# Fluid Mechanics (1)_2

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### Properties of Fluids

EXERCISE PROBLEM

1. If a certain gasoline weighs 7 KN/m3 , what are the values of its density, specific volume, and specific gravity relative to water at 150C?

a.) ρ = 𝑤𝑔 b.) ѵ = 1ρ c.) s = ws𝑤

=7 𝐾𝑁/𝑚3(1000)9.81 𝑚/𝑠2 = 713.56 𝐾𝑔/𝑚31 = 9.81 𝐾𝑁/𝑚37 𝐾𝑁/𝑚3 ρ = 713.56 Kg/m3 ѵ = 0.0014 m3/Kg s = 0.714

2. A certain gas weighs 16N/m3 at a certain temperature and pressure. What are the values of its density, specific volume, and specific gravity relative to air weighing 12N/m3?

a.) ρ = 𝑤𝑔 b.) ѵ = 1ρ c.) s = ws𝑤

=9.81 𝑚/𝑠216 𝑁/𝑚3 ѵ= 1.63 𝐾𝑔/𝑚31 s = 16 𝑁/𝑚312 𝑁/𝑚3

ρ = 1.63 Kg/m3 ѵ = 0.613 m3/Kg s = 1.33

3. If 5.30m3 of a certain oils weighs 43,860 N, calculate the specific weight, density and specific gravity of this oil.

a.) w =𝑊𝑔 b.) ρ = 𝑔𝑉𝑊 c.) s = ws𝑤 = 43.860 𝐾𝑁5.30 𝑚3 = 43860 𝑘𝑔.𝑚/𝑠2 (9.81𝑚𝑠2 )(5.30 𝑚3) = 8.28 𝐾𝑁/𝑚3 9.81 𝐾𝑁/𝑚3 w = 8.28 KN/m3 ρ = 843.58 kg/m3 s = 0.844

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4. The density of alcohol is 790 Kg/m3 . Calculate its specific weight, specific gravity and specific volume.

a.) w = ρg b.) s = ws𝑤 c.) ѵ = 1ρ

= (790 kg/m3)(9.81 m/s2) = 7.75 𝐾𝑁/𝑚39.81 𝐾𝑁/𝑚3 = 790 𝑘𝑔/𝑚31 w = 7.75 KN/m3 s = 0.79 ѵ = 0.00127 m3/kg

5. A cubic meter of air at 101.3 KPa and 150C weighs 12 N. What is its specific volume? wa= 12 N/m3 s = ρ ρs s = 12.7 𝑁/𝑚312 𝑁/𝑚3 ρa = (1.29 kg/m3)(0.94) ѵ = 1 ρ = 1 1.21 𝑘𝑔/𝑚3 s = 0.94 ρa = (1.21 kg/m3) ѵ = 0.82 m3/kg

6. At a depth of 8 km in the ocean the pressure is 82.26 MPa. Assume the specific weight on the surface to be 10.10 KN/m3 and that the average bulk modulus is 2344 MPa for that pressure range. (a) What will be the change in specific volume between at the surface and at the depth? (b) What will be the specific volume at that depth? (c) What will be the specific weight at that depth?

a.) ρ = 𝑤𝑔 = 10.10(1000)9.81 p = wh = 10.10(1000)(8000) b.) ѵ = 1ρ = 1043 𝑘𝑔/𝑚31 ρ = 1029.6 kg/m3 p = 80.80 MPa = 9.5 x 10-4 m3/kg

Δѵ = 3.3 x 10-5 m3/kg c.) w = 𝑝𝑕 = 82.26 𝑥 1068000 w = 10282. 5 N/m3

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7. To two significant figures what is the bulk modulus of water in KN/m2 at 500C under a pressure of 30 MN/m2? W = 9.689 KN/m3 Ev = -v1 𝛥𝑝 𝛥𝑣 ρ = 𝑤𝑔 = 9.6899.81 = -( 1 x 10-3)(1 x 10−3−1.012x10−330,000,000 ) ρ = 987.67 kg/ m3 = 2,500,000 Pa ѵ = 1ρ = 987.671 Bv = 2.5 x 106 Pa ѵ = 1.012 x 10-3 m3/kg

8. If the dynamic viscosity of water at 20 degree C is 1x10-3 N.s/m2, what is the kinematic viscosity in the English units?

ѵ = µ𝑝 = 1𝑥10−3 𝑘𝑔.𝑚.𝑠/𝑚2𝑠21000 𝑘𝑔/𝑚3

ѵ = 1x10-6 m2/s (3.28 𝑓𝑡1 𝑚 )2 ѵ = 1.08 x 10-5 ft2/s

9. The kinematic viscosity of 1 ft2/sec is equivalent to how many stokes? (1 stoke= 1cm2/sec).

1 inch = 2.54 cm

1 ft2/s ( 12 𝑖𝑛 21 𝑓𝑡2 )( 2.54 𝑐𝑚 21 𝑖𝑛 2 ) = 929 stokes

10. A volume of 450 liters of a certain fluids weighs 3.50 KN. Compute the mass density. (1 m3= 1000 liters).

450 liters (1000 𝐿1𝑚3 ) = 0.45 m3 ρ = 𝑔𝑉𝑊 = 9.81(0.45)3.5(1000) = 792.85 kg/m3

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11. Compute the number of watts which equivalent to one horsepower. (1 HP = 550 ft-lb/sec; 1 W = 107 dyne-cm/sec; 1 lb = 444,8000 dynes).

1 Hp = 500 𝑓𝑡−𝑙𝑏𝑠𝑒𝑐 (12 𝑖𝑛𝑓𝑡 ) (2.54 𝑐𝑚1 𝑖𝑛 ) (444,800 𝑑𝑦𝑛𝑒𝑠1 𝑙𝑏 ) 1 Hp = 7456627200 𝑑𝑦𝑛𝑒𝑠 −𝑐𝑚 𝑠𝑒𝑐 100000000 𝑑𝑦𝑛𝑒 −𝑐𝑚 /𝑠𝑒𝑐 1 Hp = 745.66 W

12. A city of 6000 population has an average total consumption per person per day of 100 gallons. Compute the daily total consumption of the city in cibic meter per second. (1 ft3 = 7.48 gallons).

100 Gallon (7.48 𝑔𝑎𝑙1 𝑓𝑡3 ) (3.28 𝑓𝑡31 𝑚3 ) = 0.379 m3 P = 6000 (0.379 m3)

P = 2274 m3

D.C. = (𝑝𝑑) = (60𝑥60𝑥242274 𝑚3) D.C. = 0.026 m3/s

13. Compute the conversion factor for reducing pounds to newtons.

32.18 𝑓𝑡𝑠2(0.3048 𝑚1 𝑓𝑡 ) (2.205 𝑚1 𝑘𝑔 ) (𝑘𝑔/𝑚21 𝑁 ) = 4.448 N

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### Principles of Hydrostatics

EXERCISE PROBLEM

1. If the pressure 3 m below the free surface of the liquid is 140 KPa, calculate its specific weight and specific gravity.

Solution:

a.) P=wh b.)

W=p/n S=W/ws

=140kPa/3m =46.67/9.81

W=46.67KN/m3 S=4.76

2. If the pressure at the point in the ocean is 1400 KPa, what is the

pressure 30 m below this point? The specific gravity of salt water is 1.03. Solution:

P=1400kPa+whs

=1400kPa+9.81(30)(1.03) P=1,703kPa

3. An open vessel contains carbon tetrachloride (s = 1.50) to a depth of 2 m and water above this liquid to a depth of 1.30 m. What is the pressure at the bottom?

Solution:

Ht=1.50(2) P=wh

=3m =9.81(4.3)

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4. How many meters of water are equivalent to a pressure of 100 KPa? How many cm. of mercury?

Solution:

a.) P=wh

b.)h=P/w=100kPa/9.81(13.6)

h=P/w=100kPa/9.81 h=0.75m

h=10.20m of water h=75cm of Hg

5. What is the equivalent pressure in KPa corresponding to one meter of air at 15®C under standard atmospheric condition?

Solution:

P=wh

=(12N/m3)(1m) P=12Pa

6. At sea level a mercury barometer reads 750 mm and at the same time on the top of the mountain another mercury barometer reads 745 mm. The temperature of air is assumed constant at 15®C and its specific weight assumed uniform at 12 N/m3. Determine the height of the mountain. Solution: P1=wsh1 ; P2=wsh2 wsh1+wh=wsh2 w(13.6)(0.745)+12h=w(13.6)(0.750) h=(13.6)[0.75-0.745](9810)/12 h=55.60m

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7. At ground level the atmospheric pressure is 101.3 KPa at 15®C.

Calculate the pressure at point 6500 m above the ground, assuming (a) no density variation, (b)an isothermal variation of density with pressure. Solution:

a.)P2=P1+wh b.)P1=P2e-gh/RT

=101.3-12(6500) =(101.3)e-9.81(6500)(287/239)

P1=23.3kPa P1=47kPa

8. If the barometer reads 755 cm of mercury, what absolute pressure corresponds to a gage pressure of 130 KPa?

Solution:

Patm=wsh

=9.81(13.6)(0.775) Patm=100.72kPa

Pabs=Patm=Pgage =100.72+130 Pabs=220.752kPa

9. Determine the absolute pressure corresponding to a vacuum of 30 cm of mercury when the barometer reads 750 mm of mercury.

Solution:

Pv=-whs Patm=whs

=-9.81(0.30)(13.6) =9.81(0.75)(13.6)

Pv=-40.02kPa Patm=100.06kPa

Pabs=Patm-Pv

=100.06-40.02 Pabs=60kPa

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10. Fig. shows two closed compartments filled with air. Gage (1) reads 210 KPa, gage (2) reds – 25 cm of mercury. What is the reading of gage (3)? Barometric pressure is 100 KPa.

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11. If the pressure in a gas tank is 2.50 atmospheres, find the pressure in KPa and the pressure head in meter of water.

Solution:

a.)P=2.5(101.3kPa) b.)P=wh

P=253.25kPa h=P/w=253.25/9.81

H=25.81m

12. The gage at the sunction side of a pump shows a vacuum of 25 cm of mercury. Compute (a) Pressure head in meter of water, (b) pressure in KPa, (c) absolute pressure in KPa if the barometer read 755 cm of mercury. Solution: a.)h=P/w=33.35/9.81 b.)Pv=-whs h=3.40m =-0.25(9.81)(13.6) Pv=-33.35kPa c.)Pabs=Patm+Pv =9.81(13.6)(0.775)-33.35 Pabs=67.38kPa

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13. Oil of specific gravity 0.80 is being pumped. A pressure gage located downstream of the pump reads 280 KPa. What is the pressure head in meter of oil?

