Heat content thermal energy storage

[scooter85]

Hi everyone!

I'm new here and just registered since I thought this might be a good place to ask.
I'm an aeronautics engineer specialized in the fields of thermodynamics and fluid mechanics, currently working as a researchers.

This is also the reason why I feel quite stupid having to ask this question... But I've been dealing with so many equations now, out of which this task is by far the easiest, but still got my brain completely confused on such a basic problem...

I want to calculate the heat content of a thermal energy storage, filled with water (liquid, no phase change). With this heat content I want to evaluate the state-of-charge of the TES. Thus I have a set of current temperatures, a set of lowest temperatures and a set of highest temperatures.

Let's simplify by dividing the storage in 3 control volumes of which I've got the temperatures. The total volume is V=1m^3. Material properties cp and rho are also quite exact in the range from 0°C to 100°C.
Now I've got three states (temperatures top to bottom):
- fully charged, temperatures, state "max": T_max = [90, 70, 50] °C
- fully discharged, temp. , state "min": T_min = [70, 30, 10] °C
- partly charged, temp. , state "now": T_now = [80, 60, 30] °C

edit: Forgot to mention that it is an "open" storage, thus mass may change due to changes in density.

To get the heat content, I'd go with
Q = sum(T * cp(T) * rho(T) * V_cell)
for all three states (charged, disch., partly ch.) with V_cell = V / 3
And then to get the remaining SOC to max/min:
SOC_max = Q_max - Q_now
SOC_min = Q_now - Q_min

BUT... I need to integrate over cp and rho between the current state and a reference state. (or at least take the mean).
And this is where I'm getting stuck. Since I need to "precalculate" the max/min heat content Q, I can't integrate, since at the point where I calculate "now"-state, I have no information about the min/max temperatures.
What should I do? The options I have considered so far:
[ul]
[li] Calculate Q with reference 0°C for all three states "max/min/now": Q = sum(T * (cp(T) + cp(0°C))/2 * (rho(T) + rho(0°C))/2 * V_cell) (currently preferred)[/li]
[li] Calculate Q with reference 0°C while on the Kelvin scale for all three states "max/min/now": Q = sum((T + 273.15) * (cp(T) + cp(0°C))/2 * (rho(T) + rho(0°C))/2 * V_cell)[/li]
[li] Calculate Q without reference for all three states "max/min/now": Q = sum(T * cp(T) * rho(T) * V_cell)[/li]
[li] Calculate Q without reference on the Kelvin scale for all three states "max/min/now": Q = sum((T + 273.15) * cp(T) * rho(T) * V_cell)[/li]
[/ul]

I guess you all know that when you start thinking about basics too much, basics become the real problem... Thus I just had to ask this.
Many thanks in advance!

Best regards,
scooter

 

[desertfox]

Hi

I might be misreading the question but why can you not evaluate the heat energy lost say between the fully charged temperature and the partially discharge temperature, so you would have say a starting temperature of 90 degrees going down to 80 degrees, I presume that rho and Cp would remain fairly constant over a 10 degree loss.
Sorry If I have misread the problem and simplified it to much.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[LittleInch]

Ok,

took me a while to work out what you are trying to do.

It seems you have a 1m3 tank of some sort with a temperature gradient of 40, 50 and 60C from fully "charged" to discharged?

Now how you develop that and maintain that temp gradient whilst taking heat out of it is beyond me, but I guess that's your problem.

Why can't you just take the average of the temperatures and use that as the heat content?

I still can't figure out what exactly you're trying to calculate though.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[scooter85]

Hi!
@desertfox:
For my calculations, cp and rho are quite significant. A 2% error is already quite much. Furthermore in my "real" calculations even larger spreads are possible, thus errors are even larger.
And this calculation has to be implemented in an optimization dealing with energy flows, thus a SOC unit in J or kWh is required.

