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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.

Answer :

Answer:

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

akindeleot

Answer:

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

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