Solution:

H=P/ws

=280/9.81(0.80) H=35.70m

14. The pressure of air inside a tank containing air and water is 20 KPa absolute. Determine the gage pressure at point 1.5 m below the water surface. Assume standard atmospheric pressure.

Solution: Pabs=20+1.5(9.81) =34.72kPa Pabs=Patm+pg 34.72=101.3=pg Pg=-66.60kPa

15. A piece of 3 m long and having a 30 cm by 30 cm is placed in a body of water in a vertical position. If the timber weights 6.5 12 KN/m3what vertical force is required to hold it to its upper end flush with the water surface? Solution: W=wV F=Wa-Ww =(9.81)(0.3x3x0.3) =2.65kN-1.756kN W=2.65kN F=0.894kN VWw=6.5(0.3x0.3x3) Vw=1.755/9.81 Vw=0.179m3 Ww=wV =0.179(9.81) Ww=1.756kN

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16. A glass tube 1.60 m long and having a diameter of 2.5 cm is inserted vertically into a tank of oil (sg = 0.80) with the open end down and the close end uppermost. If the open end is submerged 1.30 m from the oil surface, determine the height from which the oil will rise from the tube. Assume barometric pressure is 100 KPa and neglect vapor pressure. 17. A gas holder at sea level contains illuminating gas under a pressure

equivalent under a 5 cm of water. What pressure in cm of water is expected in a distributing pipe at a point of 160 m above sea level? Consider standard atmospheric pressure at sea level and assume the unit weighs of air and gas to be constant at all elevations with values of 12 N/m3and 6 N/m3respectively.

18. If the barometric pressure is 758 mm of mercury, calculate the value h of figure. Gage reads – 25 cm Hg sunction mercury h Solution: P = (13.6)(9.81)(7.08) p =wh P = 1,011.29 kpa h = p/w h = 1,011.29/9.81 h = 103.08 m

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19. The manometer of figure is tapped to a pipeline carrying oil (sg = 0.85). Determine the pressure at the center of the pipe.

mercury

75 cm oil

150 cm

20. Determine the gage reading of the manometer system of figure.

air water 20cm Gage 3m Mercury Solution: P = wsh + wsh P = (9.81) (13.6) (0.75) + (9.81) (0.85) (1.5) P = 112.6 kpa

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Solution:

P = -wsh Pg = wsh - wsh

P = - (9.81) (0.2) (13.6) Pg = 9.81 (3) – (9.81) (13.6) (0.2) P = -26.68 kpa Pg = 2.75 kpa

21. In fig. calculate the pressure at point m.

Liquid (s= 1.60) water 55 cm m 30 cm . Solution: Pm = wsh – wsh Pm = (9.81) (1.60) (0.55) – (9.81) (3) Pm = 5.70 kpa

22. In fig. find the pressure and pressure at point m ; Fluid A is oil (s= 0.90), Fluid B is carbon tetrachloride (s= 1.50) and fluid C is air.

B

C

60 cm A 45 cm

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Solution:

a) Pb = -wsh Pm = -8.829 + 0 Pb = - (9.81) (1.5) (0.6) Pm = -8.829 kpa Pb = - 8.829 kpa

23. Compute the gage and absolute pressure at point m at the fig. ; Fluids A and C is air, Fluid B is mercury.

C A m 2 cm B 6 cm Solution: Pg = - wsh Pabs = Patm + Pg Pg = - (9.81) (13.6) (0.06) Pabs = 101.3 – 10.67 Pg = - 10.67 kpa Pabs = 90.63 kpa

24. The pressure at point m is increased from 70 KPa to 105 KPa. This causes the top level of mercury to move 20 cm in the sloping tube. What is the inclination θ? Water mercury

## θ

b) h = p/w h = -8.82/9.81 h = -1.0 m

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. Solution:

P = wsh 10.5 – 26.68sin𝜃 = 0 P = (9.81) (13.6) (0.20) 26.68sin𝜃 = 10.5 P = 26.68 kpa 𝜃 = 22.6 °

25. In fig. determine the elevation of the liquid surface in each piezometer.

EL. 7 m (s= 0.75) EL. 4.5 m (s= 1.00) EL. 4.35 m EL. 2.15 m EL. 2 m (s= 1.50)

26. In fig. fluid A is water, fluid B is oil(s= 0.85). Determine the pressure difference between points m and n.

Solution: 1.02 = y – x 68 – x = z 170 – y = 68 – x Pm/w – y – 0.68 (0.85) + x = Pn/w Pm – Pn = [ ( y – x ) + ( 0.65 ) (0.85) ] 9.81 Pm – Pn = ( 1.02 + 0.578) (9.81) Pm – Pn = 15.67 kpa

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15 Solution: Pm = wsh + wsh Pm = (9.81) (0.4) (3) + (0.4) (9.81) (0.9) Pm = 14.13 kpa A 27. In fig. determine𝑝𝑚 − 𝑝𝑛. water n m 90 cm 52 cm 105 cm 65 cm 45 cm Mercury Solution: Pm/w + 1.05 – (13.6) (0.65) + 0.45 – (13.6) (0.52) – 0.38 = Pn/w Pm – Pn = [ (13.6) (0.65) – (1.05) – 1.05 - 0.45 + 0.52 (13.6) + 0.38] 9.81 Pm – Pn = 149 kpa

28. In fig. Fluid A is has a specific gravity of 0.90 and fluid B has a specific gravity of 3.00. Determine the pressure at point m.

B 12 mm. D

3 mm. D 36 cm

12 cm, D 40 cm m

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### Hydrostatic Force on Surfaces

EXERCISE PROBLEM

1. A rectangular plate 4m by 3m is emmersed vertically with one of the longer sides along the water surface. How must a dividing line be drawn parallel to the surface so as to divide the plate into two areas,the total forces upon which shall be equal?

Solution: F1 = F2 Awh1 = Awh2 (12.0)(1.50) = h(4.0)( h/2 ) 2h2 =18.0 2h = √18 h = 4.24/2 h = 2.12 m below w.s

2. A triangle of height H and base B is vertically submerged in a liquid. The base B coincides with the liquid surface.Derive the relation that will give the location of the center of pressure.

3. The composite area shown in Fig. A is submerged in a liquid with specific gravity 0.85. Determine the magnitude and location of the total hydrostatic force on one face of the area.

Solution: e = 𝐼g 𝐴𝑦 = 𝑏𝑕2 12 𝑏𝑕𝑦

hp = 𝑕 + 𝑒 F1 = wA𝑕 e = 𝑕2 12 𝑦 = 3.52 12 3.25 hp = 3.25 + 0.31 F1 = 9.81(3.5)(1.5)(3.25)(0.85) e = 0.31 m hp = 3.56 m F1 = 142.28 KN

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hp = 𝑕 + 𝑒 F1 = wA𝑕 e = 𝑕2 12 𝑦 = 1.52 12 4.25 hp = 4.25 + 0.04 F1 = 9.81(1.5)(1.5)(4.25)(0.85) e = 0.04 m hp = 4.29 m F1 =79.74 KN Ft = F1 + F2 Pt = P1 + P2 Ft = 142. 28 + 79.74 Pt h = F1 h + F2h Ft = 222.02 KN 222.02𝑕 222.02 = 142.28 3.56 + 79.76(4.29) 222.02 h = 3.83 m , below w.s

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4. The gate in fig. B is subjrcted to water pressure on one side and to air pressure on the other side. Determine the value of X for which the gate will rotate counterclockwis if the gate is (a) rectangular, 1.5m by 1.0m (b) triangular, 1.5m base and 1.0m high.

Solution: F = PA a.) F = w𝑕 A e = 𝐼 𝐴𝑦 F = 30(1.0)(1.5) F = (9.81)(x+0.5)(1.5)(1.0) e = 1 12𝑥+6 F = 45 KN F = 14.72x + 7.36 ∑𝑀𝑃1 = 0 14.42x + 7.36(0.5 + 12𝑥+61 ) = 45(0.5) 86.5x2 – 168.16x – 105.56 = 0 𝑥 =−(−168.16)± (−168.16)2(86.5)2−4 86.5 (105.56) 𝑥 = 2.40 𝑚

5. A vertical circular gate 1m in diameter is subjected to pressure of liquid of specific gravity 1.40 on one side. Thefree surface of the liquid is 2.60m above the uppermost part of the gate. Calculate the total force on the gate and the location of the center of pressure.

Solution:F = w𝑕 A e = 𝐼g 𝐴𝑦 = 𝜋 (𝑟4)2 4 𝜋𝑟2𝑦 F =9.81(1.4)(3.1)(𝜋)(0.52) e = (0.5 2) 4(3.1)

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6. A horizontal tunnel having a diameter of 3m is closed by a vertical gate. When the tunnel is (a) ½ full (b) ¾ full of wter, determine the magnitude and location of the total force.

Solution: a.) ½ full 𝑕 = 4𝑟 3𝜋

F = w𝑕 A 𝑕 = 4(1.5) 3𝜋 F = (9.81)( 𝜋(1.52) (2) )(0.64) 𝑕 = 0.64 m F = 22.15 b.) 𝑕 = 1.5+0.64 2 F = w𝑕 A 𝑕 = 1.08 m F = (9.81)(3𝜋(1.5 2) (4) )(1.08) F = 56.25 KN 𝑒 =𝐴𝑦𝐼g hp = 𝑕 + 𝑒 𝑒 =0.1098(1.5)3.53(0.64)4 hp = 0.64 + 0.25 e = 0.25 m hp = 0.89 m (below center)

7. In Fig. C is a parabolic segment submerged vertically in water.

Determine the magnitude and location of the total force on one face of the area. Solution: F = w𝑕 A F = 9.81(1.8)(23

### )

(3)(3) F = 105.95 KN 𝑒 =𝐴𝑦𝐼g = 8 3 (3)2 175 2 3 3 3 (1.8) 𝑕𝑃 = 𝑕 + 𝑒 𝑒 = 0.34 𝑚 𝑕𝑃 = 1.8 + 0.34 𝑕𝑃 = 2.14 𝑚 𝑏𝑒𝑙𝑜𝑤 𝑐𝑒𝑛𝑡𝑒𝑟

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8. A sliding gate 3m wide by 1.60m high is in a vertical position. The coefficient of friction between the gate and guides is 0.20. If the gate weighs 18KN and its upper edge is 10m below the water surface, what vertical force is required to lift it? Neglect the thickness of the gate. Solution: 𝐹 = 𝐴𝑤𝑕 𝐹𝑓 = 𝜇𝑁 𝐹 = 9.81 1.6 (10.8) 𝐹𝑓 = 0.2(508.55) 𝐹 = 508.55 𝐾𝑁 𝐹𝑓 = 101.71 𝐾𝑁 F = 508.55 KN ∑𝐹𝑣=0 𝐹 = 𝑤 + 𝐹𝑓 𝐹 = 18.0 + 101.71 𝐹 = 119.71 𝐾𝑁

9. The upper edge of a vertical rapezoidal gate is 1.60m long and flush with the water surface. The two edges are vertical and measure 2m and 3m, respectively. Calculate the force and location of the center of pressure on one side of the gate.

