Q: Given the equation $$P_{atm} = \rho g h_{max}$$, what would be the maximum height $$h_{max}$$ for mercury ($$\rho_{Hg} = 13,600 kg/m$^3) at $$standard atmospheric pressure?

$$h_{max} = \frac{101,325}{13,600 \times 9.8}$$

Q: Suppose the well is located at a high altitude where atmospheric pressure is only 80,000 Pa. What is the new maximum height $$h_{max}$$ that water can be lifted by the vacuum pump? (Use $$\rho = 1000 kg/m$^3$$, $$g = 9.8 m/s$^2.)$$

$$h_{max} = \frac{80,000}{1000 \times 9.8}$$

Q: If the liquid being pumped is oil with a density of $$850 kg/m$^3$$, what is the maximum height $$h_{max}$$ that can be lifted at sea level atmospheric pressure? (Use $$P_{atm} = 101,325 Pa$$, $$g = 9.8 m/s$^2.)$$

$$h_{max} = \frac{101,325}{850 \times 9.8}$$

Question 1

In the diagram, a vacuum pump is used to lift water from a well. What is the maximum theoretical height $$h_{max}$$ that water can be lifted by a perfect vacuum at sea level atmospheric pressure? (Assume $$P_{atm} = 101,325 Pa $$and the density of water $$\rho = 1000 kg/m$^3.)$$

Accepted Answer

$$h_{max} = \frac{P_{atm}}{\rho g}$$

Question 2

If the atmospheric pressure decreases due to a weather change, what happens to the maximum height that water can be lifted by the vacuum pump?

Accepted Answer

The maximum height decreases.

Question 3

Why is it not possible to lift water higher than approximately 10 meters using a vacuum pump at sea level?

Accepted Answer

Atmospheric pressure can only support a column of water up to a certain height.

Question 4

Given the equation $$P_{atm} = \rho g h_{max}$$, what would be the maximum height $$h_{max}$$ for mercury ($$\rho_{Hg} = 13,600 kg/m$^3) at $$standard atmospheric pressure?

Accepted Answer

$$h_{max} = \frac{101,325}{13,600 	imes 9.8}$$

Question 5

If the vacuum pump in the diagram develops a leak and air enters the system, what is the most likely immediate effect?

Accepted Answer

The maximum height the water can be lifted decreases.

Question 6

Suppose the well is located at a high altitude where atmospheric pressure is only 80,000 Pa. What is the new maximum height $$h_{max}$$ that water can be lifted by the vacuum pump? (Use $$\rho = 1000 kg/m$^3$$, $$g = 9.8 m/s$^2.)$$

Accepted Answer

$$h_{max} = \frac{80,000}{1000 	imes 9.8}$$

Question 7

Which of the following best explains why a pump placed at the bottom of the well (submerged in water) can push water to a much greater height than a vacuum pump at the surface?

Accepted Answer

A submerged pump pushes water, which is not limited by atmospheric pressure.

Question 8

If the density of the liquid being pumped is halved, what happens to the maximum height $$h_{max}$$ that can be achieved by the vacuum pump?

Accepted Answer

The maximum height doubles.

Question 9

A student claims that increasing the power of the vacuum pump will allow water to be lifted higher than 10 meters. Is this claim correct? Why or why not?

Accepted Answer

No, because the maximum height is limited by atmospheric pressure, not pump power.

Question 10

If the liquid being pumped is oil with a density of $$850 kg/m$^3$$, what is the maximum height $$h_{max}$$ that can be lifted at sea level atmospheric pressure? (Use $$P_{atm} = 101,325 Pa$$, $$g = 9.8 m/s$^2.)$$

Accepted Answer

$$h_{max} = \frac{101,325}{850 	imes 9.8}$$

Fluid Mechanics: Pumping and Air Pressure

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