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<p>Ilyas, <br>
</p>
<p> If you have not already done so, it would be wise to look at the
Type1231 documentation in ..\Tess
Models\Documentation\06-HVACLibraryMathematicalReference.pdf<br>
</p>
<p><br>
</p>
<div>1- Does the temperature of the hot water inlet to the radiator
(Twater, in) change or remain a fixed value for example 50C.</div>
<div>note that I have a system of regulation (on / off). when he
finds a flow = 0 kg / h. he will change it automatically or I have
to put a variable temperature. How to do it.<br>
<br>
I am not sure that I understand this question. Yes, the
temperature of the water entering the radiator can change. The
entering water temperature is an input to the model as is the flow
rate. The radiator model computes the amount of heat transfer to
the room as a function of the input values so it internally
handles the case when there is no water flow into the radiator.<br>
<br>
</div>
<div>2. I have a gas boiler with a yield of 0.94. I used this as the
fuel power equation (water flow * 4.19 * (50-All)) / (3600 * 0.94)
to find the power. is that correct?<br>
<br>
Again, I am not sure I understand. I would recommend that you use
a boiler model in which case you'll be asked to provide the
efficiency as a parameter (or input depending on which model you
choose). The boiler model will then compute the power that the
boiler requires.<br>
kind regards,<br>
David<br>
<br>
</div>
<br>
<div class="moz-cite-prefix">On 07/09/2018 09:13, Ilyas KHELIFA
KERFAH via TRNSYS-users wrote:<br>
</div>
<blockquote type="cite"
cite="mid:CWXP265MB0374F867F918665FAB196C93E0440@CWXP265MB0374.GBRP265.PROD.OUTLOOK.COM">
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<div>Hello;</div>
<div><br>
</div>
<div>I have questions about type 1231 (Radiator du chfage);</div>
<div><br>
</div>
<div>1- Does the temperature of the hot water inlet to the
radiator (Twater, in) change or remain a fixed value for
example 50C.</div>
<div>note that I have a system of regulation (on / off). when he
finds a flow = 0 kg / h. he will change it automatically or I
have to put a variable temperature. How to do it.</div>
<div>2. I have a gas boiler with a yield of 0.94. I used this as
the fuel power equation (water flow * 4.19 * (50-All)) / (3600
* 0.94) to find the power. is that correct?</div>
<div><br>
</div>
<div>Thank you.</div>
<div><br>
</div>
<div>Have a good day.</div>
<br>
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<pre class="moz-signature" cols="72">--
***************************
David BRADLEY
Principal
Thermal Energy Systems Specialists, LLC
22 North Carroll Street - suite 370
Madison, WI 53703 USA
P:+1.608.274.2577
F:+1.608.278.1475
<a class="moz-txt-link-abbreviated" href="mailto:d.bradley@tess-inc.com">d.bradley@tess-inc.com</a>
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