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<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>Dear Trnsys users,<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>I Have a question about the indirect evaporative
cooler type 757 from TESS library.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>I do a very simple simulation with type 757, I just
put parameters in entrance and see the results.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>If I put the same air flow rate in the primary stream
and in the secondary stream, I get coherent results, the exchange of power
between the two flows is the same. <o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>Now, if I put a little air flow rate for the
secondary stream (10 times under the primary air flow rate), the air outlet
temperature in secondary stream decreases, but the air outlet temperature in
the primary stream doesn’t change as it seems that this outlet
temperature only depends on the efficiency, which is based on the primary air
inlet dry bulb temperature and the secondary air inlet wet bulb temperature and
these two parameters does not change if we only change the secondary air flow
rate. <o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>Thus, the power change in the primary stream is not
equal to the power change in the secondary stream (calculs below). <o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>Is it a voluntary simplification of the model?
Shouldn’t the primary air outlet temperature change to balance the
exchange of power? <o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>I thank you in advance.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>Best Regards,<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>Benjamin BROTTES<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>Nota: <o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>I have these parameters : <o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>1 : Primary-stream,in<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>2 : Primary stream, out<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>3 : Secondary stream,in<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>4 : secondary stream, out<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>1 : T1 = <st1:metricconverter ProductID="30.49°C"
w:st="on">30.49°C</st1:metricconverter> ; HR1 = 38.88% ; m1 = 252kg/hr ; h1 =
57.59 kJ/kg<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>2 : (trnsys results) T2 = 27.18°C ; HR2 =
47.12% ; h2 = 54.22 kJ/kg<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>3 : T3 = <st1:metricconverter ProductID="30°C" w:st="on">30°C</st1:metricconverter>
; HR3 = 50% ; w3 = 13.31 g/kg ; <o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>4 : (trnsys results) T4 = <st1:metricconverter
ProductID="28.31°C" w:st="on">28.31°C</st1:metricconverter> ; HR4 = 57.92% ; w4
= 14.00 g/kg if m2 = 252 kg/hr<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'>
T4 = <st1:metricconverter ProductID="21.99°C" w:st="on">21.99°C</st1:metricconverter>
; HR4 = 100% ; w4 = 16.64 g/kg
if m2 = 25.2 kg/hr<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=PT-BR style='font-size:
10.0pt;font-family:Arial'>I calculate :<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=PT-BR style='font-size:
10.0pt;font-family:Arial'>Q1 = m1.(h1-h2) = 252 * (57.59-54.22) = 849 kJ/hr<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>Q2 = m2.Cp.(T3-T4) + m2.Lv.(w4-w3) <o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'> </span></font><font size=2 face=Arial><span
lang=EN-GB style='font-size:10.0pt;font-family:Arial'>=
252*1.005*(30-28.31) + 252*2432*(0.014-0.01331) = 428 + 551 =
979kJ/hr if m2 =
252 kg/hr<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'> = 25.2*1.005*(30-21.99) +
25.2*2432*(0.01664-0.01331) = 202 + 217 = 419 kJ/hr if
m2 = 25.2kg/hr which is very far from 849 kJ/hr<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span lang=EN-GB style='font-size:
10.0pt;font-family:Arial'><o:p> </o:p></span></font></p>
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