Question 9.2: An 18^'-tall facility having the dimensions 250^'×100^' has ...

The following data needed to calculate the heat loss may be obtained from a variety of sources [9].

$U_{F} =0.81 Btu/(hr)(ft^{2})(°F)$              $t_{i}=70^{\circ } F$

$U_{R} =0.20 Btu/(hr)(ft^{2})(°F)$              $t_{0}=0^{\circ } F$

Then, using Equation 9.2,

$Q_{F} =(25,000ft^{2})\left[0.81 Btu/(hr)(ft^{2})(°F)\right](70°F-0°F)$
$Q_{F} =14.175\times 10^{5} Btu/hr$

$Q_{R} =(25,000ft^{2})\left[0.20 Btu/(hr)(ft^{2})(°F)\right](70°F-0°F)$
$Q_{R} =3.5\times 10^{5} Btu/hr$

$Q_{G} =(16)(32ft^{2})\left[0.63 Btu/(hr)(ft^{2})(°F)\right](70°F-0°F)$
$Q_{G} =0.226\times 10^{5} Btu/hr$

$Q_{D} =(4)(24ft^{2})\left[1.13 Btu/(hr)(ft^{2})(°F)\right](70°F-0°F)$
$Q_{D} =0.076\times 10^{5} Btu/hr$

$Q_{W} =\left[(2)(4500ft^{2})+(2)(1800ft^{2})-(16)(32ft^{2})-(4)(24ft^{2})\right] \left[(0.39 Btu/(hr)(ft^{2})(°F)\right]$
$(70^{\circ }F-0^{\circ }F)$

Then, using Equation 9.1, the total heat loss may be calculated as follows: