Problem-setsfluid-mech.docx

Problem Sets Midterm A certain gas weighs 16𝑁⁄ 3 at a certain temperature and pressure. What 𝑚 are the values of its density, specific volume, and specific gravity relative to air weighing 12 𝑁⁄ 3 𝑚 a.) For Density 16𝑁⁄ 3 𝜌= = 𝑚𝑚 𝑔 9.81 ⁄ 2 𝑠 𝑘𝑔 𝛾

𝜌=1.63

𝑚3

b.) For Specific Volume 1

1 𝑘𝑔 𝜌 1.64 ⁄ 3 𝑚 3

V= =

V=0.61 𝑚 ⁄𝑘𝑔 c.) For Specific Gravity 𝑁 𝛾𝑠 16 ⁄𝑚3 S= = 𝑁 𝛾𝑤 12 ⁄ 3 𝑚

S=1.33

The density of alcohol is 790𝑘𝑔⁄𝑚3 . Calculate its specific weight, specific gravity and specific volume.

a.) For Specific Weight 𝑘𝑔 𝛾=𝜌. 𝑔=790 ⁄ 3 (9.81 𝑚⁄ 2 ) 𝑚 𝑠 7749.9𝑁⁄ 3 𝑚 𝛾= =7.75𝑘𝑁⁄ 3 1000 𝑚

b.) For Specific Gravity 7.75𝑘𝑁⁄ 3 S= = 𝑘𝑁 𝑚 =0.79 𝛾𝑤 9.81 ⁄ 3 𝑚 𝛾

c.) For Specific Volume 1

1 =1.27x10-3m3/kg 𝑘𝑔 𝜌 790 ⁄ 3 𝑚

V= =

At a depth of 8km in the ocean the pressure is 82.26Mpa. Assume the specific weight at the surface to be 10.10kN/m3 and that the average bulk modulus is 2344MPa for that pressure range, a.) what will be the change in specific volume bet. that at the surface and at the depth? b.) what will be the specific volume at that depth? c.) what will be specific weight at the depth?

𝑘𝑁 𝛾 10.10 ⁄𝑚3 (1000) a.) 𝜌1 = = 𝑔 9.81𝑚⁄ 2 𝑠

𝜌1 =1029.6

𝑘𝑔⁄ 𝑚3

𝜌2 =𝛾. ℎ=10.10𝑘𝑁⁄ 3 (1000)(8000) 𝑚 𝜌2 =80.80MPa

∆𝑉=3.3x10-5m3/kg 1

1 𝑘𝑔 𝜌 1043 ⁄ 3 𝑚

b.) V= =

V=9.5x10-4m3/kg 𝜌 82.26(106)

c.) 𝛾= = ℎ

8000

=1039.9N/m3

If the dynamic viscosity of water at 20℃ is 1x10-3𝑁. 𝑠⁄ 2 , what is the 𝑚 kinematic viscosity in the English unit. V=

𝑘𝑔.𝑚.𝑠⁄ 𝑚2 𝑠 2 𝑘𝑔 1000 ⁄ 3 𝑚

1𝑥10−3

𝑚 V=1x10-6

2

𝑠

3.28𝑓𝑡 2

x(

1𝑚

)

V=1.08x10-5 ft2/s

The kinematic viscosity of 1ft2/s is equivalent to how many strokes? (1 strokes= 1cm2/s). 1 inch = 2.54 cm 1𝑓𝑡 2 (12𝑖𝑛𝑐ℎ)2 (2.54𝑐𝑚)2 𝑠

x

1𝑓𝑡 2

x

1𝑖𝑛𝑐ℎ2

=929.03 strokes

A volume of 450 liters of a certain fluids weighs 3.50kN. Compute the mass density. (1m3=1000L). 450 L x 𝛾

𝜌= =

1𝑚3 1000𝐿

=0.45 m3

3.5(1000)

𝑔𝑣 9.81(0.45)

𝜌=729.84 kg/m3

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

𝑃

=

280𝑘𝑃𝑎

𝛾𝑤 .𝑆 (9.81 𝑘𝑔 )(0.80) 𝑚3

h=35.68 m

The pressure of the air inside a tank containing air and water is 20kPa absolute. Determine the gage pressure at a point 1.5m below the water surface. Assume STD. ATM. Pressure. Pabs=20+1.5(9.81) Pabs=34.72 kPa Pabs=Patm ± Pgage 34.72=101.325 ± Pgage Pgage=66.61 kPa

