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• Jorge Dominic Conde

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Problem 1 - Hydraulics If 6 m3 of soil weight 47 kn , calculate the following. a)

specific weight

47 y = 6 = 7.833 kN/m3 b)

c)

Density:

specific gravity:

7.833 sp.gr. = 9.81 = 0.80

7833 p = 9.81 = 798 kg/m3

Problem 2 - Hydraulics For the open tank with piezometers attached on the side, contains two different liquids. ➀ Find the elevation of the liquid in piezometer A. ➁ Find the pressure at the bottom of the tank. ➂ Find the elevation of the liquid in piezometer B. Solution: A B ➀ Elevation of liquid at piezometer A = 2m. El. 2.0m ➁ Pressure at the bottom of the tank: Pbottom = 9.81(0.72)(2 - 0.3) + 9.81(2.36)(0.3) Pbottom = 18.95 kPa ➂ Elevation of the liquid at piezometer B: 0 + 9.81(0.72)(1.7) + 9.81(2.36)(0.3) - 2.36(h)(9.81) = 0 h = 0.819 Elevation 0 + 0.819 = 0.819 m.

A(0.72) h

El. 0.3m

B(2.36)

Problem 3 - Hydraulics Oil of sp.gr. 0.750 flows through the nozzle shown and deflects the mercury in the U-tube gage. ➀ Determine the value of h if the pressure at A is 142 kPa. ➁ What is the pressure at B. ➂ If the diam. at A is 200 mm, compute the velocity at A if oil flows at a rate of 36000 liters per minute. Solution: ➀ Value of h: 142 + 9.81(0.75)(h + 2.75) - 13.6(9.81)(h) = 0 142 - 126.0585h + 20.233 = 0 h = 1.29 m. ➁

Pressure at B: PB = 142 + (9.81)(0.75)(2.75 + 1.29) PB = 177.72 kPa

➂

Velocity at A: Q = 36000 liters/minute Q = 0.60 m3/s Q = AV 0.60 =

π (0.2)2 V 4

V = 19.10 m/s

A 2.75 B Hg(13.6)

D C

h

Problem 4 - Hydraulics A cylinder glass tubing 2.8 cm. inside diameter and 90 cm long with one end closed is immersed vertically with the open end down into a tank of cleaning solvent (sp.gr. = 0.73) until only 5 cm. of its length remain above the liquid surface. If the barometric pressure is 1 kg/cm2 and neglecting vapor pressure, ➀ How high will the fluid rise in the tube? ➁ Compute the height of air inside the glass. ➂ What force required to maintain equilibrium. Solution: ➀ Height of fluid rise in the tube: P1 = 1 kg/cm2 P2 = P1 + wh 1000(0.73)(h) Kg m3 P2 = 1 + ( ) (100)3 m3 1003 cm3 P2 = 1 + 0.00073h kg/cm2

2.8cm 5cm 90cm h x

Cleaning solvent (S=0.73)

π V1 = 4 (2.8) 2 (90) V1 = 554.18 cm3 π V2 = 4 (2.8) 2 (h + 5) V2 = 6.158 (h + 5)

Using Boyle's Law: P1V1 = P2V2 1(554.18) = (1 + 0.00073h)(6.158)(h + 5) 554.18 6.158 = (1 + 0.00073h) (h + 5) 89.99 = h + 0.00073h2 + 5 + 0.00365h 89.99 = 0.00073h2 + 1.00365h + 5 h2 + 1374.86h - 116424.657 = 0 h = 80 cm. x = 90 - 5 - 80 x = 5 cm.

➁ Height of air inside the glass: h + 5 = 85 cm. ➂ Force required equilibrium: ρA = γ h A

to

π(0.028)2 F = 9810 0.80 4 F = 4.83 N

maintain

Problem 5 - Hydraulics The 8 ft. diam. cylinder weighs 500 lb. and rests on the bottom of a tank that is 3 ft. long. Water and oil are poured into the left and right portions of the tank to depths of 2 ft. and 4 ft. respectively. ➀ Find the horizontal component of the force that will kept the cylinder touching the tank at B. ➁ Find the vertical component of the force that will push up the cylinder. ③ Compute the force that will keep the cylinder touching the tank at B. Solution: ➀ Horizontal component of the force that will kept the cylinder touching the tank at B: P1 = γw h A 500 lb P1 = 62.4 (1) (2) (3) Oil (0.75) P1 = 374 lb. 4 2

H 2O P1

P2 = γw h A P2 = 62.4 (2) (4) (3) (0.75) P2 = 1123 lb. Ph = P2 - P1 H 2O Ph = 1123 - 374 2 Ph = 749 lb. ➁ Vertical component of the force that will push up the cylinder:

π(4)2 (3)(0.75) Fv2 = 62.4 4 = 1764 lb.

π(4)2 (60 2(4)Sin 60˚ Fv1 = 62.4 [ 360 ] (3) 2 Fv1 = 920 Fv = 1764 + 920 Fv = 2684 lb. ③ Force that will keep the cylinder touching the tank at B: FB + 500 = 2684 FB = 2184 lb. (downward)

60˚

2 2

P2

4

B Fv1

Fv2

500 lb

4

60˚

4 2 2

4

B Fv1

Oil (0.75)

Fv2

Problem 6 - Hydraulics The 6 ft. diameter cylinder weighs 5000 lb. and is 5 ft. long. ➀ ➁ ③

Determine the upward force due to the effect of oil in the left side. Compute the horizontal reaction at A. Compute the vertical reaction at B.

➀

Solution: Upward force due to the effect of oil in the left side: C

E

Pv = γw V

Oil D (0.80)

2

(π)(3) (5) Pv = 62.4(0.8) 2

A B

Pv = 3529 lb. (upward) ➁ Horizontal reaction at A: Ph = γw h A Ph = 62.4 (.80)(3)(6)(5) Ph = 4493 lb. RA = 4493 lb. to the left.

5

E

C

D

6

A

Ph

③ Vertical reaction at B: RB + Pv = 5000 RB = 5000 - 3529 RB = 1471 lb. (upward)

Pv

B RB

RA

Problem 7 - Hydraulics From the figure shown. ➀ Compute the horizontal component of the hydrostatic force. ➁ Compute the vertical component of the hydrostatic force. ③ Compute the location of the vertical component horizontally from B. Solution: ➀ Horizontal component of the hydrostatic force: Ph = γw h A Ph = 62.4 (3)(6)(1) Ph = 1123 lb.

Hinge 1 6

➁ Vertical component of the hydrostatic force: (π)(6)2 (1) Pv = 62.4 4 Pv = 1764 lb. ③ Location of the vertical component horizontally from B:

4(6) x = 3π x = 2.55 ft. from B

C

4’ Ph 1/3(6)=2 R

4r x = 3π

6’

A

6’ B Pv

x

Problem 8 - Hydraulics A wooden storage vat, 20 ft. outside diam. is filled with 24 ft. of brine sp.gr. = 1.06. 1 The wood staves are bound by flat steel bands 2 in. wide by 4 inch thick, whose allowable stress is 16,000 psi. ➀ ➁ ➂

What is the bursting pressure? What is the tensile force of steel bands. What is the spacing of the bands near the bottom of the vat, neglecting any initial stress?

