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

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Contents1FUNDAMENTAL PROPERTIES OF WATER12WATER PRESSURE AND PRESSURE FORCES73WATER FLOW IN PIPES194PIPELINES AND PIPE NETWORKS315WATER PUMPS686WATER FLOW IN OPEN CHANNELS837GROUNDWATER HYDRAULICS978HYDRAULIC STRUCTURES1099WATER PRESSURE, VELOCITY, ANDDISCHARGE MEASUREMENTS12010 HYDRAULIC SIMILITUDE AND MODEL STUDIES12611 HYDROLOGY FOR HYDRAULIC DESIGN13512 STATISTICAL METHODS IN HYDROLOGY162

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__________________________________________________________________________________Chapter 1 – Problem Solutions1.2.1E1= energy released in lowering steam temperature to100C from 110°CE1= (500 L)(1000 g/L)(10C)(0.432 cal/g∙C)E1= 2.16x106calE2= energy released when the steam liquefiesE2= (500 L)(1000 g/L)(597 cal/g)E2= 2.99x108calE3= energy released when the water temperature islowered from 100°C to 50CE3= (500 L)(1000 g/L)(50C)(1 cal/g∙C)E3= 2.50x107cal;Thus, the total energy released is:Etotal= E1+ E2+ E3=3.26x108cal1.2.2First, convert kPa pressure into atmospheres:84.6 kPa(1 atm/101.4 kPa) = 0.834 atmFrom Table 1.1, the boiling temperature is 95°CE1= energy required to bring the water temperature to95C from 15°CE1= (900 g)(95C - 15C)(1 cal/g∙C)E1= 7.20x104calE2= energy required to vaporize the waterE2= (900 g)(597 cal/g)E2= 5.37x105calEtotal= E1+ E2=6.09x105cal1.2.3E1= energy needed to vaporize the waterE1= (300 L)(1000 g/L)(597 cal/g)E1= 1.79x108calThe energy remaining(E2)is:E2= ETotal– E1E2= 2.00x108cal – 1.79x108calE2= 2.10x107calThe temperature change possible with the remainingenergy is:2.10x107cal = (300 L)(1000 g/L)(1 cal/g∙C)(T)T = 70C,making the temperatureT = 90C when it evaporates.Therefore, based on Table 1.1,P = 0.692 atm1.2.4E1= energy required to warm and then melt the iceE1= (10 g)(6C)(0.465 cal/g∙C) + 10g(79.7 cal/g)E1= 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 - T1)(1 cal/g∙C)T1= 15.0C.Now we have a mixture of water at 0°C(formerly ice) and the original 165 liters that is now at15.0C. The temperature will come to equilibrium at:1[(0.165 L)(1000 g/L)(15.0C - T2)(1 cal/g∙C)] =[(10 g)(T2- 0C)( 1 cal/g∙C)] ;T2= 14.1C

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__________________________________________________________________________________________________________________________________________________________________________1.2.5E1= energy required to melt iceE1= (5 slugs)(32.2 lbm/slug)(32F - 20F)*(0.46 BTU/lbm∙F) + (5 slugs)(32.2 lbm/slug)*(144 BTU/lbm)E1= 2.41 x 104BTU.Energy taken from the water.The resulting temperature of the water will decrease to:2.41 x 104BTU = (10 slugs)(32.2 lbm/slug)(120F –T1)(1 BTU/lbm∙F)T1= 45.2FThe energy lost by the water (to lower its temp. to45.2F) is that required to melt the ice. Now you have5 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 – T2)(1 BTU/lbm∙F)]= [(5 slugs)(32.2 lbm/slug)(T2- 32F)(1 BTU/lbm∙F)]T2= 40.8F1.2.6E1= energy required to raise the temperature to 100CE1= (7500 g)(100C – 20C)(1 cal/g∙C)E2= 6.00x105calE2= energy required to vaporize 2.5 kg of waterE2= (2500 g)(597 cal/g)E2= 1.49x106calEtotal= E1+ E2= 2.09x106calTime required =(2.09x106cal)/(500 cal/s)Time required =4180 sec = 69.7 min1.3.1F = 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∙g1.3.2SGoil= 0.976 = γoil/γ;whereγis 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/m3Also, γ = ρ∙g; or ρ = γoil/gSubstituting (noting thatN ≡ kg·m/sec2) yields,ρoil= γoil/g = (9,570 N/m3) / (9.81 m/sec2)= 976 kg/m31.3.3By definition,= W/Vol = 55.5 lb/ft3; thus,W=∙Vol = (55.5 lb/ft3)(20 ft3) =1,110 lb (4,940 N)=/g = (55.5 lb/ft3)/(32.2 ft/s2) =1.72 slug/ft3(887 kg/m3)SG=liquid/water at 4C= (55.5 lb/ft3)/(62.4 lb/ft3) =0.8891.3.4The mass of liquid can be found usingρ = γ/gandγ= weight/volume, thusγ = (47000 N – 1500 N)/(5 m3) = 9.10 x 103N/m3ρ = γ/g = (9.1 x 103N/m3)/(9.81 m/sec2);= 928 kg/m3(Note: 1 N ≡ 1 kg·m/sec2)2Specific gravity (SG) =γ/γwater at 4°CSG = (9.10 x 103N/m3)/9.81 x 103N/m3)SG = 0.928

