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

Solution Manual for Fundamentals of Hydraulic Engineering Systems, 5th Edition makes tackling textbook exercises a breeze, with clear and concise answers to every problem.

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    SOLUTIONS MANUAL

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    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

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    _________________________________________ _________________________________________ Chapter 1 – Problem Solutions 1.2.1 E 1 = energy released in lowering steam temperature to 100  C from 110°C E 1 = (500 L)(1000 g/L)(10  C)(0.432 cal/g∙  C) E 1 = 2.16x10 6 cal E 2 = energy released when the steam liquefies E 2 = (500 L)(1000 g/L)(597 cal/g) E 2 = 2.99x10 8 cal E 3 = energy released when the water temperature is lowered from 100°C to 50  C E 3 = (500 L)(1000 g/L)(50  C)(1 cal/g∙  C) E 3 = 2.50x10 7 cal; Thus, the total energy released is: E total = E 1 + E 2 + E 3 = 3.26x10 8 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 E 1 = energy required to bring the water temperature to 95  C from 15°C E 1 = (900 g)(95  C - 15  C)(1 cal/g∙  C) E 1 = 7.20x10 4 cal E 2 = energy required to vaporize the water E 2 = (900 g)(597 cal/g) E 2 = 5.37x10 5 cal E total = E 1 + E 2 = 6.09x10 5 cal 1.2.3 E 1 = energy needed to vaporize the water E 1 = (300 L)(1000 g/L)(597 cal/g) E 1 = 1.79x10 8 cal The energy remaining (E 2 ) is: E 2 = E Total – E 1 E 2 = 2.00x10 8 cal – 1.79x10 8 cal E 2 = 2.10x10 7 cal The temperature change possible with the remaining energy is: 2.10x10 7 cal = (300 L)(1000 g/L)(1 cal/g∙  C)(  T)  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 E 1 = energy required to warm and then melt the ice E 1 = (10 g)(6  C)(0.465 cal/g∙  C) + 10g(79.7 cal/g) E 1 = 825 cal. This energy is taken from the water. The resulting temperature of the water will decrease to: 825 cal = (0.165 L)(1000 g/L)(20  C - T 1 )(1 cal/g∙  C) T 1 = 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: 1 [(0.165 L)(1000 g/L)(15.0  C - T 2 )(1 cal/g∙  C)] = [(10 g)(T 2 - 0  C)( 1 cal/g∙  C)] ; T 2 = 14.1  C

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    _________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ 1.2.5 E 1 = energy required to melt ice E 1 = (5 slugs)(32.2 lbm/slug)(32  F - 20  F)* (0.46 BTU/lbm∙  F) + (5 slugs)(32.2 lbm/slug)* (144 BTU/lbm) E 1 = 2.41 x 10 4 BTU.  Energy taken from the water. The resulting temperature of the water will decrease to: 2.41 x 10 4 BTU = (10 slugs)(32.2 lbm/slug)(120  F – T 1 )(1 BTU/lbm∙  F) T 1 = 45.2  F The energy lost by the water (to lower its temp. to 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. Therefore, the final temperature of the water is: [(10 slugs)(32.2 lbm/slug)(45.2  F – T 2 )(1 BTU/lbm∙  F)] = [(5 slugs)(32.2 lbm/slug)(T 2 - 32  F)(1 BTU/lbm∙  F)] T 2 = 40.8  F 1.2.6 E 1 = energy required to raise the temperature to 100  C E 1 = (7500 g)(100  C – 20  C)(1 cal/g∙  C) E 2 = 6.00x10 5 cal E 2 = energy required to vaporize 2.5 kg of water E 2 = (2500 g)(597 cal/g) E 2 = 1.49x10 6 cal E total = E 1 + E 2 = 2.09x10 6 cal Time required = (2.09x10 6 cal)/(500 cal/s) Time required = 4180 sec = 69.7 min 1.3.1 F = m∙a; Letting a = g yields: W = m∙g, (Eq’n 1.1) Then dividing both sides of the equation by volume, W/Vol = (m/Vol)∙g; γ = ρ∙g 1.3.2 SG oil = 0.976 = γ oil /γ; where γ is for water at 4°C: γ = 9,810 N/m 3 (Table 1.2). Substituting yields, 0.977 = γ oil /9,810; γ oil = (9810)(0.976) = 9,570 N/m 3 Also, γ = ρ∙g; or ρ = γ oil /g Substituting (noting that N ≡ kg·m/sec 2 ) yields, ρ oil = γ oil /g = (9,570 N/m 3 ) / (9.81 m/sec 2 ) = 976 kg/m 3 1.3.3 By definition,  = W/Vol = 55.5 lb/ft 3 ; thus, W =  ∙Vol = (55.5 lb/ft 3 )(20 ft 3 ) = 1,110 lb (4,940 N)  =  /g = (55.5 lb/ft 3 )/(32.2 ft/s 2 ) = 1.72 slug/ft 3 (887 kg/m 3 ) SG =  liquid /  water at 4  C = (55.5 lb/ft 3 )/(62.4 lb/ft 3 ) = 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 m 3 ) = 9.10 x 10 3 N/m 3 ρ = γ/g = (9.1 x 10 3 N/m 3 )/(9.81 m/sec 2 );  = 928 kg/m 3 (Note: 1 N ≡ 1 kg·m/sec 2 ) 2 Specific gravity (SG) = γ/γ water at 4°C SG = (9.10 x 10 3 N/m 3 )/9.81 x 10 3 N/m 3 ) SG = 0.928

