Chapter ONE Properties of Fluids Exercis PDF

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Summary

A bottle consisting of a cylinder 30 cm in diameter
and 30 cm high has a neck 5 cm in diameter and 30
cm long. If the bottle, filled with air under normal
atmospheric condition, is inverted and submerged
in water until the neck is just filled with water, find
the d...

Description

CHAPTER ONE - 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.  = =

 

.

7 / 3(1000 )

ѵ=



1

c.) s = ws

7 / 3

= 9.81 / 3

= 713 .56  / 3

9.81 /2

 = 713.56 Kg/m3

1

ѵ = . 3/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.)  =

=

 

16 / 3

9.81 /2

. ѵ = ѵ= 1.63  / 3

 = 1.63 Kg/m3

1 

c.) s =



ws

16 / 3

s = 12 / 3

1

ѵ = . 3/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 =  =

43 .860  5.30  3

w = 8.28 KN/m3



b.)  =  =

c.) s =

43860  . /2

9.81 ( 2

)(5.30  3)

 = . kg/3

1

=



ws 8.28 / 3

9.81 / 3

s = 0.844

4. The density of alcohol is 790 Kg/m3 . Calculate its specific weight, specific gravity and specific volume. a.) w = g

b.) s =

= (790 kg/m3)(9.81 m/s2) w = 7.75 KN/m3

=

 ws

. ѵ = 

1

7.75 / 3

= 790  / 3

9.81 / 3

s = 0.79

1

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=

12 / 3

12.7 / 3

s = 0.94



s = s

a = (1.29 kg/m3)(0.94)

ѵ= =  1.21  / 3 1

a = (1.21 kg/m3)

1

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.)  =

 

=

10 .10 (1000 ) 9.81

 = . kg/3 m3/kg

p = wh = 10.10(1000)(8000)

. ѵ =

1

= 9.5 x 10-4

p = 80.80 MPa

Δѵ = 3.3 x 10-5 m3/kg

= 1043  / 3

1 

c.) w =





=

82 .26  106 8000

w = 10282. 5 N/m3

2

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

= -( 1 x 10-3)( 1 x 10−3−1.012 x10−3)

9.689

=

� 30,000 ,000

9.81

 = 987.67 kg/ m3

= 2,500,000 Pa

ѵ =  = 987.67

Bv = 2.5 x 106 Pa

1

1

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= =  µ

110−3  . ./ 22 1000  / 3 3.28  2 ) 1

ѵ = 1x10-6 m2/s (

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). 2.54  2 1  2

1 inch = 2.54 cm 12  2 1  2

1 ft2/s (

)(

) = 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  1 3

) = 0.45 m3

3.5(1000)

3

= 9.81 (0.45) = 792.85 kg/m

3

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

1 Hp =

 − 12  2.54  444 ,800  (  ) ( 1  ) ( 1  

7456627200

)

 − 

100000000  − /

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 (

1  3

7.48 

1 3

) (3.28  3) = 0.379 m3

P = 6000 (0.379 m3) P = 2274 m3 

2274  3

D.C. = ( ) = ( ) 60  60  24  D.C. = 0.026 m3/s

13. Compute the conversion factor for reducing pounds to newtons. 32.18

 0.3048  ) ( 2 1 

(

1 1  ) )( 2.205   / 2

= 4.448 N

4

CHAPTER TWO – 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 W=p/n =140kPa/3m W=46.67KN/m3

b.) S=W/ws =46.67/9.81 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) P=42.18kPa

5

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=10.20m of water

h=0.75m 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

6

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

7

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.

