When is ΔH = ΔU in Thermodynamics?

[Ronie Bayron]

This process of yours is better describe this way(p-v-t diagram). The two red lines are isotherms. Superheated steam @ 1 bar is raised to 1.6 bar at constant volume. This would mean one thing, you add up heat to the steam to raise its pressure at constant volume(obviously, since.transition is between 2 isotherms)
Applicable analysis would be ΔH = ΔU, since PΔV or flow work is zero.

The values @1.6 bar & 1.6959 m³/kg bar are in between highlighted values below

 

[Chestermiller]

View attachment 98379
This process of yours is better describe this way(p-v-t diagram). The two red lines are isotherms. Superheated steam @ 1 bar is raised to 1.6 bar at constant volume. This would mean one thing, you add up heat to the steam to raise its pressure at constant volume(obviously, since.transition is between 2 isotherms)
Applicable analysis would be ΔH = ΔU, since PΔV or flow work is zero.

This is not correct. The definition of ΔH is $\Delta H=\Delta U+\Delta (PV)$, not $\Delta U+P\Delta V$. So, in this case of constant volume, $\Delta H=\Delta U+V\Delta P$. The final internal energy and enthalpy are 2835.8 kJ/kg and 3107.1 kJ/kg, respectively, and the final temperature is 316.8 C. Even in the data you reported to me in your own post # 30, H and U differed from one another by $PV$ = 169.6 kJ/kg. Go back to whatever source you can find, and obtain the value of U at 1.6959 m^3/kg and 1.6 bars. You will find that the value is 2835.8 kJ/kg, as I indicated. So, for our constant volume change,

$\Delta H=3107.1-2675.8$ = 431.3 kJ/kg

$\Delta U = 2835.8 - 2506.2$ = 329.6 kJ/kg

So, $\Delta H$ and $\Delta U$ are not equal for our constant volume change (as also confirmed by my steam tables, which does give the U values).

The difference between these values of $\Delta H$ and $\Delta U$ is $V\Delta P$ = 1.6959 m^3/kg x 60 kPa = 101.7 kJ/kg

Now, Ronie, I'm counting on you to determine the internal energy at that final state so that we can finally reach consensus.

Chet

 

[Ronie Bayron]

Yes, I see the flaws now and you are right.

This is the correct equation for isochoric compression (but do you agree on this?)
₁Q ₂= ΔU

Where in Q (heat) can be equal to ΔU, ΔKE, W, ΔH but not limited

Instead of
ΔH = ΔU (erroneous eq'n by definition in this case), because equation of state for internal energy would be u = h - pv.

Question: ΔH = ΔU is there a case (an actual one) describing this equation? What insights can you infer on this simple expression?

 

[Chestermiller]

Yes, I see the flaws now and you are right.

That's it? That's all you have to say to me after all the hours of my valuable time and effort I put in trying to figure out a way of explaining this in a way that resonates with you? Are you aware that being a Mentor at Physics Forums is done strictly on a volunteer basis by people who just want to help other members? I don't get paid for this. I am very disappointed in your unappreciative response. Growing up, I was taught better manners than this.

This is the correct equation for isochoric compression (but do you agree on this?)
₁Q ₂= ΔU

Yes. This is correct in the typical case where significant KE and PE changes are absent.

Where in Q (heat) can be equal to ΔU, ΔKE, W, ΔH but not limited

The more general form of the first law applicable to a closed system is $$\Delta U+\Delta (KE)+\Delta (PE)=Q-W$$

I don't feel like answering you last question.

 

[Ronie Bayron]

That's it? That's all you have to say to me after all the hours of my valuable time and effort I put in trying to figure out a way of explaining this in a way that resonates with you? Are you aware that being a Mentor at Physics Forums is done strictly on a volunteer basis by people who just want to help other members? I don't get paid for this. I am very disappointed in your unappreciative response. Growing up, I was taught better manners than this.

Oh, my apology. Chet. Actually, we have the same effort exerted just for the sake of good and educative argument and this does turn out, best in the interest of OP and readers. I am sorry if you expected a little extra manners than the concede and I haven't been able to pamper it a little extra gratitude. Consider this the extra perhaps.
I don't know if you agree on this, technical guys almost always converse like computer algorithm.

 

[Chestermiller]

Oh, my apology. Chet. Actually, we have the same effort exerted just for the sake of good and educative argument and this does turn out, best in the interest of OP and readers. I am sorry if you expected a little extra manners than the concede and I haven't been able to pamper it a little extra gratitude. Consider this the extra perhaps.
I don't know if you agree on this, technical guys almost always converse like computer algorithm.

Apology accepted. The only thing I learned from this thread is how difficult and frustrating it is to un-teach something to someone who had originally been taught incorrectly.

Chet

 

[Ronie Bayron]

Apology accepted. The only thing I learned from this thread is how difficult and frustrating it is to un-teach something to someone who had originally been taught incorrectly.
Chet

Nope,I could not agree. Previous professors have nothing to do with this. Life is a continuous learning process, I too believe and will continuously believe I am still a babe with Thermodynamics. I acted on instincts knowing ΔH, ΔU, ΔKE, ΔPE, etc. can be converted in a form of heat. Koreans and Japs has a lay term on this as "sim, sim"