Exercise 1 Sayas Ej2

Bryan Jonas R. Sayas 11211067 | EJ2 LBYCVW1- EXERCISE 1

1. Convert the following to their equivalent pressure or pressure head value 

76 cm of mercury, S=13.6 to pounds per square inch (psi) P = 13.6(9810 N/m3)(0.76m) P = 101396.16 N/m2 =(101396.16 N/m2)(2.205416 lb/1kg)(1kg/9.81N)[1in2/(3,28ft)2][1ft2/(12in)2] =14.71 psi



10.3m of water to equivalent height of oil, S= 0.83, in meters Assume Pwater = Poil Pw = Patm + 1(9810)(10.3) Pw = 101043 Pa Poil = Patm + 0.83(9810)(h) h = 101043/(0.83x9810) h = 12.41m



40 psi to meters of water = (40 lbs/in2)( (4.448 N/1lb)[1 in2/(0.0254m)2) = 275776.55 N/m2 h = 275776.55/1(9810) h = 28.11m



300 kPa to equivalent pressure in absolute units, if local Patm = 99.79kPa 300 kPa + 99.79 kPa = 399.79 kPa



235kPa-abs to meters of oil, S=0.79 (235kPa)(1000 Pa/1 kPA) = 235000 Pa h = 235000/.79(9810) h = 30.32m

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2. The container shown has three liquids contained as shown. If the pressure gage indicates 5kPa gage, what would be the elevations of the liquid free surfaces (A, B and C) in the different limbs of the manometer? What would be the elevation of the free surface of mercury (D) in the manometer below the tank? P2 = P1 + ∑±[δ(z2 – z1) P12 = δair(2) = P10 5000 Pa = P10 A) 5000 – 1.30(9810)ha = PA ha = 5000/1.3(9810) ha = 0.39m ElA = 12.39m B) 5000 – 1.5(9810)hb = PA hb = 5000/1.5(9810) hb = 0.34m ElB = 12.34m C) 5000 – 1.7(9810)hc = PA hc = 5000/1.7(9810) hc = 0.30m ElC = 12.30m D) 5000 – 13.6(9810)hd = PA hd = 5000/13.6(9810) hd = 0.04m ElD = 12.04m 3. What if the pressure indicates –5kPa? What would be the corresponding surface Pressure free gage 12m elevations? A P2 = P1 + ∑±[δ(z2 – z1) Elev 12m P12 = δair(2) = P10 Air 5000 Pa = P10 Elev 10m A) 5000 – 1.30(9810)ha = PA ha = 5000/1.3(9810) S= 1.30 ha = 0.39m Elev 8.4m S= 1.50 4.0m

ElA = 10.39m B) 5000 – 1.5(9810)hb = PA hb = 5000/1.5(9810) hb = 0.34m

Elev 5.7m S= 1.70 4.0m

ElB = 8.74m C) 5000 – 1.7(9810)hc = PA hc = 5000/1.7(9810) hc = 0.30m ElC = 6m D) 5000 – 13.6(9810)hd = PA 2|Page

D Elev 4.0m

S= 13.6 4.0m

B C

hd = 5000/13.6(9810) hd = 0.04m ElD = 4.04m 4. Determine in (a) newtons per sq.m, (b) millibars, the increase of pressure intensity per meter of depth in fresh water Assume: h = 1m P = pgh P = pg P = 1000kg/m3 (9.81m/s2) P = 9.81 x 103 N/m2 ~ 98.1mb 9.81 x 103 N/m2, 98.1mb 5. A branch of science that deals with water at rest a) fluid mechanics

b) fluid statics

c) hydraulics

d) kinetics

6. Absolute viscosity of a fluid varies with pressure and temperature and is defined as a function of : a) density and angular deformation rate b) density and shear stress c) shear stress and density d) shear stress and angular deformation rate 7. A block of wood (s =0.65) 20 mm thick is floating in sea water (s =1.03). Find the area of a block of wood (As) in square meter which will support a man weighing 80 kg., when the top surface is just at the water surface. Ywater = (1.03)(9810) = 10104.3 ywood = (0.65)(9810) = 6376.5 10104.5 (20A) – 80(9.81)(1000) = 6376.5 (20A) A = 10.53 a) 10.53

b) 9.45

c. 11.43

d) 12.42

8. A pressure gauge, located on side of a tank at elevation 8 m, containing a liquid reads 80 kPa. Another gauge at elevation 3 reads 120kPa. Compute for the specific gravity of the liquid. 120000 = 80000 + S(9810)(5) S = 0.82 a) 0.82

b) 0.65

c) 0.72

d) 0.55

A stone weighs 468 N in air. When submerged in water, it weighs 298 N. 9. Find the volume of the stone. The volume of stone in cubic meter is 468-298 = 170N, 170/9.81 = 17.33 *kg of water displaced* V = m/d = 17.33/1000 = 0.01733 m3 a) 0.0173