Solution: ➀ Bursting pressure: 2T = F 2T = P D(S)

20’

T 24’

1/4” D T

sp.gr.=1.06

2” 1/4”

P = γw h P = 62.4 (1.06)(24) P = 1587.5 psf.

T F

T

S D

➁ Tensile force of steel bands: T = Ss A ⎛1⎞ T = 16000 (2) ⎜⎜4⎟⎟ ⎝ ⎠ T = 8000 lb. ➂ Spacing of the bands near the bottom of the vat: 2T = P D S 2(8000) = 1587.5 (20) S S = 0.504 ft. S = 6.05 inches

Problem 9 - Hydraulics A concrete dam retaining water is shown. If the specific weight of concrete is 23.5 kN/m3. ➀ ➁ ➂

Find the factor of safety against sliding. Find the factor of safety against overturning if the coeff. of friction is 0.48. Find the max. and min. pressure intensity

Solution: ➀ Factor of safety against sliding: Considering 1 meter strip: ___ P=γhA P = 9.79 (3)(6)(1) P = 176.20 kN 2(7) w1 = 2 (1)(23.5) w1 = 164.5 kN w2 = 2 (7)(1)(23.5) w2= 329 kN Ry = 164.5 + 329 Ry = 493.5 kN Factor of safety against sliding: uRy F.S. = P 0.48 (493.5) F.S. = 176.20 F.S.= 1.34

2m

7m

6m

4m 2

2

3 W2

7m

6m

W1

P 2

1.333 x R y 2

2

Pmin

Pmax 1.73 e Ry

Problem 9 – Hydraulics (Continuation) ➁

Factor of safety against overturning. O.M. = P(2) O.M. = 176.20(2) O.M. = 352.4 kN.m. R.M. = w1 (1.333) + w2 (3) R.M. = 164.5 (1.333) + 329(3) R.M. = 1206 kN.m. R.M. F.S. = O.M. 1206 F.S. = 352.4 F.S. = 3.42

➂

The max. and min. pressure intensity: ___ Ry x = R.M. - O.M. 1206 - 352.4 x= 493.5 x = 1.73 e = 2 - 1.73 e = 0.27 Ry ⎛ 6e⎞ Pmin. = B ⎜⎜1 - B ⎟⎟ ⎝ ⎠ 493.5 6(0.27) Pmin. = 4 [1 - 4 ] Pmin. = 73.4 kN/m2. R ⎛ 6e⎞ Pmax. = B ⎜⎜1 + B ⎟⎟ ⎝ ⎠ 493.5 6(0.27) Pmax. = 4 [1 + 4 ] Pmax. = 173.34 kN/m2.

Problem 10 - Hydraulics A prismatic object 200 mm thick by 200 mm wide by 400 mm long is weighed in water at a depth of 500 mm and found to be 50 N. ➀ ➁ ➂

Find its weight in air. Find its specific gravity. Find its specific weight.

Solution: ➀ Weight in air: W = BF + 50 BF = 0.2(0.2)(0.4)(9810) BF = 156.96 N W = 156.96 + 50 W = 206.96 N ➁ Specific gravity: 206.96 Sp.gr. = 156.96 Sp.gr. = 1.32 ➂

Specific weight: Sp.wt. 9.81 = 1.32 Sp.wt. = 12.94 kN/m3 Check: 206.96 3 0.2(0.2)(0.4) = 12935 N/m = 12.94 kN/m3

Problem 11 - Hydraulics ➀ ➁ ➂

What fraction of the volume of a solid object of sp.gr. 7.3 floats above the surface of a container of mercury? If the volume of the object below the liquid surface is 0.014 m3, what is the wt. of the object. What load applied vertically that would cause the object to be fully submerged?

Solution: ➀ Fraction of vol. of a solid object above the mercury: W = BF (7.3)(V)(9.81) = V1 (9.81)(13.6) V = 1.863 V1 V1 = 0.536 V V2 = V - 0.536 V V2 = 0.464 V V2 V = 0.464

V2 V1

Mercury (13.6)

➁ Wt. of object: V1 = 0.014 V = 1.863(0.014) V = 0.026 m3 W = 0.026(9.81)(7.3) W = 1.86 kN ➂

Load to cause the object to submerged: P = V2 (9.81)(13.6) P = 0.464(0.026)(9.81)(13.6) P = 1.61 kN

Problem 12 - Hydraulics A hydrometer weighs 0.002 kg has a stem at the upper end, which is 3 mm in diameter. How much deeper will it float in oil (sp.gr. = 0.78) than in alcohol having sp.gr. of 0.82? Solution: In alcohol: (sp.gr. = 0.82) BF = W (9810)(0.82)V1 = 0.002(9.81) V1 = 2.44 x 10-6 m3 V1 = 2440 mm3 In oil (sp.gr. = 0.78) BF = W 9810(0.78)V2 = 0.002(9.81) V2 = 2.56 x 10-6 m3 V2 = 2560 mm3 V = V1 – V2 V = 2560 – 2440 V = 120 mm3 π 2 (3) h = 120 4 h = 16.98 mm

h

w.s.

Alcohol (sp. gr. = 0.82)

w.s.

Oil (sp. gr. = 0.78)

Problem 13 - Hydraulics A piece of wood of sp.gr. 0.651 is 3 in. square and 5 ft. long. ➀ ➁ ➂

What is the volume of lead having a unit weight of 700 pcf that should be fastened at one end of the stick so that it will float upright in 1 ft. out of water? Determine the weight of the lead? Determine the total weight of lead and the wood.

Solution: ➀ Volume of lead: w1 + w2 = BF1 + BF2 ⎛ 3 ⎞⎛ 3 ⎞ 0.651 ⎜⎜12⎟⎟⎜⎜12⎟⎟ (5)(62.4) + 700 V ⎝ ⎠⎝ ⎠ ⎛ 3 ⎞⎛ 3 ⎞ = ⎜⎜12⎟⎟⎜⎜12⎟⎟(4) 62.4 + 62.4 V ⎝ ⎠⎝ ⎠ V = 0.00456 ft3. ➁ Weight of lead: w=VxD w = 0.00456 (700) w = 3.19 lb. ➂ Total weight:

3 3 = 3.19 + 0.651 (12 )(12 )(5)(62.4) = 15.88 lb.

Problem 14 - Hydraulics

W1 3”

3” 1’

BF1

4’

W2

Lead

BF2

A piece of wood floats in water with 50 mm projecting above the water surface. When placed in glycerine of sp.gr. 1.35, the block projects 75 mm above the liquid surface. ➀ ➁ ➂

Find the height of the piece of wood. Find the sp.gr. of wood. Find the weight of the wood if it has a cross sectional area of 200 mm x 200 mm.

Solution: ➀ Height of wood: W = 9.81(A)(h - 0.05) S h (9.81)(A) = 9.81A (h - 0.05) S h = h - 0.05 W = 9.81(1.35) A (h - 0.075) S h (9.81) A = 9.81(1.35) A (h - 0.075) S h = 1.35(h - 0.075) h - 0.05 = 1.35 (h - 0.075) 0.35h = 0.05125 h = 0.146 m.

