Question

A passive solar home has energy stored in a concrete floor of 1000 square feet. How thick should this floor be to store 150000 Btu with a temperature swing of 20 degrees F? (Compute the answer in feet. If the floor should be one tenth of a foot thick, enter 0.1) (Note: you'll need to find a value in your textbook for the energy storage capacity of concrete)

Answer 1

Answer:

Thickness is 0.086 ft

Solution:

iAs per the question:

Temperature, T = $20^{\circ}F = 36^{\circ}C$

Energy to be stored, E = 150000 Btu

Area of the concrete floor, A = $1000\ ft^{2} = 93 m^{2}$

Density of concrete, $\rho = 2400\ kg/m^{3}$

The heat energy is given by:

$Q = ms\Delta T$         (1)

Also, we know that:

1 Btu = 1055 J

$\rho = \frac{m}{V}$

$m = \rho V$

$V = A\times t$

where

$\rho$ = density

m = Mass

V = Volume

A = Area

t = thickness

s = 750 Jk/kg

Now, eqn (1) can be written as:

$Q = \rho \times A\times t\times s\Delta T$

Thickness can be written as:

$t = {Q}{\rho \times A\times s\Delta T}$

$t = {150000\times 1055}{2400 \times 1000\times 0.093\times 750\times 20\times (\frac{9}{5})}$

t = 0.02625 m

t = $0.02625\times 3.28084 ft = 0.086 ft$