Question #171582

A motor winding has a resistance of 80 ohm at room temperature of 20 deg C before switching on to a 230V supply. After 4 hours run the winding resistance is 100 ohm. Find the temperature rise if the material resistance temperature co-effiecent is 1/234.5 deg C.

Expert's answer

**Q171582**

A motor winding has a resistance of 80 ohms at room temperature of 20 deg C before switching on to a 230V supply. After 4 hours run the winding resistance is 100 ohm. Find the temperature rise if the material resistance temperature co-efficient is 1/234.5 deg C.

**Solution: **

We are given the resistance of the material at 20 ^{o}C.

We will take this as our reference resistance and temperature.

R_{ ref } = 80 ohms. T_{ ref }= 20 ^{0}C.

The effect of temperature on resistance is given by the relation

"R = R_{ ref} [ 1 + \u03b1 ( T - T _{ref} ) ]"

where R _{ref} is the reference resistance, T_{ref }is the temperature at the R _{ref}.

*α is the material resistance temperature coefficient. *

R is the resistance of the material at temperature T.

We have,

R_{ ref} = 80 ohms, T_{ ref} = 20 ^{0}C.

*α = 1/ 234.5 *^{0}*C. *

We have to find the temperature of the material when the resistance observed is 100 ohms.

So R = 100 ohms.

Substitute all this information in the formula we have

"100 ohms = 80 ohms [ 1 + \\frac{1}{234.5^0C } ( T - 20 ^0C) ]"

divide both the side by 80 ohms we have

"\\frac{100\u03a9 }{80\u03a9 } = [ 1 + \\frac{1}{234.5^0C } ( T - 20 ^0C) ]"

"1.25 = 1 + \\frac{1}{234.5^0C } ( T - 20 ^0C )"

subtract 1 from both the side we have

"1.25 -1 = 1 + \\frac{1}{234.5^0C } ( T - 20 ^0C ) - 1"

"0.25 = \\frac{1}{234.5^0C } ( T - 20 ^0C )"

multiply both the side by 234.5 ^{0}C, we have

"0.25 * 234.5 ^0C = \\frac{1}{234.5^0C } ( T - 20 ^0C ) * 234.5 ^0C"

"0.25 * 234.5 ^0C = T - 20 ^0C"

"58.625 ^0C = T - 20 ^0C"

add 20 ^{0}C on both the side we have

T = 58.63 ^{0}C + 20 ^{0}C = 78.625 ^{0}C.

which in 2 significant figure is 79 ^{ 0}C.

Hence the final temperature of the motor winding will be 79 ^{0}C.

in question, we are asked the temperature rise.

Temperature rise = Final temperature - initial temperature

= 79 ^{0}C - 20^{ 0}C

= 59 ^{0}C.

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