Question

An engineering student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A and a voltage of 110 V, a lamp that contains a 100-W bulb, an overhead light with a 60-W bulb, and various other small devices adding up to 3.00 W.
a) Assuming the power plant that supplies 110 V electricity to the dorm is 10 km away and the two aluminum transmission cables use 0-gauge wire with a diameter of 8.252 mm, estimate the total power supplied by the power company that is lost in the transmission. (Hint: Find the power needed by the student for all devices. Use this to find the current necessary to be running through the transmission cable)
(b) What would be the result is the power company delivered the electric power at 110 kV?

Answer 1

Answer:

a) $P_{L}=199.075W$

b) $P_{L}=1.991x10^{-4}W$

Explanation:

1) Notation

Power on the refrigerator: $P=IV=3Ax110V=330W$

Voltage $V=110V$

$D=8.252mm$, so then the radius would be $r=\frac{8.252}{2}=4.126mm$

$L=2x10km=20km=20000m$, representing the length of the two wires.

$\rho=2.65x10^{-8}\Omega m$, that represent the resistivity for the aluminum founded on a book

$P_L$ power lost in the transmission.

2) Part a

We can find the total power adding all the individual values for power:

$P_{tot}=(330+100+60+3)W=493W$

From the formula of electric power:

$P=IV$

We can solve for the current like this:

$I=\frac{P}{V}$

Since we know $P_{tot}$ and the voltage 110 V, we have:

$I=\frac{493W}{110V}=4.482A$

The next step would be find the cross sectional are for the aluminum cables with the following formula:

$A=\pi r^2 =\pi(0.004126m)^2=5.348x10^{-5}m^2$

Then with this area we can find the resistance for the material given by:

$R=\rho \frac{L}{A}=2.65x10^{-8}\Omega m\frac{20000m}{5.348x10^{-5}m^2}=9.910\Omega$

With this resistance then we can find the power dissipated with the following formula:

$P_{L}=I^2 R=(4.482A)^2 9.910\Omega=199.075W$

And if we want to find the percentage of power loss we can use this formula

$\% P_{L}=\frac{P_L}{P}x100$

3) Part b

Similar to part a we just need to change the value for V on this case to 110KV.

We can solve for the current like this:

$I=\frac{P}{V}$

Since we know $P_{tot}$ and the voltage 110 KV=110000V, we have:

$I=\frac{493W}{110000V}=4.482x10^{-3}A$

The cross sectional area is the same

The resistance for the material not changes.

With this resistance then we can find the power dissipated with the following formula:

$P_{L}=I^2 R=(4.482x10^{-3}A)^2 9.910\Omega=1.991x10^{-4}W$