A long, cylindrical, electrical heating element of diameter D= 10 mm, thermal conductivity k= 240 W/m·K, density rho= 2700 kg/m3, and specif

Question

A long, cylindrical, electrical heating element of diameter D= 10 mm, thermal conductivity k= 240 W/m·K, density rho= 2700 kg/m3, and specific heat cp= 900 J/kg·K is installed in a duct for which air moves in cross flow over the heater at a temperature and velocity of 27°C and 20 m/s, respectively. (a) Neglecting radiation, estimate the steady-state surface temperature when, per unit length of the heater, electrical energy is being dissipated at a rate of 2000 W/m. (b) If the heater is activated from an initial temperature of 27°C, estimate the time required for the surface temperature to come within 10°C of its steady-state value. Evaluate the properties of air at 450 K.

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Trung Dũng 5 months 2021-08-02T05:58:17+00:00 1 Answers 110 views 0

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    2021-08-02T05:59:25+00:00

    Answer:

    The steady-state surface temperature = 905.5 K

    The time required to come within 10°C of its steady-state value is 137.4 sec.

    Explanation:

    Given data

    D = 10 mm

    k = 240\frac{W}{m K}

    \rho = 2700 \frac{Kg}{m^{3} }

    C_{p} = 900 \frac{J}{kg K}

    Ambient temperature T_{o} = 27 °c = 300 K

    Heat transfer per unit length Q = 2000 \frac{W}{m}

    (a). We know that at steady state the heat transfer per unit length is given by

    Q = h (\pi d l ) (T_{s}T_{o})  —– (1)

    Since the convection heat transfer coefficient at 450 K is

    h =  105.2 \frac{W}{m^{2} K }

    Put all the values in equation (1), we get

    2000 = 105.2 (3.14 × 0.01 × 1) (T_{s} – 300)

    T_{s} = 905.5 K

    Therefore the steady-state surface temperature = 905.5 K

    (b).The time required for the surface temperature to come within 10°C of its steady-state value is given by

    \frac{T - T_{o} }{T_{s} - T_{o}  } = e^{-at} ——- (2)

    Here

    a = \frac{h A}{\rho V C}

    Put all the values in above equation we get

    a = \frac{6h}{\rho D C}

    a = \frac{(6)(105.2)}{(2700) (0.01) (900)}

    a = 0.026

    From equation (2)

    \frac{300-283}{905-300} = e^{-0.026t}

    0.028 = e^{-0.026t}

    -3.57 = -0.026 t

    t = 137.4 sec

    Therefore the time required to come within 10°C of its steady-state value is 137.4 sec.

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