contestada

) An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of material having a thermal conductivity of 5 W/m·K. Estimate the temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

Respuesta :

Answer:

Th = 40.91 C

Explanation:

We must use the equation for heat current in conduction

[tex]H = \frac{kA(Th-Tc)}{L}[/tex]

                         Where k is the thermal conductivity of the material

                                     A is the cross-sectional area

                                     (Th -Tc) is the temperature difference

                                  and L is the length of the material

Thus

[tex]A = 2\pi r*0.1 = 0.00157079 m^{2}[/tex]

[tex]50 = \frac{5*0.00157079(Th-25)}{0.0025}[/tex]

Isolating Th, we have

[tex]Th = \frac{50*0.0025}{5*0.00157079}+25[/tex]

Th = 40.91 C

Otras preguntas