0.05

h-0.05

water

0.075

➁ Sp.gr. of wood: S h = h - 0.05 S (0.146) = 0.146 - 0.05 S = 0.658 ➂

Wt. of wood: W = 9.81(0.658)(0.146)(0.2)(0.2) W = 0.038 kN W = 38 N

h-0.075

Glycerine

Problem 15 - Hydraulics An open horizontal tank 2 m high, 2 m. wide and 4 m. long is full of water. ➀ ➁ ➂ ➀

➁

➂

How much water is spilled out when the tank is accelerated horizontally at 2.45 m/sec2 in a direction parallel with its longest side? What is the force acting on the side with the greatest depth? Compute the required accelerating force. Solution: Volume of water spilled Water spilled out: θ a x=1 tan θ = g θ 2 1 2.45 tan θ = 9.81 4 1 tan θ = 4 2m x 1 1m F1 F2 4 =4 x = 1 m. 4 1(4)(2) Vol. of water spilled = 2 2m Vol. of water spilled = 4 cu. m. 1m F1 F2 Force acting on the side with the greatest depth. F1 = γw h1 A1 F1 = 9.81 (1)(2)(2) F1= 39.24 kN Required accelerating force: F2 = γw h2 A2 1 F2 = 9.81 (2 ) (1)(2) F2= 9.81 Accelerating force = 39.24 - 9.81 Accelerating force = 29.43 kN Check: F=ma W F = (g ) a [2(2)(4) - (4)] (9.81) F= (2.45) = 29.4 kN 9.81

Problem 16 - Hydraulics A 75 mm diameter pipe, 2 meter long is just filled with oil, specific gravity is 0.855 and then capped. Placed in a horizontal position, it is rotated at 27.5 rad/sec. about a vertical axis 0.5 meter from one end. ➀ What is the pressure in kPa at the far end of the pipe? ➁ What is the pressure in kPa at the other end of the pipe? ➂ What is the pressure at the midpoint of the pipe? Solution: ➀ Pressure in kPa at the far end of the pipe: P2 = γw h h = y - y1 ω 2 r2 ω y = 2g rad ω = 27.5 sec h 2 2 y (27.5) (2.5) y = 2(9.81) h1 1 y2 2 y = 240.9 m. y1 ω 2 r1 2 y1 = 2g r1 0.5m 2m (27.5)2 (0.5)2 r=2.5m y1 = 2(9.81) r =1.5 m. y1 = 9.64 h = 240.90 - 9.64 = 231.26 m. P2 = γw h P2 = 9.81(0.822) (231.26) = 1864.8 kPa ➁ Pressure in kPa at the other end of the pipe: P1 = 0 ➂ Pressure at the midpoint of the pipe: ω 2 r2 y2 = 2g (27.5)2 (1.5)2 y2 = 2(9.81) y2 = 86.73 Pm = γw h1 Pm = 9.81(0.822)(86.73 - 9.64) = 621.6 kPa

Problem 17 - Hydraulics A turbine is rated at 450 KN when the flow of water through it is 0.609 m3/s. Assuming an efficiency of 87%, what head is acting on the turbine? Solution: Power = QWE x Efficiency 450 = 0.609(9.81)E(0.87) E = 86.6 m.

Problem 18 - Hydraulics Oil of sp.gr. of 0.75 is flowing through a 150 mm pipe under a pressure of 103 kPa. If the total energy relative to a datum plane 2.40 m. below the center of the pipe is 17.9 m. kN/kN, determine the flow of oil. Solution: Energy =

V12 2g

+

P1 w

+Z

V2 103 17.9 = + + 2.40 2(9.81) 0.75(9.81) V = 5.43 m/s Q = AV π Q = (0.15)2 (5.43) 4 Q = 0.096 m3 / ss

Problem 19 - Hydraulics A 150 mm diameter jet of water is discharge from a nozzle into the air. The velocity of the jet is 36 m/s. Find the power in the jet? Solution: V2 E= 2g (36)2 E= = 66.06 2(9.81) Power = QwE Q = AV π Q = (0.15)2 (36) = 0.636 m3 / s 4 Power = 0.636(9.81)(66.06) Power = 412 kN

Problem 20 - Hydraulics For laminar conditions, what size of pipe will deliver 0.0057 m3/s of oil having viscosity of v = 6.09x10-6 m2/s. Solution: Q = AV

π 2 dV 4 0.00228 V= πd2 Rendds no. for laminar flow = 2000 Vd R= V ⎛ 0.0228 ⎞ (d) 2000 = ⎜ ⎝ πd2 ⎟⎠ 6.09x10 −6 d = 0.596 m. 0.0057 =

Problem 21 - Hydraulics A 1m. diameter new cast iron pipe C = 130 is 845 m. long and has a head loss of 1.11 m. Find the discharge capacity of the pipe according to Hazen Williams Formula? Solution: 10.64LQ1.85 HL = CD4.87 10.64(845)Q1.85 1.11 = 130(1)4.87 Q1.85 = 0.016 Q = 1.01 m3 / s

Problem 22 - Hydraulics A barrel containing water weighs 1.260 kN. What will be the reading on the scale if a 50 mm by 50 mm piece of wood is held vertically in the water to a depth of 0.60 m. Solution: BF = 0.05(0.05)(0.60)(9.81) BF = 0.015 kN Scale reading = 0.015 + 1.260 = 1.270 kN

Problem 23 - Hydraulics A piece of wood of sp.gr. 0.651 is 80 mm square and 1.5 m. long. How many Newtons of lead weighing 110 kN/m3 must be fastened at one end of the stick so that it will float upright with 0.3 m. out of water? Solution: W1 = (0.08)2(1.5)(0.651)(9.81) W1 = 0.061 kN BF1 = (0.08)2(1.2)(9.81) BF1 = 0.075 kN W2 = V2 D2 W2 = 110 V2 BF2 = 9.81V2 W1 + W2 = BF1 + BF2 0.061 + 110V2 = 0.075 + 9.81V2 100.19V2 = 0.014 V2 = 0.00014 m3 W2 = 110(0.00014) = 0.0154 kN = 15.4 N

0.08

0.08 0.3

W1

W2

BF1

BF2

1.2

Problem 24 - Hydraulics To what depth will a 2.4 m. diameter log 4.6 m. long and sp.gr. of 0.425 sink in fresh water? Solution:

r =1.2

r =1.2 w.s.

w.s.

h

W = π(1.2)2 (4.6)(0.425)(9.81) W = 1.922655(4.6)(9.81) Area of segment: (shaded section) ⎛ Sin 2θ ⎞ A = r2 ⎜ θ 2 ⎟⎠ ⎝ Try θ = 83.1667˚ θ = 1.4515 rad Sin 2θ = 0.11813 2 A = (1.2)2 ⎡⎣1.4515 - 0.11813 ⎤⎦ A = 1.92 BF = A(4.6)(9.81) BF = W 1.92(4.6)(9.81) = 1.92265(4.8)(9.81) 1.92 = 1.92265 almost equal Use ø = 83.1667˚ Depth of floatation = r - r Cos θ h = 1.2 - 1.2 Cos 83.1667˚ h = 1.057

4.6

h

Problem 25 - Hydraulics A barge with a flat bottom and square ends has a draft of 1.8 m. when fully loaded and floating in an upright position. The barge is 7.6 m. wide and 12.8 m. long and a height of 3 m. The center of gravity of the barge is 0.30 m. above the water surface if the barge is stable. ➀ ➁ ➂ ➀

Determine the distance of the metacenter above the center of buoyancy. Determine the distance of the metacenter above the barge center of gravity. What is the righting moment in water when the angle of heel is 12˚? Solution: Distance of the metacenter above the center of buoyancy I MBo = V 12.8(7.6)3 I= 12 I = 468.24

M

MBo w.s.

