Hydrau Coaching 3 PDF

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####### Problem 41 - HydraulicsThe 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...

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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.

H 2O P1

4 60˚

2 2

P2

B Fv1

Fv2

500 lb

4

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˚

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) Visit For more Pdf's Books Pdfbooksforum.com

B Fv1

Oil (0.75)

4

2

π(4)2 (60 360

4

Fv2

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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. 0.20m

0.20m

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

h BF 1

W

Con.

3m

W BF1

➂ Total weight of concrete: Wc = (0.023 + 0.0069) 23.5 Wc = 0.70265 kN Wc = 702.65 N

Con.

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3m

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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)

Steel bars

200000+Wb

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

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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 Wt. of wood= 4 (0.55) (1.833)(0.55)(9.79)

Ww

Wt. of wood = 2.34 kN BFw

L

③ Total weight of wood and concrete.

Wc

Concrete (sp. gr.=2.5)

Total weight= π (0.55) 2 (0.55)(2.5)(9.79)+2.34 4

550mm

BFc

Total weight = 5.54 kN 550mm

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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˚ = Upright Position MBo MBo = 16 D G y = MBo + 2 Bo y = 16 + 1.5 = 17.5 m.

10m

GBo=10

➁ Metacentric height. MG = MBo - GBo MG = 16 - 10 MG = 6 m.

Tilted Position

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 Visit For more Pdf's Books Pdfbooksforum.com

M

y 30˚

G x Bo

8m

D/2 D/2 BF

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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 ω 2 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

ω

x=0.10m 45˚ y

ω r=0.20

➁ Difference in pressure at a point 0.10 m from the vertical axis. ω2 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.

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y

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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.

ω2 r2 h = 2g

ω=

Vol. spilled out

30(2π) 60

1m

ω

1m h

ω = 3.14 rad/sec. (3.14)2 (1)2 2(9.81) h = 0.5 m. h=

Vol. of water spilled out:

➂ Maximum pressure at the bottom of the hemispherical at this instant. h = 1 - 0.5

π(1)2 (0.5) V= 2 V = 0.785

h = 0.5

➁ Remaining volume of water in the hemispherical bowl. 4 π r3 V = 3 2 - 0.785

P = γw h P = 9.81(0.5) P = 4.91 kPa

2 V = 3 π (1)3 - 0.785 V = 1.31 m3

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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 2g + γ w + ZA = 2g + γw + ZB + HE + HL 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 Visit For more Pdf's Books Pdfbooksforum.com

B

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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 1 2 3

Diam. 900 mm 600 mm 450 mm

Length 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 = 2.52 m. hf2 = (0.6)5 hf3 = 23 + hf1 hf3 – hf1 = 23 Q 3 = Q2 – Q 1 Q3 = 0.56 – Q1 hf3 - hf1 = 23 0.0826(0.0175)(1200)Q32 0.0826(0.0208)(1500)Q12 = 23 (0.9)5 (0.45)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

hf1

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 = 0.018 m Hf1 = (0.9)5 Elev. B = 300 + 0.018 + 2.52 = 302.538 m.

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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

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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

2 HL H 1

➁ Head loss of orifice: V12 1 HL = 2g Cv2 - 1

[

[

1 HL = 8 (0.98)2 HL = 0.33 m.

h=8m

Orifice

] - 1]

③ Head of orifice: H = 0.33 + 8 H = 8.33 m.

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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 h 2=5.6 h2 = 5.6 B 5.6 + 2 = h + 5.5 h = 2.1 m.

2m

A (0.80)

h 4m

B (1.0)

5.5 2m

2m C

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C (1.60)

h

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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

σ=

p=γ h (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 Visit For more Pdf's Books Pdfbooksforum.com

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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. B

Solution: p=γ hA

2T1

p = 9.81(1.5)(3)(1.8)

2T1 + 2T2 = 79.46

p = 79.46 kN

2(13.24) + 2T2 = 79.46

∑M A = 0

T2 = 26.48

1(79.46) = 2T1(3)

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

T1 = 13.24

3m P 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 A 1

p1 = 9.81(h/2)(4)h p1 = 19.62 h

h1

2

p2 = γ w h2 A2 ⎛ h + 4⎞ (4 - h)(4) p2 = 9.81⎜ ⎟ 2 ⎝ ⎠

h h2

4m w.s. 4-h

p2 = 19.62(4 + h)(4 - h) p1 = p 2 19.62h2 = 19.62(4 + h)(4 - h) h = 2.828 m. Visit For more Pdf's Books Pdfbooksforum.com

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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 P

3m

e 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 Solution: Consider 1-meter length of dam. P = γ wh 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

W1 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 Visit For more Pdf's Books Pdfbooksforum.com

12m

P 4m

W2 Heel

Toe

4m 7m 8m

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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) Cos 3.58˚ x = 0.037 m 0.037 Sin 3.58˚ = MG MG = 0.594 m. metacentric height

4050x =

θ

w.s.

3m

M

50

Gx

θ

θ θ

3/cosθ 3600 θ

B BF=4050 tons 225

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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? θ

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

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:

CA 2g Qave =

Qave ...