10. How far below the water surface is it necessary to immerse a vertical plane surface, 1m square, two edges of which are horizontal, so that the center of pressure will be located 2.50cm below the center of gravity? Solution: 𝑒 = 𝑕2 12

### 𝑕

0.025 = 12 12 𝑕 − 0.5 𝑕 = 2.83 m

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11. The gate shown in fig. D is hinged at B and rest on a smooth surface at A. If the gate is 1.60m wide perpendicular to the paper, find BH and BH

Solution: 𝜃 = 𝑡𝑎𝑛−1(3 2) 𝐹 = 𝐴𝑤𝑕 𝜃 = 56.31° 𝐹 = 9.81 3.61 1.6 (2.8) 𝐹 = 158.66 𝐾𝑁 𝐵𝑉= 𝐹𝑠𝑖𝑛𝜃 𝐵𝑕 = 𝐹𝑠𝑖𝑛𝜃 𝐵𝑉= 158.66𝑠𝑖𝑛56.31° 𝐵𝑕 = 158.66𝑠𝑖𝑛56.31° 𝐵𝑉= 132.01 𝐾𝑁 𝐵𝑉= 80.70 𝐾𝑁

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12. In fig. E gate AB is 2m wide perpendicular to the paper. Determine FH to hold the gate in equilibrium.

Solution: 𝐴 = 3.2 2 = 6.4 𝑒 =𝐴𝑦𝐼g 𝐹 = 𝐴𝑤𝑕 𝑒 =12𝑦𝑕2 𝐹 = 9.81 1.21 (6.4) 𝑒 = 3.20 12(1.26) 𝐹 = 77.85 𝐾𝑁 𝑒 = 0.22 𝑚 𝑥 =3.2−0.482 − 0.53 ∑𝑀𝑣= 0 𝑥 = 1.38 𝑚 3.20𝐹𝑕− 1.38 77.85 = 0 𝐹𝑕 = 42.50 𝐾𝑁

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16. A triangular gate having a horizontal base of 1.30m and an altitude of 2m is inclined 45o from the vertical with the vertex pointing upward. The base of the gate is 2.60m below the surface of oil (s=0.80). What normal force must be applied at the vertex of the gate to open it?

17. What depth of water will cause the rectangular gate of Fig. I to fall? Neglect weight of the gate.

Solution: 𝑕 =𝑠𝑖𝑛60°0.5𝑕 𝑒𝑞. 1 𝑒 =𝐴𝑦𝐼g 𝑒 =12𝑦𝑕2 𝑒 = 𝑕 𝑠𝑖𝑛60° 12(𝑠𝑖𝑛60°0.5𝑕) 𝑒 = 0.19𝑕 𝑒𝑞. 3 ∑𝑀𝑣= 0 𝐹 𝑠𝑖𝑛 60𝑕 −𝑠𝑖𝑛 600.5𝑕 + 0.19𝑕𝑠𝑖𝑛 60 = 22.5(5.0) 5.95𝑕3= 112.5(5.0) 𝑕 = 18.91 𝑕 = 2.66 𝑚 𝐹 = 𝐴𝑤𝑕 𝐹 = 9.81 𝑕 𝑠𝑖𝑛60° 2.6( 0.5𝑕 𝑠𝑖𝑛60°) 𝐹 = 17.0𝑕2 𝑒𝑞. 2

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24

18. Determine the horizontal and vertical components of the total force on the gate of Fig. J. The width of the gate normal to the paper is 2m. Solution: A1 = AAOBC A2 = ( 1 2)(6)(6)(c0s30°) 𝐴1 60° = 𝐻(6)2 360°

A2 = 15.59 𝑚2 A1= 18.85 𝑚2 𝐹 = 𝐴𝑤𝑕 A = A1 – A2 Fh= 9.81(6)(3)(2) FV = 9.81(3.26)(2) A = 18.85 – 15.59 Fh = 353.16 KN FV = 63.96 KN A = 3.26 𝑚2

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19. The corner of floating body has a quarter cylinder AB having a length normal to the paper of 3m. Calculate the magnitude and location of each of the components of the force on AB. Fig. K.

Solution:

𝐹𝑕 = 𝐴𝑤𝑕 𝐹𝑣= 𝐹𝑕𝑐0𝑠𝜃

𝐹𝑕 = 9.81 1.5 3 (1.03) 𝐹𝑣= 147.48 𝑐𝑜𝑠30° 𝐹𝑕 = 147.78 𝐾𝑁 𝐹𝑣= 128.56 𝐾𝑁

20. The cylindrical gate of Fig. L is 3m long. Find the total force on the gate. What is the minimum weight of the gate to maintain equilibrium of the system?

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21. The gate if Fig. O is 3m long. Find the magnitude and location of the horizontal and vertical components of the force on the gate AB. Solution: 𝐹𝑕 = 𝐴𝑤𝑕 𝐹𝑕 = 9.81 1.06 3 (2.12) 𝐹𝑕 = 66.14 𝐾𝑁 𝐴𝐴𝐵𝐶 = 0.88 + 3 2.12 2 𝐴𝐴𝐵𝐶 = 4.11 𝑚2 𝐴𝑠𝑒𝑐𝑡𝑜𝑟 = 𝜋𝑟 2𝜃 360°= 𝜋(3)2(45°) 360° 𝐴𝑠𝑒𝑐𝑡𝑜𝑟 = 3.58 𝑚2

22. A pyramidal object having a square base (2m on a side) and 1.50m high weighs 18KN. The base covers a square hole (2m on a side) at the bottom of a tank. If water stands 1.50m in the tank, what force is necessary to lift the object off the bottom? Assume that atmospheric pressure acts on the water surface and underneath the bottom of the tank.

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23. The hemesphirical dome of Fig. P surmounts a closed tank containing a liquid of specific gravity 0.75. The gage indicates 60KPa. Determine the tension holding the bolts in place.

Solution: 𝑃 = 𝑤𝑠𝑕 𝑇 = 𝑤𝑉𝑠 60 = 9.81 0.75 𝑕 𝑇 = 9.81 39.23 (0.75) 𝑕 = 8.15 𝑚 𝑇 = 288.63 𝐾𝑁 𝑉 = 𝜋𝑟2𝑕 −4𝜋𝑟3 6 𝑉 = 𝜋 1.5 2(8.15) −4𝜋(1.5)3 6 𝑉 = 39.23𝑚3

24. Fig. Q shows semi-conical buttress. Calculate the components of the total force acting on the surface of this semi-conical buttress.

Solution: 𝐴 =𝜋𝑟32𝑕 𝐹𝑕 = 𝑤𝑕 𝐴 𝐴 =𝜋 0.15 2(3) 3 𝐹𝑕 = 9.81 1.463 (7.07) 𝐴 = 7.07 𝑚2 𝐹 𝑕 = 101.47 𝐾𝑁 𝑕 = 𝑦𝑝+ 𝑒(𝜋 1.5 4 4 ) 𝐹𝑉 = 𝑤𝑉 𝑕 = 1.3 + 0.163 𝐹𝑉 =1.3 3 1.3 1.5 + 𝜋 1.5 2(3) 3 𝑕 = 1.463 𝑚 𝐹𝑉 = 0.12 9.81 𝐹𝑉 = 7.8 𝐾𝑁

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28

25. In Fig. R a circular opening is closed by a sphere. If the pressure at B is 350KPa absolute, what horizontal force is exerted by the sphere on the opening? Solution: 𝐴 = 𝜋𝑟2 𝐹 𝑕 = 𝑤𝑕 𝐴 𝐴 = 𝜋(0.125)2 𝐹 𝑕 = 9.81 178.39 0.099 (0.71) 𝐴 = 0.099 𝑚2 𝐹 𝑕 = 7.8 𝐾𝑁 𝑃 = 𝑤𝑕 𝐴 350 = 9.81 0.20 𝐴 𝐴 = 178.39 𝑚2

26. Calculate the force required to hold the cone of Fig. S in position. Solution: 𝑃2 = 𝑃1+ 𝑤𝑠𝑕 𝑊 = 𝜋𝑟2𝑃 𝑃2 = 3.5 − 9.81 0.8 1.5 𝑊 = 𝜋(0.375)2(0.8)(9.81)(2.5) 𝑃2 = −8.26 𝐾𝑝𝑎 𝑊 = 8.66 𝐾𝑁 𝐹1= 𝐴𝑃 𝑇2= 𝑤𝑕 1 3𝜋𝑟 2 𝐹1= 𝜋𝑟2𝑃 𝑇 2= 9.81(0.8) 13𝜋(0.375)2 𝐹1= 𝜋(0.375)2(8.26) 𝑇 2= 1.16 𝐾𝑁 𝐹1= 3.65 𝐾𝑁 ∑𝐹𝑉 = 0 𝐹 + 𝑃2+ 𝐹1= 𝑊 𝐹 = 8.66 − 3.65 − 1.16 𝐹 = 3.85 𝐾𝑁

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27. A steel pipe having a diameter of 15cm and wall thickness of 9,50mm has an allowable stress of 140,000KPa. What is the maximum allowable internal pressure in the pipe?