@littleinch
Yeah, the temperature gradient is correct. And in reality, it is even higher. The error with using average temperatures would be too large, since cp and rho are not linear.
I'll try to reformulate it like this: Temperatures are measured each minute. I called this "now"-state. And for each "now" state I have to calculate the remaining heat capacity before the TES is either full charged ("max"-state) or discharged ("min"-state).

 

[LittleInch]

Compared to how you're maintaining that sort of temperature gradient thought 2% is nothing....

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[scooter85]

Yes, that may be true. But let's consider it as part of a high-resolution thermal simulation (yeah, errors apply here as well), so the resolution of the sensor array is very high.
And introducing an additional 2% error is not applicable.
Besides the simulation the gradient can be also measured in a "real" storage, but of course measurement errors and spatial resolution errors are >>2% in this case.

 

[IRstuff]

I don't quite get your equations, specific heat is defined in J/kg-K, so the calculation of rho(T)*V_cell, should be rho(T)*V_cell(T), since the actual mass doesn't change over temperature; therefore, you should be able to calculate the mass of each cell at the start and assume conservation of mass, throughout.

SOC_max and SOC_min essentially take into account the reference state already, since subtraction subtracts out the reference state values in Q_max, Q_min, and Q_now. For any given temperature of a volume, there can only be one value of Q, since there can only be a single value of cp(T) for any temperature

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[scooter85]

@IRStuff:
The storage has several pipe connectors, so that there is water in- and out-flowing. Thus the mass may change, depending on the density.
Considering the SOC: Ok, thanks. So should I just take the basic equation without reference state: Q = sum(T * cp(T) * rho(T) * V_cell) ?

 

[LittleInch]

If you've got several pipe connections with flow in and out you're not going to get a reliable fixed temperature gradient.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Compositepro]

There are many details missing. Thermal gradients in liquid tanks are generally small due to convection or even slight agitation. It seems more like you are analyzing a gravel bed. Then there are flow details like, is flow of any stream reversing or not? For example, if hot air flows around a tank it makes a great deal of difference whether the flow is top to bottom or reverse or horizontal. To get the greatest storage capacity you want to maximize the temperature change of most of the fluid. To do this you would use hot air flowing downward around the tank so that the the top heats first and the bottom heats last, but at the end the whole tank is uniformly hot and at the hot air temperature. To extract the stored heat, the airflow will be bottom to top of the tank so that the top of the tank cools last, but in the end, the tank is at a uniformly cool temperature equal to the cold air inlet temperature.

Generally it is far more practical to use a pump and heat exchanger and use valves to control whether liquid is drawn from the top of the tank and returned to the bottom, or the opposite, depending on whether the liquid is being heated or cooled. Efficiency can be greatly increased by having a rolling rubber diaphragm in the tank to prevent mixing in the tank. Gravel beds do not have any issue with mixing of hot and cold.

 

[scooter85]

I made a small drawing to depict what I want to do:

Sometimes a small drawing can help to reactivate the brain after a hot and long day... So it's going the be the first way, using ref. temperature 0°C.

@LittleInch: Trust me, I am. In my "real" storage, which is deployed in 7 real field tests, I've got an average stable gradient in the mixing zone of more than 200K/m. Only the vertical position of the gradient is moving, depending on the state of charge. Lowest gradients, in times of large disturbances, are in the range of 100-150K/m.

@compositepro: Thanks for the details! I'm (well, "we"...) are using flow diffusors to avoid turbulence. This helps maintaining very large gradients, even though peak flows are quite high (45K temperature spread at 120kW -> roughly 0.64kg/s, with the real storages having volumes of 2m^3 to 4m^3).
But for my question, this does not really matter. I am just looking at one point in time, assuming a nearly stationary state or transient state. I just need the SOC for this specific moment I am looking at.

 

[Compositepro]

It seems that with what you have described you cannot even get to a 50% accuracy with your state-of-charge calculation, let alone 2%. This just an intelectual exercise that will help you realize that. This explains your confusion.