# I Confusion: Internal energy u vs enthalpy h

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1. Dec 1, 2016

### Imolopa

I've been given the following relations where as I understand it subindex 2 equals subindex e and subindex 1 equals subindex i:

***EDIT***

More accurately
subindex 1; initial state of a control mass
subindex 2: end state of a control mass (end state is simply state 2 in the problem at hand)

subindex CV for work W and heat Q denotes what work W and work Q passes through the boundaries of what is defined as the control volume CV in the situation.

***END EDIT***

u2 = he
and
u 1 = h i

Relation 1:
1(Q - W)2 = 1Q2 -1W2 = m [ {u2 - u1 } + ( 1/2 ) * [ { v 2 2 } - { v 1 2 } ] + g {Z2 - Z1} ]

Relation 2:
(Q - W)CV = QCV - WCV = ∑ m e [ ( he + ( 1/2 ) * { v e 2 } + gZe ] - ∑ m i [ ( hi + ( 1/2 ) * { v i 2 } + gZi ]

Now in thermodynamic tables we have different columns for internal energy u and enthalpy h with different values, iow. h and u are different for the same state..

However the above relations seem to contradict the previous statement:

So what is correct? And what is happening here?

Last edited: Dec 1, 2016
2. Dec 1, 2016

### Staff: Mentor

Please describe in words the kinds of systems relation 1 and 2 respectively apply to, and tell the different way in which W in each equation is defined.

3. Dec 1, 2016

### Imolopa

I believe my edit in orginal post describes the situation a bit better now.

I can't edit the original post anymore, so adding here:

subindex e denotes the exit point at the control volume CV (where the working fluid exits, or more precisely the exit state of the working fluid in terms of enthalpy h, velocity v and height Z)
subindex i denotes the enter point at the control volume CV

m denotes the mass flow

4. Dec 1, 2016

### Staff: Mentor

I'm still having trouble understanding what you are asking. Are you asking why one equation uses enthalpy and the other equation used internal energy? Or are you asking what the enthalpies and internal energies mean in your tables? You do understand that the values in the tables are independent of whether you are considering an open system or a closed system, correct? You still haven't answered the questions I asked in my previous post.

5. Dec 1, 2016

### Imolopa

Yes Im wondering why one equation uses enthalpy and the other internal energy.

I'm not sure I'm answering some of your questions, but as I understand it:
The first equation is the energy equation that refers to an instantaneous process, where the mass flow is steady state, the whole equation is equal to the derivative of the energy (in other words the instantaneous change in energy).

The second equation refers to a transient process and the rightmost part of the equation is equal to the energy equation: E2 - E1

6. Dec 1, 2016

### Imolopa

I believe I found the answer: when the difference in volume and pressure is zero the difference in internal energy will equal the difference in enthalpies

7. Dec 1, 2016

### Staff: Mentor

This is not correct. Also, see my next response.

8. Dec 1, 2016

### Staff: Mentor

Actually, you have it backwards. The first relation applies to a closed system, in which no mass enters or leaves. The quantity m in the equation is the (constant) mass in the system. W includes all the work being done by the system on the surroundings.

The 2nd relation applies to an open system, in which mass is entering and/or leaving the system (control volume). It applies to steady state, where the internal energy within the system itself is not changing with time. Wcv is not the entire work being done. The total work is subdivided into two parts, the work done to force material into and out of the control volume plus all other work (the latter called shaft work or Wcv). The part of the work done to force material into and out of the control volume is combined with the internal energy of the material entering and exiting the control volume to give the enthalpy of the material entering and exiting the control volume. Per unit mass, this is hi and he. So the enthalpies present in the 2nd relation include part of the work (in addition to the entering internal energy).

I suggest you go back to your textbook and study more carefully the manner in which the open system version of the first law of thermodynamics is derived.