs)2]1/2 = 825 lbs θ = tan-1 (FV/FH) = 18.4˚ 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/ γ = [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 =   Vol = (0.9)(9790 N/m3)[(31.2 m – 0.75 m) – (1/2)π(1.0 m)2] (2.0 m)(8.0 m) FV = 4.07 x 106 N ↑; 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 45˚) = (40 ft)(sin 45˚) = 28.3 ft. Now, FV=   Vol = (9790 N/m3)[(1/2)(4/3)π(1m)3] = 6530 N __________________________________________ FH =   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 =   h  A = (1.03)(9790 N/m3)(3.875 m)· [(1.75 m)(1m)] = 68.4 kN← (per unit length) FV =   Vol = (1.03)(9790 N/m )[(3 m)(1.75 m)(1 m) 3 + π/4(1.75 m) ] (1 m) = 77.2 kN↑; 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) – (π/8)(40ft)2](33 ft) = 3,410 ft3 2.6.6 FH is the total pressure on the vertical projection of the FV =   Vol = (62.3 lb/ft3)(3410 ft3) = 2.12 x 105 lb; curved surface ABC. The total force is FH    h  A = (62.3lb/ft3)(2ft)[(4ft)(1ft)] = 498 lbs→ 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: θ = tan-1 (FV/FH) = 14.4˚ Vol = (Aquaarter circle + Arectangle – Aquarter circle )(unit length) Vol = (Arectangle)(unit length) = (4ft)(2ft)(1ft) = 8 ft FV    Vol = (62.3 lb/ft3)[8 ft3] = 498 lbs↓ 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    h  A = (62.3 lb/ft3)(7.0 ft)[(8.0 ft)(1.0 ft)] FH    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 → (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) + (53.1˚/360˚)π(10 ft)2 – (1/2)(8 ft)(6 ft)](1 ft) = 34.3 ft3 FV    Vol = (62.3 lb/ft )(34.3 ft ) = 2,140 lb; Thus, 3 3 yP    I0 (1 ft )(12.4 ft )3 / 12 y  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    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 θ = tan-1 (FV/FH) = 31.5˚ Since all pressure vectors pass through the center of the gate radius, the resultant force will do the same. ___________________________________________ 2.6.10 FVquadrant    Vol = (62.3 lb/ft3)[(π/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 =   h  A = (9.79 kN/m )(4 m)[π(1 m) ] = 123 kN The cone unplugs when ∑Fy = 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 =   h  A = (9.79 kN/m3)(4m)(8m2) = 313 kN FATop    Vol = (9790 N/m3)[π(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    Vol = [π(0.15m)2(0.3m) + (π/3) (0.15m)2(0.3m) – (π/3) (0.05m)2(0.1m)](9790 N/m3) = 274 N FVS =   Vol = (9.79 kN/m3)[π(1m)2/2](4m) = 61.5 kN↓ 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    Vol = (0.8)(9790 N/m3) [π(0.05m)2(1.5m) + θ = tan (FV/FH) = 11.1˚ -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. (π/3) (0.05m)2(0.1m)] = 94.3 N The weight of the cone is unknown. Solving yields γ: ∑Fy = 208 – 274 – 94.3 + (γcone)(π/3) (0.15m)2(0.3m) = 0; γcone = 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/γ = (8,500 N/m2)/[(0.8)(9790 N/m3)] = 1.09 m B = γ∙Vol = (1.26)(9790 N/m3)[π(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    Vol = (0.8)(9790 N/m3) [π(0.05m)2(1.09m) + (π/3) (0.05m)2(0.1m)] = 69.1 N 2.8.2 ∑Fy = 208 – 274 – 69.1 + (γcone)(π/3) (0.15m)2(0.3m) = 0; Neutral buoyancy Weight (W) = Buoyant force (B) γcone = 19,100 N/m3; S.G. = 1.95 __________________________________________ Also, B = wt. of water displaced = γ∙Vol ; therefore, W = B = (62.4 lb/ft3)(4/3)π[(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    h  A = (0.9)(9790 N/m3)(6 m)[(1 m)(1.41 m)] γb = wt/Vol = 12lb/{(4/3)π[(4.3/12)ft]3} = 62.3 lb/ft3 FHALeft = 74.5 kN Finally, S.G. = γb/γ = 62.3 lb/ft3/62.4 lb/ft3 = 1.0 __________________________________________ FHARight    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    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) – π/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)+π/4(0.707m) (1.0m)] the y-direction occurs with W = B. Therefore, FVABottom = 40.8 kN W = γsphere [(4/3)π(0.14m)3] + γbuoy[π(0.25m)2(1.5m)] FVBBottom = (1.5)(9790N/m3)[(0.707m)(1m)(5m)+π/4(0.707m)2(1.0m)] W = (13.5γ)[0.0115m3] + (S.G.)