# How can a non flowing fluid have kinetic energy?

I am currently studying engineering thermodynamics from Cengel and Boles and there it is given that total energy of non flowing fluid per unit mass basis is given as $$e=u+\frac{v^2}{2}+gz$$.But if the fluid is non flowing how can it have macroscopic kinetic energy$$\frac{v^2}{2}$$?

• Does Cengel & Boles say the fluid is moving or stationary? Or is the expression given to cover all conditions and you have to make the correct assumptions? Jul 25, 2019 at 19:09
• That's is the definition of the specific energy. Depends on the thermodynamic conditions one or more parameter can take zero value.
– user14407
Jul 25, 2019 at 20:20
• "... the total energy of a flowing fluid …" "But if the fluid is non flowing ..." - see your problem there? But if the fluid is not flowing, $v = 0$ and its macroscopic kinetic energy per unit mass is $v^2/2 = 0$. So what exactly is the difficulty? Jul 25, 2019 at 22:25
• @alephzero It was a typo.That was given as energy of a non flowing fluid.I can upload the snapshot but it may violate copyright laws. Jul 26, 2019 at 4:55
• @AtulGautam uploading an image of that particular part will not violate copyright, but if you upload the whole book then you may well have an issue... The are hundreds of questions with snapshots of particular formulae, expressions, questions and diagrams on here.... Jul 26, 2019 at 7:38

A non-flowing fluid can have kinetic energy if it is moving as a rigid body. For example, a full water bottle (no room for flow inside) is thrown through the air without rotation.

However, this is not likely the intent of the author for this situation.

• This is coordinate dependent. If the coordinate system is stationary with respect to the controle volume, then yes your answer is applicable. If we choose the coordinate system attached to the CV, then the kinetic energy is zero.
– user14407
Jul 26, 2019 at 13:12
• Yes of course, but the question was how can it happen. Jul 26, 2019 at 17:39

The general equation for the specific energy of a fluid is $$e=u+\frac {v^2}{2}+gz$$. It doesn't matter that the question states that the fluid is non-moving. It's always best to start with the general equation and then the fact that the fluid is non-moving is one of the conditions you apply to that equation.
I think this is made clear in the comments of Solar Mike, user14407 and alephzero.

If you begin by altering a general equation before applying it (at least write it down!) you may run into difficulties later. (Also if this is a test question, removing the $$\frac {v^2}{2}$$ part of the formula is giving you a hint toward the answer.)