|
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
AG Corporate
Technology CT
PS 3 Günther-Scharowsky-Str.
1 91058
Erlangen, Deutschland Tel.:
+49 (9131) 7-24724 Fax:
+49 (9131) 7-21339 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 No virus
found in this incoming message. |
Attachment:
Tank_Heating1.tpf
Description: Binary data