Long cylindrical steel rods (rho = 7833 kg/m3 and cp = 0.465 kJ/kg·°C) of 8 cm diameter are heat-treated by drawing them at a velocity of 2 m/min through an oven maintained at 900°C. If the rods enter the oven at 30°C and leave at a mean temperature of 540°C, determine the rate of heat transfer to the rods in the oven.

The heat transfer rate Q' is 311.26 kJ / s.

Explanation:

To determine the rate of heat transfer Q we will need the total heat transferred Q and the time it takes to transfer it t = 60 s. To determine the heat transfer Q we will use the energy balance equation. The only non-zero term in the equation is the internal energy change ΔU and to calculate it we will need the total mass of rod that is passing through the furnace m, the given capacity c = 0.465 kJ / kg°C, and the initial T1 = 30°C and final T2 = 540°C temperature. To determine the mass m we will need the volume of rod in the furnace V and the rod density ρ = 7833 kg / m^3. For volume we need the diameter of the rod d = 8 cm and the length of the piece of rod l = 2 m. Before the calculation we need to express the diameter in m^2 units.

d = 8 cm = 8 / 100 = 0.08 m

m = (πd^2 / 4) l ρ

m = (π (0.08) ^2 / 4) (2) (7833)

m = 78.75 kg.

The heat transfer Q is

Q = m c (T2 - T1)

= (78.75) (0.465) (540 - 30)

= 18676 kJ

The heat transfer rate Q' is

Q' = Q / t

= 18676 / 60

Q' = 311.26 kJ / s.