2016-03-08-ETC-infiltration.ipynb

Kernel: Python 3 (Anaconda)

What is the heating load due to infiltration for the ETC?

How does this compare to our estimate of the conductive loss?

Assume

• 0.5 air changes per hour

• Volume of ETC 800 cubic meters

• Density of air 1.2 kg per cubic meter

• Heat capacity of air 1 joule per gram per kelvin

• 25 C inside and 5 C outside

$$q_{inf} = \rho_{air} c_{air} n V \Delta T$$$$\frac{1.2 kg}{m^3} \frac{1 joule}{gram kelvin}$$

1.2 * 1 * 0.5 * 800 * 20 * 1000 / 3600

2666.6666666666665

from pint import UnitRegistry
u = UnitRegistry()

density = 1.2 * u.kg / u.meter**3
volume = 800 * u.meter**3
heat_capacity = 1 * u.joule / u.gram / u.kelvin
Delta_T = 20 * u.kelvin
air_changes = 0.5 / u.hour

q_inf = density * heat_capacity * air_changes * volume * Delta_T


print('This is my ETC report')
print('ETC volume = ', volume)
print('ETC infiltration loss =', q_inf.to(u.kilowatt))

This is my ETC report
ETC volume =  800 meter ** 3
ETC infiltration loss = 2.666666666666667 kilowatt

from pint import UnitRegistry
u = UnitRegistry()

density = 1.2 * u.kg / u.meter**3
volume = 800 * u.meter**3
heat_capacity = 1 * u.joule / u.gram / u.kelvin
Delta_T = 20 * u.kelvin
air_changes = 0.5 / u.hour

q_inf = density * heat_capacity * air_changes * volume * Delta_T

print('infiltration', q_inf.to(u.watt))

infiltration 2666.666666666667 watt

How does this compare to other thermal gains or losses?

We estimated a UA product for the building insulation of about 150 watts per kelvin.  If we multiply this by the temperature difference, we get the heating load for the building.

UA = 150 * u.watt / u.kelvin
UA * Delta_T