Engineering
23.01.2022 05:58
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) An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole

) 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.
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nyandog50
nyandog50
4,5(18 marks)

Th = 40.91 C

Explanation:

We must use the equation for heat current in conduction

H = \frac{kA(Th-Tc)}{L}

                         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

A = 2\pi r*0.1 = 0.00157079 m^{2}

50 = \frac{5*0.00157079(Th-25)}{0.0025}

Isolating Th, we have

Th = \frac{50*0.0025}{5*0.00157079}+25

Th = 40.91 C

RIZGT500
RIZGT500
4,4(81 marks)

B. By shivering

Explanation:

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