Heat in a const. pressure/volume calculation

[guideonl]

Hi All,
In thermodynamics close system there are two const. processes in which we can calculate the heat:
const. pressure
const. volume
I took 2 simple examples for these processes to test the results according the manufacturer data:
For const. pressure I choose 1 liter 1.5 KW teapot
For const. volume I choose 200 liter 2.5 KW water heater (for shower)
I calculated heats & time in two ways:

For const. pressure I used the data: T1=15 deg C, T2=100 deg C, P=1atm, Cp(avg)=1.875KJ/Kg K, m=1Kg, and the relation q=Cp(T2-T1), thus for 1Kg mass I got Q=159.375KJ. I devided the result by the electric power (1.5KW) and got the time needed to boil the water (1.77 min).
The second way I have calculated was using the 1st thermo. law dq=du (no work), where u2=uf(sat@100degC)=419.06KJ/Kg, & u1~uf(@15degC)=62.98 KJ/Kg assuming the liquid properties could be estimated as the sat prop. @ that temp. Thus, for 1Kg mass I got Q=356.08 KJ. I (again) devided the result by the electric power and got the time needed to boil the water (3.95 min) which makes sense more than the prev. result (about half time shorter).

I did the same calc. procedure for the next example:
For const. volumeI used the data: T1=15 deg C, T2=60 deg C, P=5atm, Cv(avg)=1.42 KJ/Kg K, m=200Kg, and the relation q=Cv(T2-T1), thus for 200Kg mass I got Q=11,360 KJ. I devided the result by the electric power (2.5KW) and got the time needed to heat the water (1.26 hr) makes sense.
The second way I have calculated was using the 1st thermo. law dq=du (no work), where u2~uf(sat@60degC)=251.16KJ/Kg, & u1~uf(@15degC)=62.98 KJ/Kg assuming the liquid properties could be estimated as the sat prop. @ that temp. Thus, for 200Kg mass I got Q=37,636 KJ. I (again) devided the result by the electric power and got the time needed to boil the water (4.18 hr) which does'nt make sense than the prev result.

If I summerize the results:
At const. pressure process, using q=Cp(T2-T1) eq. gives the real and makes sense result.
At const. volume process, using q=(u2-u1) eq. gives the real and better result than q=Cv(T2-T1) eq.

I hope the problems are clear, and I'll be thankfull if somebody will explain why the results are different (when using 1st thermo. law dq=du vs. the process eqs. Cp;Cv(T2-T1)).

 

[PeterDonis]

For const. pressure I choose 1 liter 1.5 KW teapot

Why do you think the teapot will be at constant pressure as it is heated?

For const. volume I choose 200 liter 2.5 KW water heater (for shower)

Why do you think the shower water heater will heat a constant volume?

 

[PeterDonis]

I hope the problems are clear

Not really. Your post is very hard to read.

 

[PeterDonis]

If I summerize the results:
At const. pressure process, using q=Cp(T2-T1) eq. gives the real and makes sense result.
At const. volume process, using q=(u2-u1) eq. gives the real and better result than q=Cv(T2-T1) eq.

It seems like your summary is backwards. For the constant pressure case, you say the q(u2 - u1) result "makes sense more". For the constant volume case, you say the q(u2 - u1) result "doesn't make sense".

 

[guideonl]

Hi Peter,
Sorry for my EN, I'll try to explain:

I assumed the teapot will be at constant pressure as it is heated because it is open to the atmospheric pressure, and neglating the small mass amount evaporized during heating.

I assumed the shower water heater will heat a constant volume while there is no water use from the heater until reaching 60 degC. At this time (no mass enter/leave the heater, thus could be assumed as a closed system).

For the summary, yes you are right, I wrote it wrong. It should be:
At const. pressure process, using q=(u2-u1) eq. gives the real and makes sense result.
At const. volume process, using q=Cv(T2-T1) eq. gives the real and better result than q=(u2-u1) eq.

 

[russ_watters]

Cp(avg)=1.875KJ/Kg K,

Where did you get that Cp? Shouldn't it be 4.186?

 

[PeterDonis]

I assumed the teapot will be at constant pressure as it is heated because it is open to the atmospheric pressure, and neglating the small mass amount evaporized during heating.

Ok. So what will happen to the volume of the water in the teapot as it is being heated?

I assumed the shower water heater will heat a constant volume while there is no water use from the heater until reaching 60 degC. At this time (no mass enter/leave the heater, thus could be assumed as a closed system).

Ok.

For the summary, yes you are right, I wrote it wrong.

Ok, good.

At const. pressure process, using q=(u2-u1) eq. gives the real and makes sense result.
At const. volume process, using q=Cv(T2-T1) eq. gives the real and better result than q=(u2-u1) eq.

What is u? You seem to think it's internal energy (since you say "no work"), but your answer to my first question above (about the volume of water in the teapot) should show you that that's not the case, because "no work" is not correct for that case.

Also, as @russ_watters pointed out, your Cp value is way off. With a correct value of Cp, you should find that the two calculation methods for the constant pressure case give approximately the same answer. The reason why is connected to the correct definition of what you are calling u.

