What would happen if the same pump were used on the Moon?
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The same centrifugal pump has a flow rate of 100 cubic meters, a head of 100 meters, and a rotational speed of 2950 rpm on Earth. If this pump is used on the Moon, its rotation speed is also 2950 rpm. What will the parameters of the pump be? The head should be about 6 times that of Earth, right?
The head remains constant; head is defined as the energy gained per unit weight of liquid after passing through the pump, as the parameters related to gravitational acceleration have been canceled out. The volumetric flow rate remains unchanged, the mass flow rate remains unchanged, the weight flow rate decreases, and the shaft power decreases as well, because there is a factor of gravitational acceleration in the formula for shaft power. The outlet pressure remains unchanged, but the effect that this pressure can produce is greatly amplified.
The head is approximately equal to the square of the linear velocity at the outer circumference of the impeller divided by 2g. g has become 1/6 of its value on Earth. Does the head remain unchanged?
The Foam Super Edition really thinks ahead, hahaha
What the moderator said makes some sense as well, but I still think the head pressure remains constant; I need to find more evidence for this (it seems my foundation in theory isn’t solid enough; I’ve handed it all back to the teacher;P)
This post was last edited by 3983596_FPPZ on 2018-9-25 at 10:28. The moderator is right; the head is the energy difference per unit weight. Due to the term “unit weight,” g is included in the denominator, which helps to define this concept.:lol So, on Earth the head is 100 meters, while on the Moon it is 600 meters. However, on Earth these figures are nominal values. On the Moon, a head of 600 meters allows water to be raised to a vertical height of approximately 600 meters, which is equivalent to the potential energy associated with 100 meters on Earth. Therefore, the motor power and the absolute pressure at the pump outlet remain the same as those on Earth.
If possible, if there is an atmospheric pressure (otherwise cavitation would occur due to the lack of pressure), then W=ρ_liquid·g·H/Q. Here, W remains constant; the density of the liquid does not change, the flow rate does not change, and the pressure does not change either. What changes is the relationship between the head and the acceleration due to gravity – in other words, the head is 6 times greater, but the outlet pressure remains unchanged. The head corresponding to a pressure of 0.1 MPa (1 kilogram), as we often say, is approximately 10 meters. The conversion formula here is ρgH, with g being based on a value of 9.81. Therefore, all the pump can do is reach the pump outlet; what happens beyond the outlet is determined by the gravitational accelerations of the Earth and the Moon. Just for discussion. Thank you to the original poster for giving everyone these thought-provoking questions.