0

There is an engine pumping liquid, for example, water, through a pipe. One way to figure out the engine's power is to times the force with the velocity of the liquid. The other way is to calculate the kinetic energy in the liquid per unit of time. Which is correct?

There is a similar question Work power energy contradiction in hose pipe problem [closed]. But it was closed due to the check-my-work and homework policy. Some answers on the Internet support the first way: times the force with the velocity.

However, some answers on physics.stackexchange.com support the second way: the engine power is equal to the kinetic energy.

Which is correct?

I didn't give any example in this post because I didn't want this post to get the same fate as Work power energy contradiction in hose pipe problem [closed] although I think this post has educational value. If you want to find examples, click the link. However, welcome to give examples. Examples are always the best way to demonstrate the concept and also a method we learn things from the fundamentals. This is true not only in the physics field.

Response from knzhou

knzhou is the author of the picked answer of Power of water pump. knzhou claimed the power of the engine should equal to the kinetic energy in the liquid per unit of time.

The physically correct answer is 400 W. As for what was marked "correct" on that test question, I have no idea -- it depends on whether the test writer knew what they were doing. Actually, many JEE prep problems are incorrectly posed, and don't have any correct answers.

Backup for the similar question

Work power energy contradiction in hose pipe problem [closed] has been deleted. I failed to cache the page with archive.org. To let others see the example, I copy-pasted that question from the page cached by Google.

  • Author: Neel Narlawar

Question: An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?

Approach 1:

Force = vdm/dt = 2(2×100) = 400

Power= Force × Velocity = 800W

800 is the answer given in most of the books out there.

Approach 2:

Change in kinetic energy of water = 1/2 * m * v^2

Rate = 1/2 * m/t * v^2 = 1/2 * 200 * 4 = 400W

I am unable to find an error in either of the approaches so can someone tell me what's wrong in the second approach? Both the concepts seem correct, the former being related to the usual force power relation while the second uses kinetic energy work relation. Is there a problem in my understanding of the concept? Thank you!

Note: I have tried looking at other sources for clarification but nothing seems to answer my question properly.

IvanaGyro
  • 267
  • This question is conceptual and cannot be construed as "homework-type". You need not worry about it being closed due to that policy. Also, is there a reason why you think it has to be one method and not the other, as opposed to both being valid methods? – joseph h Jan 04 '23 at 22:50
  • If we define engine power in one way, there should be only one correct answer. It's not possible that we input the same energy but get different velocity of liquid if other settings are the same. – IvanaGyro Jan 05 '23 at 02:13
  • 1
    Mechanical power is change in kinetic energy per unit time. We also know that mechanical power is the product of force and velocity. Both are valid. That someone tried to do a calculation using both these definitions and got two different results does not change this. It just means that the calculations were done wrong. – joseph h Jan 05 '23 at 02:29
  • Sorry, I can't understand what you mentioned. Can you explain more clearlier? – IvanaGyro Jan 05 '23 at 02:37
  • I mean you can use both approaches: kinetic energy/time or $F.v$ Both methods will give the same results IF the calculations are done correctly. In the first question you linked, you did the calculations incorrectly and that’s why you get two different answers. Also, you should merge all your accounts into one. Go to meta.physics.stackexchange.com to see how. Cheers – joseph h Jan 05 '23 at 22:38
  • I only have one account. The other is not me. Did you mean other answers on the Internet are also wrong? – IvanaGyro Jan 07 '23 at 00:47
  • No. Like I said, the first question you linked was done wrong, and that's why there are two different answers. I had a look and the other answers you linked, and they appear correct. – joseph h Jan 07 '23 at 03:31
  • https://www.toppr.com/ask/question/an-engine-pumps-water-through-a-hose-pipe-water-passes-through-the-pipe-and/ – IvanaGyro Jan 08 '23 at 06:00
  • https://byjus.com/question-answer/an-engine-pumps-water-through-a-pipe-water-passes-through-the-pipe-and-leaves-it/ – IvanaGyro Jan 08 '23 at 06:00
  • These two have the same answers. Did you mean they are wrong? Can you point out clearer what concept is wrong? I cannot understand what you said. – IvanaGyro Jan 08 '23 at 06:02
  • I think the steps in the comment above and the step in this https://physics.stackexchange.com/a/700942/203452 are different. The later claim 1/2Fv is correct Fv is wrong. The former said Fv is correct. – IvanaGyro Jan 08 '23 at 06:07
  • "These two have the same answers." Yes, they do. Using $P=F\cdot v$ they both get 800 W. "Did you mean they are wrong?" No, they are right. "I think the steps in the comment above and the step in this are different." They may use different methods, but all get the correct answer $P= \frac 12 \rho A v^3$. "The later claim 1/2Fv is correct Fv is wrong. The former said Fv is correct." I'm not sure how you came to this conclusion. The question was why there is a missing $\frac 12$ factor on the equation. Cheers. – joseph h Jan 09 '23 at 02:05
  • Again, the very first question you linked above in your OP is what contains a wrong answer. Cheers. – joseph h Jan 09 '23 at 02:08
  • I can accept that OP contains a wrong answer, but why is the answer wrong? Sorry, I am so stupid that I cannot understand. I was not the opposite of your opinion. I just cannot understand. In this, there is an equation $P= \rho A v^3$, which is different from this. Why 800W can be the result of $P= \frac 12 \rho A v^3$? – IvanaGyro Jan 10 '23 at 03:21
  • If you consider momentum as energy of a body then force is equivalent to power but in spatial term. Power is energy exchanged in temporal sense, and kinetic energy is equivalent to power because only it can exchanged by a body. Also kinetic energy is average force if acceleration is square of speed. – Neil Libertine Jan 17 '23 at 10:09

0 Answers0