Along (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

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prevailing winds carry air toward the mountain range. as the air rises up over a mountain range, the air cools, water vapor condenses, and clouds form. on this side of the mountains, called the windward side, precipitation falls in the form of  rain  or snow. this dry air produces a  rain shadow.