Sol’n:

Sa = T/t FB = PiD

T = Sat = 14,000(0.0095) Pi = FB/D = 2T/D = (2(1330)/1000)/0.15

T = 1330 kN/m Pi = 17.73 Mpa

28. A pipe carrying steam at a pressure of 7,000KPa has an inside diameter of 20cm. If the pipe is made of steel with an allowable stress of

400,000Kpa, what is the factor safety if the wall thickness is 6.25mm? Sol’n: S = PD′/2t f c = 𝐷 ′ 𝐷 = 0.714 𝑚 0.2 D’ = 2𝑠𝑡𝑃 = 2(400,000)(0.00625 )7000 fc = 3.60 D’ = 0.714 m

29. A 60 cm cast iron main leads from a reservoir whose water surface is at EL. 1590m. In the heart of the city the main is at EL. 1415m. What is the stress in the pipe wall if the the thickness of the wall is 12.5mm and the external soil pressure is 520Kpa? Assume static condition.

Sol’n: ∆EL = 𝐸𝐿1 - 𝐸𝐿2 S = 𝑃𝑑2𝑡 − 𝑃𝑠 = 1716.75−3202(0.0125) = 1590 – 1415 S = 28,709 kPa = 175 m = 28.7 MPa P = wh = 9.81 (175m) P = 1716.75 kPa

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30. Compute the stress in a 90cm pipe with wall thickness of 9.50mm if water fills under a head of 70m.

Sol’n:

FB = PiD T = FB/2 = 618.03/2 = 309.01 kN/m

= whD Sa = T/t = 309.01/0.0095 = 9.81(70)(0.9) Sa = 32,527 kPa FB = 618.03 kN/m

31. A wood stave pipe, 120cm in inside diameter, is to resist a maximum water pressure of 1,200KPa. If the staves are bound by steel flat bands (10cm by 2.50cm), find the spacing of the bands if its allowable stress is 105MPa. Sol’n: FB = PiD T = FB/2 = 1200kPa(1.2) = 1440/2 = 1440 kN/m T = 720 kN/m S = SaAH/T = (105(2.5))/0.72 S = 36.46 cm

32. A continuous wood stave pipe is 3m in diameter and is in service under a pressure head of 30m of water. The staves are secured by metal hoops 2.50cm in diameter. How far apart should the hoops be spaced in order that the allowable stress in the metal hoop of 105MPa be not exceeded? Assume that there is an initial tension in the hoops of 4.50KN due to cinching.

33. A vertical cylindrical container, 1.60m diameter and 4m high, is hel

together by means of hoops,one at the top and the other at the bottom. A liquid of specific gravity 1.40 stands 3m in the container. Calculate the tension in each hoop.

Sol’n:

F = wAh ∑MCD = 0 ∑Mab = 0

= 9081(1.4)(3)(1.6)(1.5) 4(2TU) = 1F 4(2TL) = 3F F = 98.9 kN TU = 12.40 kN TL= 37.09 kN

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34. A masonry dam has trapezoidal section: one face is vertical, width at the top is 60cm and at the bottom is 3m. The dam is 7m high with the vertical face subjected to water pressure. If the depth of water is 5m, where will the resultant force intersect the base? Determine the distribution of pressure along the base, (a) assuming there is no uplift pressure; (b) assuming that the uplift pressure varies uniformly from full hydrostatic at the heel to zero at the toe. Specific weight of masonry is 23.54KN/m3. Sol’n: a.)G_1 = wVs ∑R.M = G_1 + G_2 = 266.95 + 316.38 = 23.54(0.6)(7(1) = 583.33 kN.m = 98.87 kN ∑O.M = F_1 = 204.38 kN.m G_2 = 23.54(0.5)(7)(2.4)(1) = 197.74 kN x = (∑R.M-EO.M)/RV = (583.33-204.38)/296.61 F1 = 1/2wh^2 x = 1.28 m (from toe) = 0.5(9.81)(5^2) e = b/2 – x = 3/2 – 1.28 = 122.63 kN e = 0.22 Moment Forces: G_1 = 2.7(98.87) = 266.95 kN.m Smax = Rv/b (1 + 6e/b) = 296.61/3(1 + (6(0.22))/3) G_2 = 1.6(197.74) = 316.38 kN.m Smax = 142.38 kPa F_1 = 1/3 (5)(122.63) = 204.38 kN.m Smin = 296.61/3(1 - (6(0.22))/3) ∑FV = G_1 + G_2 Smin = 55.38 kPa = 98.87 + 197.74 RV = 296.61 kN ∑FH = F_1 = 122.63 kN

b.)U_1 = 1/2 whb x = (583.33-351.54)/223.03 = 1.04 m (from toe) = (1/2)(9.81)(5)(3)(1) e = 3/2 – 1.04

U_1 = 73.58 kN e = 0.46

Moment forces: Smax = 223.03/3 (1 + 6x0.46/3)

U_1 = 2/3 (3)(73.58) = 147.16 kN Smax = 142.74 kPa RV = G_1 + G_2 - U_1 = 98.87 + 197.74 – 73.58 Smin = 223.03/3 (1 - 6x0.46/3) RV = 223.03 kN Smin = 5.95 kPa ∑R.M = 583.33 kN.m ∑O.M = 204.38 + 147.16 = 351.54 kN.m

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35. The masonry dam of Problem 40 has its inclined face subjected to pressure due to a depth of 5m of water. If there is no uplift pressure , where will the resultant intersect the base? Specific weight to concrete is 23.54KN/m3 Soln: a/5 = 2.4/7 ∑R.M = W_1 + W_2+ W_3 a = 1.71 m = 23.91 + 316.38 + 266.95 = 607.24 kN.m W_1 = wV ∑O.M = 204.38 kN.m = 9.81(0.5)(5)(1.71)(1) = 41.94 kN x = (607.24-204.38)/338.55 W_2 = 1/2 (2.4)(1)(7)(23.54) x = 1.19 m =197.74 kN W_3 = 0.6(7)(1)(23.54) = 98.87 F = 1/2 (9.81)(5^2) = 122.63 kN Moment Forces: W_1= 0.57(41.94) = 23.91 kN.m W_2 = 1.6(197.74) = 316.38 kN.m W_3 = 2.7(98.87) = 266.95 kN.m F_1 = 1/3 (5)(122.63) = 204.38 kN.m RV = W_1+W_2+W_3 = 41.94 + 197.74 + 98.87 RV = 338.55 kN RH = F = 122.63 kN

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36. A masonry dam of trapezoidal cross section, with one face vertical has thickness pf 60cm at the top, 3.70m at the base, and has height of 7.40m. what is the depth of water on the vertical face if the resultant intersect the base at the downstream edge of the middle third? Assume that the uplift pressure varies uniformly from full hydrostatic at the heel to zero at the toe.

Soln: G_1 = 104.52 kN ∑R.M = G_1+G_2-U G_2 = 270 kN = 355.37 + 558.9 – 44.77h F = 1/2 (9.81)h^2 ∑O.M = 1.635h^3 = 4.905h^2 U = 1/2wh(3.7) x = (∑R.M- ∑O.M)/Rv = 1/2 (9.81)h(3.7) 1.23 = ((355.37+558.9-44.77h)- U = 18.15h1.635h^3)/(374.52-18.15h) h = 5.83 m Moment Forces: G_1 = 3.4 (104.52) = 355.37 kN.m G_2 = 2.07(270) = 558.9 kN.m F = 1/3 h(4.905h^2) = 1.635h^3 kN.m U = 2/3 (3.7)(18.15h) = 44.77h kN.m ∑Fv = G_1+G_2-U = 104.52 + 270 – 18.15h Rv = 374.52 – 18.15h

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37. A concrete dam is triangular in cross section and 30 m high from the horizontal base. If water reaches a depth of 27 m on the vertical face, what is the minimum length of the base of the dam such that the

resultant will intersect the base within the middle third? What minimum coefficient of friction is required to prevent sliding? Determine the pressure distribution along the base.

Soln: a.) G = wV RVx = ∑R.M - ∑O.M = 23.54(1/2)(30)(1)B 353.1B(B/3) = 235.4B^2 – 32181.75 G = 353.1B (235.4 – 117.7)B^2 = 32181.75 B = 16.54m F = 1/2 (9.81)(〖27〗^2) F = 3575.75 kN Moment Force: G = 2/3 B(253.1B) G = 235.4B^2 F = 1/3 (27)(3575.75) = 32181.75 kN.m b.) G = 235.4B^2 µ = RH/RV = 235.4(〖16.54〗^2) = 3575.75/5840.27 G = 64398.76 kN.m µ = 0.61 RV = 353.1B = 353.1(16.54) RV = 5840.27 kN c.) x = 1/3 (16.54) = 5.51 S = 5840.27/16.54 (1+6x2.76/16.54) S = 706.20 kPa e = 16.54/2-5.51 e = 2.76

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44. The section of masonry dam is shown in Fig. U. If the uplift pressure varies uniformly from full hydrostatic at the heel to full hydrostatic at the toe, but acts only 2/3 of the area of the base, find: (a) the location of the resultant, (b) factor safety against overturning, (c) factor of safety against sliding if the coefficient of friction between base andfoundation is 0.60.

Soln: a.) 𝐺1 = 5(8)(1)w ∑Fv = 𝐺1+ 𝐺2+ 𝐺3 + 𝐺4 + 𝐺5+ 𝐺6− 𝑈1− 𝑈2 = 40w = (40+25+60+252+176.4+4.18-73.67-56.67)w 𝐺2 = 12 (5)(10)(1)w Rv = 427.24w = 25w 𝐺3 = 12 (5)(10)(1)(2.4)w ∑Fh = 𝐹1− 𝐹2 = 60w = (162-12.5)w 𝐺4 = 5(21)(1)(2.4)w Rh = 149.5w = 252w 𝐺5 = 12 (7)(21)(1)(2.4)w Moment Forces: = 176.4w 𝐺1 = 14.5(40w) = 580w 𝐺6 = 12 (1.67)(5)(1)w 𝐺2 = 15.33(25w) = 383.25w = 4.18w 𝐺3 = 13.67(60w) = 820.2w 𝐹1 = 12 (182)w 𝐺4 = 9.5(252w) = 2394w = 162w 𝐺5 = 4.67(176.4w) = 823.79w 𝐹2 = 12 (52)w 𝐺6 = 0.56(4.18w) = 2.34w = 12.5w 𝑈1 = 11.33(73.67w) = 834.67w 𝑈1 = 12 (17)[23 (176.58 – 49.05)] 𝑈2 = 8.5(56.67w) = 481.70w = 73.67w 𝐹1 = 6(162w) = 972w 𝑈2 = 23 (17)(49.05) 𝐹2 = 1.67 (12.5w) = 20.88w = 56.67w ∑R.M = (580+383.28+820.2+2394+823.79+2.34+20.88)w = 5024.46w ∑O.M = 𝐹1+ 𝑈1+ 𝑈2 = (972+834.68+481.7)w = 2288.38w