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31.3.5The force exerted on the tank bottom is equal to theweight of the water body (Eq’n 1.2).F = W = mg = [ρ(Vol)] (g);ρ found in Table 1.2920 lbs = [1.94 slugs/ft3(π ∙(1.25 ft)2∙d)] (32.2 ft/sec2)d = 3.00 ft(Note:1 slug = 1 lb∙sec2/ft)___________________________________________1.3.6Weight of water on earth = 8.83 kNFrom E’qn (1.1):m = W/g = (8,830 N)/(9.81 m/s2)m = 900 kgNote: mass on moon is the same as mass on earthW (moon) = mg = (900 kg)[(9.81 m/s2)/(6)]W(moon) = 1,470 N_________________________________________1.3.7Density is expressed as ρ = m/Vol, and even thoughvolume changes with temperature, mass does not.Thus,(ρ1)(Vol1) = (ρ2)(Vol2)= constant; orVol2= (ρ1)(Vol1)/(ρ2)Vol2= (999 kg/m3)(100 m3)/(996 kg/m3)Vol2= 100.3 m3(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 10-1ftlb_________________________________________1.3.9(1 N/m2) [(1 m)/(3.281 ft)]2[(1 ft)/(12 in)]2·[(1 lb)/(4.448 N)] =1.450 x 10-4psi1.4.1(a)Note that:1 poise = 0.1 Nsec/m2.Therefore,1 lb·sec/ft2[(1 N)/(0.2248 lb)]·[(3.281 ft)2/(1 m)2] =47.9 Nsec/m2[(1 poise)/(0.1 Nsec/m2)] = 478.9 poiseConversion:1 lbsec/ft2= 478.9 poise(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 stokesConversion:1 ft2/sec = 929.0 stokes_________________________________________1.4.2[(air)/(H2O)]0C= (1.717x10-5)/(1.781x10-3)[(air)/(H2O)]0C= 9.641x10-3[(air)/(H2O)]100C= (2.174x10-5)/(0.282x10-3)[(air)/(H2O)]100C= 7.709x10-2[(air)/(H2O)]0C= (1.329x10-5)/(1.785x10-6)[(air)/(H2O)]0C= 7.445[(air)/(H2O)]100C= (2.302x10-5)/(0.294x10-6)[(air)/(H2O)]100C= 78.30Note: The ratio of the viscosity of air to water increaseswith temperature. Why? Because the viscosity of airincreases with temperature and that of water decreaseswith temperature magnifying the effect. Also, thevalues of kinematic viscosity () for air and water aremuch closer than those of absolute viscosity. Why?_________________________________________1.4.320C= 1.002x10-3Nsec/m2;20C= 1.003x10-6m2/s(1.002x10-3Nsec/m2)∙[(0.2248 lb)/(1 N)]∙[(1 m)2/(3.281 ft)2] =2.092x10-5lbsec/ft2(1.003x10-6m2/s)[(3.281 ft)2/(1 m)2] =1.080x10-5ft2/s

