Question

Two thousand (2,000) pigs in the growing-finishing stage (60 kg) are housed at a temperature (T_i) of 15 C. The indoor relative humidity (RH_i) should not exceed 70%. The outdoor air conditions are T_o = -10 C and RH_o = 50%. The UA value for the structure is 10,050 J/s.C. Will supplemental heat (Q_supp) be required for this structure, if so how much (J/s)?

Answer 1

Relative humidity = actual water vapor pressure /saturated water vapor pressure.*100%

As temperature goes up, saturated vapor pressure goes up as well, but relative humidity goes down

The temperature reaches equilibrium Teq by heat exchange

Q inside lost=Q outside gained

Mass of air inside*specific heat of air*(Ti-Teq)= Mass of air inside*specific heat of air*(Teq-To)

(Ti-Teq)= Teq-To

Teq=Ti+To/2=15+10/2=25/2=12.5C

RH(inside)=Vapour pressure of water inside/saturated vapour pressure *100

RH(outside)=Vapour pressure of water outside/saturated vapour pressure *100

RH(inside)/ RH(outside)= saturated vapour pressure (o)/ saturated vapour pressure(i)

Relative humidity is the percentage of water vapour the air at a given temperature can hold. This is given by the Clausius–Clapeyron equation, which rises exponentially with temperature doubling approx every 10degrees C.

Clausius–Clapeyron equation,

lnP=-H vap/R *(1/T)

For every 1C fall in temp=sat . vapour pressure decreases by 1/5 times or RH increases 5 times

So if the temperature inside changes from 15 C to 2.5C (12.5C) ,then the saturated vapour pressure also changes by 12.5* 1/5 =2.5 times ie RH would increase by 12.5* 5 times and exceed the value of 70 %

Decrease in temperature inside=15-2.5=12.5C

So heat lost = 10,050 J/s.C*12.5C=125625 J/S=125.625 KJ/s

supplemental heat (Q_supp) be required=125.625 KJ/s