When is ΔH=ΔU

  • #26
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OK. Fasten your seat belt.

I have 1kg of superheated water vapor at 1 bar and 100 C in a closed rigid container. This is going to be the initial condition. From your steam tables, what is the volume of the container? What is the specific volume of the water vapor, the specific internal energy, and specific enthalpy?

Chet
Yes according to spiraxsarco.com container should have volume of 1.69612 m3per kg of steam and all the rest are given on the attached table of values.
 

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  • #27
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Yes according to spiraxsarco.com container should have volume of 1.69612 m3per kg of steam and all the rest are given on the attached table of values.
Good. The two steam table references I have both give H = 2676.2 kJ/kg. So we are pretty much in agreement on that. But I asked you to provide the internal energy U from your steam tables under these conditions, and there is no value given in your table. So what is the value from your source? We can't continue until you provide this.

Chet
 
  • #28
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Good. The two steam table references I have both give H = 2676.2 kJ/kg. So we are pretty much in agreement on that. But I asked you to provide the internal energy U from your steam tables under these conditions, and there is no value given in your table. So what is the value from your source? We can't continue until you provide this.

Chet
Yes, let us use this, perhaps.
@ 1 bar, 100 deg C

v = 1.6959 m3/kg
u = 2506.2 kJ/kg
h = 2675.8 kJ/kg
 

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  • #29
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Yes, let us use this, perhaps.
@ 1 bar, 100 deg C

v = 1.6959 m3/kg
u = 2506.2 kJ/kg
h = 2675.8 kJ/kg
Excellent!! That summarizes the initial state of our system. Now, to get to our final state, we are going to gradually add heat to the 1 kg of vapor in our rigid container until its pressure is 1.5 bars. From your steam tables, what is the final temperature, volume, internal energy, and enthalpy of our vapor?

Chet
 
  • #30
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Excellent!! That summarizes the initial state of our system. Now, to get to our final state, we are going to gradually add heat to the 1 kg of vapor in our rigid container until its pressure is 1.5 bars. From your steam tables, what is the final temperature, volume, internal energy, and enthalpy of our vapor?

Chet
Let us just make it 1.6 bars to avoid numerous interpolation since its readily available on the table. Again, we recall initial state
@ initial state 1 bar, 1000C (we'll include entropy)
v = 1.6959 m3/kg
u = 2506.2 kJ/kg
h = 2675.8 kJ/kg
s = 7.3611 kJ/kg-K
@ Final State 1.6 bar, (s = 7.3611 kJ/kg-K) given process does not lost heat, ideally.
Values of T is in between 4000C and 5000C
 

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  • #31
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Let us just make it 1.6 bars to avoid numerous interpolation since its readily available on the table. Again, we recall initial state
@ initial state 1 bar, 1000C (we'll include entropy)
v = 1.6959 m3/kg
u = 2506.2 kJ/kg
h = 2675.8 kJ/kg
s = 7.3611 kJ/kg-K

@ Final State 1.6 bar, (s = 7.3611 kJ/kg-K) given process does not lost heat, ideally.
Values of T is in between 4000C and 5000C

View attachment 98056
I asked for the results at 1.5 bar, and that's what I want. Also, if the mass of water vapor is 1 kg, and the volume of the rigid container hasn't changed, then has the specific volume of the vapor changed between the initial and final states? Also, regarding your statement that "given process does not lost heat," what did you think I meant when I said "we are going to gradually add heat to the 1 kg of vapor in our rigid container?"

Chet
 
  • #32
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I asked for the results at 1.5 bar, and that's what I want. Also, if the mass of water vapor is 1 kg, and the volume of the rigid container hasn't changed, then has the specific volume of the vapor changed between the initial and final states? Also, regarding your statement that "given process does not lost heat," what did you think I meant when I said "we are going to gradually add heat to the 1 kg of vapor in our rigid container?"