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    3 1.3.5 The force exerted on the tank bottom is equal to the weight of the water body (Eq’n 1.2). F = W = mg = [ρ(Vol)] (g); ρ found in Table 1.2 920 lbs = [1.94 slugs/ft 3 (π ∙(1.25 ft) 2 ∙ d )] (32.2 ft/sec 2 ) d = 3.00 ft (Note: 1 slug = 1 lb∙sec 2 /ft) ___________________________________________ 1.3.6 Weight of water on earth = 8.83 kN From E’qn (1.1): m = W/g = (8,830 N)/(9.81 m/s 2 ) m = 900 kg Note: mass on moon is the same as mass on earth W (moon) = mg = (900 kg)[(9.81 m/s 2 )/(6)] W(moon) = 1,470 N _________________________________________ 1.3.7 Density is expressed as ρ = m/Vol, and even though volume changes with temperature, mass does not. Thus, (ρ 1 )(Vol 1 ) = (ρ 2 )(Vol 2 ) = constant; or Vol 2 = (ρ 1 )(Vol 1 )/(ρ 2 ) Vol 2 = (999 kg/m 3 )(100 m 3 )/(996 kg/m 3 ) Vol 2 = 100.3 m 3 (or a 0.3% change in volume) ____________________________________________ 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 2 ) [(1 m)/(3.281 ft)] 2 [(1 ft)/(12 in)] 2 · [(1 lb)/(4.448 N)] = 1.450 x 10 -4 psi 1.4.1 (a) Note that: 1 poise = 0.1 N  sec/m 2 . Therefore, 1 lb·sec/ft 2 [(1 N)/(0.2248 lb)]·[(3.281 ft) 2 /(1 m) 2 ] = 47.9 N  sec/m 2 [(1 poise)/(0.1 N  sec/m 2 )] = 478.9 poise Conversion: 1 lb  sec/ft 2 = 478.9 poise (b) Note that: 1 stoke = 1 cm 2 /sec. Therefore, 1 ft 2 /sec [(12 in) 2 /(1 ft) 2 ]· [(1 cm) 2 /(0.3937 in) 2 ] = 929.0 cm 2 /sec [(1 stoke)/(1 cm 2 /sec)] = 929.0 stokes Conversion: 1 ft 2 /sec = 929.0 stokes _________________________________________ 1.4.2 [  (air)/  (H 2 O)] 0  C = (1.717x10 -5 )/(1.781x10 -3 ) [  (air)/  (H 2 O)] 0  C = 9.641x10 -3 [  (air)/  (H 2 O)] 100  C = (2.174x10 -5 )/(0.282x10 -3 ) [  (air)/  (H 2 O)] 100  C = 7.709x10 -2 [  (air)/  (H 2 O)] 0  C = (1.329x10 -5 )/(1.785x10 -6 ) [  (air)/  (H 2 O)] 0  C = 7.445 [  (air)/  (H 2 O)] 100  C = (2.302x10 -5 )/(0.294x10 -6 ) [  (air)/  (H 2 O)] 100  C = 78.30 Note: The ratio of the viscosity of air to water increases with temperature. Why? Because the viscosity of air increases with temperature and that of water decreases with temperature magnifying the effect. Also, the values of kinematic viscosity (  ) for air and water are much closer than those of absolute viscosity. Why? _________________________________________ 1.4.3  20  C = 1.002x10 -3 N  sec/m 2 ;  20  C = 1.003x10 -6 m 2 /s (1.002x10 -3 N  sec/m 2 )∙[(0.2248 lb)/(1 N)]∙ [(1 m) 2 /(3.281 ft) 2 ] = 2.092x10 -5 lb  sec/ft 2 (1.003x10 -6 m 2 /s)[(3.281 ft) 2 /(1 m) 2 ] = 1.080x10 -5 ft 2 /s