γ[0.295m3] 3 2 FVBBottom = 57.7 kN WCylinder = (2.0)(9790N/m3)[π(0.707m)2(1.0m)] = 30.7 kN Thus, FV = 0.2 kN W = 0.155γ + 0.295(S.G.)γ B = γ[(4/3)π(0.14m)3] + γ[π(0.25m)2(1.33m)] = 0.273γ 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 γA(H)(w)(L) + γB(1.5∙H)(w)(L) = γ (2∙H)(w)(L) γA + (2γA)(1.5) = γ(2); γA (1 + 3) = γ (2); γA = 0.5γ; and since γB = 2∙γA = γ __________________________________________ 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 θ) = 109 lb/sin θ; B = 109 lb/sin θ; The buoyant force acts at the center of the submerged portion. W = 165 lb ∑Mhinge = 0, and assuming the rod is homogeneous, (109 lb/sin θ)(3.5 ft/tan θ) – (165 lb)[(6 ft)(cos θ)] = 0 When the anchor is lifted off the bottom, equilibrium in the y-direction occurs (∑Fy = 0). Therefore, T (sin60˚) + B = W; where T = anchor line tension Noting that tan θ = (sin θ/cos θ) and dividing by cos θ (sin θ)2 = 0.385; sin θ = 0.621; θ = 38.4˚ __________________________________________ B = (62.3 lb/ft3)π(0.75 ft)2(1.2 ft) = 132 lb 2.8.9 W = (SG)(62.3 lb/ft3)π(0.75 ft)2(1.2 ft) = 132(SG) The center of gravity (G) is given as 6 ft up from the Substituting: 244(sin60°) + 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  GB ; where Io is the waterline Vol 2.8.7 GM  MB  GB  The hydrostatic force on the gate: F    h  A moment of inertia about the tilting axis. Chopping off F = (9790 N/m3)(1.5m)[(1m/sin45˚)2] = 29.4 kN a circle with a 30 ft diameter. Thus, I0 y Inclined location from water surface: y P  Ay yP  (1m / sin 45)(1m / sin 45) /12 3 (1m / sin 45)(1m / sin 45)(1.5 / sin 45)  (1.5 / sin 45) yp = 2.20 m. The perpendicular distance from hinge is: y’ = (2m/sin45˚) – 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     30 ft  / 64 I0  GB   8.0 ft = 10.0 ft; Vol  15 ft 2 28 ft  4 Note: Vol is the submerged volume and a positive sign is used since G is located below the center of buoyancy. M  W  GM  sin  M =[(1.03)(62.3 lb/ft3)(π)(15ft)2(28ft)](10ft)(sin θ) B = γ∙Vol = (9790 N/m3)(1/2)(4/3)π(R)3 = 20.5(R)3 kN M = 1.11 x 106 ft-lb (for a heel angle of 5˚) ∑Mhinge = 0, ignoring the buoy weight M = 2.21 x 106 ft-lb (for a heel angle of 10˚) 29.4kN(0.628 m) + 6kN(0.5m) – [20.5(R)3kN](1m) = 0 M = 3.29 x 106 ft-lb (for a heel angle of 15˚) 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  MB  GB  I0  GB ; where Io is the waterline Vol moment of inertial about the tilting axis. Chopping off (Vols)(SGs)(γ) + (Volp)(SGp)(γ) = (Vols)(γ) + (Volp’)(γ) the tube at the waterline and looking down we have a [(4/3)π(0.25m)3](1.4)+[π(0.125m)2(2m)](0.62) = circle with a 36 ft diameter. Thus, [(4/3)π(0.25m)3]+[π(0.125m)2(h)]; GM  0.0916m3 + 0.0609m3 = 0.0654m3 + (0.0491m2)h h = 1.77 m; Find “B” using the principle of moments. [(Vols)(γ)+(Volp’)(γ)](hb)=(Vols)(γ)(h+0.25m)+(Volp’)(γ)(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 “G” 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  36 ft 4 / 64  GB   4.0 ft = 5.93 ft; Vol  / 436 ft 2 42 ft  Note: Vol is the submerged volume and a positive sign is used since G is located below the center of buoyancy. M  W  GM  sin  M = [(1.02)(62.3 lb/ft3)(π/4)(36 ft)2(42 ft)](5.93 ft)(sin 4˚) M = 1.12 x 106 ft∙lb (for a heel angle of 4˚) __________________________________________ 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)π(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    I0 12m 4.8m 3 / 12  0.3m = 0.660 m;  GB  12m 4.8m 2m  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  W  GM  sin  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 15˚) gravity, then GM = 0 and the righting moment is M  W  GM  sin  = 0. With no righting moment, the block will not be stable. is used since G is located above the center of buoyancy. M = 198 kN∙m; The distance G can be moved is d = GM (sin θ) = (0.66 m)(sin 15˚) = 0.171 m 18