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15 February, 06:07

A cylindrical resistor element on a circuit board dissipates 0.15 W of power in an environment at 40°C. The resistor is 1.2 cm long, and has a diameter of 0.3 cm. Assuming heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this resistor dissipates during a 24-h period; (b) the heat flux on the surface of the resistor, in W/m2; and (c) the surface temperature of the resistor for a combined convection and radiation heat transfer coefficient of 9 W/m2 ·K.

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  1. 15 February, 06:16
    0
    A) 12,960 J

    B) 1178.9 W/m³

    C) 171⁰C

    Explanation:

    A)

    The energy balance equation of the process is "Rate of heat dissipated from the element = Rate of heat energy coming out of the system".

    Q = Q' t

    where Q' = 0.15 W, and t = 24 hour = 86,400sec

    Q = 0.15W x 86,400sec = 12,960J

    B)

    heat flux is q' = Q'/A

    Where A = πDL + 2 (πD²/4) given that L = 0.3cm, D = 1.2cm.

    Q' = 0.15 W

    substituting all parameters into the bellow equation.

    q' = Q'/[πDL + 2 (πD²/4) ]

    q' = 1178.9W/m³

    C)

    Surface temperature q' = h (Tₐ-Tₙ) where:

    h = combined heat transfer coefficient, Tₐ = Surface Temperature, Tₙ = Surrounding Temperature.

    substituting for q' = 1178.9W/m³, h = 9 W/m2 ·K, Tₙ = 40°C in the above equation.

    Tₐ = 171°C
  2. 15 February, 06:30
    0
    A) 3.6 Wh

    B) q' = 1178.93 W/m^2

    C) Ts = 171 °C

    Explanation:

    Q' = 0.15

    D = 0.3cm = 0.003m

    L = 1.2cm = 0.012m

    T∞ = 40°C and hcomb = 9W/m^2. k

    A) The amount of heat this resistor dissipates during a 24 hr period;

    Q = Q' ∆t = 0.15 x 24 = 3.6Wh

    B) The heat flux on the surface of the resistor;

    q' = Q'/A

    Now we will find the surface area of the resistor;

    A = 2 (πD^2) / 4 + πDL

    A = (π/4) (0.003^2) + π (0.003 x 0.012) = 1.272 x 10^-4 m^2

    Therefore, q' = 0.15 / (1.272 x 10^-4) = 1178.93 W/m^2

    C) The surface temperature of the resistor;

    Ts = T∞ + (q'/hcomb)

    Ts = 40 + (1178.93/9)

    Ts = 40 + 131 = 171 °C
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