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    4 1.4.4 _________________________________________ Using Newton’s law of viscosity (Eq’n 1.2):  =  (dv/dy) =  (Δv/Δy)  = (1.00 x 10 -3 N  sec/m 2 )[{(4.8 – 2.4)m/sec}/(0.02 m)]  = 0.12 N/m 2 1.4.5 From Eq’n (1.2):  =  (  v/  y) = _________________________________________  = (0.0065 lb  sec/ft 2 )[(1.5 ft/s)/(0.25/12 ft)]  = 0.468 lb/ft 2 F = (  )(A) = (2 sides)(0.468 lb/ft 2 )[(0.5 ft)(1.5 ft)] F = 0.702 lb 1.4.6 Summing forces parallel to the incline yields: T shear force = W(sin15  ) =  A =  (  v/  y)A  y = [(  )(  v)(A)] / [(W)(sin15  )]  y = [(1.52 N  sec/m 2 )(0.025 m/sec)(0.80m)(0.90m)]/ [(100 N)(sin15  )]  y = 1.06 x 10 -3 m = 1.06 mm ___________________________________________ 1.4.7 Using Newton’s law of viscosity (Eq’n 1.2):  =  (dv/dy) =  (  v/  y)  = (0.04 N  sec/m 2 )[(15 cm/s)/[(25.015 – 25)cm/2]  = 80 N/m 2 F shear resistance =  A = (80 N/m 2 )[(  )(0.25 m)(3 m)] F shear resistance = 188 N 1.4.8 v = y 2 – 3y, where y is in inches and v is in ft/s v = 144y 2 – 36y, where y is in ft and  is in ft/s Taking the first derivative: dv/dy = 288y – 36 sec -1 _________________________________________  =  (dv/dy) = (8.35 x 10 -3 lb-sec/ft 2 )( 288y – 36 sec -1 ) Solutions: y = 0 ft,  = -0.301lb/ft 2 y = 1/12 ft,  = -0.100 lb/ft 2 ; y = 1/6 ft,  = 0.100 lb/ft 2 y = 1/4 ft,  = 0.301 lb/ft 2 ; y = 1/3 ft,  = 0.501 lb/ft 2 1.4.9  = (16)(1.00x10 -3 N  sec/m 2 ) = 1.60x10 -2 N  sec/m 2 Torque = dA r dF r R R       0 0 ) (     R dA y v r 0 ) )( )( (  Torque = _________________________________________    R dr r y r r 0 ) 2 )( 0 ) )( ( )( )( (    Torque =   R dr r y 0 3 ) ( ) )( )( 2 (    Torque =          4 ) 1 ( 0005 . 0 sec) / 65 . 0 )( sec/ 10 60 . 1 )( 2 ( 4 2 2 m m rad m N  Torque = 32.7 N  m 1.4.10  =  /(dv/dy) = (F/A)/(  v/  y); Torque (T) = Force∙distance = F∙R where R = radius Thus;  = (T/R)/[(A)(  v/  y)]  = ) )( )( )( 2 ( ) / )( )( )( 2 ( / 3     h R y T y R h R R T       =          rpm rad rpm ft ft ft ft lb 60 sec / 2 ) 2000 )( ) 12 / 6 . 1 (( ) ) 12 / 1 )(( 2 ( ] ) 12 / 008 . 0 )[( 10 . 1 ( 3    = 7.22x10 -3 lb  sec/ft 2

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    5 _________________________________________ 1.5.1 The concept of a line force is logical for two reasons: 1) The surface tension acts along the perimeter of the tube pulling the column of water upwards due to adhesion between the water and the tube. 2) The surface tension is multiplied by the tube 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/m 3 )(1.0x10 -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. 1.5.3 For capillary rise, apply Equation 1.3: h = [(4)(  )(sin θ)] / [(  )(D)] But sin 90˚ = 1,  = 62.3 lb/ft 3 (at 20˚C), and σ = 4.89x10 -3 lb/ft (from inside book cover) thus, D = [(4)(  )] / [(  )(h) ]; for h = 1.5 in. D = [(4)(4.89x10 -3 lb/ft)] / [(62.3 lb/ft 3 )(1.5/12)ft] D = 2.51 x 10 -3 ft = 3.01 x 10 -2 in.; thus, for h = 1.5 in., D = 2.51x10 -3 ft = 0.0301 in. for h = 1.0 in., D = 3.77x10 -3 ft = 0.0452 in. for h = 0.5 in., D = 7.54x10 -3 ft = 0.0904 in. 1.5.4 Condition 1: h 1 = [(4)(  1 )(sin  1 )] / [(  )(D)] h 1 = [(4)(  1 )(sin30  )] / [(  )(0.8 mm)] Condition 2: h 2 = [(4)(  2 )(sin  2 )] / [(  )(D)] h 2 = [(4)(0.88  1 )(sin50  )] / [(  )(0.8 mm)] h 2 /h 1 = [(0.88)(sin50°)] / (sin30  ) = 1.35 alternatively, h 2 = 1.35(h 1 ), 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 10 4 N/m 3 h =[(4)( 8.28 x 10 -2 N/m)(sin 35)] / [(1.00 x 10 4 N/m 3 )(0.012m)] h = 1.58x10 -3 m = 0.158 cm _________________________________________ 1.5.6  P R  P = P i – P e (internal pressure minus external pressure)  F x = 0;  P(  )(R 2 ) - 2  (R)(  ) = 0  P = 2  /R

