The components of an electronic system dissipating 180 W are located in a 1-m-long horizontal duct whose cross section is 16 cm × 16 cm. The components in the duct are cooled by forced air, which enters at 27 oC at a rate of 0.65 m3 /min. Assuming 85 percent of the heat generated inside is transferred to air flowing through the duct and the remaining 15 percent is lost through the outer surfaces of the duct, determine (a) the exit temperature of air and (b) the highest component surface temperature in the duct. As a first approximation assume flow is fully developed in the channel. Evaluate properties of air at a bulk mean temperature of 35 oC. Is this a good assumption?

c) The average temperature for air equal to 35°C is a good assumption because the air temperature at the inlet will increase due to the result in the heat gain produced by the duct and whose surface is exposed to a flow of hot.

Explanation:

a) The properties of the air at 35°C:

p = density = 1.145 kg/m³

v = 1.655x10⁻⁵m²/s

k = 0.02625 W/m°C

Pr = 0.7268

cp = 1007 J/kg°C

a) The mass flow rate of air is equal to:

[tex]m=\rho *V = 1.145*0.65=0.7443kg/min=0.0124kg/s[/tex]

The exit temperature is:

[tex]T=T_{i} +\frac{Q}{m*c_{p} } =27+\frac{0.85*180}{0.0124*1007} =39.25[/tex]°C

b) The mean fluid velocity is:

[tex]V_{m} =\frac{V}{A} =\frac{0.65}{0.16*0.16} =25.4m/min=0.4232m/s[/tex]

The hydraulic diameter is:

[tex]D_{h} =\frac{4A}{p} =\frac{4*0.16*0.16}{4*0.16} =0.16m[/tex]

The Reynold´s number is:

[tex]Re=\frac{VD_{h} }{v} =\frac{0.4232*0.16}{1.655x10^{-5} } =4091.36[/tex]

Assuming fully developed turbulent flow, the Nusselt number is:

[tex]Nu=0.023Re^{0.8} *Pr^{0.4} =0.023*4091.36^{0.8} *0.7268^{0.4} =15.69[/tex]

[tex]h=\frac{k*Nu}{D_{h} } =\frac{0.02625*15.69}{0.16} =2.57W/m^{2} C[/tex]

The highest component surface temperature is:

[tex]T=T_{e} +\frac{\frac{Q}{A} }{h} =39.2+\frac{0.85*\frac{180}{4*0.16*1} }{2.57} =132.22[/tex]°C