A long cylindrical steel rod is heat treated in an oven that is 7 meters long and is maintained at a temperature of 900°C. The rod is 10-cm in diameter and is drawn at a velocity of 3 m/min. The rod enters the oven at 30°C and leave at 700°C. The rod has a density of 7800 kg/m³ and an average heat capacity of 0.465 kJ/kg °C. Calculate the rate of heat transfer to the rods in the oven.

Q' = 954.28 KJ/s

Explanation:

We are given;

Density of rod = 7800 kg/m³

Length of steel rod; L= 3m

Diameter; D = 10cm = 0.1m

Initial temperature;T1 = 30°C

Exit temperature; T2 = 700°C

Average heat capacity; c = 0.465 kJ/kg °C

Area = πD²/4 = π x 0.1²/4 = 0.0025π

Now,we know that volume = Area x Length

We also know that density = mass/volume.

Thus, mass = density x volume

So, mass = 7800 x 0.0025π x 3 = 183.78 kg

Formula for heat transfer is;

Q = m•c•(T2 - T1)

Q = 183.78 x 0.465(700 - 30)

Q = 57256.66 KJ

Rate of heat transfer is given as;

Q' = Q/t

Question says velocity at 3m/minutes. So, for every 3m it's i minute.. Thus, t = 1 minute = 60 seconds

Thus, Q' = 57256.66/60 = 954.28 KJ/s

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