A vertical circular gate 1m in ∅ is subjected to pressure of liquid of specific gravity 1.40 on one side. The pipe surface of the liquid is 2.60m above the upper most part of the gate. Calculate the total force on the gate and the location of the center of pressure. P=𝛾𝑤 .ħ.A P=9.81(1.4)(3.1)𝜋(0.5)2 P=33.44 kN

𝐼𝑔 (0.5)2

e= =

=0.02m

𝐴ȳ 4(3.1)

A horizon tunnel having a diameter of 3m is closed by a vertical gate. When 1

3

2

4

the tunnel is a.) Full, b.) full of water, determine the magnitude and location of the total force. 1

P=𝛾𝑤 .ħ.A=9.81[

a.) Full 2

𝜋(1.5)2 2

] (0.64)=22.19 kN

4𝑟 4(1.5)

ħ= = 3𝜋

3𝜋

ħ=0.64m 1.5+0.64

b.) ħ=

=1.08m

2 𝐼𝑔 0.1098(1.5)4

e= = 𝐴ȳ

3.53(0.64)

e=0.25m

P=𝛾𝑤 .ħ.A=9.81[

3𝜋(1.5)2 4

] (1.08)=56.17 kN

hp=ħ+e=0.64+0.25=0.89m

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. 𝐼𝑔

8(3)(3)2 175 2(3)(3) (1.8) 3

P=𝛾𝑤 .ħ.A

e= =

P=9.81(1.8)(2/3)(3)(3)

hp=ħ+e=1.8+0.34=2.14m

P=105.95 kN

∴ Location of P at 2.14m fr. the ws.

𝐴ȳ

=0.34m

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. P=𝛾𝑤 .ħ.A

Pf=𝜇𝑁

P=9.81(1.6)(10.8)

Pf=0.2(508.55)

P=508.55kN

Pf=101.71kN

ƩFv=0 Pv=𝛾𝑤 +Pf Pv=18.0+101.71 Pv=119.71kN

The upper edge of a vertical trapezoidal gate is 1.60m long and plush w/the water surface. The two edges are vertical and measure 2m & 3m, respectively. Calculate the force and location of the center of pressure on one side of the gate.

How far below the water surface is it necessary to immerse a vertical plane surface, 1m square, two edges of w/c are horizontal so that the center of pressure will be located 2.50m below the center of gravity? ℎ2

e= 12 ħ

0.025=

12 12ħ

ħ=2.83m

− 0.5

The width of the Fig. G normal to the paper is 3m. What vertical force must be applied at “a” to prevent collapse when h=6m? Neglect weight of the gate, what is stress in strut bc?

The gate of Fig. H is hinged at A and rest on a smooth surface face at B. The gate is circular having a ∅ of 3m. Determine the value of the vertical force P that will open the gate at B.

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

0.5ℎ

sin 60

𝐼𝑔 ℎ2

→eq. 1

e= =

=

𝐴ȳ 12ȳ

ℎ sin 60 0.5ℎ 12(sin 60)

P=𝛾𝑤 .ħ.A P=9.81(

ℎ

sin 60

) (2.6) (

0.5ℎ

sin 60

) → eq. 2

ƩMA=0 P(

ℎ

sin 60

−

0.5ℎ sin 60

5.95h3=112.5 h=2.66m

+

0.19ℎ sin 60

)=22.5(0.5)

→ eq. 3

End AB of Fig. N has a section in the shape of a quadrant if the tank has a length of 3m. Determine the total force acting on the end AB.

The hemisphere dome of Fig. P surmounts a close tank containing a liquid of S.G.=0.75 the gate indicates 60kPa. Determine the tension holding the bolts in place. P=𝛾𝑤 .S.h

T=𝛾.V.S=9.81(39.23)(0.75)=288.63kN

60kPa=9.81kN/m3(0.75)(h) h=8.15 V=𝜋𝑟 2 .h-

4𝜋𝑟 3 6

V=𝜋(1.5)2 (8.15) − V=39.23m3

4𝜋(1.5)3 6

An open cylindrical wood stave tank contains three liquids with specific gravities 2, 3, and 4, respectively. The depth of the bottom liquid is 2m, while the other two has a depth of 1m each. If the diameter of the container is 2m determine the tension in the top and bottom hoops which are holding the container in place.