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41.4.4_________________________________________Using Newton’s law of viscosity (Eq’n 1.2):=(dv/dy) =(Δv/Δy)= (1.00 x 10-3Nsec/m2)[{(4.8 – 2.4)m/sec}/(0.02 m)]= 0.12 N/m21.4.5From Eq’n (1.2):=(v/y) =_________________________________________= (0.0065 lbsec/ft2)[(1.5 ft/s)/(0.25/12 ft)]= 0.468 lb/ft2F = ()(A) = (2 sides)(0.468 lb/ft2)[(0.5 ft)(1.5 ft)]F = 0.702 lb1.4.6Summing 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 10-3m = 1.06 mm___________________________________________1.4.7Using Newton’s law of viscosity (Eq’n 1.2):=(dv/dy) =(v/y)= (0.04 Nsec/m2)[(15 cm/s)/[(25.015 – 25)cm/2]= 80 N/m2Fshear resistance=A = (80 N/m2)[()(0.25 m)(3 m)]Fshear resistance= 188 N1.4.8v = y2– 3y,where y is in inches and v is in ft/sv = 144y2– 36y,where y is in ft andis in ft/sTaking the first derivative:dv/dy = 288y – 36 sec-1_________________________________________=(dv/dy) = (8.35 x 10-3lb-sec/ft2)( 288y – 36 sec-1)Solutions:y = 0 ft,= -0.301lb/ft2y = 1/12 ft,= -0.100 lb/ft2; y = 1/6 ft,= 0.100 lb/ft2y = 1/4 ft,= 0.301 lb/ft2; y = 1/3 ft,= 0.501 lb/ft21.4.9= (16)(1.00x10-3Nsec/m2) = 1.60x10-2Nsec/m2Torque =dArdFrRR00)(RdAyvr0))()((Torque =_________________________________________Rdrryrr0)2)(0))(()()((Torque =Rdrry03)())()(2(Torque =4)1(0005.0sec)/65.0)(sec/1060.1)(2(422mmradmNTorque = 32.7 Nm1.4.10=/(dv/dy) = (F/A)/(v/y);Torque (T) = Force∙distance = F∙R where R = radiusThus;= (T/R)/[(A)(v/y)]=))()()(2()/)()()(2(/3hRyTyRhRRT=rpmradrpmftftftftlb60sec/2)2000)()12/6.1(())12/1)((2(])12/008.0)[(10.1(3= 7.22x10-3lbsec/ft2

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5_________________________________________1.5.1The concept of a line force is logical for two reasons:1)The surface tension acts along the perimeter of thetube pulling the column of water upwards due toadhesion between the water and the tube.2)The surface tension is multiplied by the tubeperimeter, a length, to obtain the upward force usedin the force balance developed in Equation 1.3 forcapillary rise.1.5.2To 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)(1.0x10-3m)]_________________________________________D = 0.0131 m = 1.31 cmNote: 50° was used instead or 40° because it producesthe largest D. A 40° angle produces a smaller error.1.5.3For capillary rise, apply Equation 1.3:h = [(4)()(sin θ)] / [()(D)]Butsin 90˚ = 1,= 62.3 lb/ft3(at 20˚C),andσ = 4.89x10-3lb/ft(from inside book cover)thus,D = [(4)()] / [()(h)]; for h = 1.5 in.D = [(4)(4.89x10-3lb/ft)] / [(62.3 lb/ft3)(1.5/12)ft]D = 2.51 x 10-3ft = 3.01 x 10-2in.;thus,for h = 1.5 in.,D = 2.51x10-3ft = 0.0301 in.for h = 1.0 in.,D = 3.77x10-3ft = 0.0452 in.for h = 0.5 in.,D = 7.54x10-3ft = 0.0904 in.1.5.4Condition 1:h1= [(4)(1)(sin1)] / [()(D)]h1= [(4)(1)(sin30)] / [()(0.8 mm)]Condition 2:h2= [(4)(2)(sin2)] / [()(D)]h2= [(4)(0.881)(sin50)] / [()(0.8 mm)]h2/h1= [(0.88)(sin50°)] / (sin30) =1.35alternatively,h2= 1.35(h1), about a 35% increase!_________________________________________1.5.5Capillary rise (measurement error) is found usingEquation 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 10-2N/mand= (9750)(1.03) = 1.00 x 104N/m3h =[(4)( 8.28 x 10-2N/m)(sin 35)] /[(1.00 x 104N/m3)(0.012m)]h = 1.58x10-3m = 0.158 cm_________________________________________1.5.6PRP = Pi– Pe(internal pressure minus external pressure)Fx= 0;P()(R2) - 2(R)() = 0P = 2/R

