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Time Needed for Cooling by Radiation

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A solid cube of side 30 cm at an initial temperature of 1000 K is kept in vacuum at absolute zero temperature. Calculate the time required to cool it to 500 K. The material have the following properties:

Density = 2700 kg/m^{3}

Specific heat = 0.9 kJ/kg.K

Emissivity, ε = 0.1

Stefan-Boltzmann constant, σ = 5.67 x 10^{-8} W/m^{2}.K^{4}.

Calculations:

The given problem is an unsteady state problem. We shall start with the heat balance equation, with the following assumptions:

(i) There is no temperature distribution within the solid

(ii) The surrounding is at absolute zero temperature throughout the cooling period.

Rate of heat flow of out of the solid of volume V through the boundary surfaces A = rate of decrease of internal energy of the solid of volume V

Volume of cube = 0.30^{3} m^{3} = 0.027 m^{3}

Surface area of cube = 6 x 0.3 x 0.3 = 0.54 m^{2}

Aεσ(T(t)^{4} - T_{∞}
^{4}) = -ρC_{P}VdT(t)/dt

Where T(t) is the temperature of the solid at time t and T_{∞} is the temperature of the surrounding.

Since T_{∞}
is 0,

0.54 x 0.1 x 5.67 x 10^{-8} x T(t)^{4} = -2700 x 900 x 0.027 x dT(t)/dt

0.46665 x 10^{-13} T(t)^{4} = -dT(t)/dt

Integrating the above equation,

∫ dT(t)/T(t)^{4} = -0.46665 x 10^{-13} x ∫
dt

-2.1429 x 10^{13 }x [-1/(3T(t)^{3})] = t + C

2.1429 x 10^{13}/(3T(t)^{3}) = t + C

Substituting the initial condition eliminates the constant:

T(t) = 1000 at t = 0 sec

C = 7143.11

Time needed to cool down to 500 K:

2.1429 x 10^{13} / (3 x 500^{3}) = t + 7143.11

t = 50002 sec = 13 hour 54 min

Time needed for the solid to cool to 500 K = 13 hour and 54 minutes

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Last Modified on: 01-May-2024

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