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Fluids

Problems

(See below for answers)

1. Calculate the pressure created by ocean water (density = 1.025 g/cm 3), at a depth of

11.0 Km.2. Oil having a Reynold's number of 2500, specific gravity of 0.86, and viscosity at 20

degrees Celsius of 084 poise flows through a 5 inch diameter pipe.

a. What is the maximum average speed of the oil without turbulence?

b. How would the average speed change if the temperature of the oil was to rise?

3. The pressure in a tightly closed building is the same as outside, 988 mbars. The

windows in this building are 1.2 m by 2.15 m. If a 23 m/s gust suddenly blows across

the face of this building

a. What pressure difference across these windows does the wind create?

b. What force acts on each window?

4. A tennis ball has a density of 0.084 g/cc and a diameter of 3.8 cm. What force isrequired to submerge the ball in water?

5. Describe the relationship between fluid flow through a hole in a tank and the area of

the hole.

6. Water (62 lb per cubic foot) is flowing into a 1 ft diameter pipe entrance at 100 ft/sec

and 200 lb per square foot. Neglecting friction, what is the velocity and pressure at the 2

ft diameter exit?

7. A hot-air balloon consists of a basket and a 2.18x103 m3 envelope having a

combined weight of 2.45 kN. What should be the temperature of the air used to inflate

the envelope to provide a net lift of 2.67 kN? The surrounding air is 20C, has a weight

of 11.41 N/m3, has an average molecular mass of 0.028 kg/mole, and is at a pressure of

1.0 atm.

8. Air pressure is 1x105 N/m

2, air density is 1.3 kg/m

3. How fast must air be blown

across the top of a straw rising 0.10 m above the water in a glass, to make the water

rise half way up the straw?

9. At what depth will a submersible experience 10.0 N per square millimeter pressure?

10. What lift does Bernoulli's principle predict for a wing of area 78 m2 if the air

passes over the top surface at 260 m/s and the bottom surface at 150 m/s?

11. In liquid A a body floats with 9/10th of its volume immersed, while in liquid B it floats

with 3/5th of its volume immersed. Compare the densities of the liquids.12. At what depth will a diver experience 4 atmospheres pressure?

13. A 10 cm radius water main connects a hilltop reservoir to houses in a valley. The

supply line to each house is 1cm radius. If only one house has water running, the flow is

0.25 L/s. What is the velocity of the water in the main and in the supply pipe? Where is

the pressure highest?

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14. The average velocity of flow in a river is 1.1 m/s where it is 0.5 m deep and 5 m

wide. (A) What is its flow rate? (B) Another part of the river it is 2 m wide and 1 m deep.

What are the flow rate and the average velocity?

15. A thin hollow sphere of mass 0.500 kg and diameter 0.180 m is filled with alcohol (=

806 kg/m3). Find the acceleration of the sphere after it is released under water.

16. A helium balloon tied by light string to an armrest in a stationary train hovers in the

still air. When the train accelerates forward, what does the balloon do?

17. An empty ship is moored to a loading dock then filled with Styrofoam blocks. Does

the ship float higher in the water or sink deeper?

18. An air bubble has a diameter of 1.0 cm at a depth of 18.0 m. What will the bubble's

diameter at the surface? The temperature change is negligible.

Answers

1. Given the density of ocean water is uniform at 1.025 g/cm3= 1025 kg/m

3, the weight

of water is 1.005 x 104N/m3.

The pressure of water at this depth (the weight per square meter) would be(1.005 x 10

4N/m

3)(11.0 x 10

3m) = 1.105 x 10

8N/m

2= 1.105 x 10

5kPa

(about 16000 psi)

For solutions to all the problems on this page clickhere.

7.

The lifting force required = 2.45 kN + 2.67 kN = 5120 N

weight of 2.18x103m3air at 20C = 11.41 N/m3* 2.18x103m3= 24874 N

Archimedes' principle informs us of buoyancy:

Lifting force = [weight of 2.18x103m

3air at 20C] - [weight of 2.18x10

3m

3hot air]

Substituting calculated values,

5120 N = 24874 N - [weight of 2.18x103m3hot air]

weight of 2.18x103m3hot air = 19754 N

From the ideal gas law, PV is proportional to WT

Since P and V are constant,

W1T1= W2T2

or

(W1/ W2)T1= T2

(24874 N / 19754 N)(293 K) = 369 K

365 K = 96C

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