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Q82PE

Expert-verifiedFound in: Page 736

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Nichrome wire is used in some radiative heaters. **

**(a) Find the resistance needed if the average power output is to be \(1.00\;kW\) utilizing \(120 - V\)\({\rm{AC}}\). **

**(b) What length of Nichrome wire, having a cross-sectional area of \(5.00{\rm{ }}m{m^2}\), is needed if the operating temperature is \({500^o}C\) ? **

**(c) What power will it draw when first switched on?**

- The resistance needed if the average power output is \(1.00\;kW\)utilizing \(120 - V\)\({\rm{AC}}\)is, \(R = 14.4{\rm{ }}\Omega \).
\(L = 60\;m\) is length of Nichrome wire, having a cross-sectional area of \(5.00{\rm{ }}m{m^2}\), is needed if the operating temperature is \({500^\circ }C\).

- The amount of power drawn by the Nichrome wire when first switched on is \({P_0} = 1.2\;kW\).

**The average power delivered by an **\({\rm{AC}}\)** voltage source to a purely resistive load is,**

\({P_{avg}} = {I_{rms}}\Delta {V_{rms}}\)

**Because the potential difference across a resistor is given by **\({V_{rms}} = {I_{rms}}R\)**, we can express the average power delivered to a resistor as,**

\(\begin{aligned}{c}P = {I^2}_{rms}R\\ = \frac{{{{\left( {\Delta {V_{rms}}} \right)}^2}}}{R}...(1)\end{aligned}\)

**The energy delivered to a resistor by electrical transmission appears in the form of internal energy in the resistor.**

**The resistivity of a conductor varies approximately linearly with temperature according to the expression,**

\(\rho = {\rho _0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)...(2)\)

**Where **\({\rho _0}\)** is the resistivity at some reference temperature **\({T_0}\)** , and **\(\alpha \)** is the temperature coefficient of resistivity.**

**The electrical resistance **\(R\) **of a resistive circuit element depends on its geometric structure (its length \(L\) and cross-sectional area \(A\)) and on the resistivity \(\rho \) of the material of which it is made,**

\(R = \rho \frac{L}{A}...(3)\)

**The resistance of a conductor varies approximately linearly with temperature according to the expression,**

\(R = {R_0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)...(4)\)

**Where **\({R_0}\)** is the resistance at some reference temperature **\({T_0}\)** , and **\(\alpha \)** is the temperature coefficient of resistivity.**

- The average power output of the heater is: \({P_{avg}} = \left( {1.00{\rm{ }}kW} \right)\left( {\frac{{1000{\rm{ }}W}}{{1{\rm{ }}kW}}} \right) = 1000{\rm{ }}W\).
- The rms voltage across the heater is: \(\Delta {V_{rms}} = 120{\rm{ }}V\).
- The cross-sectional area of the Nichrome wire is: \(A = \left( {500{\rm{ }}m{m^2}} \right)\left( {\frac{{1{\rm{ }}m}}{{{{10}^6}{\rm{ }}m{m^2}}}} \right) = 5.00 \times {10^{ - 6}}{\rm{ }}{m^2}\).
- The operating temperature of the heater is: \(T = {500^o}C\).
- The resistivity of Nichrome at temperature \({T_0} = {20.0^o}C\) is: \({\rho _0} = 1.00 \times {10^{ - 6}}{\rm{ }}\Omega \cdot M\).
- The temperature coefficient of resistivity for Nichrome is: \(\alpha = 4 \times {10^{ - 4}}^o{C^{ - 1}}\).

a.

The average power output of the heater is found from Equation \((1)\),\({P_{avg}} = \frac{{{{\left( {\Delta {V_{rms}}} \right)}^2}}}{R}\)

Solving for \(R\), it is obtained,

\({\rm{R = }}\frac{{{{\left( {{\rm{\Delta }}{{\rm{V}}_{{\rm{rms}}}}} \right)}^{\rm{2}}}}}{{{{\rm{P}}_{{\rm{avg}}}}}}\)

Entering the values for \(\Delta {V_{rms}}\) and \({P_{avg}}\), gives,

\(\begin{aligned}{}R = \frac{{{{\left( {120{\rm{ }}V} \right)}^2}}}{{1000{\rm{ }}W}}\\ = 14.4{\rm{ }}\Omega \end{aligned}\)

Therefore, the value for the resistance obtained is \(R = 14.4{\rm{ }}\Omega \).

b.

The resistivity of Nichrome is expressed as a function of temperature from Equation \((2)\),

\(\rho = {\rho _0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)\)

Where \({\rm{\rho }}\) is the resistivity of the Nichrome wire at the operating temperature \(T\). Substitute numerical values,

\(\begin{aligned}{}\rho = \left( {1.00 \times {{10}^{ - 6}}{\rm{ }}\Omega \cdot m} \right)\left( {1 + \left( {4 \times {{10}^{ - 4o}}{C^{ - 1}})\left( {{{500}^o}C - {{20.0}^o}C} \right)} \right.} \right.\\ = 1.192 \times {10^{ - 6}}{\rm{ }}\Omega \cdot m\end{aligned}\)

The resistance of the Nichrome wire at the operating temperature \(T\) is then found from Equation \((3)\),

\(R = \rho \frac{L}{A}\)

Solve for \(L\),

\(L = \frac{{RA}}{\rho }\)

Entering the values for \(R\), \(A\), and \(\rho \), it is obtained,

\(\begin{aligned}{}L = \frac{{(14.4{\rm{ }}\Omega )\left( {5.00 \times {{10}^{ - 6}}\;{m^2}} \right)}}{{1.192 \times {{10}^{ - 6}}{\rm{ }}\Omega \times m}}\\ = 60\;m\end{aligned}\)

Therefore, the value of the length is obtained as \(L = 60\;m\).

c.

The resistance of the Nichrome wire is found as a function of temperature from Equation \((4)\) as,

\(R = {R_0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)\)

Where \({R_0}\) is the resistance of the wire at room temperature \({T_0} = {20.0^\circ }C\) , which is the temperature of the wire when first switched on.

Rearrange and solve for \({R_0}\)as,

\({R_0} = \frac{R}{{1 + \alpha \left( {T - {T_0}} \right)}}\)

Substitute numerical values as,

\(\begin{aligned}{}{R_0} = \frac{{14.4{\rm{ }}\Omega }}{{1 + \left( {4 \times {{10}^{ - 4}}^\circ {C^{ - 1}}} \right)\left( {{{500}^\circ }C - {{20.0}^\circ }C} \right)}}\\ = 12.1{\rm{ }}\Omega \end{aligned}\)

The power drawn by the heater when it is first turned on is found from Equation \((1)\) as,

\({P_0} = \frac{{{{\left( {\Delta {V_{rms}}} \right)}^2}}}{{{R_0}}}\)

Entering the values for \(\Delta {V_{rms}}\) and \({R_0}\), it is obtained as,

\(\begin{aligned}{}{P_0} = \frac{{{{(120\;V)}^2}}}{{12.1{\rm{ }}\Omega }}\\ = 1.2 \times {10^3}\;W\\ = \left( {1.2 \times {{10}^3}\;W} \right)\left( {\frac{{1\;kW}}{{1000\;W}}} \right)\\ = 1.2\;kW\end{aligned}\)

Therefore, the value for the Power obtained is \({P_0} = 1.2\;kW\).

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