Chet
O you're so fast. Haven't got enough time to think and edit my post.
Anyhow, isentropic process was mistake to describe what you want.- since you add up heat gradually and heat is not by some sort of chemical potential, inside a perfect adiabatic enclosure in which case can also be true at conditions you describe.

I realized, that this is isochoric process. It can not be found on the table fixated @ final v = 1.6959 m3/kg which remains as is, no matter what. Assuming water vapor behaves an ideal gas above regions of the saturated vapor curve, we can find T using PV =mRT or T = PV/mR; where R- is specific gas constant of water of course. This holds true unless, you rebutted.
Do you have problem with this?


I asked for the results at 1.5 bar, and that's what I want
It's not so essential to prove your point. Let's just cut the chase. Is that okay with you?

 
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  • #33
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O you're so fast. Haven't got enough time to think and edit my post.
Anyhow, isentropic process was mistake to describe what you want.- since you add up heat gradually and heat is not by some sort of chemical potential, inside a perfect adiabatic enclosure in which case can also be true at conditions you describe.

I realized, that this is isochoric process. It can not be found on the table fixated @ final v = 1.6959 m3/kg which remains as is, no matter what. Assuming water vapor behaves an ideal gas above regions of the saturated vapor curve, we can find T using PV =mRT or T = PV/mR; where R- is specific gas constant of water of course. This holds true unless, you rebutted.
Do you have problem with this?



It's not so essential to prove your point. Let's just cut the chase, okay with you?
Well, we said we were going to do it strictly by using your steam tables (to make sure I'm not trying to pull a fast one). Later, we can compare with the ideal gas law, but, for now I'd like to stick with the steam tables. Besides, if you assume an ideal gas, you still will have to deal with the problem of determining the internal energy and the enthalpy at the final state.

There are steam tables online that have the properties evaluated at 1.5 bar, if you want to avoid that interpolation (and I deliberately chose 1.5 bar because one of the specific volume values listed in the table at 1.5 bar is almost exactly our specific volume, so you wouldn't have to do any interpolating). However, if you can't find a steam table with 1.5 bar in it, I will settle for your using 1.6 bar. But then you will have to interpolate with respect to the specific volume to get the temperature, internal energy and enthalpy. Your choice.
 
  • #34
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Well, we said we were going to do it strictly by using your steam tables (to make sure I'm not trying to pull a fast one). Later, we can compare with the ideal gas law, but, for now I'd like to stick with the steam tables. Besides, if you assume an ideal gas, you still will have to deal with the problem of determining the internal energy and the enthalpy at the final state.

There are steam tables online that have the properties evaluated at 1.5 bar, if you want to avoid that interpolation (and I deliberately chose 1.5 bar because one of the specific volume values listed in the table at 1.5 bar is almost exactly our specific volume, so you wouldn't have to do any interpolating). However, if you can't find a steam table with 1.5 bar in it, I will settle for your using 1.6 bar. But then you will have to interpolate with respect to the specific volume to get the temperature, internal energy and enthalpy. Your choice.
Do you have any to recommend? So what's the value of u and h @ 1.5 bars & v = 1.6959 m3/kg
 
  • #35
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Do you have any to recommend? So what's the value of u and h @ 1.5 bars & v = 1.6959 m3/kg
I got one by googling "steam tables," but it was a PDF file, and I don't know how to get the actual URL. There are also steam tables in thermo books I have (that are consistent with the PDF table). In Moran et al, they give v = 1.695 m^3 at 280 C, with u = 2778.6 kJ/kg and h = 3032.8 kJ/kg. If you are comfortable using these values, we can continue.

Chet
 
  • #36
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upload_2016-4-2_23-52-2.png

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
upload_2016-4-3_0-12-21.png
 
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  • #37
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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
 
  • #38
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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?
 
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  • #39
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4,298
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.
 
  • #40
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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.:wink:
I don't know if you agree on this, technical guys almost always converse like computer algorithm.
 
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  • #41
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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.:wink:
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
 
  • #42
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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"
 
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