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36 x = ∑𝑅.𝑀− ∑𝑂.𝑀𝑅𝑣 = 5024.46−2288.38 𝑤427.24𝑤 x = 6.40 m (from toe) b.) F.S. vs. Overturning = ∑𝑂.𝑀∑𝑅.𝑀 = 2024.46𝑤2288.38𝑤 = 2.20 c.) F.S vs. Sliding = µ𝑅𝑣𝑅𝑕 = 0.6(427.24)𝑤149.5𝑤 = 1.70

45. Shown in Fig. V is an overflow dam. If there is no uplift pressure, determine the location of the resultant.

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37 Soln: G_1 = 2(3)(1)(9.81) Moment Force: = 58.86kN G_1 = 6.5(58.86) = 382.59 kN.m G_2 = 1/2 (3)(6)(1)(9.81) G_2 = 7(88.29) = 618.03 kN.m = 88.29 kN G_3 = 6(211.86) = 1271.16 kN.m G_3 = 1/2 (3)(6)(1)(23.54) G_4 = 4(38.24) = 156.96 kN.m = 211.86 kN G_5 = 4(282.48) = 1129.92 kN.m G_4 = 2(2)(1)(9.81) G_6 = 2(211.86) = 423.72 kN.m = 39.24 kN G_7 = 0.67(39.24) = 26.29 kN.m G_5 = 2(6)(1)(23.54) F_1 = 2.4(294.3) = 706.32 kN.m = 282.48 kN F_2 = 1.33(78.48) = 104.38 kN.m G_6 = 1/2 (3)(6)(23.54) = 211.36 kN G_7 = 1/2 (2)(4)(1)(9.81) = 39.24 kN ∑R.M = 382.59+618.03+1271.16+156.96+1129.92 +423.72+26.29+104.38 F_1 = Awh = 4113.05 kN.m = 6(1)(9.81)(5) = 294.3 kN ∑O.M = F_1 = 706.32 kN.m F_2 = 1/2 (9.81)(4^2) = 78.48 kN ∑Fv = G_1+G_2+G_3+G_4+G_5+G_6+G_7 = 58.86+88.29+211.86+39.24+282.48+211.86+39.24 Rv = 931.83 kN ∑Fh = F_1-F_2 x = (∑R.M- ∑O.M)/Rv = (4113.05- 706.32)/931.83

= 294.3 – 78.48 x = 3.66 m (from the toe) Rh = 215.82 kN

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46. The base of a solid metal cone (Sp. Gr. 6.95) is 25 cm in diameter. The altitude of the cone is 30 cm. If placed in a basin containing mercury (Sp. Gr. 13.60) with the apex of the cone down, how deep will the cone float? ∑Fy=0] Given :

𝐹𝑏=W d=25cm.

(wV)displaced mercury = (wV)cone r=12.5cm=0.125 (9.81)(13.60)Vm = 9.81(6.95)Vcone Vcone=4.9087x10−3 13.60Vm = 6.95𝜋𝑟 2𝑕 3 Vm = 6.95𝜋 0.125 2 (0.30) 3(13.60) Vm = 2.50x10−3𝑚3

47. If a metal sphere 60 cm in diameter weighs 11,120 N in the air, what would be its weight when submerged in (a) water? (b) mercury? Sol’n: a.) b.) FB = 9.81 (4/3 πr^3)W_hy = 11,120 – 9.81(13.6)(4/3)(π)(〖0.3〗^2) = 9.81(4/3)(〖0.3〗^2) = -3976 N FB = 1.11 kN W = 11.12 – 1110 = 10.01 N 𝑉𝑐𝑜𝑛𝑒 𝑉𝑚

0.30 𝑥

3

3

### =

(0.30)3𝑉𝑚 𝑉𝑐𝑜𝑛𝑒 x = 3 0.30 4.91 𝑥 103(2.50𝑥10−3−3) x = 0.24m x = 24cm

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48. A rectangular solid piece of wood 30 cm square and 5 cm thick floats in water to depth of 3.25 cm. How heavy an object must be placed on the wood (Sp. Gr. 0.50) in such a way that it will just be submerged? Given: dept=3.25cm Req. F=? 30cm 5cm Fb=w w.s. wv'=wsv s=v’/v Fb s= (30) (30) (3.25) w.s. 4500 S= 0.65// ans. F=Fb-W Fb F=wv-wsv

49. A hollow vessel in the shape of paraboloid of revolution floats in fresh water with its axis vertical and vertex down. Find the depth to which it must be filled with a liquid (Sp. Gr. 1.20) so that its vertex will be submerged at 45 cm from the water surface.

Solution: 𝐹𝑏=W 9.81Vd = 9.81(1.20)Vp Vd = 1.20Vp W W F=wv(1-s) F= (9.81)(4500)(1-0.65) F=15.45 N By Similar Solids: 𝑉𝑝 𝑉𝑑=

𝑎 0.45

### )

3 𝑉𝑝 1.20𝑉𝑑= 𝑎3 0.453 1 1.20= = 𝑎3 0.453 a = 0.42m a = 42cm

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50. A barge is 16 m long by 7 m wide 120 cm deep, outside dimensions. The sides and bottom of the barge are made of timber having thickness of 30 cm. The timber weighs 7860 N/cu.m. If there is to be freeboard of 20 cm in fresh water how many cubic meters of sand weighing 15700 N/cu.m may be loaded uniformly into the barge?

Sol’n: V_t=V_o-V_i F_b-W_t-W_s = 0 = (16)(7)(1.2) – (15.4)(16.4)(0.9) 9810(16)(7)(1) – 7860(45.7) – 15700 = 45.7 m^3 V_s = 0 V_s = 47.10 m^3

51. A brass sphere (Sp. Gr. 8.60) is placed in a body of mercury. If the diameter of the sphere is 30 cm (a) what minimum force would be required to hold it submerged in mercury? (b) what is the depth of flotation of the sphere when it is floating freely?

hg.s. F F=Fb-W F=wSmVs-wSsVs F=wVs(Sm-Ss) F=(9.81) (3/4)(3.14)(0.15)^3(13.60-8.6) F=693.43N//ans. y Fb W Fb

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41 V=4/3(3.14)(r)^3 V=4/3(3.14)(15)^3 V=14,137.17cm^3 Fb=w V’=3.14/3 D^2(3r-D) wSmV’=wSsV 8939.68 =3.14/3(y^2)((3x15)-y) V’=8.60(14137.17) 13.60 y=17.10cm V’=8939.68cm^3

52. A spherical balloon weighs 3115 N. How many newton of helium have to be put in the balloon to cause it to rise, (a) at sea level? (b) at an

elevation of 4570 m? Soln: W = Fb – Fh W = ρ_agV - ρ_hgV [W = V(ρ_ag - ρ_hg)]1/(〖(ρ〗_a g - ρ_h g)) V = w/(g(ρ_a-ρ_h)) = 3115/(9.81(1.29-0.179)) V = 286.1 m^3

53. The Sp. Gr. of rock used as concrete aggregate is often desirable to know. If a rock weighed 6.15 N in the air and 3.80 N when submerged in water, what would be the specific gravity of the rock?

Soln: W = W_a - W_w S = 6.15/(9810(2.4 x 〖10〗^(-4))) W = 6.15 – 3.8 S.g = 2.62 (9810)V = 2.35 V = 2.4 x 〖10〗^(-4) m^3 S.W = W_a/V = (6.15 N)/(2.4 x 〖10〗^(-4) ) S.W = 25625 N/m^3

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42

54. A piece of wood weighs 17.80 N in air and piece of metal weighs 17.80 in water. Together the two weighs 13.35 N in water. What is the specific gravity of the wood?

Solution:

Wwo=17.80N (air) Wm=17.80N(water)

WT=Wwo+Wm ; WT=13.85N Wwo=17.80(air)

55. A sphere 1.0 in diameter floats half submerged in tank of liquid (Sp. Gr. 0.80) (a) what is the weight of the sphere? (b) What is the minimum weight of the anchor (Sp. Gr. 2.40) that will require to submerge the sphere completely? Given: Find: Sa = 7.40 Sliquid = 0.80 W s Vs = 4/3π^3 Wa = 4/3π(0.53)^3 = 0.52m³ A.) W=fb =WsLVs/2 Ws=9.81Kn/m3(0.80)(0.52m3)/2 Ws=2.05KN B.) Wa=Fba+Fbs–W where: Va=Wa/Wsa =WslVa+WslVs-WsVs SS=Ws/wVs Wwo=17.80-FB 13.35N=17.80-FB+17.80 FB=22.25N (Displaced Water) Gs=Wwo/FB =17.80N/22.25N Gs=0.80

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43 Wa=w[0.80xwa/w2.40]+[0.80x0.32m3]-[2.04kn/9.81kn/m3] Wa=0.33wa+4.08kn-2.04kn Wa-0.33wa=2.04kn-4.08kn 0.67wa/0.67=2.04/0.67 Wa= 3.50kn

56. Fig. Z shows a hemispherical shell covering a circular hole 1.30 m in diameter at the vertical side of a tank. If the shell weighs 12,450 N, what vertical force is necessary to lift the shell considering a friction factor of 0.30 between the wall and the shell?

57. An iceberg has a specific gravity of 0.92 and floats in salt water (Sp. Gr. 1.03). If the volume of ice above the water surface is 700 cu.m, what is the total volume of the iceberg?

given: find: Si=0.92 Vt

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44 W=WvƩfv=0] vt=v1+v2 W=WsiVt W=Fb v2=vt-v1 Fb=Wssw wsivt=wssw(vt-v1) =Wsssv2 Sivt=wswvt-sswv1 sswv1=vt(ssw-si)

58. A concrete cube 60 cm on each edge (Sp. Gr. 2.40) rests on the bottom of a tank in which sea water stands to a depth of 5 m. The bottom edges of the block are sealed off so that no water is admitted under the block. Find the vertical pull required to lift the block.

