[TRNSYS-users] Auxiliary Heater and Stratified Storage Tank

Fri Dec 18 07:14:02 PST 2009

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

 <mailto:alan.kiff at icax.co.uk> francois.badinier at icax.co.uk

 <http://www.icax.co.uk> 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 

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Vorstand: Peter Löscher, Vorsitzender; Wolfgang Dehen, Heinrich Hiesinger,
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Registergericht: Berlin Charlottenburg, HRB 12300, München, HRB 6684;
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