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61.6.1Pi= 1 atm = 14.7 psi. and Pf= 220 psiFrom Equation (1.4):Vol/Vol = -P/EbVol/Vol = -(14.7 psi – 220 psi)/(3.2x105psi)Vol/Vol = 6.42 x 10-4= 0.0642%(volume decrease)/= -Vol/Vol =-0.0642%(density increase)_________________________________________1.6.2m = W/g = (7,490 lb)/(32.2 ft/s2) = 233 slug= m/Vol = (233 slug)/(120 ft3)= 1.94 slug/ft3Vol = (-P/Eb)(Vol)Vol = [-(1470 psi –14.7 psi)/(3.20x105psi)](120 ft3)Vol = -0.546 ft3new= (233 slug)/(120 ft3– 0.546 ft3)= 1.95 slug/ft3Note: The mass does not change._________________________________________1.6.3Surface pressure:Ps= 1 atm = 1.014 x 105N/m2Bottom pressure:Pb= 1.61 x 107N/m2From Equation (1.4):Vol/Vol = -P/EbVol/Vol = [-(1.014 x 105- 1.61 x 107)N/m2](2.2x109N/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/m3Specific weight at the bottom:γb= (9,810 N/m3)(1.00727) =9,880 N/m3Note: These answers assumes thatEbholds constant forthis great change in pressure.1.6.4Pi= 30 N/cm2= 300,000 N/m2= 3 barP = 3 bar – 30 bar = -27 bar = -2.7x105N/m2Amount of water that enters pipe =VolVolpipe= [()(1.50 m)2/(4)]∙(2000 m) = 3530 m3Vol = (-P/Eb)(Vol)Vol = [-(-2.7x105N/m2)/(2.2x109N/m2)]*(3530 m3)Vol = 0.433 m3Water in the pipe is compressed by this amount. Thus,thevolume of H2O that enters the pipe is 0.433m3

