# Bernoulli equation on moving cart

Let’s say I have moving cart with velocity U .

And water from pipe flows at velocity V with respect to ground

The question asks to use Bernoulli equation , pressure is atmospheric in the top

If I use Bernoulli equation relative to the cart I get hight h different than that when being on ground . Why ?

V3 = 2V relative to Cart Using Bernoulli equation I can easily get the hight

But when moving to the ground frame and using Bernoulli I get different answer

Which one is true ?  • Added , my question is , should I say h =( 2V + U )^2 /2g Jan 31, 2021 at 20:57
• I should calculate h , but if I move to the ground frame I get different answer Jan 31, 2021 at 20:57
• I wrote the relataive velocities and conservation of mass gives the this velocity from the exit pipe Jan 31, 2021 at 20:59
• Write out the Bernoulli eq. and substitute each case (U=0, U ≠ 0), carefully equating the in-cart condition at a known point (surface is a good one) with the exit jet condition. Pay particular attention to the cart velocity term. Jan 31, 2021 at 21:11
• @Mather -- why should V3 = 2V ?? Jan 31, 2021 at 22:39

I'm going to post this as an answer even though it isn't one.

I was hoping the OP would post the actual substitutions into the Bernoulli equation to support their observation that you measure a different height of water if you watch the car roll past compared with sitting on the car. (which is ridiculous, of course).

Notice that so far, three of the four items of influence on the car have been accounted for. The forth one, the force in red, hasn't appeared yet. If the cart velocity is zero, the force is not doing any work on the system. But if the cart is moving, then there is work being done on the cart system.

Now if the fluid volume in the cart is steady and the velocity of the cart is steady, that work is being done on the fluid transiting the cart, and must be accounted for.

Credit due to the OP for sticking with this question, because as posed, it is a really subtle and well disguised paradox involving one moving part.

• The force is not known we need to calculate it in the second sub- question . So it’s unlikely to use it in the first place since it comes after this question . Feb 1, 2021 at 6:18
• Also the question says that there is no energy loss Feb 1, 2021 at 6:23
• I got the mark I was right you can’t use Bernoulli from the ground frame only from the cart frame and I got Almost full mark in this question ( 28/30 ) Feb 9, 2021 at 6:11

I don't know where you went wrong, but the key will be that all velocities must be measured from the same frame of reference. If your frame of reference is the cart, then velocities into & out of the cart need to be expressed relative to it. If the reference is the ground, velocities are calculated that way, but you need to add/subtract cart speed to pipe velocities. But the reference doesn't change the math. Moving toward a pipe will be a higher velocity of fluid flowing in or a lower velocity of fluid flowing out. No matter the reference, the velocities through the pipe and into the cart must be the same. Find where your model makes them different and that's your error.

• No error in my calculation , all speed relative to the car are right . But Bernoulli equation is not the same . Feb 4, 2021 at 15:13
• Meaning that Bernoulli equation is not right because it’s not invariant when moving from one reference frame to another Feb 4, 2021 at 15:15
• @Mather, no, by definition your numbers can't be correct, or else they would be the same. I can't read your notation so I can't figure it out for you. But engineering is just not dependent on the frame of reference. The physics in play is on the materials, not on the methods of measurement. Feb 4, 2021 at 20:36
• But this is not model I designed it’s simply an exam question in fluid mechanics Feb 4, 2021 at 20:54
• The relative speed is correct given the speed in the ground frame in the picture Feb 4, 2021 at 20:55