[TRNSYS-users] Auxiliary Heater and Stratified Storage Tank

Fri Dec 18 07:29:45 PST 2009

Hello Francois,

thank you for your answer!

It was very helpfull and I also found the solution for the strange results that I've got. I tried all, except setting the "cold side flow rate" to 0, and that was the reason for the low energy in the tank.

Mit freundlichen Grüßen / Kind regards
Christopher Haluf
Intern

Siemens AG
Corporate Technology
CT PS 3
Günther-Scharowsky-Str. 1
91058 Erlangen, Deutschland
Tel.: +49 (9131) 7-24724
Fax: +49 (9131) 7-21339
mailto:christopher.haluf at siemens.com

Siemens Aktiengesellschaft: Vorsitzender des Aufsichtsrats: Gerhard Cromme; Vorstand: Peter Löscher, Vorsitzender; Wolfgang Dehen, Heinrich Hiesinger, Joe Kaeser, Barbara Kux, Hermann Requardt, Siegfried Russwurm, Peter Y. Solmssen; Sitz der Gesellschaft: Berlin und München, Deutschland; Registergericht: Berlin Charlottenburg, HRB 12300, München, HRB 6684; WEEE-Reg.-Nr. DE 23691322

________________________________
Von: Francois Badinier [mailto:francois.badinier at icax.co.uk]
Gesendet: Friday, December 18, 2009 4:14 PM
An: Haluf, Christopher; trnsys-users at cae.wisc.edu
Betreff: RE: [TRNSYS-users] Auxiliary Heater and Stratified Storage Tank

Hi Christopher

I found several things to correct in your model to find the solution:

1.       The main thing is that you are withdrawing some heat from your tank at the same time as you are heating it, because your "cold side flow rate" input for the Tank is not set to 0.

2.       In your pump, if you want an ideal pump, set the conversion coefficient to 0, so that the pump does not heat the fluid by its inefficiencies.

3.       In your auxiliary heater, you'll have to increase your "boiling temperature" so that it continues heating when you reach 100C (because you reach 100 C before the end of the 10h)

4.       If you play with the time step (more accurate when smaller), you'll see that what I called Q_to_Tank tends to the 360,000 kJ.

5.       However, I also plotted the "internal energy change", which is exactly what you want to see, it is equal to the "Q_to_Tank_kJ" up to the point where the temperature of the tank reaches 100 C, and then the Tank cannot accept any more energy, so you'll see that for the last minutes of your simulation the "internal energy change" flattens out.

I attached the tpf so you can see all that.

Regards,

François Badinier
Development Engineer
ICAX Ltd
1 Hatfield House
Baltic Street West
London EC1Y OST
francois.badinier at icax.co.uk<mailto:alan.kiff at icax.co.uk>
www.icax.co.uk<http://www.icax.co.uk>

From: Haluf, Christopher [mailto:christopher.haluf at siemens.com]
Sent: 18 December 2009 09:07
To: trnsys-users at cae.wisc.edu
Subject: [TRNSYS-users] Auxiliary Heater and Stratified Storage Tank

Dear TRNSYS-Users,

I'm modelling an easy system that consists of a pump (Type 3b), an auxiliary heater (Type 6) and a stratified storage tank (Type 4a). I've attached my .tpf-file, so you can comprehend what I'm writing.

I'm doing this simulation because I want to get familiar with the behavior of a tank while it is loaded. I set the losses of each component to 0 and the efficiencies to 1, so that I have a loss-free system.

Concerning the heater, I set the "maximum heating rate" to 10 kW and the "set point temperature" to 100°C. I defined 3 nodes besides the top and the bottom in the tank, so I have 5 nodes. I set the "initial temperature" for each node to 20°C. The volume of the tank is 1m³ and I disabled the internal heaters, so that all of the heat is provided by Type6.

What I want to do, is to check if all of the energy which is provided by the Heater gets into the tank. Therefore, I run the simulation for 10 hours. Now, I calculate the energy provided by the heater which is 10 kW * 36.000 s = 360.000 kJ. In order to control if all of this energy got into the tank, I take the highest and the lowest value of the average tank temperature which is the delta T. Cp for water is 4.19 kJ/kg*K and the mass of the water is 1.000kg. Now, I can calculate the internal energy of the tank (1000 kg * 4.19 kJ/kg*K * (74.2°C-20°C) = 227.098 kJ). This value is definitely much lower than the energy that is provided by the heater...

Another strange thing is, that the average tank temperature begins with a value of 0°C. I don't understand this, because I set the initial temperature for all nodes to 20°C, so I think that the average tank temperature should have the same value...

Could someone of you please have a look at my simulation and see if there is something wrong. And if there is all right, could you please explain the bahvior of the tank? It would be very helpfull for me!

Thank you in advance!

Mit freundlichen Grüßen / Kind regards
Christopher Haluf
Intern

Siemens AG
Corporate Technology
CT PS 3
Günther-Scharowsky-Str. 1
91058 Erlangen, Deutschland
Tel.: +49 (9131) 7-24724
Fax: +49 (9131) 7-21339
mailto:christopher.haluf at siemens.com

Siemens Aktiengesellschaft: Vorsitzender des Aufsichtsrats: Gerhard Cromme; Vorstand: Peter Löscher, Vorsitzender; Wolfgang Dehen, Heinrich Hiesinger, Joe Kaeser, Barbara Kux, Hermann Requardt, Siegfried Russwurm, Peter Y. Solmssen; Sitz der Gesellschaft: Berlin und München, Deutschland; Registergericht: Berlin Charlottenburg, HRB 12300, München, HRB 6684; WEEE-Reg.-Nr. DE 23691322

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