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7Chapter 2 – Problem Solutions2.2.1a)The depth of water is determined using therelationship between weight and specific weight.W =γ·Vol =γ(A·h);where h = water depth14,700 lbs = (62.3 lb/ft3)[π·(5 ft)2](h);h = 3.00 ftb)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/ft3)(3 ft)= 187 lb/ft22.2.2___________________________________________The absolute pressure includes atmospheric pressure.Therefore,Pabs= Patm+ (water)(h)5(Patm);Thus from Eq’n 2.4:h = 4(Patm)/waterh = 4(1.014 x 105N/m2)/(1.03)(9790 N/m3)h = 40.2 m (132 ft)2.2.3waterat 30C = 9.77 kN/m3(from Table 1.2)Pvaporat 30C = 4.24 kN/m2(from Table 1.1)Patm= Pcolumn+ PvaporPatm= (8.7 m)(9.77 kN/m3) + (4.24 kN/m2)Patm= 85.0 kN/m2+ 4.24 kN/m2= 89.2 kN/m2The percentage error if the direct reading is used andthe 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%2.2.4Since mercury has a specific gravity of 13.6, the waterheight 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+ PatmPabs= 0.408 m + 10.3 m = 10.7 m of waterPabs= (10.7 m)(9790 N/m3) =1.05 x 105N/m22.2.5Since force equals pressure on the bottom times area:P = γ∙h = (62.3 lb/ft3)(10 ft) = 623 lb/ft2___________________________________________Fbottom= P∙A = (623 lb/ft2)(100 ft2)= 6.23 x 104lbsPressure varies linearly with depth. So the averagepressure on the sides of the tank occur at half the depth.Pavg= γ∙h = (62.3 lb/ft3)(5 ft) = 312 lb/ft2Now,Fside= Pavg∙A = (312 lb/ft2)(100 ft2)= 3.12 x 104lbs2.2.6Pressure and force on the bottom of both containers is:P = (water)(h) = (9790 N/m3)(10 m) = 97.9 kN/m2Also,F= P∙A = (97.9 kN/m2)(2 m)(2 m)= 391 kNThis may be confusing since the water weights aredifferent. To clarify the situation, draw a free bodydiagram 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.)W1W2P1·A1Note the three vertical forces will be opposed at thebottom.W1is the weight of the water column above it.W2is the weight of water in the lower portion of thecontainer. But there is also an upward pressure forcefrom the water on the underside of the L indent (P =γh). This is opposed by an equal and opposite pressureon the water below. The resulting force isP1·A1whichequalsγ·h·A1orγ·Vol. In other words, the force equalsthe weight of the imaginary column of displaced waterabove it. Hence, the force on the bottom opposing theseforce components is equal to the force on the bottom ofthe water column without the indent.2.2.7_________________________________________seawater= (SG)() = (1.03)(62.3 lb/ft3) = 64.2 lb/ft3Ptank= (seawater)(h) = (64.2 lb/ft3)(18 ft)(1 ft2/144 in2)Ptank= 8.03 psi = 5.54 x 104N/m2(Pascals)2.2.8Pbottom= Pgage+ (liquid)(1.4 m);andliquid= (SG)(water) = (0.85)(9790 N/m3) = 8,320 N/m3Pbottom= 3.55x104N/m2+ (8320 N/m3)(1.4 m)Pbottom= 4.71 x 104N/m2The pressure at the bottom of the liquid column can bedetermined two different ways which must be equal.(h)(liquid) = Pgage+ (liquid)(1 m)h = (Pgage)/(liquid) + 1mh=(3.55x104N/m2)/8320 N/m3+ 1 m =5.27 m2.2.9oil= (SG)(water) = (0.85)( 62.3 lb/ft3) = 53.0 lb/ft3P10ft= Pair+ (oil)(12 ft)orPair= P10ft- (oil)(12 ft)Pair= 25.7 psi (144 in2/ft2) – (53.0 lb/ft3)(12 ft)Pair= 3.06 x 103lb/ft2(21.3 psi);Gage pressurePabs= Pgage+ Patm= 21.3 psi + 14.7 psi___________________________________________Pabs= 36.0;Absolute pressure2.2.10∑Fx= 0;(Px)(Δy) – (Ps)(Δs)(sin θ) = 0since(Δs∙sin θ) = Δy,Ps= Px∑Fy= 0;and accounting for weight yields:(Py)(Δx) – (Ps)(Δs)( cos θ) – (Δx∙Δy/2)(Δz)(γ) = 0since(Δs∙cos θ) = Δx and (Δy)(Δx)(Δz)0;Ps= PyHence, the pressure is the same in alldirections at the same depth, thus omnidirectional.___________________________________________2.2.11The mechanical advantage in the lever increases theinput 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/cm2Psystem= 180 kN/m2= 180 kPaFrom Pascal’s law, the pressure at the input pistonshould equal the pressure at the two output pistons.The force exerted on each output piston is:Pinput= Poutputequates to: 18 N/cm2= Foutput/250 cm2Foutput=(18 N/cm2)(250 cm2)(1 kN/1000 N)= 4.50 kN

Solution Manual for Fundamentals of Hydraulic Engineering Systems, 5th Edition - Page 12 preview image