(1)

(2)

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

8

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/m3 what 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

VW w=6.5(0.3x0.3x3)

Ww=wV

Vw=1.755/9.81

=0.179(9.81)

Vw=0.179m3

Ww=1.756kN 9

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/m3 and 6 N/m3 respectively. 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

10

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

Solution: P = wsh + wsh P = (9.81) (13.6) (0.75) + (9.81) (0.85) (1.5) P = 112.6 kpa 20.

e.

air water

20cm Gage

3m

Mercury

11

Solution: P = -wsh

Pg = wsh - wsh Pg = 9.81 (3) – (9.81) (13.6) (0.2)

P = - (9.81) (0.2) (13.6) 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

m 12

Solution: a) Pb = -wsh

Pm = -8.829 + 0

Pb = - (9.81) (1.5) (0.6)

Pm = -8.829 kpa

b) h = p/w

Pb = - 8.829 kpa

h = -8.82/9.81

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 m

B

6 cm

Solution: Pg = - wsh

Pabs = Patm + Pg Pabs = 101.3 – 10.67

Pg = - (9.81) (13.6) (0.06) 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 13

. Solution:

10.5 – 26.68sin� = 0

P = wsh

26.68sin� = 10.5

P = (9.81) (13.6) (0.20)

� = 22.6 °

P = 26.68 kpa

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

Pm/w – y – 0.68 (0.85) + x = Pn/w

170 – y = 68 – x

Pm – Pn = [ ( y – x ) + ( 0.65 ) (0.85) ] 9.81 Pm – Pn = ( 1.02 + 0.578) (9.81) Pm – Pn = 15.67 kpa

14

27. In fig. determine −  . 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

12 cm, D

36 cm

40 cm

m

Solution: sh + wsh A

= (9.81) (0.4) (3) + (0.4) (9.81) (0.9)

CHAPTER THREE – 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=

g

 2 12

 

2 12



e= =

=

3.5 2 12

3.25

e = 0.31 m

hp = + 

F1 = wA

hp = 3.25 + 0.31

F1 = 9.81(3.5)(1.5)(3.25)(0.85)

hp = 3.56 m

F1 = 142.28 KN

16

e=

g

= 

 2

 12



e= =

12  2

1.5 2 12

4.25

e = 0.04 m

hp = + 

F1 = wA

hp = 4.25 + 0.04

F1 = 9.81(1.5)(1.5)(4.25)(0.85)

hp = 4.29 m

F1 =79.74 KN

Ft = F1 + F2

Pt = P1 + P2

Ft = 142. 28 + 79.74

Pt h = F1 h + F2h 222 .02 

Ft = 222.02 KN

222.02

=

142 .28 3.56 + 79.76 (4.29) 222.02

h = 3.83 m , below w.s

17

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

F = 30(1.0)(1.5)

F = (9.81)(x+0.5)(1.5)(1.0)

F = 45 KN

∑1 = 0

14.42x + 7.36(0.5 +

e=

e=

 

1

12+6

F = 14.72x + 7.36

1 12+6

) = 45(0.5)

86.5x2 – 168.16x – 105.56 = 0 =

−(−168 .16)±(−168 .16 )2 −486 .5 (105 .56 ) 2(86.5)

 = 2.40 

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=

 ( 4 )2 4 =   2 

g

F =9.81(1.4)(3.1)()(0.52)

e=

F = 33.44 KN

e = 0.02m (below the center) 18

(0.52 ) 4(3.1)

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 =

=

= 0.64 m

b.)

=

4

F = wA

3

 (1.52 )

4(1.5) 3

F = (9.81)(

(2)

)(0.64)

F = 22.15

1.5+0.64

F = wA

2

= 1.08 m

=

=

g 

0.1098(1.5) 3.53(0.64)

3 (1.52 )

F = (9.81)(

(4)

)(1.08)

F = 56.25 KN

4

e = 0.25 m

hp = + 

hp = 0.64 + 0.25 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 2 F = 9.81(1.8)( )(3)(3) 3

F = 105.95 KN

83 (3)2 g 175  = = 2 3 3 (1.8) 3

 = 0.34  19

 = + 

 = 1.8 + 0.34

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:

 = 

 = 9.811.6(10.8)

 = 508.55 

F = 508.55 KN

 = �

 = 0.2(508.55)

 = 101.71 

∑=0

 =  + 

 = 18.0 + 101.71  = 119.71 

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:

0.025 =

− 0.5

=

12 12



= 2.83 m

20

 12 2



Solution: a) Pb = -wsh

Pm = -8.829 + 0

Pb = - (9.81) (1.5) (0.6)

Pm = -8.829 kpa

Pb = - 8.829 kpa

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

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 m

B

6 cm

Solution: Pg = - wsh
...