Solution: W1=23.54(0.6x7xd) ;d=1 W1=98.868 W2=197.736[1/2(24)(7)d] X=(RM-OM)/RV W2=197.736 =(583.3692-204.375)/296.604 F=δhA X=1.28m =9.81(2.5)(5) F=122.625 e=b/2-x Rx=122.625d S=Ry/b(1±6e/b) Ry=296.604d S=142.36992kPa RM=98.868(3-0.3)+197.766[2/3(2.4)] RM=583.3692kN.m OM=122.625(5/3) OM=204.375kN.m vt=sswv1/ssw-si =1.03(700)/1.03-0.92 Vt=6554.55m3

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59. A 15 cm by 15 cm by 7 m long timber weighing 6280 N/cu.m is hinged at one end and held in horizontal position by an anchor at the other end as shown in Fig. AA. If the anchor weighs 23450 N/cu.m, determine the minimum total weight it must have

Solution: Vt=(0.15)2(7) Wa=WaVa Vt=0.1575m2 Wa=23540Va Va=Wa/23540 Wt=WtVt =(0.1575)(62.80) Wt=989.1 Fbt=9810(0.1575) Fba=WVa Fbt=1545.075 Fba=9810Va Mh= 3.5Fbt+7Fba=3.5Wt+Wa 3.5(1545.075)+7(9810) Va=3.5(989.1)+7(23540)Va Va=0.02m3 Wa=WaVa =23540N/m3(0.02m3) Wa=470.8N

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46

60. A cylinder weighing 445 N and having a diameter of 1.0 m floats in salt water (Sp. Gr. 1.03) with its axis vertical as in Fig. BB. The anchor consist 0f 0.0280 cu.m of concrete weighing 23450 N/cu.m. What rise in the tide r will be required to lift the anchor off the bottom?

Solution: Wa=23540(0.280) Fba=9810(1.05)(0.280) Wa=659102N Fba=2829.204N Wo=445N Fbc=9810(1.03)p(0.5)2(0.3+r) Fbc=2380.769+7935.89866 ∈Fv=0 Fba+Fbc=Wa+Wc 2829.204+2380.769+7935.898r=6591.2+445 r=0.23m ; 23cm

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47

0.11x2-1.1x+1.375=0

X=1.46m

𝑥 =−𝑏 ± 𝑏2− 4𝑎𝑐 2𝑎

61. A timber 15 cm square and 5 m long has a specific gravity of 0.50. One end is hinged to the wall and the other is left to float in water (Fig. CC). For a=60 cm, what is the length of the timber submerged in water? Solution: Wt=9.81(0.5)(5)(0.15)2 Wt=0.5518125kN Fb=9.81(0.15)2(x) Fb=0.220725x Mh=0 2.5cosθWt=(5-0.5x)cosθFb 2.5(0.5518125)=(5-0.5x)(0.220725x) 1.375=1.1x-0.11x2

62. A metal block 30 cm square and 25 cm deep is allowed to float on a body of liquid which consist of 20 cm layer of water above a layer of mercury. The block weighs 18,850 N/cu.m. What is the position of the upper level of the block? If a downward vertical force of 1110 N is

applied to the centroid of the block, what is the new position of the upper level of the block?

Solution: a.) Fbm=wV Fbw=9.81(0.20)(009) =9.81(13.6)(0.09)(0.05-x) Fbw=0.17658kN Fbm=0.600372-12.00744x W=18.85(0.09)(0.25) W=0.424125kN

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48 ∈Fv=0 Fbw+Fbm=W 0.17658+0.600372-12.000744x=0.424125 X=0.0294m X=2.94cm b.) Fbm=9.81(13.6)(0.09)(0.25-x) W=0.4241225 Fbm=3.00186-12.00744x Wv=1.11kN Fbw=9.81(0.09)(x) Fbw=0.8829x ∈Fv=0 0.8829x+3.00186-12.00744x=0.4241225+1.11kN X=0.132m H=0.20m-0.132m H=0.068m H=68cm

63. Two spheres, each 1.2 m diameter, weigh 4 and 12 KN, respectively. They are connected with a short rope and placed in water. What is the tension in the rope and what portion of the lighter sphere produces from the water? What should be the weight of the heavier sphere so that the lighter sphere will float halfway out of the water?

Solution: 𝐹𝐵𝐻 = 9.81 4𝜋 (0.603 3)𝐹𝐵𝐻 = (9.81)(12)(4𝜋(0.60)3 3) 𝐹𝐵𝐻 = 8.8759 kN𝐹𝐵𝐻 = 4.4379 kN T = 𝑊𝑆𝐻 - 𝐹𝐵𝐻 T = Wss - 𝐹𝐵 T =12 kN - 8.8759 kN T = 4 kN - 4.4379 kN T = 3.12 kN T = 0.4379 kN ∑Fy=0] 𝐹𝐵𝑆 = 𝑊𝑆𝑆 - T 9.81Vss = 4 + 3.12 Vss = 0.72579𝑚3

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𝑊𝐻 = T + 𝐹𝐵𝐻 𝑊𝐻 = 0.44 kN + 8.88 kN

Vs = 𝜋3𝐷2(3𝑟 − 𝐷)𝑊𝑆𝐻 = 9.32 kN 0.72579 = 0.60π𝐷2 - 0.33π𝐷3 D = 0.85m --- by trial & error X = 1.20 – D

X = 0.35m

68. If the specific gravity of a body is 0.80, what proportional part of its total volume will be submerged below the surface of a liquid (Sp. Gr. 1.20) upon which it floats?

Solution: 𝐹𝐵 = 𝑊𝐵 (𝑤𝑉)𝑆𝐿 = (𝑤𝑉)𝑆𝐵 (9.81)(1.20)𝑉𝑆𝐿 = (9.81)(0.80)𝑉𝑇 (1.20)𝑉𝑆𝐿 = (0.80)𝑉𝑇 𝑉𝑆𝐿 = 23𝑉𝑇 2

3of the total Volume

69. A vertical cylinder tank, open at the top, contains 45.50 cu.m of water. It has a horizontal sectional area of 7.40 sq.m and its sides are 12.20 m high. Into its lowered another similar tank, having a sectional area of 5.60 sq.m and a height of 12.20 m. The second tank is inverted so that its open end is down, and it is allowed to rest on the bottom of the first. Find the maximum hoop tension in the outer tank. Neglect the thickness of the inner tank.

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70. A small metal pan of length of 1.0 m, width 20 cm and depth 4 cm floats in water. When a uniform load of 15 N/m is applied as shown in Fig. DD, the pan assumes the figure shown. Find the weight of the pan and the magnitude of the righting moment developed.

Solution: 𝑉𝑃 = 0.04m (0.20m) (1m) 𝑉𝑃 = 8x10−3𝑚3𝑉 2 = 𝑉1= 4x10−3𝑚3 Θ = 𝑡𝑎𝑛−1 (0.04 0.20) Θ = 11.31𝑜 𝑊𝑃 + T = 𝐹𝐵 𝑊𝑃= 𝐹𝐵- T 𝑊𝑃 = 9810(4x10−3) – 15 𝑊𝑃 = 39.24 - 15 𝑊𝑃 = 24.24 N

71. A ship of 39,140 KN displacement floats in sea water with the axis of symmetry vertical when a weight of 490 KIN is mid ship. Moving a weight 3 m toward one side of the deck cause a plumb bob, suspended at the end of a string 4 m long, to move 24 cm. Find the metacentric height. Given: W=39140kn TanΦ=0.24/4 C=Wx Φ=3.43˚ 490(3)=39140x SinΦ=X/MG T = 𝐹𝑙 15𝑚𝑁 = 1𝑚𝐹 F = 15 X=MGSinΦ 490(3)=39140(MGSin3.43˚) MG=0.63m

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72. A rectangular scow 9.15 m wide by 15.25 m long and 3.65 m high has a draft of 2.44 m in sea water. Its center of gravity is 2.75 m above the bottom of the scow. (a) Determine the initial metacentric height. (b) If the scow tilts until one of the longitudinal sides is just at the point of

submergence, determine the righting couple or the overturning couple. Soln:

a.) b.)

GB_o = 2.75 – 1.22 tanθ = 1.21/4.575

= 1.53 m θ = 14.81°

MG = MB_o - GB_o MB_o = B^2/12D(1 + 〖tan〗^2θ/2)

= 2.86 – 1.53 = 〖9.15〗^2/(12(2.46))(1 + 〖tan〗 MG = 1.33 m^(2(14.81))/2) MB_o = 2.96 m ∑Fv = 0 ; FB = W FB = wV = 9.15(15.25)(2.44)(9.81)(1.03) FB = W W = 3490.23 kN RM = W(MGsinθ) = 3490.23(1.34sin14.81°) RM = 1257.7 kN.m

73. A cylindrical caisson has an outside diameter of 6 m and floats in fresh water with its axis vertical. Its lower end is submerged to a depth of 6 m below the water surface. Find: (a) the initial metacentric height; (b) the righting couple when the caisson is tipped through an angle of 10 degrees.

Soln:

a.) b.)

MB_o = I/V ; I=(π(6^2))/(12(4)) ; V=(π(6)(6))/4 MG = MB_o+GB_o = 0.375 + 0.5

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74. A rectangular scow 9.15 m wide by 15.25 m long has a draft of 2.44 m in fresh water. Its center of gravity is 4.60 m above the bottom. Determine the height of the scow if, with one side just at the point of submergence, the scow is in unstable position.

75. A rectangular raft 3 m wide 6 m long has a thickness of 60 cm and is made of solid timbers (Sp. Gr. 0.60). If a man weighing 890 N steps on the edge of the raft at the middle of one side, how much will the original water line on that side be depressed below the water surface?