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    6 1.6.1 P i = 1 atm = 14.7 psi. and P f = 220 psi From Equation (1.4):  Vol/Vol = -  P/E b  Vol/Vol = -(14.7 psi – 220 psi)/(3.2x10 5 psi)  Vol/Vol = 6.42 x 10 -4 = 0.0642% (volume decrease)  /  = -  Vol/Vol = -0.0642% (density increase) _________________________________________ 1.6.2 m = W/g = (7,490 lb)/(32.2 ft/s 2 ) = 233 slug  = m/Vol = (233 slug)/(120 ft 3 ) = 1.94 slug/ft 3  Vol = (-  P/E b )(Vol)  Vol = [-(1470 psi –14.7 psi)/(3.20x10 5 psi)](120 ft 3 )  Vol = -0.546 ft 3  new = (233 slug)/(120 ft 3 – 0.546 ft 3 ) = 1.95 slug/ft 3 Note : The mass does not change. _________________________________________ 1.6.3 Surface pressure: P s = 1 atm = 1.014 x 10 5 N/m 2 Bottom pressure: P b = 1.61 x 10 7 N/m 2 From Equation (1.4):  Vol/Vol = -  P/E b  Vol/Vol = [-(1.014 x 10 5 - 1.61 x 10 7 )N/m 2 ] (2.2x10 9 N/m 2 )  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: γ s = 9,810 N/m 3 Specific weight at the bottom: γ b = (9,810 N/m 3 )(1.00727) = 9,880 N/m 3 Note : These answers assumes that E b holds constant for this great change in pressure. 1.6.4 P i = 30 N/cm 2 = 300,000 N/m 2 = 3 bar  P = 3 bar – 30 bar = -27 bar = -2.7x10 5 N/m 2 Amount of water that enters pipe =  Vol Vol pipe = [(  )(1.50 m) 2 /(4)]∙(2000 m) = 3530 m 3  Vol = (-  P/E b )(Vol)  Vol = [-(-2.7x10 5 N/m 2 )/(2.2x10 9 N/m 2 )]*(3530 m 3 )  Vol = 0.433 m 3 Water in the pipe is compressed by this amount. Thus, the volume of H 2 O that enters the pipe is 0.433m 3

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    7 Chapter 2 – Problem Solutions 2.2.1 a) The depth of water is determined using the relationship between weight and specific weight. W = γ· Vol = γ (A · h); where h = water depth 14,700 lbs = (62.3 lb/ft 3 )[ π·( 5 ft) 2 ](h); h = 3.00 ft 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: P = γ∙h = (62.3 lb/ft 3 )(3 ft) = 187 lb/ft 2 2.2.2 ___________________________________________ The absolute pressure includes atmospheric pressure. Therefore, P abs = P atm + (  water )(h)  5(P atm ); Thus from Eq’n 2.4: h = 4(P atm )/  water h = 4(1.014 x 10 5 N/m 2 )/(1.03)(9790 N/m 3 ) h = 40.2 m (132 ft) 2.2.3  water at 30  C = 9.77 kN/m 3 (from Table 1.2) P vapor at 30  C = 4.24 kN/m 2 (from Table 1.1) P atm = P column + P vapor P atm = (8.7 m)(9.77 kN/m 3 ) + (4.24 kN/m 2 ) P atm = 85.0 kN/m 2 + 4.24 kN/m 2 = 89.2 kN/m 2 The percentage error if the direct reading is used and the vapor pressure is ignored is: Error = (P atm - P column )/(P atm ) Error = (89.2 kN/m 2 – 85.0 kN/m 2 )/(89.2 kN/m 2 ) Error = 0.0471 = 4.71% 2.2.4 Since mercury has a specific gravity of 13.6, the water height can be found from: h water = (h Hg )(SG Hg ) h water = (30 mm)(13.6) = 408 mm = 40.8 cm of water _________________________________________ Also, absolute pressure is: P abs = P gage + P atm P abs = 0.408 m + 10.3 m = 10.7 m of water P abs = (10.7 m)(9790 N/m 3 ) = 1.05 x 10 5 N/m 2 2.2.5 Since force equals pressure on the bottom times area: P = γ∙h = (62.3 lb/ft 3 )(10 ft) = 623 lb/ft 2 ___________________________________________ F bottom = P∙A = (623 lb/ft 2 )(100 ft 2 ) = 6.23 x 10 4 lbs Pressure varies linearly with depth. So the average pressure on the sides of the tank occur at half the depth. P avg = γ∙h = (62.3 lb/ft 3 )(5 ft) = 312 lb/ft 2 Now, F side = P avg ∙A = (312 lb/ft 2 )(100 ft 2 ) = 3.12 x 10 4 lbs 2.2.6 Pressure and force on the bottom of both containers is: P = (  water )(h) = (9790 N/m 3 )(10 m) = 97.9 kN/m 2 Also, F = P∙A = (97.9 kN/m 2 )(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.)