C.G.

0.3 m

3m

1.8 m

Bo 7.6 m

V = 7.6(12.8)(1.8) V = 175.10 468.24 175.10 MBo = 2.67 m. MBo =

12.8 m

Distance of the metacenter is 2.67 m. above the center of buoyancy. ➁

➂

7.6 m

Distance of the metacenter above the center of gravity of the barge Distance = 2.67 – 0.3 – 0.9 = 1.47 m. Righting moment in water when the angle of heel is 12˚ x = 1.47 Sin 12˚ x = 0.3056 BF = W BF = 7.6(1.8)(12.8)(9.81) BF = 1717.77 kN Righting moment = BF(x) Righting moment = 1717.77(0.3056) Righting moment = 584.95 kN.m.

1.47 1.2 1.8

M

W

12˚ C.G.

x 0.3

Bo

0.9

7.6

BF

0.9 m 0.9 m

Problem 26 - Hydraulics A tank is 1.5 m. square and contains 1.0 m. of water. How high must its sides be if no water is to be spilled when the acceleration is 4 m/s2 parallel to a pair of sides? Solution: a tan θ = g

y

y tan θ = 0.75 4 y = 9.81 0.75 y = 0.306 m. Its sides must be = 0.306 + 1 = 1.306 m. high (deep)



 1m

1.5 m

1.5 m

Problem 27 - Hydraulics A rectangular tank 6 m. long by 1.8 m. deep and 2.10 m. wide is filled with water accelerated in the direction of its length at a rate of 1.52 m/sec2. How many liters of water are spilled out? Solution: a tan θ = g h a = 6 g 6(1.52) h= 9.81 h = 0.93 Vol. spilled out: 0.93(6) V= (2.10) 2 V = 5.859 m3 V = 5859 liters

h



6m

1.8 m

Problem 28 - Hydraulics

A rectangular tank 6 m. long by 1.8 m. deep is 2.1 m. wide contains 0.90 m. of water. Find the unbalanced force necessary to accelerate the liquid mass in the direction of the tank length to accelerate 2.45 m/s2. Solution: F = ma m=

W g

m=

(6)(2.1)(0.90) (9810) 9.81

m = 11340 kg F = 2.45(11340) F = 27783 N

Problem 29 - Hydraulics A cubic tank is filled with 1.5 m. of oil, sp.gr. 0.752. ➀ Find the force acting on the side of the tank when the acceleration is 4.9 m/s2 vertically upward. ➁ Find the force acting on the side of the tank when the acceleration is 4.9 m/s2 vertically downward. ➂ Determine the pressure at the bottom of the tank when the acceleration is 9.81 m/s2 vertically downward.

a=4.9 m/s2

Solution: ➀ Force acting on the side of the tank when the acceleration is 4.9 m/s2 vertically upward. ⎛ a⎞ PA = γ h ⎜ 1 + ⎟ g⎠ ⎝

Oil

⎛ 4.9 ⎞ PA = 0.752(9.81)(1.5) ⎜ 1 + 9.81⎟⎠ ⎝ PA = 16.59 kPa

PA

16.59 F= (1.5)(1.5) = 18.67 kN 2

PB = 5.54 kPa F=

5.54 (1.5)(1.5) = 6.23 kN 2

A a=4.9 m/s2

➁ Force acting on the side of the tank when the acceleration is 4.9 m/s2 vertically downward ⎛ a⎞ PB = γ h ⎜ 1 - ⎟ ⎝ g⎠

⎛ 4.9 ⎞ PB = 0.752(9.81)(1.5) ⎜ 1 ⎝ 9.81⎟⎠

Oil

F

PB

1.5 m

B a=9.81 m/s2

➂ Pressure at the bottom of the tank when the acceleration is 9.81 m/s2 vertically downward ⎛ a⎞ PC = γ h ⎜ 1 - ⎟ ⎝ g⎠

Oil

PC = 0.752(9.81)(1.5)(1 - 1) PC = 0

1.5 m

PC

C

1.5 m

Problem 30 - Hydraulics An open cylindrical tank, 1.8 m. and 0.9 m. in diameter contains water to a depth of ¾ of its height. If the cylinder rotates about its geometric axis. ➀ What constant angular velocity can be attained without spilling any water? ➁ What is the pressure at the center of the bottom of the tank when ω = 6 rad/sec? ➂ What is the pressure at the bottom wall of the tank when ω = 6 rad/sec? Solution: ➀ Constant angular velocity can be attained without spilling any water ω2 r2 y= 2g

y=0.90

0.45

0.45 0.45

1.8 1.35

ω 2 (0.45)2 0.90 = 2(9.81) ω = 9.34 rad / sec.

D

C

0.90

➁ Pressure at the center of the bottom of the tank

ω2 r2 y= 2g 2

2

(6) (0.45) 2(9.81) y = 0.372 m. y h = 1.35 2 0.372 h = 1.35 2 h = 1.164

y

y=

0.45

3/2 3/2

1.8 1.35

h

PC = 9.81(1.164)

C

PC = 11.42 kPa

➂ Pressure at the bottom wall of the tank H = 1.35 + 0.186 H = 1.536 PD = 9.81(1.536) PD = 15.07 kPa

0.45

y

1.35

H

D

y/2 y/2=0.372/2

Problem 31 - Hydraulics A closed cylindrical tank 1.8 m. high and 0.9 m. in diameter contains 1.4 m. of water. When the angular velocity is constant at 20 rad/sec, how much of the bottom of the tank is uncovered? 

