Question

a) What current is needed to transmit\({\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{ MW}}\)of power at\({\rm{10}}{\rm{.0 kV}}\)? (b) Find the resistance of\({\rm{1}}{\rm{.00 km}}\)of wire that would cause a\({\rm{0}}{\rm{.0100\% }}\)power loss. (c) What is the diameter of a\({\rm{1}}{\rm{.00 km}}\)long copper wire having this resistance? (d) What is unreasonable about these results? (e) Which assumptions are unreasonable, or which premises are inconsistent?

Short answer

1. Current needed to transmit the given power is\({\rm{10}}{\rm{.0 kA}}\).
2. Resistance of wire is\({\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ \Omega }}\).
3. Diameter of the resistance wire is\({\rm{0}}{\rm{.468\;m}}\).
4. The diameter of a wire is unreasonably large.
5. The power loss is extremely low for a transmission line at this voltage.

Step-by-step solution

Definition of power, electrical resistance and relation between power and potential difference.

Power: The power of a process is the amount of some type of energy converted into a different type divided by the time interval\({\rm{\Delta t}}\)in which the process occurred:

\({\rm{P = }}\frac{{{\rm{\Delta E}}}}{{{\rm{\Delta t}}}}\) ….. (1)

The SI unit of power is the watt (W).

Because the potential difference across a resistor is given by\({\rm{\Delta V = IR}}\)we can express the power delivered to a resistor as

\({\rm{P = }}{{\rm{I}}^{\rm{2}}}{\rm{R}}\)

\({\rm{P = }}\frac{{{{{\rm{(\Delta V)}}}^{\rm{2}}}}}{{\rm{R}}}\) ….. (2)

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

Electrical resistance: 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\({\rm{\rho }}\)of the material of which it is made:

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

Given data and required data.

The power output of the transmission line is:

\(\begin{aligned}{}{{\rm{P}}_{{\rm{out }}}} &= \left( {{\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{MW}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{\;W}}}}{{{\rm{1MW}}}}} \right)\\ &= {\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;W}}\end{aligned}\)

The voltage at which the power is transmitted is:

\(\begin{aligned}{}{\rm{\Delta V}} &= \left( {{\rm{10}}{\rm{.0kV}}} \right)\left( {\frac{{{\rm{1000\;V}}}}{{{\rm{1kV}}}}} \right)\\ &= {\rm{10}}{\rm{.0 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;V}}\end{aligned}\)

(\({\rm{0}}{\rm{.0100\% }}\)of power is dissipated in the transmission lines.)

(a) Calculation of current using the power output of transmission:

The power output of the transmission line is found from equation (1):

\({P_{out}} = I\Delta V\)

Solve the above equation for I:

\(I = \frac{{{P_{out}}}}{{\Delta V}}\)

Substitute numerical values:

\(\begin{aligned}{}I &= \frac{{1.00 \times {{10}^8}\;W}}{{10.0 \times {{10}^3}\;V}}\\ &= 10.0 \times {10^3}\;A\\ &= \left( {10.0 \times {{10}^3}\;A} \right)\left( {\frac{{1{\rm{ }}kA}}{{{{10}^3}\;A}}} \right)\\ &= 10.0{\rm{ }}kA\end{aligned}\)

Therefore, current needed to transmit the given power is \({\rm{10}}{\rm{.0 kA}}\).

(b) Calculation of resistance using the power dissipated:

The power dissipated\({{\rm{P}}_{{\rm{dissipated }}}}\)in the transmission lines is found from equation (2). But we are given that\({{\rm{P}}_{{\rm{dissipated }}}}\)is equal to\({\rm{0}}{\rm{.0100\% }}\)of the power transmitted\({{\rm{P}}_{{\rm{out }}}}\).

So

\(\left( {0.0100\% \pi } \right){P_{out }} = {I^2}R\)

Solve the above expression for R as below.

\({\rm{R}} = \frac{{\left( {{\rm{0}}{\rm{.0100\% }}} \right){{\rm{P}}_{{\rm{out }}}}}}{{{{\rm{I}}^{\rm{2}}}}}\)

Entering the values for\({{\rm{P}}_{{\rm{out }}}}\)and I, and you obtain:

\(\begin{aligned}{}{\rm{R}} &= \frac{{\left( {{\rm{0}}{\rm{.0100\% }}} \right)\left( {{\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;W}}} \right)}}{{\left( {{\rm{10}}{\rm{.0 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;A}}} \right)}}\\ &= {\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ \Omega }}\end{aligned}\)

Therefore, resistance of wire is \({\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ \Omega }}\).

(c) Calculation of the diameter of the resistance wire:

The resistance of the wire is found from equation (3):

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

Here, the cross-sectional area of the wire is\(\frac{{{\rm{\pi }}{{\rm{D}}^{\rm{2}}}}}{{\rm{4}}}\). Therefore,

\({\mathop{\rm R}\nolimits} = \frac{{4rL}}{{p{D^2}}}\)

Solve the above equation for D:

\({\rm{D}} = \sqrt {\frac{{{\rm{4\rho L}}}}{{{\rm{\pi R}}}}} \)

Entering the values\({\rm{\rho ,L,}}\)and\({\rm{R}}\), you have

\(\begin{aligned}{}{\rm{D}} &= \sqrt {\frac{{{\rm{4}}\left( {{\rm{1}}{\rm{.72 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{ \Omega }} \cdot {\rm{m}}} \right)}}{{{\rm{\pi }}\left( {{\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ \Omega }}} \right)}}} \\ &= {\rm{0}}{\rm{.468\;m}}\end{aligned}\)

Therefore, the diameter of a resistance wire is \({\rm{0}}{\rm{.468\;m}}\).

(d) Define unreasonable about these results:

The diameter of the wire is unreasonably large. A wire with such a diameter would be too heavy for use.

(e) Define the assumptions are unreasonable, or which premises are inconsistent:

The power loss is extremely low for a transmission line at this voltage.

For example, a \({\rm{160\;km}}\) transmission line at \({\rm{345 kV}}\) (which is much greater than the voltage in our problem) carrying \({\rm{100 MW}}\) of power can have losses of \({\rm{4}}{\rm{.2\% }}\).