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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
Von: Francois Badinier [mailto:francois.badinier@icax.co.uk] Gesendet: Friday, December 18, 2009 4:14 PM An: Haluf, Christopher; trnsys-users@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 From: Haluf,
Christopher [mailto:christopher.haluf@siemens.com] 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
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Technology CT PS
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