Your Cv value is also way off. However, even with a correct Cv value, the two calculation methods will not give the same answer. Again, the reason why is connected to the correct definition of what you are calling u.

 

[Chestermiller]

What is u? You seem to think it's internal energy (since you say "no work"), but your answer to my first question above (about the volume of water in the teapot) should show you that that's not the case, because "no work" is not correct for that case.

Also, as @russ_watters pointed out, your Cp value is way off. With a correct value of Cp, you should find that the two calculation methods for the constant pressure case give approximately the same answer. The reason why is connected to the correct definition of what you are calling u.

Your Cv value is also way off. However, even with a correct Cv value, the two calculation methods will not give the same answer. Again, the reason why is connected to the correct definition of what you are calling u.

I think it is pretty clear that the OP recognizes that u is the internal energy per unit mass. After all, he used the u values taken straight out of the steam tables in his calculation. And since liquid water is nearly incompressible and has a nearly constant heat capacity over the temperature range of interest, it was valid in his alternate calculation to use $\Delta u=C_v\Delta T$. The only real mistake that the OP made was in using an incorrect value for the heat capacity.

 

[guideonl]

Hi Russ & Chestermiller
Yes, you are right. The Cp(avg) should be 4.186 ! It gives the correct result (and make sense) to the 1st case (teapot) for the two ways of calculation I noted above.
My mistake probably because I've found a formula which Cp (T), for water in the temp. range 375-1360 K , and I used the formula without noticing the incorrect results acheived.

For the 2nd case (water heater), after using the correct Cp=4.186, Cv=3.724 I got two different new results for the calculated heat (at the two ways of calculations), which reveal to heating time of ~3.7-4.2 hr (which does not makes sense).

u is the internal energy [KJ/Kg]

Thank you for your help, and if you can find the problem in the 2nd case it would be helpful.

 

[PeterDonis]

I think it is pretty clear that the OP recognizes that u is the internal energy per unit mass. After all, he used the u values taken straight out of the steam tables in his calculation.

Those values aren't internal energy, they're enthalpy. Enthalpy is $u + p v$, where $u$ is the internal energy, $p$ is the pressure, and $v$ is the volume (all per unit mass).

However, I think those values are at saturation pressure, not atmospheric pressure (at least that's what I find when I look them up), so they aren't really applicable to either of the cases described by the OP (except for the 100 C endpoint of the first case, where saturation pressure equals atmospheric pressure). The constant pressure case is at atmospheric pressure throughout, and the constant volume case will have pressure increasing from whatever the supply pressure is of his water main, which should be significantly higher than atmospheric.

liquid water is nearly incompressible

Nearly, but not quite. The difference between the constant pressure and constant volume cases--the larger value of $Q$ (the heat added from the external source) required to cause the same temperature change--is due to the additional work required for the thermal expansion of the water. This shows up as $C_p$ being larger than $C_v$.

 

[PeterDonis]

The Cp(avg) should be 4.186 !

Actually it's a little larger over the range 15 C to 100 C; I estimate about 4.21.

For the 2nd case (water heater), after using the correct Cp=4.186

You should not be using $C_p$ for the constant volume case; you should be using $C_v$. The average value of $C_v$ over a given temperature range is a bit lower than the average value of $C_p$. I estimate it at about 4.08 over the range 15 C to 60 C.

 

[PeterDonis]

heating time of 4.18 hr (which does not makes sense)

Why not? (The correct value using $C_v$ instead of $C_p$ will be a little smaller, but still of the rough order of magnitude of 4 hours.)

 

[guideonl]

Why not? (The correct value using $C_v$ instead of $C_p$ will be a little smaller, but still of the rough order of magnitude of 4 hours.)

Actualy it takes about 1.5 hr

 

[PeterDonis]

Actualy it takes about 1.5 hr

Actually meaning what? You've measured it? In a controlled experiment? And you've verified that the numbers you gave in the OP match the numbers in the actual experiment?

Please show your work.

 

[guideonl]

calculated Cv =3.724 KJ/Kg K

 

[PeterDonis]

calculated Cv =3.724 KJ/Kg K

Calculated how? Please show your work.

 

[guideonl]

Cv=Cp-R=4.186-.462 R=8.314/18 KJ/Kg K

 

[PeterDonis]

Cv=Cp-R

Why do you think this is true for liquid water? (Hint: it isn't.)

 

[guideonl]

Oh, I'm sorry, Is it for ideal gases only? I did'nt know it. Is there any other relation?
I must go sleep.. 00:30 ...I have to work at 06:30.
Thanks for meanwhile

 

[PeterDonis]

Is it for ideal gases only?

Yes.

Is there any other relation?

I'm not aware of any general relation between specific heats for liquids or solids. You just have to look up the values that have been measured empirically.

 

[Chestermiller]

For an incompressible liquid, Cp=Cv

 

[PeterDonis]

For an incompressible liquid, Cp=Cv

Water is not incompressible. The actual values of Cp and Cv, which you can find online, make that clear. It is fairly close to being incompressible, but "fairly close" is not the same as "exactly".