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b) 0.0165

c) 0.045

d) none of above

10. Find the specific weight of the stone in kN per cubic meter. The specific weight of the stone is m = 468/9.81 = 47.71kg p = 47.71/0.0173 = 2757.8 y = (2757.8)(9.81) = 27054 ~ 27.05kN a) 27.05 b) 23.9 c) 24.5 d) none of above 11. It is the pressure measured above absolute zero; also the sum of atmospheric pressure and gage pressure. a) absolute b)absolute pressure c)absolute d) absolute volume viscosity temperature 12. The term used as the resultant vertical force exerted on a body by a static fluid in which is submerged or floating is: a)reaction b)buoyancy c)hydrostatic force d)inertial force

13. If force, length and time are selected as the three fundamental dimensions, the units of mass in the SI system would be written as: a) FT2/L

b) FL/T2

c)Nm/s2

d) Ns2/m

14. Calculate the pressure at a depth of 10m in a liquid with specific gravity of 0.8 P = (0.8)(9810)(10) P = 78480 ~78.5kPa a)78.5kPa b) 62.7 kPa

c) 89.4 kPa

d) 98.3kPa

15. How many meters of water are equivalent to 75cm mercury (Hg)? 0.75 = Hh20 (1/13.6) H = 10.2m a) 8.5m b) 10.2 m

c) 9.4 m

d) 6.3 m

16. Oil with S= 0.86 is being transported in a pipe. Calculate the pressure if a U-tube manometer reads 240mm Hg (mercury. The oil in the manometer is depressed 125 mm below the pipe centreline. Phg = 0.24(13.6)(9810) = 32019.84 Pa Poil = 0.125(0.86)(9810) = 1054.575 Pa PA = Phg – Poil = 30.96kPa a)68.5kPa

b) 30.96 kPa

c) 49.4 kPa

d) 28.3kPa

17. An object is constructed of a material lighter than water. It weighs 50N in air and a force of 10N is required to hold it under water. What is its density, specific weight and specific gravity. 18. The mass density is : (50/60)(100) =833.33 kg/m2 a) 832.5 kg/m2

b) 742.5 kg/m2 c) 539.5 kg/m2 a) 972.5 kg/m2

19. The specific weight is: y = pg = (832.5)(9.81) = 8166.83 N/m3

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a) 9178 N/m3

b) 9123 N/m3 c) 7882 N/m3

d) 8167 N/m3

20. The specific gravity is: S = 832.5/1000 = 0.8325 a) 0.7784 b) 0.9084

c) 0.8325

d) 0.6785

21. Whenever the absolute pressure is less than the atmospheric pressure, the pressure is said to be: a) atmospheric

b) vacuum

c) positive gage d) absolutely positive.

22. This property can be thought of as the internal stickiness of a fluid. The rate of deformation of a fluid is directly linked to this parameter. a) specific weight

b) Viscosity

c) roughness coefficient d) compressibility

The pressure at the bottom of the river section 9.2 m deep was determined to be 97kPa. It was noticed however that the river has two layers of water having two different densities. The lighter one has a specific gravity of 1.02 and 7 m deep. 23. What is the specific weight of the denser liquid? 97000 = Po + 1.02(9810)(7) + s(9810)(2.2) S = 1.25 a) 1.75

b) 1.55

c) 1.45

d) 1.25

The gage pressure at a depth of 6 meters in an open tank of liquid is 47 kPa 24. What is the specific weight in Newton per cubic meter? 6y = 47000 y = 7833 N/m3 a) 7833

b) 8910

c) 7650

d) 9234

A cylindrical container, having a diameter of 4 meters and a height of 3 meters is inverted vertically with the open mouth facing down and submerged in water. 25. When the water has risen 30 cm inside the cylinder, how deep in meters is the open end (mouth) from the free surface? Vcylinder = (3.14)(22)(3) = 37.68 Vh2o = (3.14)(22)(0.30) = 3.768 Vair = 37.68 – 3.768 = 33.912 P1 = 101325 Pa P2 = (101325 x 37.68)/33.912 = 112583.33 – 101325 = 11258.33 Pa Pmouth = Psubmerged + Ph2o = 11258.33 + 9810(0.3) = 14201.33 Pa h = 14201.33/9810 h=1.448 a) 1.43

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b) 1.67

c) 1.53

d) 1.23