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Q68P

Expert-verifiedFound in: Page 769

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**The copper windings of a motor have a resistance of ${\mathbf{50}}{\mathit{\Omega}}$ ****at ${\mathbf{20}}{\mathbf{\xb0}}{\mathbf{C}}$**** when the motor is idle. After the motor has run for several hours, the resistance rises to ${\mathbf{58}}{\mathit{\Omega}}$****.What is the temperature of the windings now? Ignore changes in the dimensions of the windings. (Use Table 26-1.)**

The temperature of windings now is $57\xb0C$.

a) Resistance of the windings, ${R}_{0}=50\Omega $

b) Initial temperature of the motor,${T}_{0}=20\xb0C$

c) New resistance,$R=58\Omega $

d) Coefficient of expansion,$\alpha =4.3\times {10}^{-3}{K}^{-1}$

e) Resistivity of the copper material, $\rho =1.69\times {10}^{-8}\Omega .m$

**By using the equation 26-17, the parameter, initial temperature is selected as a reference temperature and the resistivity at that temperature. Using the relationship between the temperature, resistance, and resistivity, we can find the temperature.**

Formulae:

The resistivity relation due to change in resistance due to temperature for resistivity being directly proportional to resistance,$\rho -{\rho}_{0}=\rho \alpha (T-{T}_{0})$ (i)

Rearranging the resistivity relation of equation (i), we can get the value of the temperature of the windings of the motor now as follows:

$T=\frac{\left(\rho -{\rho}_{0}\right)}{\rho \alpha}+{T}_{0}\phantom{\rule{0ex}{0ex}}=\left(\frac{\rho}{{\rho}_{0}}-1\right)\frac{1}{\alpha}+{T}_{0}\phantom{\rule{0ex}{0ex}}=\left(\frac{R}{{R}_{0}}-1\right)\frac{1}{\alpha}+{T}_{0}\left(\because \frac{\mathrm{\rho}}{{\mathrm{\rho}}_{0}}=\frac{\mathrm{R}}{{\mathrm{R}}_{0}},\mathrm{for}\mathrm{area}\mathrm{and}\mathrm{length}\mathrm{being}\mathrm{constant},\mathrm{R}=\frac{\mathrm{\rho L}}{\mathrm{A}}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{58}{50}-1\right)\frac{1}{4.3\times {10}^{-3}}+20\left(\because \mathrm{substituting}\mathrm{the}\mathrm{given}\mathrm{values}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{58}{50}-1\right)\frac{1}{4.3\times {10}^{-3}}+20\phantom{\rule{0ex}{0ex}}=0.037\times {10}^{3}+20\phantom{\rule{0ex}{0ex}}=37+20\phantom{\rule{0ex}{0ex}}=57\xb0C$

Hence, the value of the now temperature is $57\xb0C$.

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