One surface of a semi-infinite copper slab is subjected to a constant heat flux of 2000 BTU/h ft2. If the initial temperature is 50oF, determine the surface temperature and the temperature at a depth of 2.5 ft, after 55 min.

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

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