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    8 ___________________________________________ (Solution 2.2.6 cont.) W 1 W 2 P 1 · A 1 Note the three vertical forces will be opposed at the bottom. W 1 is the weight of the water column above it. W 2 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 = γ h). This is opposed by an equal and opposite pressure on the water below. The resulting force is P 1 · A 1 which equals γ· h · A 1 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 _________________________________________  seawater = (SG)(  ) = (1.03)(62.3 lb/ft 3 ) = 64.2 lb/ft 3 P tank = (  seawater )(  h) = (64.2 lb/ft 3 )(18 ft)(1 ft 2 /144 in 2 ) P tank = 8.03 psi = 5.54 x 10 4 N/m 2 (Pascals) 2.2.8 P bottom = P gage + (  liquid )(1.4 m); and  liquid = (SG)(  water ) = (0.85)(9790 N/m 3 ) = 8,320 N/m 3  P bottom = 3.55x10 4 N/m 2 + (8320 N/m 3 )(1.4 m) P bottom = 4.71 x 10 4 N/m 2 The pressure at the bottom of the liquid column can be determined two different ways which must be equal. (h)(  liquid ) = P gage + (  liquid )(1 m) h = (P gage )/(  liquid ) + 1m h = (3.55x10 4 N/m 2 )/8320 N/m 3 + 1 m = 5.27 m 2.2.9  oil = (SG)(  water ) = (0.85)( 62.3 lb/ft 3 ) = 53.0 lb/ft 3 P 10ft = P air + (  oil )(12 ft) or P air = P 10ft - (  oil )(12 ft) P air = 25.7 psi (144 in 2 /ft 2 ) – (53.0 lb/ft 3 )(12 ft) P air = 3.06 x 10 3 lb/ft 2 (21.3 psi); Gage pressure P abs = P gage + P atm = 21.3 psi + 14.7 psi ___________________________________________ P abs = 36.0; Absolute pressure 2.2.10 ∑F x = 0; (P x )(Δy) – (P s )(Δs)(sin θ) = 0 since (Δs∙sin θ) = Δy, P s = P x ∑F y = 0; and accounting for weight yields: (P y )(Δx) – (P s )(Δs)( cos θ) – (Δx∙Δy/2)(Δz)(γ) = 0 since (Δs∙cos θ) = Δx and (Δy)(Δx)(Δz)  0; P s = P y Hence, the pressure is the same in all 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, F input = (9)(50 N) = 450 N; The pressure developed in the system is therefore: P system = F/A = (450 N)/(25 cm 2 ) = 18 N/cm 2 P system = 180 kN/m 2 = 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: P input = P output equates to: 18 N/cm 2 = F output /250 cm 2 F output =(18 N/cm 2 )(250 cm 2 )(1 kN/1000 N) = 4.50 kN

