Question

a) What current is needed to transmit\({\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{ MW}}\)of power at\({\rm{480 V}}\)? (b) What power is dissipated by the transmission lines if they have a\({\rm{1}}{\rm{.00 \Omega }}\)resistance? (c) What is unreasonable about this result? (d) Which assumptions are unreasonable, or which premises are inconsistent?

Short answer

1. Current needed to transmit the given power is\({\rm{2}}{\rm{.08 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{\;A}}\).
2. Power dissipated by the transmission line is\({\rm{43}}{\rm{.3 GW}}\).
3. Power dissipated in transmission is huge.
4. The voltage is too low.

Step-by-step solution

Definition of power and relation between potential difference & power:

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\({\rm{\Delta V = IR}}\)it can be expressed into 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.

Given data and required data.

The power output of the transmission line is:

\(\begin{aligned}{c}{P_{out{\rm{ }}}} &= \left( {1.00 \times {{10}^2}{\rm{ }}MW} \right)\left( {\frac{{{{10}^6}\;W}}{{1{\rm{ }}MW}}} \right)\\ &= 1.00 \times {10^8}\;W\end{aligned}\)

The voltage at which the power is transmitted,\({\rm{\Delta V}} = {\rm{480\;V}}\).

The resistance of the transmission lines is:\({\rm{R}} = {\rm{1}}{\rm{.00\Omega }}\).

In part (a), we are asked to determine the current needed to transmit a power \({{\rm{P}}_{{\rm{out }}}}\)at voltage \({\rm{\Delta V}}\).

(a) Calculation of current using the power output of transmission line.

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

\({{\rm{P}}_{{\rm{out}}}}{\rm{ = I\Delta V}}\)

Solve the above equation for I:

\({\rm{I = }}\frac{{{{\rm{P}}_{{\rm{out}}}}}}{{{\rm{\Delta V}}}}\)

Substitute numerical values in the above expression.

\(\begin{aligned}{}{\rm{I}} &= \frac{{{\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;W}}}}{{{\rm{480\;V}}}}\\ &= {\rm{2}}{\rm{.08 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{\;A}}\end{aligned}\)

Therefore, current needed to transmit the given power is \({\rm{2}}{\rm{.08 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{\;A}}\).

(b) Calculation of power dissipated.

The power dissipated within the transmission lines before reaching its destination is found from equation (2):

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

Entering values for I and R, and you obtain:

\(\begin{aligned}{}P &= {\left( {2.08 \times {{10}^5}\;A} \right)^2}\left( {1.00{\rm{ }}\Omega } \right)\\ = 4.34 \times {10^{10}}\;W\\ &= \left( {4.34 \times {{10}^{10}}\;W} \right)\left( {\frac{{1{\rm{ }}GW}}{{{{10}^9}\;W}}} \right)\\ &= 43.3{\rm{ }}GW\end{aligned}\)

Hence, Power dissipated by the transmission line is \(43.3{\rm{ }}GW\).

 Step 5: (c) Unreasonable about result:

The power dissipated within the transmission lines is huge. It is even greater than the power output of the transmission lines.

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

The voltage is too low. Such a transmission line in real life would have much higher voltage and low current so that the power dissipated in the line is reduced.