Find: RM V’=(9.15m)(2.75m)(15.25) W=wV =383.73m3=(9.81kn/m3)(383.73m3) W=3764.39kn TanΦ=1.85/4.575 V=1/2(15.25m)(1.85m)(4.573) Φ=222.02˚ V=64.54m3 Mbo=VL/V’SinΦ Gbo=2.30-1.375m =64.54m3(6.1)/383.73m3(Sin22.02˚) Gbo=0.925m Mbo=2.74m MG=Mbo-Gbo RM=w(MGSinΦ) =2.74m-0.925m =(3764.39kn)(1.815)sinΦ22.02˚ MG=1.815m RM=1242.60kn.m

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### Accelerated Liquids in Relative Equilibrium

EXERCISE PROBLEM

1. A car travelling on a horizontal road has a rectangular cross section, 6m long by 2.40m wide by 1.50m high. If the car is half full of water, what is the maximum acceleration it can undergo without spilling any water? Neglecting the weight of the car, what force is required to produce maximum acceleration? Given: d=0.75 m L=6 m Wide=2.4 m H=1.5 m Solution: W = Vw W = . 75 6 2.4 9.81 = 105.95 kN d = La 2g𝐹 = 𝑚𝑎 = 𝑊𝑎 𝑔 a = 2dg L 𝐹 = 105.95 2.45 9.81 a =2 0.75 9.81 6 𝐹 = 26.46 𝑘𝑁 𝑎 = 2.45 𝑚 𝑠2

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54 1.20 m 0.90 m Sg = 22 000 N/m 3 0.60 m a = 9.81 m/s2

### F

2. A cylindrical bucket is accelerated upward with an acceleration of gravity. If the bucket is 0.60m in diameter and 1.20m deep, what is the force on the bottom of the bucket if it contains 0.90m depth of wet concrete whose specific weight is 22,000 N/m3? Solution: 𝐹 = 𝑤𝑕𝐴 1 + 𝑎𝑔 𝐹 = 22 1.20 𝜋. 60 2 4 (1 + 9.81 9.81) 𝐹 = 11.20 𝑘𝑁

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3. A rectangular car is 3m long by 1.5m wide and 1.5m deep. If the friction is neglected and the car rolls down a plane with an inclination of water surface if the car contained 0.60m depth of water when the car was horizontal? Given: Find: θ tanθ= macos 20° mg −masin 20° =m(g−asin 20°)m(acos 20°) = g−asin 20°acos 20° Consider: tanα =0.91.5 a= 9.81 tan(30.96°) =30.96° a=5.885 m/s2 tanα =ag

tan(30.96°) = 𝑎𝑔 tan 𝜃 = 9.81− 5.885 sin 20° 5.885 cos 20° = 5.537.79 = 35.35°

3m θ 0.9 0.6 1.5 1.5 1.5 20° W REFн REFv 20° W REFн REFv 20° W REFн REFv a 20° a av aн W=mg θ REFн = maн =masin20° REFv = mav =macos20° F F mg masin20° macos20°

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4. An open tank, 9.15m long is supported on a car moving on a level track and uniformly accelerated from rest to 48km/hr.When at rest the tank was filled with water to within 15cm of its top. Find the shortest time in which the acceleration may be accomplished without spilling over the edge. Given: VF=48 Km/s= 13.33m/s Find: t Solution: tanθ = 4.5750.15 = 1.878° tanθ =𝑔𝑎 tan(1.878°)=9.81𝑎 a = 9.81 tan(1.878°) a= 0.322 m/s2 a=𝑉𝑓−𝑉𝑡 𝑜 t= 13.330.322 =41.44 s θ 15 cm a h 4.575 4.575 15 cm

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5. A rectangular tank, 60cm long and containing 20 cm of water is given an acceleration of a quarter of the acceleration of gravity along the length. How deep will the water be at rear end? At the front end? What is the pressure force at the rear end if it is 45 cm wide?

Given: Find : hF, hr, F Solution: tanθ = 𝑎 g = 1 4𝑔 2= 1 4 tanθ=_x_ 0.30 1 =_x_ 4 0.30 x= 0.075m or 7.5 cm hr=20+7.5 =27.5 cm hF=20-7.5 =12.5cm F=Awh =(0.45)(0.275)(9.81)(0.275/2) =0.16692 KN =166.92 N F x x 30 cm 30 cm a= 1 g 4 20 cm

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W.S.

600

3.0 m 1.3o m

6. Figure GG shows a container having a width of 1.50 m. Calculate the total forces on the ends and bottom of the container when at rest and when being accelerated vertically upward at 3m/s2?

Given: a=3m/s2 w= 1.50m Solution: 𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 𝑤𝑎𝑕 𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 𝑤𝑕𝐴(1 +𝑎𝑔) 𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 9.81 1.3 1.5 (0.65)𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 9.81 1.3 1.5 (0.65)(1 + 3 9.81) 𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 12.43 𝑘𝑁𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 16.23 𝐾𝑁 𝐹𝑟𝑖𝑔𝑡 𝑕 𝑒𝑛𝑑 = 𝑤𝑎𝑕 𝐹𝑟𝑖𝑔𝑡 𝑕 𝑒𝑛𝑑 = 𝑤𝑕𝐴(1 +𝑎𝑔) 𝐹𝑟𝑖𝑔𝑡 𝑕 𝑒𝑛𝑑 = 9.81 1.5 1.5 (0.65)𝐹𝑟𝑖𝑔𝑡 𝑕 𝑒𝑛𝑑 = 9.81 1.5 1.5 (0.65)(1 + 3 9.81) 𝐹𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 = 14.35 𝑘𝑁𝐹𝑟𝑖𝑔𝑡 𝑕 𝑒𝑛𝑑 = 18.74 𝑘𝑁 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 𝑤𝑎𝑕 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 𝑤𝑕𝐴(1 +𝑎𝑔) 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 9.81 1.5 1.3 1.3 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 9.81 1.5 1.3 (1.3) (1 + 3 9.81) 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 57.40 𝑘𝑁 𝐹𝑏𝑜𝑡𝑡𝑜𝑚 = 74.95 𝑘𝑁

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7. A closed rectangular tank 1.20m high by 2.40 m long by 1.50 m wide is filled with water and the pressure at the top is raised to 140 Kpa.

Calculate the pressures in the corners of this tank when it is accelerated horizontally along its length at 4.60m/s2?

Given: Find: P1 , P2 Solution: h=p = 1.40 = 14.2712 m w 9.81 tanθ = a = y_ = 4.6_ g 2.4 9.81 y= 1.125 m P1 = wh1 = 9.81 (1.2 + 14.2712 + 1.125) = 162.81 Kpa P2 = wh2 = 9.81(1.2 +14.2712) = 151.77 Kpa y h θ 1.20 m P1 P2 2.40 m a=4.60m/s

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8. A pipe 2.50 cm in diameter is 1.0 m long and filled with 0.60m water., what is the pressure at the other end of the pipe when it is rotating at 200 RPM? Given: Find: P Solution: y1 = w2x12 2g =(20/3л)2 (0.4)2 2(9.81) =3.577 m y2== w2x22 2g =(20/3л)2 (1)2 2(9.81) =22.357 m h2=22.357-3.577 =18.78 m P=wh2 =9.81(18.78) = 184.23 Kpa h2 0.6 X1=0.4 y1 y2 w= 200rpm =20/3 л rad/s x2= 1m

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9. An open vertical cylindrical tank 0.60 m in diameter and 1.20 m high is half full of water. If it is rotated about its vertical axis so that the water just reach the top, find the speed of rotation. What will then be the maximum pressure in the tank? If the water were 1.0 m deep, what speed will cause the water to just reach the top? What is the depth of the water at the center? a. 𝑦 =𝑤 2𝑥2 2𝑔 1.2 =𝑤 2 0.3 2 2 9.81 𝑤 = 16.17𝑟𝑎𝑑 𝑠 𝑃 = 𝑤𝑕 𝑃 = 9810(𝑤 2𝑟2 2𝑔 ) 𝑃 = 9810(16.17 2× 0. 32 2 × 9.81 ) 𝑃 = 11.80 𝑘𝑃𝑎 𝑦 =𝑤2𝑥2 2𝑔 0.4 =𝑤2 0.3 2 2 9.81 𝑤 = 9.34𝑟𝑎𝑑 𝑠 𝑑 = 𝐻 2 + 𝑦 2 𝑑 = 0.6 + 0.2 𝑑 = 0.8 𝑚

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10. If the tank of problem 9 is half full of oil (sp. Gr. 0.75) what speed of rotation is necessary to expose one-half of the bottom diameter? How much oil is lost in attaining this speed?

Given: D=0.6 m H=1.2 m 1.2 + 𝑑 =𝜔2𝑔2𝑥2 eq.1 𝑑 =𝜔2 𝑥 2 2 2𝑔 eq.2 By equating eq 1 and eq 2 𝜔 = 18.68𝑟𝑎𝑑 𝑠 𝑑 = 0.40 𝑚

11. The U-tube of figure HH is given a uniform acceleration of 1.22 m/ s2 to the right. What is the depth in AB and the pressures at B, G and D?

Given: Find: Solution: 𝑦 𝑧 = 𝑎 𝑞 𝑦 0.30= 1.22 9.81 𝑦 = 0.04𝑚 B A H G C D 30 cm 30 cm 30 cm y h1 h2 H3 𝑃𝐵 = 45.36 𝑘𝑃𝑎 𝑃𝑔 = 𝑤𝑕2 𝑃𝑔 = 40.02 𝑘𝑃𝑎 𝑃𝑑 = 𝑤𝑕3 𝑃𝑑 = 37.40 𝑘𝑃𝑎 𝑃𝐵 = 9.81 13.6 (. 34) 𝑃𝑔 = 9.81 13.6 (. 30) 𝑃𝑑 = 9.81 13.6 (. 26)

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𝑃𝐵 = 𝑤𝑕1

12. The U-tube of figure HH is rotated about an axis through HG so that the velocity at B is 3m/s. What are the pressures at B and G?

Given: Find: PR, PG Solution : PR= wh1 =9.81(3.6)(0.30) =40.02 Kpa Y2 = 𝑉2 2𝑔 = 3 2 2 (9.81) =0.459 m PG= wh2 =9.81(13.6)(0.459-0.30)

=21.18 Kpa it is below the point Therefore, PG = -21.18 Kpa 30 cm 30 cm H G C A D B 30 cm Figure HH 30 cm 30 cm H G A D B Figure HH C 30 cm

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13. The U-tube of figure HH is rotated about HG. At what angular velocity does the pressure at G become zero gage? What angular velocity is required to produce a cavity at G?

y1 = 𝑤12𝑥 12 2𝑔 ; in LL1 w= 𝑦12𝑔 𝑥2 = 0.30 2 (9.81) 0.302 w1=8.09 rad/s in LL1; h=Patm 𝑤 y2=0.30 +0.759 w2 = 𝑦12𝑔 𝑥2 = 9.81 (13.6)101.3 =1.059 w2 = 1.059 2 (9.81) (0.30)2 30 cm 30 cm H G C A D B Figure HH 0.30 m 0.30m Patm/w w 0.30,0.30 0.30 m 0.30, y2 Y2 LL1 LL2

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14. The tank of problem 9 is covered with a lid having a small hole at the center and filled with water. If the tank is then rotated about its vertical axis at 8rad/s, what is the pressure at any point of circumference of the upper cover? Of the lower cover?

