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A heavy-duty electrical resistor is 2cm in diameter by 16cm long. 5 amps of current through it heats the resistor to 100°C, and produces 6 volts across its terminals. Assuming the room air is at 20°C and radiation is negligible, what is the convective heat transfer coefficient?

User Asafge
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1 Answer

4 votes

Answer:

The convective coefficient is 37.3 W/m²K.

Step-by-step explanation:

Use Newton’s law of cooling to determine the heat transfer coefficient. Assume there is no heat transfer from the ends of electric resistor. Heat is transferred from the resistor curved surface.

Step1

Given:

Diameter of the resistor is 2 cm.

Length of the resistor is 16 cm.

Current is 5 amp.

Voltage is 6 volts.

Resistor temperature is 100°C.

Room air temperature is 20°C.

Step2

Electric power from the resistor is transferred to heat and this heat is transferred to the environment by means of convection.

Power of resistor is calculated as follows:

P=VI


P=6*5

P= 30 watts.

Step3

Newton’s law of cooling is expressed as follows:


Q=h* \pi DL(T_(r)-T_(\infty))

Here, h is the convection heat coefficient and
\pi DL is the exposed surface area of the resistor.

Substitute the values as follows:


30=h* \pi (2cm)((1m)/(100cm))(16cm)((1m)/(100cm))(100-20)


h=(30)/(0.8042)

h = 37.3 W/m²K.

Thus, the convective coefficient is 37.3 W/m²K.

User Sean Dunford
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