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    9 2.4.1 The gage pressure at the bottom of the tank is equal to the pressure due to the liquid heights. Thus, (h Hg )(SG Hg )(  ) = (5h)(  ) + (h)(SG oil )(  ) h = (h Hg )(SG Hg )/(5 + SG oil ) h = (1.43 in.)(13.6)/(5 + 0.80) = 3.35 inches ____________________________________________ 2.4.2 _________________________________________ Note the equal pressure surface; P 7 = P 8 or, (h water )(  ) = (h oil )(  oil ) = (h oil )(  )(SG oil ), thus h water = (h oil )/(SG oil ) = (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, P left = P right (1 m)(  ) = (h)(  Hg ) + (0.6 m)(  oil ) (1 m)(  ) = (h)(SG Hg )(  ) + (0.6 m)(SG oil )(  ); h = [1 m – (0.6 m)(SG oil )]/(SG Hg ) h = [1m – (0.6 m)(0.79)]/(13.6) = 0.0387m = 3.87 cm _________________________________________ 2.4.4 Equal pressure surface at Hg-H 2 O interface: Thus, (3 ft)(  Hg ) = P + (2 ft)(  ) where  Hg = (SG Hg )(  ) (3 ft)(13.6)(62.3 lb/ft 3 ) = P + (2 ft)(62.3 lb/ft 3 ); P = 2,420 lb/ft 2 = 16.8 psi Pressure can be expressed as a fluid height: h = P/  Hg = (2420 lb/ft 2 )/[(13.6)(62.3 lb/ft 2 )] h = 2.86 ft of Hg (or 34.3 inches) 2.4.5 Note the equal pressure surface: P 1 = P 2 Therefore, P A + (y)(  ) = (h)(  Hg ) P A + [(1.34/12)ft](  ) = [(1.02/12)ft](  Hg ) P A = [(1.02/12)ft](13.6)(62.3 lb/ft 3 ) - [(1.34/12)ft](62.3 lb/ft 3 ) P A = 65.1 lb/ft 2 = 0.452 psi _________________________________________ 2.4.6 A surface of equal pressure surface can be drawn at the mercury-water meniscus. Therefore, P pipe + (h 1 )(  ) = (h 2 )(  Hg ) = (h 2 )(SG Hg )(  ) ___________________________________________ P pipe + (0.18 m)(9790 N/m 3 ) = (0.60 m)(13.6)(9790 N/m 3 ) P pipe = 7.81 x 10 4 N/m 2 (Pascals) = 78.1 KPa 2.4.7 Use the swim-through-technique : Start at the end of the manometer open to the atmosphere (P gage = 0). Then “swim through” adding pressure when “swimming” down and subtracting when “swimming” up until you reach the pipe (P pipe ). 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 ) = P pipe The specific weight of air is negligible when compared to fluids, so that term in the equation can be dropped. P pipe = 0 - (0.66 m)(SG ml )(  ) – (0.58 m)(SG oil )(  ) P pipe = 0 - (0.66 m)(0.8)(9790 N/m 3 ) – (0.58 m)(0.82)( 9790 N/m 3 ) P pipe = - 9.83 kN/m 2 (kPa) Pressure can be converted to height (head) of any liquid through P =  ∙h. Thus, h pipe = (-9,830 N/m 2 )/(9790 N/m 3 ) = -1.00 m of water

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    10 2.4.8 A surface of equal pressure surface can be drawn at the mercury-water meniscus. Therefore, P + (h 1 )(  ) = (h 2 )(  Hg ) = (h 2 )(SG Hg )(  ) P + (0.5 ft)(62.3 lb/ft 3 ) = (1.5 ft)(13.6)(62.3 lb/ft 3 ) P = 1,240 lb/ft 2 = 8.61 psi When the manometer reading rises or falls, volume balance is preserved for constant density. Therefore, Vol res = Vol tube or A res ∙h 1 = A tube ∙h 2 h 1 = h 2 (A tube /A res ) = h 2 [(D tube ) 2 /(D res ) 2 ] _________________________________________ h 1 = (4 in.)[(1 in.) 2 /(5 in.) 2 ] = 0.16 in. 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: P A + (5.33 ft)(  oil ) - (1.67ft)(  ct ) - (1.0 ft)(  ) = P B _________________________________________ P A - P B = (62.3 lb/ft 3 ) [(1.0 ft) + (1.67 ft)(1.6) – (5.33)(0.82)] P A - P B = -43.5 lb/ft 2 = -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 P 1 . The computations are: P 2 + (Δh)(ρ 1 ∙g) + (y)(ρ 1 ∙g) + (h)(ρ 2 ∙g) - (h)(ρ 1 ∙g) - (y)(ρ 1 ∙g) = P 1 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, P 1 – P 2 = (Δh)(ρ 1 ∙g) + (h)(ρ 2 ∙g) - (h)(ρ 1 ∙g) 2.4.10 – cont. When the manometer reading (h) rises or falls, volume balance must be preserve in the system. Therefore, Vol res = Vol tube or A res ∙(Δh) = A tube ∙ h Δh = h (A tube /A res ) = h [(d 2 ) 2 /(d 1 ) 2 ]; substituting yields _________________________________________ P 1 – P 2 = h [(d 2 ) 2 /(d 1 ) 2 ] (ρ 1 ∙g) + (h)(ρ 2 ∙g) - (h)(ρ 1 ∙g) P 1 – P 2 = h∙g [ρ 2 - ρ 1 + ρ 1 {(d 2 ) 2 /(d 1 ) 2 }] P 1 – P 2 = h∙g [ρ 2 - ρ 1 {1 - (d 2 ) 2 /(d 1 ) 2 }] 2.4.11 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 E A in the resulting equation. 20 kN/m 2 + (37 m - E A )(9.79 kN/m 3 ) - (35m - E A )(1.6)(9.79 kN/m 3 ) - (5 m)(0.8)(9.79 kN/m 3 ) = -29.0 kN/m 2 E A = 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 P B . 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)(  ) = P A ; P A = 269∙γ 0 + (46)(0.8)(  ) + (20)(13.6)(  ) – (40)(γ) = P B P B = 269∙γ; Therefore, P A = P B and h = 0