Solution: ω2 r2 y= 2g (20)2 (0.45)2 y= 2(9.81) y = 4.13 m. ω 2 x12 y1 = 2(9.81) (20)2 x12 y1 = = 20.387 x12 2(9.81) ω 2 x 22 y2 = 2g (20)2 x 22 y2 = = 20.387 x 22 2(9.81) Vol. of air is constant since the tank is closed. π x 22 y 2 π x12 y1 V= (after rotation) 2 2 V = π(0.45)2 (0.40) (before rotation) V = 0.2545 m3

x2 0.4 m

Air

y 1.8 m

1.4 m

x1

y1 0.90

 x2 0.4 m

Air

y2

x1 y

y2=1.8 + y1

Problem 32 - Hydraulics An open vessel of water accelerates up a 30˚ plane at 3.66 m/s2. What is the angle the water surface makes with the horizontal? Solution: a x = 3.66 Cos 30˚



a x = 3.17 Cot θ = tan 30˚ +

g ax

Cot θ = tan 30˚ +

9.81 3.17

a

30˚

Cot θ = 3.672 θ = 15˚14'

Problem 33 - Hydraulics A vessel partly filled with water is accelerated horizontally at a constant rate. The inclination of the water surface is 30˚. What is the acceleration of the vessel? Solution: a tan θ = g a 9.81 a = 5.66 m / s 2 tan 30˚ =

Problem 34 - Hydraulics

An open tank of water accelerates down a 15˚ inclined plane at 16.1 ft.sec2. What is the slope of the water surface? Solution:

g Cot θ = tan 15˚ + ax a x = 16.1 Cos 15˚ a x = 15.55 32.2 Cot θ = - tan 15˚ + 15.55 Cot θ = 1.80279 tan θ = 0.5546956671 θ = 29˚01'

Problem 35 - Hydraulics An open cylindrical tank 4 ft. in diameter and 6 ft. deep is filled with water and rotated about its axis at 60 rpm. ➀ How much liquid is spilled? ➁ How deep is the water at the x-axis? ➂ At what speed should the tank be rotated in order for the center of the bottom of the tank to have zero depth of water? Solution: ➀ Vol. of liquid is spilled ω2 r2 y= 2g (2π)2 (2)2 y= 2(32.2) y = 2.452

=60 rpm

y 6 ft h

π(2)2 (2.452) Vol. spilled out = = 15.4 ft 3 2 4

➁ Depth of the water at the x-axis h = 6 – 2.452 h = 3.548 ➂ Speed that the tank be rotated about its axis to have zero depth of water ω2 r2 y= 2g ω 2 (2)2 6= 2(32.2) ω = 9.83 rad / sec.

Problem 36 - Hydraulics A closed vessel 1 m. in diameter is completely filled with water. If the vessel is rotated at 1200 rpm, what increase in pressure will occur at the top of the tank at the circumference? Solution: 1200(2π) ω= 60

=1200 rpm

ω = 40π

y

ω2 r2 y= 2g (40π)2 (0.5)2 y= 2(9.81) y = 201.22 p =γ y p = 9.81(201.22) p = 1973 kPa

1m

Problem 37 - Hydraulics An open vessel 18 in. in diameter and filled with water is rotated about its vertical axis at such a velocity that the water surface 4 in. from the axis makes an angle of 40˚ with the horizontal. Compute the speed of rotation. Solution: ω2 r2 y= 2g dy = tan 40˚ dx dy ω 2 (2)r = dx 2(g)

r=

4 ft. 12

ω 2 (2)(4 /12) tan 40˚ = 2(32.2)

x=4

40˚

ω = 9 rad / sec.

Problem 38 - Hydraulics The 2 m. diameter impeller of a closed centrifugal water pumped is rotated at 1500 rpm. If the casing is full of water, what pressure head is developed by rotation? Solution:

ω2 r2 y= 2g 1500(2π) ω= = 50π rad/sec. 60 (50π)2 (1)2 y= 2(9.81) y = 1258 m.

Problem 39 - Hydraulics An unbalanced vertical force of 268 N upward accelerates a volume of 0.044 m3 of water. If the water is 0.90 m. deep in a cylindrical tank, what is the force acting on the bottom of the tank? Solution: W = VD

W = 0.044(9.81) W = 0.43164 kN W = 431.64 N m F= a g W F= a g 431.64 a 9.81 a = 6 m/s 2 268 =

W = π r 2 h (9.81) 431.64 = π r 2 (0.9)(9.81) r = 0.12 m. ⎛ a⎞ P = γ h ⎜1 + ⎟ g⎠ ⎝ ⎛ 6 ⎞ P = 9.81(0.9) ⎜ 1 + 9.81⎟⎠ ⎝ P = 14.23 kPa Force acting on the bottom of the tank: F = PA F = 14.23(π)(0.12)2 F = 0.6437 kN F = 643.7 N

Problem 41 - Hydraulics The 8 ft. diam. cylinder weighs 500 lb. and rests on the bottom of a tank that is 3 ft. long. Water and oil are poured into the left and right portions of the tank to depths of 2 ft. and 4 ft. respectively. ➀ Find the horizontal component of the force that will kept the cylinder touching the tank at B. ➁ Find the vertical component of the force that will push up the cylinder. ③ Compute the force that will keep the cylinder touching the tank at B.

Solution: ➀ Horizontal component of the force that will kept the cylinder touching the tank at B: P1 = γw h A P1 = 62.4 (1) (2) (3) P1 = 374 lb.

500 lb Oil (0.75)

2

P2 = γw h A P2 = 62.4 (2) (4) (3) (0.75) P2 = 1123 lb. Ph = P2 - P1 Ph = 1123 - 374 Ph = 749 lb.

H2O P1

4 60˚

2 2

P2

B Fv1

Fv2

500 lb

➁ Vertical component of the force that will push up the cylinder: 2 π(4) Fv2 = 62.4 4 (3)(0.75) = 1764 lb.

H2O

4

60˚

4 2 2

Fv1 = 62.4

[

]

2(4)Sin 60˚ (3) 2

Fv1 = 920 Fv = 1764 + 920 Fv = 2684 lb. ③ Force that will keep the cylinder touching the tank at B: FB + 500 = 2684 FB = 2184 lb. (downward)

B Fv1

Oil (0.75)

4

2

π(4)2 (60 360

4

Fv2

Problem 43 - Hydraulics A concrete block with a volume of 0.023 cu.m. is tied to one end of a wooden post having dimensions of 200 mm x 200 mm by 3 m. long and placed in fresh water. Weight of wood is 6.4 kN/m3 and that of concrete is 23.5 kN/m3. ➀ Det. the length of the wooden post above the water surface. ➁ Det. the volume of additional concrete to be tied to the bottom of the post to make its top flush with the water surface. ➂ Det. the total weight of concrete to make its top flush with the water surface.

Solution: ➀ Length above the water surface: W + Wc = BF1 + BF2 3(0.20)(0.20)(6.4) + 0.023(23.5) = 0.2(0.2) h (9.81) + 0.023(9.81) h = 2.76 Length above water surface = 3 - 2.76 Length above water surface = 0.24 m. ➁ Additional vol. of concrete: W + W1 = BF1 + BF2 0.20(0.20)(3)(6.4) + (0.023 + V) 23.5 = 0.2(0.2)(3)(9.81) + (0.023 + V)9.81 0.768 + 0.5405 + 23.5V = 1.772 + 0.22563 + 9.81V 13.69V = 0.09433 V = 0.0069 m3 ➂ Total weight of concrete: Wc = (0.023 + 0.0069) 23.5 Wc = 0.70265 kN Wc = 702.65 N

0.20m

0.20m

h BF1

3m

W

Con.

BF1

3m

W

Con.

BF2

Problem 44 - Hydraulics A rectangular barge weighing 200000 kg is 14 m long, 8 m. wide and 4.5 m deep. It will transport to Manila 20 mm diameter; 6 m. long steel reinforcing bars having a density of 7850 kg/m3. Density of salt water is 1026 kg/m3. ➀ ➁ ➂

Det. the draft of the barge on sea water before the bars was loaded. If a draft is to be maintained at 3 m., how many pieces of steel bars could it carry? What is the draft of the barge when one half of its cargo is unloaded in fresh water?

