[TRNSYS-users] Type 757 indirect evaporative cooler

[BROTTES Benjamin 227974]

Dear Trnsys users,

 

I Have a question about the indirect evaporative cooler type 757 from TESS library.

 

I do a very simple simulation with type 757, I just put parameters in entrance and see the results.

 

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. 

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. 

Thus, the power change in the primary stream is not equal to the power change in the secondary stream (calculs below). 

 

Is it a voluntary simplification of the model? Shouldn't the primary air outlet temperature change to balance the exchange of power? 

 

 

I thank you in advance.

 

Best Regards,

Benjamin BROTTES

 

Nota: 

I have these parameters : 

 

1 : Primary-stream,in

2 : Primary stream, out

3 : Secondary stream,in

4 : secondary stream, out

 

1 : T1 = 30.49°C ; HR1 = 38.88% ; m1 = 252kg/hr ; h1 = 57.59 kJ/kg

2 : (trnsys results) T2 = 27.18°C ; HR2 = 47.12% ; h2 = 54.22 kJ/kg

3 : T3 = 30°C ; HR3 = 50% ; w3 = 13.31 g/kg ; 

4 : (trnsys results) T4 = 28.31°C ; HR4 = 57.92% ; w4 = 14.00 g/kg       if m2 = 252 kg/hr

                            T4 = 21.99°C ; HR4 = 100% ; w4 = 16.64 g/kg          if m2 = 25.2 kg/hr

 

I calculate :

Q1 = m1.(h1-h2) = 252 * (57.59-54.22) = 849 kJ/hr

Q2 = m2.Cp.(T3-T4) + m2.Lv.(w4-w3) 

 =  252*1.005*(30-28.31) + 252*2432*(0.014-0.01331) = 428 + 551 = 979kJ/hr           if m2 = 252 kg/hr

 =  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

 

 

 

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