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Apr 16 Discussions Hydraulic equilibrium Pressure and depth

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Apr 16 Discussions. Hydraulic equilibrium Pressure and depth. Q1: Pressure. - PowerPoint PPT Presentation

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Page 1: Apr 16 Discussions

Apr 16 Discussions

Hydraulic equilibrium

Pressure and depth

Page 2: Apr 16 Discussions

Q1: Pressure

• Which of the following situations represents the greatest pressure?(Recall that P=F/A)A) the weight of 10 tons on 1.0 square meterB) the weight of 100 kg on 100 square centimetersC) the weight of 1.0 kg on 1.0 square centimeterD) the force of your pencil (say, 1 Newton) on its point

Page 3: Apr 16 Discussions

Q1: Pressure

• Which of the following situations represents the greatest pressure?(Recall that P=F/A)

D) the force of your pencil (say, 1 Newton) on its point:

P = F/A = (1 N)/(1 mm)2=106 Pa?

Page 4: Apr 16 Discussions

P1: Hydraulic lift• Because of your physics knowledge, you are

asked to advise on the design of a hydraulic lift in a car repair garage that must be able to hold a 3 ton vehicle. The piston that supports the vehicle platform has a diameter of 30 cm and a mass of 1 ton. A pump pushes fluid into the piston’s cylinder through a 3 cm diameter pipe.

• What is the minimum pressure that needs to be exerted on the bottom of the piston to support the vehicle?

• What is the minimum force that the piston on the pump needs to exert on the fluid entering the pipe to hold the vehicle in place?

F

W

Use g=10 m/s2

Page 5: Apr 16 Discussions

Hydraulic lift-2

• The force the large piston supports is its own weight plus the weight of the vehicle (4 ton=4000 kg):Wtot=mtot g = (4000 kg)(10 m/s2)=40 kNP = Wtot/A = (40 kN)/( (0.15 m)2)=5.7x105 Pa

• The small piston needs to be able to sustain this pressure. The force necessary is:F = P•Apipe= (5.7x105 Pa)( (0.015 m)2)=400 N

• Another way to see this is that since the pipe has 1/100 of the cross-sectional area of the large piston, the necessary force is 1/100 of the total weight (40 kN)/100 = 0.4 kN = 400 N. Note that this is about 90 lb.

Page 6: Apr 16 Discussions

P2: Pressure below the surface • You are designing a deep sea exploration vessel. From P201

you recall that pressure increases as you go beneath the surface of the ocean, the pressure of the atmosphere at sea level is approximately 100 kPa, and the density of water is 1025 kg/m3.

• What pressure does your vessel need to withstand if it is to reach at the deepest part of the ocean (about 10 km deep)?

• Metals such as titanium have compressive strengths over 500 MPa, but metals are not transparent and you want your vessel to have a window. Glass has a compressive strength of approximately 50 MPa. What do you recommend about providing the ability to pressurize your vessel?

Page 7: Apr 16 Discussions

Pressure below the surface-2

• The pressure is

gd+Po=(1025 kg/m3)(10 m/s2)(10000m)=108 N/m2=100 MPa

(about 1000 atm), since every 10 meter of depth is equivalent to an extra atmosphere-equivalent of pressure.

• Your window will not be able to withstand this unless you pressurize your vessel to over 50 MPa (so the pressure difference is less than 50 MPa). It would be best to do this gradually on the way down (you don’t want it to be 50 MPa internally at the surface, since your window would blow out!)

Page 8: Apr 16 Discussions

P3: Water Supply 1

• Water supplies are pressurized for several reasons, but to provide sufficient flow, they need to be pressurized. A city is to have a 24-inch (60 cm) diameter supply line that is to provide 20 million gallon/day of flow (3.75 liter = 1gallon).

• What flow velocity is needed?• According to Bernoulli’s Law, what is the minimum

pressure drop between the storage tank and the entrance to the pipe needed to produce this velocity?

Page 9: Apr 16 Discussions

Water supply-1

• The volume flow rate=20x106 gal/day•(3.75 liter/gal)•(1 m3/1000 liter) =75x103 m3/day•(1 day/86400 s)=0.87 m3/s=A•vflow=(0.30 m)2•vflow

vflow=3.1 m/s

• The pressure must obey: Ptank=Ppipe+(1/2)v2

• The pressure drop is: Ptank-Ppipe=(1/2)v2=(1/2)(1000 kg/m3)(3.1 m/s)2

=5000 Pa

Page 10: Apr 16 Discussions

Water supply-2

• The end of the city water supply pipe is 60 meter above the entrance to the pipe. A pump provides the necessary pressure. What pressure is required at the exit of the pump for the water to reach city, if the pressure at the city end needs to be at least 200 kPa?

Page 11: Apr 16 Discussions

Water supply-2

• The end of the city water supply pipe is 60 meter above the entrance to the pipe. A pump provides the necessary pressure. What pressure is required at the exit of the pump for the water to reach city, if the pressure at the city end needs to be at least 200 kPa?

• Ppump+ghpump=Pcity+ghcity

Ppump=Pcity+g(hcity-hpump)=(200 kPa)+(1000 kg/m3)(10 m/s2)(60 m)=800 kPa