Solution: ➀

200000 kg

Draft of empty barge on sea water: 200000 = 14(8)(d)(1026) d = 1.74 m.

4.5 d

➁

No. of bars loaded: 8(14)(3)(1026) = 200000 + Wb Wb = 144736 kg

8m

π 2 Wb = 4 (0.02) (6) N (7850) π 2 (0.02) 144736 = 4 (6)(7850) N N = 9782 bars

Steel bars

200000+Wb

3m

➂

Draft of barge on fresh water when one half of its cargo is unloaded: 1 200000 + 2 (144736) = 8(14)(d)(1000) d = 2.43 m.

8m

200000+1/2Wb

d 8m

Problem 45 - Hydraulics A wooden pole (sp.gr. = 0.55) 550 mm. in diameter, has a concrete cylinder (sp.gr. = 2.5) 550 mm. long and of same diameter attached to one end. Unit weight of water is 9.79 kN/m3. ➀ Determine the min. length of pole for the system to float vertically in static equilibrium. ➁ Determine the weight of wood. ③ Determine the total weight of wood and concrete.

Solution: ➀ Min. length of pole: ww + wc = BFw + BFc A( L) (0.55)(9.79) + A (0.55)(2.5)(9.79) = A (L + 0.55)(9.79) 0.55 L + 1.375 = L + 0.55 0.45 L = 0.825 L = 1.833 m.

Wood (0.55)

L

Concrete (sp. gr.=2.5)

550mm

550mm

➁ Weight of wood. π 2 (0.55) Wt. of wood= 4 (1.833)(0.55)(9.79)

Ww

Wt. of wood = 2.34 kN BFw

L

③ Total weight of wood and concrete. Total weight= π4 (0.55) 2 (0.55)(2.5)(9.79)+2.34

Total weight = 5.54 kN

Wc 550mm

Concrete (sp. gr.=2.5)

BFc 550mm

Problem 46 - Hydraulics If the center of gravity of a ship in the upright position is 10 m. above the center of gravity of the portion under water, the displacement being 1000 metric tons, and the ship is tipped 30˚ causing the center of buoyancy to shift sidewise by 8 m. ➀ Find the location of the metacenter from the bottom of the ship if its draft is 3 m. ➁ Find the metacentric height. ③ What is the value at the moment in kg.m. Solution: ➀ Location of the metacenter from the bottom of the ship if draft is 3 m. 8 Sin 30˚ = MB Upright Position o MBo = 16 D G y = MBo + 2 Bo y = 16 + 1.5 = 17.5 m. GBo=10

➁ Metacentric height. MG = MBo - GBo MG = 16 - 10 MG = 6 m. ③ Value at the moment in kg.m. x Sin 30˚ = MG x = (Sin 30˚) (6) x = 3 m. 1000 kg BF = 1000 metric tons x metric tons BF = 1,000,000 kg Moment = 1,000,000 (3) Moment = 3,000,000 kg.m

10m

Tilted Position

M

y 30˚

G x Bo

8m

D/2 D/2 BF

Problem 47 - Hydraulics A cylindrical vessel 400 mm in diameter and filled with water in rotated about its vertical axis with speed such that the water surface at a distance of 100 mm from the vertical axis makes an angle of 45˚ with the horizontal. ➀ Find the speed of rotation in rpm. ➁ Find the difference in pressure at a point 0.10 m from the vertical axis and at the vortex of the water surface. ➂ How far is the vortex of the water surface from the top of the vessel.

Solution : ➀ Speed of rotation in rpm. ω 2 x2 y= 2g dy ω2 dx = 2 g 2x dy tan 45˚ = dx = 1 ω2 (2)(0.1) 1 = 2 (9.81) ω = 9.9 rad/sec 9.9(60) ω= = 94.58 rpm 2π ➁ Difference in pressure at a point 0.10 m from the vertical axis. ω 2 r2 y = 2g (9.9)2 (0.1)2 y = 2(9.81) y = 0.05 m. P = γw h P = 9.81(0.05) P = 0.49 kPa ➂ Distance of the vortex of the water surface from the top of the vessel. ω2 r2 (9.9)2 (0.2)2 y = 2g = 2(9.81) = 0.20 m.

ω x=0.10m 45˚ y

ω r=0.20

y

Problem 48 - Hydraulics A hemispherical bowl having a radius of 1 m. is full of water. If the hermispherical bowl is made to rotate uniformly about the vertical axis at the rate of 30 rpm. ➀ Determine the volume of water that is spilled out. ➁ Determine the remaining volume of water in the hemispherical bowl. ➂ Determine the maximum pressure at the bottom of the hemispherical at this instant.

Solution: ➀ Volume of water that is spilled out.

ω 2 r2 h = 2g

ω=

Vol. spilled out

30(2π) 60

1m

ω

1m h

ω = 3.14 rad/sec. (3.14)2 (1)2 h = 2(9.81) h = 0.5 m. Vol. of water spilled out:

π(1) (0.5) 2 V = 0.785 2

V=

➁ Remaining volume of water in the hemispherical bowl. 4 π r3 V = 3 2 - 0.785 2 V = 3 π (1)3 - 0.785 V = 1.31 m3

➂ Maximum pressure at the bottom of the hemispherical at this instant. h = 1 - 0.5 h = 0.5 P = γw h P = 9.81(0.5) P = 4.91 kPa

Problem 49 - Hydraulics Two reservoirs A and B have elevations of 250 m and 100 m respectively. It is connected by a pipe having a diameter of 25 mmø and a length of 100 m. A turbine is installed at point in between reservoirs A and B. If C = 120, compute the following if the discharge flowing in the pipe is 150 liters/sec. El. 250m A

150m El. 100m

T

250 mm ø

➀ Head loss of pipe due to friction. ② The head extracted by the turbine. ③ The power generated by the turbine.

Solution: ➀ Headloss of pipe: 10.64LQ1.85 hf = 1.85 4.87 C D Q = 0.15 m3/s 10.64(100)(0.15)1.85 hf = (120)1.85(0.25)4.87 hf = 3.87 m. ② Head extracted by the turbine: VA2 PA VB2 PB + + Z = + ZB + HE + HL A 2g + 2g γw γw 0 + 0 + 250 = 0 + 0 + 100 + HE + 3.87 HE = 250 - 103.87 HE = 146.13 m. ③ Power generated by the turbine: Power = QWE Power = 0.15(9810)(146.13) Power = 215030 watts Power = 215.03 kW

B

Problem 50 - Hydraulics Reservoirs A, B and C are connected by pipelines 1, 2 and 3 respectively which meets at the junction D. The elevation of reservoir A is 300 m, while that of C is 277 m. Reservoir B is higher than reservoir A. The rate of flow out of reservoir B is 560 liters/sec. Pipes

Diam.

Length

1 2 3

900 mm 600 mm 450 mm

1500 m 450 m 1200 m

Friction factor "f" 0.0208 0.0168 0.0175

hf 2 El. 300

A Q2

➀ Compute the discharge flowing in or out of reservoir A. ➁ Compute the discharge flowing towards reservoir C. ③ Compute the elevation of reservoir B.