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    _________________________________________ _________________________________________ _________________________________________ ___________________________________________ 2.5.1 A h F     = (62.3 lb/ft 3 )[4(2ft)/(3 π )]∙[( 1 /2) π (2ft) 2 ] F = 332 lb (Note: y h  in Table 2.1)   )] [4(2ft)/(3 )] 3 ][4(2ft)/( (2ft) [(1/2) 72 / ) 2 ( 64 9 2 4 2 0           ft y y A I y P y p = 0.853 ft (depth to center of pressure) The center of pressure (0.853 ft) is deeper than the centroid ( ft y 848 . 0  ) of the gate. 2.5.2 A h F     = (9790 N/m 3 )[(9 m)/2]∙[(9 m)(1 m)] F = 3.96 x 10 5 N per meter of length     m m m m m m y y A I y P 5 . 4 ) 5 . 4 ( ) 1 )( 9 ( 12 / ) 9 )( 1 ( 3 0     y p = 6.0 m (depth to the center of pressure) In summing moments about the toe of the dam (∑M A ) , the weight acts to stabilize the dam (called a righting moment) and the hydrostatic force tends to tip it over (overturning moment). ∑M A = (Wt.)[(2/3)(4.5 m)] – (F)(3 m) = [1/2 (3 m)(9 m)∙(1 m)](2.78)(9790 N/m 3 )∙(3 m) – (3.96 x 10 5 N)∙(3 m) = -8.57 x 10 4 N-m ∑M A = 8.57 x 10 4 N-m (overturning; dam is unsafe) 2.5.3 The hydrostatic force and its locations are: A h F     F = (9790 N/m 3 )[(5 m)(sin 45˚)](π)(0.5 m) 2 F = 2.72 x 10 4 N = 27.2 kN          m m m m y y A I y P 5 ) 5 ( 4 / ) 1 ( 64 / ) 1 ( 2 4 0   5.01 m 2.5.4 The hydrostatic force and its locations are: A h F     = (γ)(h)[π(D) 2 /4]     h h D D y y A I y P     ) ( 4 / ) ( 64 / ) ( 2 4 0   ; y p = D 2 /(16h) + h (depth to the center of pressure) Thus, summing moments: ∑ M hinge = 0 ; P(D/2) – F(y p – h) = 0 P(D/2) – {(γ)(h)[π(D) 2 /4][D 2 /(16h)]} = 0; P = (1/32)(γ)(π)(D) 3 2.5.5 The hydrostatic force and its locations are: A h F     = (62.3 lb/ft 3 )(1.5 ft)(9 ft 2 ) F = 841 lb     ) 5 . 1 ( 9 12 / ) 3 )( 3 ( 2 3 0 ft ft ft ft y A I y y P    ; y p – 𝒚̿ = 0.500 ft If the gate was submerged by 10 m (to top of the gate): F A h     = (62.3 lb/ft 3 )(11.5 ft)(9 ft 2 ) = 6,450 lb     ) 5 . 11 ( 9 12 / ) 3 )( 3 ( 2 3 0 ft ft ft ft y A I y y P    ; y p – 𝒚̿ = 0.0652 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, 11          5 . 6 ) 5 . 6 ( ) 3 )( 2 ( 12 / ) 3 )( 2 ( 3 0 ft ft ft ft ft y y A I y P 6.62 ft

