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While deriving the Bernoulli's equation, we write the change gravitational potential energy as $mg(h' - h)$ , say where $m$ is the mass and $h'$ and $h$ are the two heights. Why we don't consider the centre of mass in this case? I mean why we don't have this term written as $\frac{mg(h' - h)}{2}$. I feel I am having some problem in understanding some concept.

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    $\begingroup$ The center of mass of what? $\endgroup$ – PiKindOfGuy Feb 27 at 12:09
  • $\begingroup$ The change of height is from h to h', not h to the average height of h and h'. The difference of 13 and 17 (mean is 15) is 4, not 2. $\endgroup$ – helpme Feb 27 at 12:15
  • $\begingroup$ @helpme in what places do we use the centre of mass concept? I need some examples. $\endgroup$ – Akash Roy Feb 27 at 12:21
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    $\begingroup$ @AkashRoy: in what places do we use the centre of mass concept? The attractive force (gravity) between two uniform bodies, e.g. For Bernoulli, $\Delta h$ is the difference in height along a flowline. $\endgroup$ – Gert Feb 27 at 14:26
  • $\begingroup$ @Gert , I want to give you a situation . If we consider a U-shaped limb in which the left limb is filled with water upto a height say $h$ and the other has a height say $h1$. The limb has a valve in between to restrict any kind of flow of water. Say $h$>$h1$ . Then what should be the initial potential energy of the water in left limb? Considering $A$ as cross sectional area and the U-shaped limb is uniform. Assuming density is $d$. Say I will be opening the valve after sometime and the level equalises $\endgroup$ – Akash Roy Feb 27 at 14:42
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In the derivation of Bernoulli's equation we consider a small element of a fluid so that we can assume that all fluid particles in the element has same pressure, velocity etc. The potential of this small element is the one to be considered in Bernoulli's equation, the height of the element itself is neglected.

You asked what the potential energy in the left limb was, we cannot use Bernoulli's equation on such a large element we use it ideally between 2 points ; so you can determine potential by integrating the potential of each small element but you cannot apply this potential into the Bernoulli's equation.

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