Question

2. An electrical heater 200mm long and 15mm in diameter is inserted into a drilled hole normal to the surface of a large block of material having a thermal conductivity of 5W/m·K. Estimate the temperature reached by the heater when dissipating 25 W with the surface of the block at a temperature of 35 °C.

Answer 1

Answer: 50.63° C

Explanation:

Given

Length of heater, L = 200 mm = 0.2 m

Diameter of heater, D = 15 mm = 0.015 m

Thermal conductivity, k = 5 W/m.K

Power of the heater, q = 25 W

Temperature of the block, = 35° C

T1 = T2 + (q/kS)

S can be gotten from the relationship

S = 2πL/In(4L/D)

On substituting we have

S = (2 * 3.142 * 0.2) / In (4 * 0.2 / 0.015)

S = 1.2568 / In 53.33

S = 1.2568 / 3.98

S = 0.32 m

Proceeding to substitute into the main equation, we have

T1 = T2 + (q/kS)

T1 = 35 + (25 / 5 * 0.32)

T1 = 35 + (25 / 1.6)

T1 = 35 + 15.625

T1 = 50.63° C

Answer 2

Answer:

The final temperature is 50.8degrees celcius

Explanation:

Pls refer to attached handwritten document