Respuesta :
Answer:
Thickness is 0.086 ft
Solution:
iAs per the question:
Temperature, T = [tex]20^{\circ}F = 36^{\circ}C[/tex]
Energy to be stored, E = 150000 Btu
Area of the concrete floor, A = [tex]1000\ ft^{2} = 93 m^{2}[/tex]
Density of concrete, [tex]\rho = 2400\ kg/m^{3}[/tex]
The heat energy is given by:
[tex]Q = ms\Delta T[/tex] (1)
Also, we know that:
1 Btu = 1055 J
[tex]\rho = \frac{m}{V}[/tex]
[tex]m = \rho V[/tex]
[tex]V = A\times t[/tex]
where
[tex]\rho[/tex] = density
m = Mass
V = Volume
A = Area
t = thickness
s = 750 Jk/kg
Now, eqn (1) can be written as:
[tex]Q = \rho \times A\times t\times s\Delta T[/tex]
Thickness can be written as:
[tex]t = {Q}{\rho \times A\times s\Delta T}[/tex]
[tex]t = {150000\times 1055}{2400 \times 1000\times 0.093\times 750\times 20\times (\frac{9}{5})}[/tex]
t = 0.02625 m
t = [tex]0.02625\times 3.28084 ft = 0.086 ft[/tex]
