Re: [TRNSYS-users] rel. humidity type 56: where is the difference

["Diego A. Arias" <daarias@wisc.edu>]

Dear Uwe,

Let me rephrase your problem:
You are specifying two inlet streams:
- m_dot_a = Vol*ach*rho_a
- m_dot_w = 0.3[kg/h]
Where Vol = room volume, ach = air change, rho_a = air density at inlet 
conditions (P=1atm,T=293.15K).

The steady state humidity ratio is:
w = m_dot_w / m_dot_w

Depending on the correlation that you use to calculate the thermodynamic 
properties you will get slightly different results:
- Following Sonntag, Borgnakke and VanWylen (Fundamentals of 
Thermodynamics) theory, assuming ideal mixtures:
W = 0.622*P_v/P_a
RH = W*P_a/(0.622*P_g)
where P_g = Saturation pressure at room temperature, and P_a and P_v are 
the partial pressures for air and vapor.
You get as result RH = 49.01%. Basically, the same result that you are 
getting from TRNSYS.

- If you use EES, which uses air-water vapor mixture (psychometric) 
properties using thermodynamic data from the built-in AIR and STEAM 
property relations from Hyland and Wexler ( “Formulations for the 
Thermodynamic Properties of the Saturated Phases of H2O from 173.15 K to 
473.15 K, ASHRAE Transactions, Part 2A,Paper 2793 (RP-216), (1983).) , 
you will get as result RH = 48.82%

Best regards,

Diego

Uwe Meinhold wrote:

> Dear TRNSYS User
> I checked a simple steady state problem in trnsys15.
> I have a room: 50 m^3 with an air change rate of 0.7 per hour and a 
> source of 0.3 kg/h water. The outside and inside temperature is 20 °C 
> and the outside relative humidity is 0 (or 0.000000001%).
> The result of TRNSYS type 56 is 49.0263 %
> but the result of my calculation is 49.572% (see enclosed pdf file).
> The differences comes with the source inside the room. That means I go 
> confirm with the psych - routine.
> Now my question is what makes the difference?
> Thanks in advance
>
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--

Diego A. Arias
TRNSYS Coordinator
Solar Energy Laboratory
University of Wisconsin - Madison
1500 Engineering Drive
Madison, WI 53706