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14 March, 07:30

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.

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  1. 14 March, 07:47
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    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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