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92.4.1The gage pressure at the bottom of the tank is equalto 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____________________________________________2.4.2_________________________________________Note the equal pressure surface;P7= P8or,(hwater)() = (hoil)(oil) = (hoil)()(SGoil),thushwater= (hoil)/(SGoil) = (61.5 cm)(0.85) = 52.3 cm2.4.3The mercury-water meniscus (left leg) will be lowerthan the mercury-oil meniscus based on the relativeamounts of each poured in and their specific gravity.Also, a surface of equal pressure can be drawn at themercury-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_________________________________________2.4.4Equal pressure surface atHg-H2Ointerface: 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);P = 2,420 lb/ft2= 16.8 psiPressure can be expressed as a fluid height:h = P/Hg= (2420 lb/ft2)/[(13.6)(62.3 lb/ft2)]h=2.86 ft of Hg (or 34.3 inches)2.4.5Note the equal pressure surface:P1= P2Therefore,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)PA= 65.1 lb/ft2= 0.452 psi_________________________________________2.4.6A surface of equal pressure surface can be drawn at themercury-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 104N/m2(Pascals) = 78.1 KPa2.4.7Use theswim-through-technique: Start at the end ofthe manometer open to the atmosphere(Pgage= 0).Then“swim through” adding pressure when “swimming”down and subtracting when “swimming” up until youreach the pipe(Ppipe).Remember to jump across themanometer at surfaces of equal pressure (ES). Thus,0 - (0.66 m)(ml) + [(0.66 +y+ 0.58)m](air)– (0.58 m)(oil) = PpipeThe specific weight of air is negligible when comparedto fluids, so that term in the equation can be dropped.Ppipe= 0 - (0.66 m)(SGml)() – (0.58 m)(SGoil)()Ppipe= 0 - (0.66 m)(0.8)(9790 N/m3)– (0.58 m)(0.82)( 9790 N/m3)Ppipe= - 9.83 kN/m2(kPa)Pressure can be converted toheight (head) of any liquid through P =∙h. Thus,hpipe= (-9,830 N/m2)/(9790 N/m3)=-1.00 m of water

Solution Manual for Fundamentals of Hydraulic Engineering Systems, 5th Edition - Page 13 preview image

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102.4.8A surface of equal pressure surface can be drawn at themercury-water meniscus. Therefore,P + (h1)() = (h2)(Hg) = (h2)(SGHg)()P + (0.5 ft)(62.3 lb/ft3) = (1.5 ft)(13.6)(62.3 lb/ft3)P = 1,240 lb/ft2= 8.61 psiWhen the manometer reading rises or falls, volumebalance is preserved for constant density. Therefore,Volres= VoltubeorAres∙h1= Atube∙h2h1= h2(Atube/Ares) = h2[(Dtube)2/(Dres)2]_________________________________________h1= (4 in.)[(1 in.)2/(5 in.)2] =0.16 in.2.4.9Using the “swim through” technique, start at pipeAand“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 psi2.4.10Using the “swim through” technique, start atP2and“swim through” the manometer, adding pressure when“swimming” down and subtracting when “swimming”up until you reachP1. The computations are:P2+ (Δh)(ρ1∙g) + (y)(ρ1∙g) + (h)(ρ2∙g) - (h)(ρ1∙g) -(y)(ρ1∙g) = P1where y is the vertical elevation difference between thefluid surface in the left hand reservoir and the interfacebetween the two fluids on the right side of the U-tube.Simplifying,P1– P2= (Δh)(ρ1∙g) + (h)(ρ2∙g) - (h)(ρ1∙g)2.4.10 – cont.When the manometer reading (h) rises or falls, volumebalance must be preserve in the system. Therefore,Volres= VoltubeorAres∙(Δh) = Atube∙ hΔh = h (Atube/Ares) = h [(d2)2/(d1)2];substituting yields_________________________________________P1– P2= h [(d2)2/(d1)2] (ρ1∙g) + (h)(ρ2∙g) - (h)(ρ1∙g)P1– P2= h∙g [ρ2- ρ1+ ρ1{(d2)2/(d1)2}]P1– P2= h∙g [ρ2- ρ1{1 - (d2)2/(d1)2}]2.4.11Using the “swim through” technique, start at the righttank where pressure is known and “swim through” thetanks and pipes, adding pressure when “swimming”down and subtracting when “swimming” up until youreach the known pressure in the left tank. Remember tojump across the pipe at the surfaces of equal pressure(ES). Solve forEAin the resulting equation.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/m2EA= 30 m___________________________________________2.4.12Using the “swim through” technique, start at both endsof the manometers which are open to the atmosphereand thus equal to zero gage pressure. Then “swimthrough” the manometer, adding pressure when“swimming” down and subtracting when “swimming”up until you reach the pipes in order to determinePAandPB. With those pressures, you may solve for thevalue of h. Remember to jump across the manometer atsurfaces of equal pressure (ES). The computations are:0 + (23)(13.6)() - (44)() = PA;PA= 269∙γ0 + (46)(0.8)() + (20)(13.6)() – (40)(γ) = PBPB= 269∙γ;Therefore,PA= PBandh = 0