 

[Chestermiller]

Water is not incompressible. The actual values of Cp and Cv, which you can find online, make that clear. It is fairly close to being incompressible, but "fairly close" is not the same as "exactly".

Speaking as an engineer, it’s close enough for me in most situations.

 

[PeterDonis]

Speaking as an engineer, it’s close enough for me in most situations.

At 100 C, the figures I get from the "Engineering Toolbox" website [1] are Cp = 4.22, Cv = 3.77 (in kJ/kg-K). That's not very close.

[1] https://www.engineeringtoolbox.com/specific-heat-capacity-water-d_660.html

 

[Chestermiller]

At 100 C, the figures I get from the "Engineering Toolbox" website [1] are Cp = 4.22, Cv = 3.77 (in kJ/kg-K). That's not very close.

[1] https://www.engineeringtoolbox.com/specific-heat-capacity-water-d_660.html

The value you give for Cv seems way out of line with the steam tables. From the steam tables, both Cp and Cv are 4.22 at 100 C. It looks like you took Cp and subtracted R to get Cv; this is valid only for an ideal gas.

 

[PeterDonis]

The value you give for Cv seems way out of line with the steam tables.

Do you have a reference?

It looks like you took Cp and subtracted R to get Cv

I didn't do anything except pull numbers from the website I linked to.

 

[russ_watters]

Speaking as an engineer, it’s close enough for me in most situations.

This is why I am an engineer.

At 100 C, the figures I get from the "Engineering Toolbox" website [1] are Cp = 4.22, Cv = 3.77 (in kJ/kg-K). That's not very close.

It's around 10% but if you average over the ranges of the OP's problems it is 6% and less. It's good enough for fixing the OP's conundrum of a more than 100% error or sizing a water heater, even if not for a scientific paper. Let's not let the perfect be the enemy of the good.

[edit]
How about we split the difference: The OP's methods gloss-over the fact that volume and Cp aren't quite constant as temperatures vary, but they are close enough to resolve the apparent inconsistency in results reported by the OP.

 

[PeterDonis]

It's around 10% but if you average over the ranges of the OP's problems it is 6% and less.

Yes, agreed. The site I linked to gives much smaller differences at 60 C and 15 C.

 

[Chestermiller]

Do you have a reference?
I didn't do anything except pull numbers from the website I linked to.

I stand corrected. In evaluating Cp and Cv, I merely took the derivatives of h and u, respectively, with respect to T for saturated liquid water. Apparently, for Cv, this is not a very good approximation as temperature increases. The actual general thermodynamic equation for the difference for molar heat capacities Cp-Cv is:
$$C_p-C_v=\frac{\alpha^2}{\beta}Tv$$where $\alpha$ is the volume expansivity and $\beta$ is the isothermal compressibility:

$$\alpha=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_P$$and$$\beta=-\frac{1}{v}\left(\frac{\partial v}{\partial P}\right)_T$$When I substitute tabulated values of these parameters for liquid water at 1 atm. into this equation, I get results in agreement with the table that Peter provided.

I was unaware that the coefficient of volume expansion for liquid water increased so much, even at a low temperature like 100 C.

 

[guideonl]

Well guys,
It was interesting and important lesson for me.
Thanks

 

[guideonl]

Hi guys,
I am back, because I found properties tables in the web, which aprove my 1st Cp value (1.875KJ/Kg K).
russ_watters asked where did I find this value?
Actully I got this value from calculation a formula for Cp(T), but recently I found it also in a textbook tables. So, now I am confused .. from one side I've got the same value as in the table, but from the other side I wonder if 300 deg K water could be considered as an ideal gas (water steam is appeared in the table under the title of ideal gas !)
Hereby the tables I used, see table A-2 (a) last row (also Cv value for that Temp is noted).

 

[Chestermiller]

Hi guys,
I am back, because I found properties tables in the web, which aprove my 1st Cp value (1.875KJ/Kg K).
russ_watters asked where did I find this value?
Actully I got this value from calculation a formula for Cp(T), but recently I found it also in a textbook tables. So, now I am confused .. from one side I've got the same value as in the table, but from the other side I wonder if 300 deg K water could be considered as an ideal gas (water steam is appeared in the table under the title of ideal gas !)
Hereby the tables I used, see table A-2 (a) last row (also Cv value for that Temp is noted).

That's the heat capacity of water vapor, and you are dealing with liquid water.

 

[guideonl]

That's the heat capacity of water vapor, and you are dealing with liquid water.

Sorry, I don't understand. The table's values are for 300 deg K ! Does it mean that the pressure is not atmospheric but lower than the atmospheric pressure? (to be at the vapor region associated to vapor at that temp.)

 

[Chestermiller]

Sorry, I don't understand. The table's values are for 300 deg K ! Does it mean that the pressure is not atmospheric but lower than the atmospheric pressure? (to be at the vapor region associated to vapor at that temp.)

Are you saying you can’t have liquid water at 100 C?

 

[PeterDonis]

Are you saying you can’t have liquid water at 100 C?

He said 300 K, which is about 27 C. I assume he's looking at saturation pressure values and mistaking the vapor one for the liquid one.