Given: Solution: y=𝑤2𝑥2 2𝑔 =8 2 0.32 2(9.81) =0.294 m PU= wh1 = 9.81(0.294) = 2.88 Kpa PL = wh2 = 9.81 (1.20 +0.294) = 14.65 Kpa h2=1.20m y 0.30 m 0.30 m h2

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15. The tank of problem 9 contains 0.60 m of water covered by 0.30m of oil (sp. Gr. 0.75). What speed of rotation will cause the oil to reach the top? What is then the pressure at any point on the circumference of the bottom? Given : Find: w, PB Solution: y1 = w12x12 ; in LL1 2g w =y2g x2 = (0.6)(2)(9.81) ` (0.30)2 w = 11.44 rad/s PB= woilh2 + whwh1 = 9.81(0.75)(0.3) + 9.81(0.9) PB = 11.04 Kpa oil water 0.3 m 0.3 m h2 h1 1.20 m

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16. The tube of figure II is rotated about axis AB. What angular velocity is required to make the pressures at B and C equal? At that speed where is the location of the minimum pressure along BC?

Given: Y2 Find: w,z Y1 Solution: w Tan 45° = 0.3x = 0.3m Y = w2x2 2g W= y22g x2 = 2 (9.81) 0.3 = 8.08 rad s 0.30 = y3 + z ; y3 = 0.3 – z h = y1+ 0.3-z = w2z2 2𝑔 +0.3-z = 8.092𝑧2 2 (9.81) +0.3 – z = 3.336z2 –z +0.3 P = wh (𝑑𝑃𝑑𝑥 ) = w(3.336z2 – z + 0.3) = 0 = w (6.672z – z) = 0 Z = 0.15 m B 45° ° x h C A z z Z 30c m

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17. A vessel 30 cm in diameter and filled with water is rotated about its vertical axis with such a speed that the water surface at a distance of 7.50cm from its axis makes an angle of 45 degrees with the horizontal. Determine the speed of rotation.

tan 45 = 𝐹𝑐 𝑤 = 𝑊 𝑔𝑤 2𝑥 𝑊 tan 45= 𝑤 2 𝑔 𝑥 𝑤 = 𝑔𝑡𝑎𝑛 45 𝑥 𝑤 = 9.81 tan 45 0.075 𝑤 = 11.44 𝑟𝑎𝑑 𝑠

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18. A cylindrical vessel, 0.30 m deep, is half filled with water. When it is rotated about its vertical axis with the speed of 150 RPM, the water just rises to the rim of the vessel. Find the diameter of the vessel.

Given: Find : D w = 150 rpm = 5πrad s Solution: Y1 = 𝑤2𝑥2 2𝑔 X = 𝑦𝑤22𝑔 = 0.3 2 (9.81)(5𝜋)2 = 0.154 m D = 2x = 2(0.154) = 0.3089 m = 30.89 cm 0.30 x

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19. A conical vessel with vertical axis has an altitude of 1m and is filled with water. Its base, 0.60m in diameter, is horizontal and uppermost. If the vessel is rotated about its axis with a speed of 60RPM, how much water will remain in it?

Given : Find: Vr w = 60 rpm = 2πrad s Solution: y = 𝑤 2𝑥2 2𝑔 = (2𝜋)2(0.3)2 2(9.81) = 0.1811 m Vr = Vcone – Vpar = 1 3𝜋 (𝑟 ) 3𝑕 −1 2𝜋𝑟 2𝑦 = 1 3𝜋 0.3 3(1) −1 2𝜋 0.3 2(0.1811) Vr = 0.060 m3 1m y 0.3 0.3

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20. A cylindrical bucket, 35 cm deep and 30 cm in diameter, contains water to a depth of 30 cm. A man swings this bucket describing a circle having a diameter of 2.15 m. what is the minimum speed of rotation that the bucket can have without permitting water to spill?

Given: Find : w Solution: r= R – 0.30 2 = 1.075 – 0.15 = 0.925 m w = Fc mg = mwr w = 𝑔𝑟 = 9.81 0.925 w = 3.26 𝑟𝑎𝑑𝑠 R = 1.075 S w

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21. If the water which just fills a hemispherical bowl of 1.0m radius be made to rotate uniformly about the vertical axis of the bowl at the rate of 30 RPM, determine the amount of water that will spill out?

Given: Find : Vspill

Solution: y = w2x2 2g = (π)2(9.81)2(1)2 = 0.503 m Vspill = 1 2πr 2y = 1 2π 1 2(0.503) Vspill = 0.79 m3 y 1.0 W = 30 RPM = πrads

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22. The open cylindrical tank of figure JJ is rotated about its vertical axis at the rate of 60 RPM. If the initially filled with water, how high above the top of the tank will water rise in the attached piezometer?

Given: Solution: y1 = w2x12 2g = (2π)2(9.81)2(0.65)2 = 0.850 m Y2 = w2x22 2g = (2π)2(9.81)2(1)2 = 2.012 m h= y2 – y1 = 2.012 – 0.850 = 1.16 m 1.30 m 1 m 1.30 cm Figure JJ

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23. A closed cylindrical tank with axis vertical, 2m high and 0.60m in

diameter is filled with water, the intensity of pressure at the top being 140 Kpa. The metal making up the side is 0.25 cm thick. If the vessel is rotated about its vertical axis at 240 RPM, compute (a) total pressure on the side wall, (b) total pressure against the top, (c) maximum intensity of hoop tension in pascals.

Given: Solution: h = 𝑝 𝑤 = 140 9.81= 14.271 m y = w2x2 2g = (8π)2(0.3)2 2(9.81) = 2.897 m

a.) Fside = Awh

= 𝜋 2 0.6 9.81 14.271 + 2.897 + 1 = 671.90 KN b.) Ftop = Awh =𝜋 0.3 2 9.81 14.27 + 2.879 = 47.62 KN c.) P = wh = 9.81(2.9 +14.27 + 2) = 188.044 Kpa FB = PD = 2T T =188.04(0.6) 2 = 56.41 Kpa S = 𝑇𝑡 = 56.41 2.5𝑥10−3 =22565.34 Kpa y 0.6 w 2m Ftop Fside = 240 rpm = 8πrads

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24. A small pipe, 0.60m long, is filled with water and capped at both ends. If placed in a horizontal position, how fast must it be rotated about a vertical axis, 0.30m from one end, to produce maximum pressure of 6,900 Kpa? Given: Solution: h = 𝑝 𝑤 = 6900 9.81 = 703.36 m h = y2 – y1 = w2(9.81)2(0.9)2 - w2(9.81)2(0.3)2 703.3639 = 9𝑤2 218 - 𝑤2 218 703.3639 = 4𝑤 2 109 w= 138.44 rad s 0.3 0.6 0.3 + 0.6 = 0.9 Y2 w h= wp

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25. A vertical cylindrical tank 2m high and 1.30m in diameter, two thirds full of water, is rotated uniformly about its axis until it is on the point of overflowing. Compute the linear velocity at the circumference. How fast will it have to rotate in order that 0.170 m3 of water will spill out?

Given: Solution: y = w2x2 2g w = y2g x2 = 1.33 2 (9.81)(0.65)2 = 7.869 rad s V = wx = 7.869(0.65) =5.11 m s Vspill = 1 2πr 2𝑕 0.170 = 1 2π(0.65) 2h h = 0.256 m w = y2g x2 = (1.33+0.256) 2 (9.81)(0.65)2 = 8.59 rad s y 0.67 0.67 w 2m 1.3 h With spillage 1.33 w 1.3 Vspill = 0.170m3

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26. A steel cylinder, closed at the top, is 3m high and 2m in diameter. It is filled with water and rotated about its vertical axis until the water

pressure is about to burst the sides of the cylinder by hoop tension. The metal is 0.625 cm thick and its ultimate strength is 345 Mpa. How fast must the vessel be rotated?

2𝑇 = 𝑃𝐷 𝑇 = 𝑃 2 2 (1) 𝑇 = 𝑃 𝑇 = 𝑆𝑡 𝐴𝑡 𝑇 = 345 1000 (6.25 × 10−3) 𝑇 = 2156.25 𝑘𝑁 𝑇 = 𝑃 𝑃 = 𝑦 9.81 + wh 2156.25 =𝑦 9.81 + 9.81 (3) 𝑦 = 216.80 𝑚 𝑦 =𝑤2𝑟2 2𝑔 216.80 = 𝑤 2𝑥 12 2(9.81) 𝑤 = 65.22 𝑟𝑎𝑑 𝑠

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27. A conical vessel with axis vertical and sides sloping at 30 degrees with the same is rotated about another axis 0.60 m from it. What must be the speed of rotation so that water poured into it will be entirely discharged by the rotative effect?

Given: Solution: tanθ = w2x2 2g w = gtanθ x = 9.81tan(60)0.6 w = 5.32 rad s 30 ° 30 ° 60° w 0.6 m

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### Principles of Hydrodynamics

EXERCISE PROBLEM

1. A fluid flowing in a pipe 30cm in diameter has a uniform velocity of 4m/s. the pressure at the center of the pipe is 40kpa and the elevation of pipes centerline above an assumed datum is 4.5m. compute the total energy per unit weight of the flowing fluid if

(a)oil (sp.gr. 0.80) (b)gas(w=8.50N/m3) GIVEN: a) E = V2 +P + Z b) E = V2 + P + Z oil (sp.gr. 0.80) 2g w 2g w gas (w=8.50N/m3) = (4)2 + 40__ + 4.5 = (4)2 + 40__ + 4.5 Z = 4.5m 2(9.81) (9.81).8 2(9.81) V = 4m/s E= 10.41 J/N E = 4.7 J/N

2. A liquid of specific gravity 1.75 flows in a 6cm pipe. The total energy at appoint in the flowing liquid is 80 J/N. the elevation of the pipe above a fixed datum is 2.60m and the pressure in the pipe is 75kpa. Determine the velocity of flow and the power available at the point.

GIVEN: E = V2 +P + Z P = QwE Sp.gr = 1.7 2g w P = AVwES P= 75kpa V2 = E – P +Z P= 𝜋(0.06)2 (37.85)(80)(1.75) Z= 2.6m 2g w 4 V2 = 80 – 75__ +26 P= 147 kW 2g (9.81)(1.75) V = 37.85m/s

4. A city requires a flow of 1.5m3/s for its water supply. Determine the diameter of the pipe if the velocity of flow is to be 1.80m/s.

GIVEN: Q = AV : A= Q/V Q = 1.5m3/s 𝜋d2 = Q V = 1.80m/s 4 V d2 = 4Q 𝜋V d2 = 4(1.5)

References

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