Solution Manual for Fundamentals of Hydraulic Engineering Systems, 5th Edition - Page 14 preview image

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______________________________________________________________________________________________________________________________________________________________________2.5.1AhF= (62.3 lb/ft3)[4(2ft)/(3π)]∙[(1/2)π(2ft)2]F = 332 lb(Note:yhin Table 2.1))][4(2ft)/(3)]3][4(2ft)/((2ft)[(1/2)72/)2(6492420ftyyAIyPyp= 0.853 ft(depth to center of pressure)The center of pressure (0.853 ft) is deeper than thecentroid (fty848.0) of the gate.2.5.2AhF= (9790 N/m3)[(9 m)/2]∙[(9 m)(1 m)]F = 3.96 x 105Nper meter of lengthmmmmmmyyAIyP5.4)5.4()1)(9(12/)9)(1(30yp= 6.0 m(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 rightingmoment) 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/m3)∙(3 m) –(3.96x 105N)∙(3 m) = -8.57 x 104N-m∑MA= 8.57 x 104N-m(overturning; dam is unsafe)2.5.3The hydrostatic force and its locations are:AhFF = (9790 N/m3)[(5 m)(sin 45˚)](π)(0.5 m)2F = 2.72 x 104N = 27.2 kNmmmmyyAIyP5)5(4/)1(64/)1(2405.01 m2.5.4The hydrostatic force and its locations are:AhF= (γ)(h)[π(D)2/4]hhDDyyAIyP)(4/)(64/)(240;yp= D2/(16h) + h(depth to the center of pressure)Thus, summing moments:∑ Mhinge= 0 ;P(D/2) – F(yp– h) = 0P(D/2) – {(γ)(h)[π(D)2/4][D2/(16h)]} = 0;P = (1/32)(γ)(π)(D)32.5.5The hydrostatic force and its locations are:AhF= (62.3 lb/ft3)(1.5 ft)(9 ft2)F = 841 lb)5.1(912/)3)(3(230ftftftftyAIyyP;yp–𝒚̿= 0.500 ftIf the gate was submerged by 10 m (to top of the gate):FAh= (62.3 lb/ft3)(11.5 ft)(9 ft2)= 6,450 lb)5.11(912/)3)(3(230ftftftftyAIyyP;yp–𝒚̿= 0.0652 ftThe force increases tremendously as depth increases.The distance between the centroid and the center ofpressure becomes negligible with increasing depth.2.5.6The center of pressure represents the solution to bothparts of the question. Thus,115.6)5.6()3)(2(12/)3)(2(30ftftftftftyyAIyP6.62ft

Solution Manual for Fundamentals of Hydraulic Engineering Systems, 5th Edition - Page 15 preview image

