Thermo - Difference between Potential and P/rho

[LT72884]

Ok, I am stuck on understanding when to use a part of an equation and when not to.

e = (P/rho) + (Ke/m) + (Pe/m) where P is pressure, Ke and Pe are Kinetic and potential energy divided by mass.

Ok, if we look at P/rho and P is defined as rho(g)(h) then the rho's divide out and we are left with the exact same thing as Pe/m or just gh.

My question. How do i know when to use P/rho or Pe/m in a question? yes, i end up with same answer, but my procedure is incorrect.

Ex
Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power. The free surface of the upper reservoir is 45 m higher than that of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine mechanical power that is converted to thermal energy during this process due to frictional effects.

My procedure is as follows:

e(mechanical) = P/rho. i chose this because there is no change in velocity or PE. granted there is a height difference but P/rho yields (gravity)(height) which is the same as Pe/m.

i then plug e(mech) into `E = `me which yeilds (0.03m3/sec)(9.81m/s2)(45m) = 13.2 kW after a conversion.

here is my issue. i end up with same answer even though i WAS NOT SUPPOSED TO USE (P/rho) I was SUPPOSED TO USE (Pe/m). Yes, i got lucky but why is my approach wrong? To me, in my mind, i said to myself " P/rho is for fluids and is a flow rate. Water is a fluid and it is flowing at some rate into the turbo, producing some amount of power in kj/kg of water"

Thanks. I am just trying to understand how to decifer the problems correctly

 

[Chestermiller]

There is a change in potential energy, as you pointed out. P is the pressure, and it is the same at the surface of the upper reservoir as at the surface of the lower reservoir (atmospheric pressure) So the change in P/rho is zero between these two points. The flow velocities are also zero at the reservoir surfaces.

Your mistake is assuming that the change in P/rho is the same as gz, which it is not (unless the system is hydrostatic).

 

[LT72884]

There is a change in potential energy, as you pointed out. P is the pressure, and it is the same at the surface of the upper reservoir as at the surface of the lower reservoir (atmospheric pressure) So the change in P/rho is zero between these two points. The flow velocities are also zero at the reservoir surfaces.

Your mistake is assuming that the change in P/rho is the same as gz, which it is not (unless the system is hydrostatic).

I see what you are saying, and the numbers come out exatly the same regardless if i use Pe or P/rho. Thats what is throwing me off. I have tried it both ways with exact same numbers. The solutions manual did it a different way and ended up same numbers haha.

Im not trying to contradict you or argue. Just trying to learn and connect the dots

 

[Chestermiller]

I see what you are saying, and the numbers come out exatly the same regardless if i use Pe or P/rho. Thats what is throwing me off. I have tried it both ways with exact same numbers. The solutions manual did it a different way and ended up same numbers haha.

Im not trying to contradict you or argue. Just trying to learn and connect the dots

The answer is coming out correct because you made an error by replacing $g\Delta z$ by $\frac{\Delta P}{\rho}$. In this problem $\Delta P/\rho$ is zero. So, basically, you made two compensating errors.

 

[LT72884]

The answer is coming out correct because you made an error by replacing $g\Delta z$ by $\frac{\Delta P}{\rho}$. In this problem $\Delta P/\rho$ is zero. So, basically, you made two compensating errors.

So what you are saying is... i got lucky haha.

Ok, i need to re-read your post and the question and follow it the right way.

So what IS P/rho used for then?

Thanks

 

[Chestermiller]

So what you are saying is... i got lucky haha.

Ok, i need to re-read your post and the question and follow it the right way.

So what IS P/rho used for then?

Thanks

It is used in cases where the pressure actually is different between the two points you are considering.