Solution:

➀ Rate of flow at A: 0.0826 f L Q2 hf = D5 0.0826(0.0168)(450)(0.56)2 hf2 = = 2.52 m. (0.6)5 hf3 = 23 + hf1 hf3 – hf1 = 23 Q3 = Q2 – Q1 Q3 = 0.56 – Q1 hf3 - hf1 = 23 0.0826(0.0175)(1200)Q32 0.0826(0.0208)(1500)Q12 = 23 (0.45)5 (0.9)5 94Q32 - 4.36Q12 = 23 94(0.56 – Q1)2 - 4.36Q12 = 23 94(0.3136 - 1.12Q1+Q12) - 4.36Q12 = 23 29.4784 - 105.28Q1+94Q12 - 4.36Q12 = 23 89.64Q12 - 105.28Q1 + 6.4787 = 0 Q12 - 1.17Q1 + 0.0723 = 0 Q1 = 0.065 m3/s = 65 liters/sec ➁ Rate of flow towards reservoir C: Q3 = 0.56 – Q1 Q3 = 0.56 - 0.065 = 0.495 Q3 = 495 liters/ec

B

hf 1

1

2

23

Q1

D

hf 3

Q3 El. 277 3

C

③ Elevation of B: Elev. of B = 300 + hf1 + hf2 0.0826(0.0208)(1500)(0.065)2 Hf1 = = 0.018 m (0.9)5 Elev. B = 300 + 0.018 + 2.52 = 302.538 m.

Problem 51 - Hydraulics A valve is suddenly closed in a 200 mmø pipe. The increase in pressure is 700 kPa. Assuming that the pipe is rigid and the bulk modulus of water is 2.07 x 109 N/m2. ➀ ➁ ➂

Compute the celerity of the pressure wave. Compute the velocity of flow. If the length of the pipe is 650 m. long, compute the water hammer pressure at the valve if it is closed in 3 sec.

Solution: ➀

Celerity of the pressure wave: C=

EB ρ

2.07 x 109 1000 C = 1438.75 m/s C=

➁

Velocity of flow. Increase in pressure: Ph = ρCV 700000 = 1000(1438.75)V V = 0.486 m/s

➂

Water hammer pressure when it is closed in 3 sec: 2L t= C 2(650) t = 1438.75 t = 0.904 sec 0.904 Ph = 3 (700) Ph = 210.83 kPa

Problem 52 - Hydraulics A sharp edge orifice, 75 mm in diameter lies in a horizontal plane, the jet being directed upward. If the jet rises to a height of 8 m. and the coefficient of velocity is 0.98. ➀ Determine the velocity of the jet. ➁ Determine the head loss of the orifice. ③ Determine the head under which the orifice is discharging neglecting air resistance.

Solution:

➀ Velocity of the jet: V22 = V12 - 2g h 0 = V12 - 2(9.81)(8) V12 2g = 8 V1 = 12.53 m/s ➁ Head loss of orifice: V12 1 HL = 2g C 2 - 1 v

[

[

1 HL = 8 (0.98)2 HL = 0.33 m. ③ Head of orifice: H = 0.33 + 8 H = 8.33 m.

] - 1]

2 HL H

h=8m

Orifice

1

Problem 58 - Hydraulics If the viscosity of oil (sp.gr. = 0.85) is 15.4 poises, compute the kinematic viscosity in m2/s. Solution: Absolute viscosity = 15.14(0.1) = 1.514

Kinematic viscosity =

1.514 = 0.00178 m2 / s 0.85(1000)

Problem 59 - Hydraulics If the viscosity of oil having a sp.gr. of 0.75 is 500 centipoise, compute the absolute viscosity in Pa.S. Solution: Absolute viscosity = 500(10)-3 = 0.5 Pa.S.

Problem 60 - Hydraulics Liquid A, B, and C in the container shown have sp.gr. of 0.80, 1.0 and 1.60 respectively. Determine the difference in elevation of the liquid B and C in each piezometer tube. Solution: For liquid C: 2(0.80) + 4(1) + 2(1.6) – 1.60h1 = 0 h1 = 5.5 For liquid B: 1(0.8) + 4(1) – h2 = 0 h2=5.6 h2 = 5.6 B 5.6 + 2 = h + 5.5 h = 2.1 m. 2m C

2m h

A (0.80)

4m 5.5

B (1.0) 2m

C (1.60)

h

Problem 61 - Hydraulics A water tank 3 m. in diameter and 6 m. high is made from a steel having a thickness of 12 mm. If the circumferential stress is limited to 5 MPa, what is the minimum height of water to which the tank may be filled? Solution: pD 2t p(3)(1000) 5= 2(12) p = 0.04 MPa p=γ h σ=

(1000)2 0.04 = 981 h h = 4.08 m

Problem 62 - Hydraulics A pressure vessel 320 mm in diameter is to be fabricated from steel plates. The vessel is to carry an internal pressure of 4 MPa. What is the required thickness of the plate if the vessel is to be spherical with an allowable stress of 120 MPa? Solution: pD σ= 2t 4(320) 120 = 4t t = 2.67 say 3 mm

Problem 63 - Hydraulics A cylindrical tank, having a vertical axis is 1.8 m. in diameter and 3 m. high. Its sides are held in position by means of two steel hoops, one at the top and one at the bottom. Find the ratio of the tensile stress at the bottom to that of the top. Solution: p =γ hA

2T1

p = 9.81(1.5)(3)(1.8) p = 79.46 kN ∑M A = 0

2T1 + 2T2 = 79.46

1(79.46) = 2T1 (3)

2(26.48) Ratio = =2 2(13.24)

T1 = 13.24

B

3m

2(13.24) + 2T2 = 79.46

P

T2 = 26.48

1m

A

2T2

Problem 64 - Hydraulics A vertical surface 4 m. square has its upper edge horizontal and on the water surface. At what depth must a horizontal line a drawn on this surface so as to divide it into two parts on each of which the total pressure is the same? Solution: p1 = γ h1 A1

p1 = 9.81(h/2)(4)h p1 = 19.62 h2 p 2 = γ w h2 A 2 ⎛ h + 4⎞ p 2 = 9.81⎜ (4 - h)(4) ⎝ 2 ⎟⎠ p 2 = 19.62(4 + h)(4 - h) p1 = p 2 19.62h2 = 19.62(4 + h)(4 - h) h = 2.828 m.

4m

h1

h w.s. 4-h

h2

Problem 65 - Hydraulics A vertical rectangular gate 2 m. wide and 3 m. high, hinged at the top, has water on one side. What force applied at the bottom of the gate at an angle of 45˚ with the vertical is required to open the gate w/hen the water surface is 1.5 m. above the top of the gate? w.s.

Solution: P =γw hA

P = 9.81(3)(2)(3) P = 176.58 KN 2(3)3 Ig = = 45 12 Ss = Ay = 2(3)(3) = 18 e=

Ig Ss

=

4.5 = 0.25 18

1.5 m

Hinge

y =h=3m

A 1.5 m

F

3m

e

P

1.5 m 45˚

∑MA = 0 F Sin 45˚(3) = 176.58(1.5 + 0.25) F = 145.67 KN

Problem 66 - Hydraulics A trapezoidal dam having a total height of 20 m. on the vertical side has a width of 2 m. at the top and 8 m. at the bottom. The height of water in the vertical side of the dam is 12 m. Neglecting hydrostatic uplift; determine the factor of safety against overturning. Assume concrete weighs 23.5 KN/m3. 2m W1

Solution: Consider 1-meter length of dam.