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122.5.7AhFleft= (9790 N/m3)(0.5 m)[(1.41m)(3m)]Fleft= 20.7 kN(where A is “wet” surface area)mmmmmmyyAIyP705.0)705.0()41.1)(3(12/)41.1)(3(30yp= 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 mAhFright= (9790 N/m3)(h/2 m)[(h/sin45˚)(3m)]Fright= 20.8∙h2kNhhhhyyAIyP705.0)705.0()41.1)(3(12/)41.1)(3(30yp= (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)mThe force due to the gate weight: W = 20.0 kNMoment arm of this force from hinge: X = 0.707 mSumming moments about the hinge yields:∑Mhinge= 0(20.8∙h2)[2m – (0.474∙h)] – 20.7(1.53) – 32.3(0.707) = 0h = 1.40 m(gate opens when depth exceeds 1.40 m)_________________________________________2.5.8Logic dictates that the center of pressure should be atthe pivot point (i.e., the force at the bottom check blockis zero).As water rises above h = 2 feet, the center ofpressure will rise above the pivot point and open thegate. Below h = 2 feet, the center of pressure will belower than the pivot point and the gate will remainclosed. For a unit width of gate, the center of pressure isftftftftftftyyAIyP6)6()8)(1(12/)8)(1(30yp= 6.89 ft(vertical distance from water surface to thecenter of pressure)Thus, the horizontal axis of rotation (0-0’) should be10 ft – 6.89 ft = 3.11 ft above the bottom of the gate.2.5.9AhF= (9790 N/m3)(2.5 m)[(π){(1.5)2–(0.5)2}m2]F = 1.54 x 105N = 154 kNmmmmmmyyAIyP5.2)5.2(})5.0()5.1{(})1()3{(64/22440yp= 2.75 m(below the water surface)___________________________________________2.5.10The total force from fluids A and B can be found as:AhFA= (γA)(hA)∙[π(d)2/4]AhFB= (γB)(hB)∙[π(d)2/4]For equilibrium, forces are equal, opposite, & collinear.FA= FB;(γA)(hA)∙[π(d)2/4] = (γB)(hB)∙[π(d)2/4]hA= [(γB)/(γA)](hB)___________________________________________2.5.11AhF= (62.3 lb/ft3)[(20 ft)]∙[(10 ft)(6 ft)]F = 7.48 x 104lbs(Horizontal force on gate)Summing vertical forces where T is the lifting force:∑ Fy= 0;T – W – F(Cfriction) = 0T = 6040 + 74,800(0.2)= 21,000 lbs (lifting force)___________________________________________2.5.12The hydrostatic force on the cover and its locations are:AhF=(9790 N/m3)(1.5m)[(10m)(5m)] = 734 kNmmmmmmyyAIyP5.2)5.2()5)(10(12/)5)(10(30= 3.33 m∑ Mhinge= 0;(734 kN)(3.33 m) – W(2 m) = 0;W = 1,220 kN

Solution Manual for Fundamentals of Hydraulic Engineering Systems, 5th Edition - Page 16 preview image

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132.5.13F hA= (62.3 lb/ft3)[(d/2)ft]∙[{(d/cos30)ft}(8ft)]F = 288∙d2lbsyAyIyP0(8)(d/ cos30) (d/ 2cos30)3(8)(d/ cos30) /12yP(d/ 2cos30)yp= [(0.192∙d) + 0.577∙d)] ft = 0.769∙dThus, summing moments:∑ Mhinge= 0(288∙d2)[(d/cos30)-0.769d] – (5,000)(15) = 0d = 8.77 ftA depth greater than this will make the gate open.Any depth less than this will make it close.___________________________________________2.5.14The force and location on the vertical side of the gate:F hA= (9790 N/m3)[(h/2)m][(hm)(1 m)]F = (4.90 x 103)h2N (per meter of gate width)(1)(h) (h/ 2)(1)(h) /1230hAyIyyP/ 2= (2/3)hmThe force upward on the bottom of the gate::F = p∙A = (9790 N/m3)(h)[(1m)(1m)]F = (9.79 x 103)hN (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 103)h2][h - (2/3)h] – [(9.79 x 103)h](0.5) = 0h = 1.73 m2.6.1The resultant force in the horizontal direction is zero(FH= 0)since equal pressures surround the viewingwindow in all directions. For the vertical direction, theforce equals the weight of the water above the window.FV Vol= (1.03)(62.3 lb/ft3)[π(1 ft)2(6 ft) –(1/2)(4/3)π(1 ft)3]= 1,080 lbs___________________________________________2.6.2Obtain the horizontal component of the total hydrostaticpressure force by determining the total pressure on thevertical projection of the curved gate.FH hA= (0.82)(9790 N/m3)(5m)[(8 m)(2 m)]FH= 6.42 x 105N = 642 kN →The vertical component of the hydrostatic force equalsthe weight of the water column above the curved gate.VolFV =(0.82)(9790N/m3)[(4m)(2m)+π/4(2m)2](8m)FV= 7.16 x 105N = 716 kN ↓The total force isF= [(642 kN)2+ (716 kN)2]1/2=962 kNθ= 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 theyact), then the resultant must also pass through point A.__________________________________________2.6.3Obtain the horizontal component of the total hydrostaticpressure force by determining the total pressure on thevertical 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 lbsθ= tan-1(FV/FH)= 18.4˚

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