Answer:
a) surface temperature is 70.196 °F
b) temperature at a depth of 2.5 ft, after 55 min is 58.426 °F
Explanation:
We know that the surface of semi-infinite copper slab is given as;
T - Ti = (2q₀√(∝t/π))/KA exp( -x²/4∝t) -q₀x/KA ( 1 - erf x/2√∝t)
where x is the depth of slab
∝ is the thermal diffusivity
t is time
A is area of the slab
k is the thermal conductivity
Ti is the initial temperature
Now for COPPER
thermal diffusivity ∝ is 11.23 × 10⁻⁵ m²/s = 120.88 × 10⁻⁵ ft²/s
thermal conductivity K = 386 W/m°C = 223.176 BTU/h.ft².°F
so
a) determine the surface temperature
Given that;
initial temperature Ti = 50°F and q₀ = 2000 BTU/h.ft², t = 55m = 55 × 60 = 3300
AT x = 0 ( surface )
so we substitute our values
T - 50 = [(2×2000√(120.88 × 10⁻⁵ × 3300)/π)) / 223.176] × esp( 0)
T - 50 = [(4000 × 1.1268) / 223.176] × 1
T - 50 = 20.1955
T = 50 + 20.1955
T = 70.196 °F
therefore the surface temperature is 70.196 °F
b)
at a depth of 2.5 ft i.e x = 2.5 ft
Error Function, exf = ( 2.5 / 2×√( 120.88 × 10⁻⁵ × 3300)
= erf( 0.62586 )
Error for function table erf( 0.62586 ) = 0.623897
now from our initial equation
T - Ti = (2q₀√(∝t/π))/KA exp( -x²/4∝t) -q₀x/KA ( 1 - erf x/2√∝t)
T - 50 = (2 × 2000√( 120.88×10⁻⁵×3300/π))/223.176 exp( 2.5²/ 4×120.88 × 10⁻⁵ × 3300)
= 2000×2.5 / 223.176 ×( 1 - 0.623897)
T = 58.426 °F
temperature at a depth of 2.5 ft, after 55 min is 58.426 °F