P =γwh A P = 9.81(6)(12)(1) P = 706.32 KN W1 = 2(20)(1)(23.5) W1 = 940 KN 8(20) W2 = (1)(235) 2 W2 = 1880

w.s.

20m

RM = W1 (7) + W2 (4) RM = 940(7) + 1880(4) RM = 14100 KN.m. OM = P(4) OM = 706.32(4) OM = 2825.28 KN.m. RM 14100 F.S. = = = 4.99 OM 2825.28

12m

P 4m

W2 Heel

Toe

4m 7m 8m

Problem 67 - Hydraulics A tank with vertical sides is 1.5 m. square, 3.5 m. depth is filled to a depth of 2.8 m. of a liquid having a sp.gr. of 0.80. A cube of wood having a sp.gr. of 0.60 measuring 1 m. on an edge is placed on the liquid. By what amount will the liquid rise on the tank? 0.80

Solution: Weight of cube = buoyant force (1)(1)(1)(0.60)(0.81) = (1)(1)(d)(0.80)(9.81) d = 0.75 m. (1)(1)(0.75 - x) = [1.5(1.5) - 1(1)] x 0.75 - x = 1.25x 2.25x = 0.75 x = 0.333 m.

w.s.

y 0.40-y

0.4 0.4

2.5

1.4

Problem 68 - Hydraulics A ship of 4000 displacement floats in seawater with its axis of symmetry vertical when a weight of 50 tons is midship. Moving the weight 3 m. towards one side of the deck causes a plumb bob, suspended at the end of a strong 3600 mm long to move 225 mm. Find the metacentric height. 4000

Solution: 225 Sin θ = 3600 θ = 3.58˚ ∑M G = 0

50(3) 4050x = Cos 3.58˚ x = 0.037 m 0.037 Sin 3.58˚ = MG MG = 0.594 m. metacentric height

3m

M

θ

w.s.

50

Gx

θ

θ θ

3/cosθ 3600 θ

B

BF=4050 tons 225

Problem 69 - Hydraulics An open vessel of water accelerates up a 30˚ plane at 2 m/s2. What is the angle that the water surface makes with the horizontal? Solution: ah = 2 Cos 30˚ ah = 1.73 m/s2 av = 2 Sin 30˚ av = 1 m/s2 ah tan ø =g + a v 1.73 tan ø = 9.81 + 1 ø = 9.09˚

av

θ

a 30˚ ah

W 30˚

(Wah)/g θ

(Wav)/g

θ

W+W(1)/g

W(1.73)/g

Problem 70 - Hydraulics A cylindrical vessel 2 m. in diameter and 3 m. high has a rounded circular orifice 50 mm in diameter at the bottom. If the vessel is filled with water, how long will it take to lower the water surface by 2 m. Assume C = 0.60 2m

Solution:

Qave =

Qave Qave

CA 2g

(

h1 + h2 2

)

⎛ π⎞ 0.60 ⎜ ⎟ (0.05)2 2(9.81) ⎝ 4⎠ = 2 = 0.0143 m3 /s

(

3+ 1

)

2m 3m 1m

Vol. of water removed = π(1)2 (2) Vol. of water removed = 2π m3 V 2π t= = = 440.7 sec. Q 0.0143

Orifice 50 mm ¿

Problem 61 - Hydraulics A rectangular channel having a width of 3 m. carries water flowing at a rate of 20 m3/s. If the depth of water in the channel is 1.2 m., compute the critical kinetic energy. Solution: Vc2 Critical kinetic energy = 2g Vc2/2g

Critical depth : q=

20 = 6.67 m3 /s/m 3

dc =

3

Emin. dc

q2 g

2 (6.67) dc = 3 = 1.65 m 9.81 Vc2 + dc = Emin 2g

Emin Emin Emin

3 = dc 2

Vc2 + 1.65 = 2.475 2g

3 = (1.65) 2 = 2.475

Vc2 = 0.825 2g

Problem 62 - Hydraulics A trapezoidal channel has a bottom width of 6 m. and side slopes of 2 hor. to 1 vertical. When the depth of flow is 1.2 m., the flow is 30.40 m3/s. ➀ Compute the specific energy. ➁ Compute the slope of channel if n = 0.014. ③ Compute the average shearing stress at the boundary.

V2 Specific energy = +d 2g

1.2

1

2.4

1.2

2

1 2

6

(2.02)2 E= + 1.2 = 1.41 m. 2(9.81)

➁ Slope of channel:

1 V = R 2/3 S1/2 n P = 2.683(2) + 6 P = 11.366 A R= P 10.08 R= = 0.887 11.36 (0.887)2/3 S1/2 2.02 = 0.014 S = 0.00094

③ Ave. shearing stress: τ = γw R S τ = 9.81(0.887)(0.00094) τ = 0.0082 kPa τ = 8.2 N/m2

3

Q = AV 20.40 = 10.08 V

10.8 6

2.4

2. 68

Solution: ➀ Specific energy: (10.8 + 6) A= (1.2) 2 A = 10.08 m2

Problem 63 - Hydraulics A circular concrete sewer 1.5 m. in diameter and flowing half full has a slope of 4 m per 5 km. Determine the discharge if n = 0.013. Solution: π(0.75)2 A= 2 A = 0.884 m2

P = π(0.75) P = 2.356 A R= P 0.884 R= 2.356 R = 0.375 m 4 S= 5000 S = 0.0008 1 V = R 2/3 S1/2 n 1 V= (0.375)2/3 (0.0008)1/2 0.013 V = 1.131 m/s Q = AV Q = 0.884(1.131) Q = 1 m3/sec. Q = 1000 liters/sec.

r =0.75

Problem 64 - Hydraulics The cross section of a right triangular channel is shown with a coefficient of roughness n = 0.012. If the rate of flow = 4 m3/s. ➀ Calculate the critical depth. ➁ Calculate the critical velocity. ➂ Calculate the critical slope.

dc

dc 45˚ 45˚

Solution: ➀ Critical depth: 2d d A = c c = d2c 2 B = 2 dc Q2 A3 = g B

B=2dc

➂ Critical slope: 1 Vc = R 2/3 S1/2 c n

2 3 (4)2 (dc ) = 9.81 2dc

P = 2 2 dc

d5c = 3.26

P = 2 2 (1.267)

dc = 1.267m.

P = 3.58 ➁ Critical velocity:

Vc = g

A B

A = (1.267)2 A = 1.605 m

2

B = 2(1.267) B = 2.534 Vc =

9.81(1.605) 2.534

Vc = 2.49 m / s

R=

A P

R=

1.605 3.58

R = 0.448 1 2/3 1/2 R Sc n 1 2.49 = (0.448)2/3 S1/2 c 0.012 Sc = 0.0026 Vc =

dc

2dc