A long (i. e., L >> D) stainless steel rod 6.4 mm in diameter is initially at a uniform temperature of 25°C and is suddenly immersed in a liquid at 150°C with h = 120 W/m2 °C. Calculate the time necessary for the center of the rod to reach 120°C.

68.4 seconds

Explanation:

Assume that the temperature of the stainless steel rod will reach 120 C in t seconds. According to lumped - capacity analysis use the following expression to determine the value of t

T - T∞/T₀ - T∞ = e^ - (hA/ρcV) t, where T is the required temperature, T∞ is the surface temperature, T₀ is initial temperature of the rod, h is the convection heat transfer coefficient, A is surface are, ρ is density of stainless steel, c is specific head and V is the volume of the rod.

Area of rod = π x d x L and Volume of rod is (π/4) x d² L

T - T∞/T₀ - T∞ = e^ - (h x π x d x L/ρc (π/4 x d² x L)) t

T - T∞/T₀ - T∞ = e^ - (4h / (ρ x c x d)) t, d is the radius of the log and L is the length of rod

The value of c and ρ for stainless steel can be obtained from the table of properties of metals

Substitute 120 C for T, 150 C for T∞, 25 C for T₀, 7817 kg/m³ for ρ, 460 J/kg. C for c, 6.4 x 10⁻³ m for d, 120 W/m². C for h in the above equation

T - T∞/T₀ - T∞ = e^ - (4h/ρ x c x d)) t

120 - 150/25 - 150 = e ^ - (4 x 120/7817 x 460 x 6.4 x 10⁻³) t

0.24 = e^-0.02086t

Take natural log on both sides

㏑0.24 = - 0.02086t

T = 68.4 seconds

Thus, the temperature of the stainless steel will reach 120 C in 68.4s