Question

A cylindrical electric heater is located at the centre of a room. The heater has an outer diameter of 0.05 m, length of 0.6 m

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Surface area of the cylindrical heater, A = \pi d L = \pi * 0.05*0.6 = 0.09425 m^2

Heat transfer coefficient between the heater and surrounding, h = 35 W/m2 K

Emissivity of the heater, \varepsilon = 0.97

Surface temperature of the heater wall, Ts = 105 oC

Surrounding temperature, Ta = 20 oC

Room wall temperature, Tw = 20 oC

Stefan - Boltzmann constant, \sigma = 5.67 * 10^{-8} Wm^{-2} K^{−4} \sigma = 5.67 * 10^{-8} Wm^{-2} K^{-4}

Solution:

Heating power that should be produced = heat lost by the convection + heat lost by the radiation

  Q = hA(T_s - T_a) + \epsilon\sigma A(T_s^4 - T_w^4)

  = (35*0.09425)(105 - 20) + (0.97*5.67*10^{-8}*0.09425)((105+273)^4 - (20+273)^4)

=280.39+ 67.63

=348.02 W(ans)

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