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    12 2.5.7 A h F left     = (9790 N/m 3 )(0.5 m)[(1.41m)(3m)] F left = 20.7 kN (where A is “wet” surface area)     m m m m m m y y A I y P 705 . 0 ) 705 . 0 ( ) 41 . 1 )( 3 ( 12 / ) 41 . 1 )( 3 ( 3 0     y p = 0.940 m (inclined distance to center of pressure) Location of this force from the hinge (moment arm): Y’ = 2 m – 1.41 m + 0.940 m = 1.53 m A h F right     = (9790 N/m 3 )(h/2 m)[(h/sin45˚)(3m)] F right = 20.8∙h 2 kN     h h h h y y A I y P         705 . 0 ) 705 . 0 ( ) 41 . 1 )( 3 ( 12 / ) 41 . 1 )( 3 ( 3 0 y p = (0.940∙h)m; Moment arm of force from hinge: Y” = 2 m – (h/sin 45)m + (0.940∙h)m = 2m – (0.474∙h)m 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: ∑M hinge = 0 (20.8∙h 2 )[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) _________________________________________ 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     ft ft ft ft ft ft y y A I y P 6 ) 6 ( ) 8 )( 1 ( 12 / ) 8 )( 1 ( 3 0     y p = 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. 2.5.9 A h F     = (9790 N/m 3 )(2.5 m)[(π){(1.5) 2 –(0.5) 2 }m 2 ] F = 1.54 x 10 5 N = 154 kN     m m m m m m y y A I y P 5 . 2 ) 5 . 2 ( } ) 5 . 0 ( ) 5 . 1 {( } ) 1 ( ) 3 {( 64 / 2 2 4 4 0         y p = 2.75 m (below the water surface) ___________________________________________ 2.5.10 The total force from fluids A and B can be found as: A h F A     = (γ A )(h A )∙[π(d) 2 /4] A h F B     = (γ B )(h B )∙[π(d) 2 /4] For equilibrium, forces are equal, opposite, & collinear. F A = F B ; (γ A )(h A )∙[π(d) 2 /4] = (γ B )(h B )∙[π(d) 2 /4] h A = [(γ B )/(γ A )](h B ) ___________________________________________ 2.5.11 A h F     = (62.3 lb/ft 3 )[(20 ft)]∙[(10 ft)(6 ft)] F = 7.48 x 10 4 lbs (Horizontal force on gate) Summing vertical forces where T is the lifting force: ∑ F y = 0; T – W – F(C friction ) = 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: A h F     =(9790 N/m 3 )(1.5m)[(10m)(5m)] = 734 kN     m m m m m m y y A I y P 5 . 2 ) 5 . 2 ( ) 5 )( 10 ( 12 / ) 5 )( 10 ( 3 0     = 3.33 m ∑ M hinge = 0; (734 kN)(3.33 m) – W(2 m) = 0 ; W = 1,220 kN

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    13 2.5.13 F    h  A = (62.3 lb/ft 3 )[( d /2)ft]∙[{( d /cos30)ft}(8ft)] F = 288∙ d 2 lbs y Ay I  y P  0     (8)( d / cos30  ) ( d / 2cos30  ) 3 (8)( d / cos30  ) /12 y P   ( d / 2cos30) y p = [(0.192∙ d ) + 0.577∙ d )] ft = 0.769∙ d Thus, summing moments: ∑ M hinge = 0 (288∙ d 2 )[( d /cos30)-0.769 d ] – (5,000)(15) = 0 d = 8.77 ft A depth greater than this will make the gate open. Any depth less than this will make it close. ___________________________________________ 2.5.14 The force and location on the vertical side of the gate: F    h  A = (9790 N/m 3 )[( h /2)m][( h m)(1 m)] F = (4.90 x 10 3 ) h 2 N (per meter of gate width)     (1)( h ) ( h / 2) (1)( h ) /12 3 0 h Ay I   y  y P  / 2 = (2/3) h m The force upward on the bottom of the gate:: F = p∙A = (9790 N/m 3 )( h )[(1m)(1m)] F = (9.79 x 10 3 ) h N (per meter of gate width) This force is located 0.50 m from the hinge. Summing moments about the hinge; ∑ M h = 0 [(4.90 x 10 3 ) h 2 ][h - (2/3) h ] – [(9.79 x 10 3 ) h ](0.5) = 0 h = 1.73 m 2.6.1 The resultant force in the horizontal direction is zero ( F H = 0 ) since equal pressures surround the viewing window in all directions. For the vertical direction, the force equals the weight of the water above the window. F V    Vol = (1.03)(62.3 lb/ft 3 )[π(1 ft) 2 (6 ft) – (1/2)(4/3)π(1 ft) 3 ] = 1,080 lbs ___________________________________________ 2.6.2 Obtain the horizontal component of the total hydrostatic pressure force by determining the total pressure on the vertical projection of the curved gate. F H    h  A = (0.82)(9790 N/m 3 )(5m)[(8 m)(2 m)] F H = 6.42 x 10 5 N = 642 kN → The vertical component of the hydrostatic force equals the weight of the water column above the curved gate. Vol F V    = (0.82)(9790N/m 3 )[(4m)(2m)+π/4(2m) 2 ](8m) F V = 7.16 x 10 5 N = 716 kN ↓ The total force is F = [(642 kN) 2 + (716 kN) 2 ] 1/2 = 962 kN θ = tan -1 (F V /F H ) = 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. F H =   h  A = (62.3 lb/ft 3 )(4 ft)[π(1 ft) 2 ] = 783 lbs ← F = [(783 lbs) 2 + (261 lbs) 2 ] 1/2 = 825 lbs θ = tan -1 (F V /F H ) = 18.4˚

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