Problem 87 - College Physics

Question

A power plant transmits 150 \(\mathrm{kW}\) of power to a nearby town, through wires that have total resistance of 0.25\(\Omega\). What percentage of the power is dissipated as heat in the wire if the power is transmitted at (a) 220 \(\mathrm{V}\) and \((\mathrm{b}) 22 \mathrm{kV} ?\)

Step-by-Step Solution

Answer

At 220 V, 77.43% of the power is lost as heat. At 22 kV, only 0.0078% is lost.

Step 1: Calculate Current at 220 V

First, we need to calculate the current when the power is transmitted at 220 V. Using the formula for power, \( P = IV \), where \( I \) is the current and \( V \) is the voltage, solve for \( I \):\[ I = \frac{P}{V} = \frac{150,000 \, \text{W}}{220 \, \text{V}} = 681.82 \, \text{A} \]

Step 2: Calculate Power Dissipated at 220 V

The power dissipated in the wires as heat is given by \( P_{\text{heat}} = I^2R \), where \( R \) is the resistance. Substituting the known values, \[ P_{\text{heat}} = (681.82)^2 \times 0.25 = 116,137.82 \, \text{W} \]

Step 3: Find Percentage Power Loss at 220 V

To find the percentage of power dissipated, use the equation \( \frac{P_{\text{heat}}}{P} \times 100 \% \): \[ \frac{116,137.82}{150,000} \times 100\% = 77.43\% \] So, 77.43% of the power is lost as heat when transmitted at 220 V.

Step 4: Calculate Current at 22 kV

Now, calculating the current for a transmission at 22 kV. Use the same power formula, \( I = \frac{P}{V} = \frac{150,000 \, \text{W}}{22,000 \, \text{V}} = 6.82 \, \text{A} \).

Step 5: Calculate Power Dissipated at 22 kV

Calculate the power dissipated with \( I^2R \): \[ P_{\text{heat}} = (6.82)^2 \times 0.25 = 11.63 \, \text{W} \]

Step 6: Find Percentage Power Loss at 22 kV

Calculate the percentage lost: \[ \frac{11.63}{150,000} \times 100\% = 0.0078\% \] Thus, 0.0078% of the power is lost as heat when transmitted at 22 kV.

Key Concepts

Electrical Resistance

Electrical resistance is a fundamental concept in understanding how electricity flows through materials. It is the opposition a material offers to the flow of electric current. Think of it like a narrow pipe restricting water flow. The larger the pipe (or conductor), the easier it is for current to pass. It's measured in ohms (\(\Omega\)). Resistance affects power transmission because it causes some of the electrical energy to convert into heat, leading to power loss. Every conductor, like the wires used to transmit power from a plant to a town, has some resistance. In our problem, the total resistance of the transmission line is \(0.25\Omega\). This resistance directly influences how much power is lost when electricity is transmitted across the line.

Power Loss

Power loss in electrical systems often refers to the energy dissipated in the form of heat due to resistance. When electric current passes through a conductor with resistance, not all the energy makes it to the destination. Some of it converts to heat and is lost.The formula to find power loss due to resistance is:

• \(P_{\text{heat}} = I^2R\)
• Where \(I\) is the current in amperes and \(R\) is the resistance in ohms.

This formula tells us that power loss is proportional to the square of the current. Thus, higher currents result in significantly more power being lost as heat. Our exercise calculates power loss at two different voltage levels, which drastically changes how much power is lost. The aim is to transmit power with minimal loss, which is more effectively done at higher voltages. In the given problem, the power loss at 220 V was very high at 77.43%, while at 22 kV, it was significantly lower, only 0.0078%.

Current Calculation

Calculating current is essential for understanding power transmission. The formula used here is derived from the basic power equation:

• \( P = IV \)
• Where \(P\) is power in watts, \(I\) is current in amperes, and \(V\) is voltage in volts.

To find the current, the formula rearranges to:

• \( I = \frac{P}{V} \)

In our exercise, first, with a voltage of 220 V, the current came out to be 681.82 A. Such a high current leads to a substantial power loss due to the equation \(I^2R\).
When the voltage increased to 22 kV, the current reduced significantly to 6.82 A. This lower current results in a much lower power loss, demonstrating why high-voltage transmission is preferred for reducing losses over long distances.

Voltage Levels

Voltage levels play a crucial role in power transmission efficiency. The higher the voltage, the lower the current needed to transmit the same amount of power. This is because of the relationship expressed in the equation:

• \( P = IV \)
• For a constant power \((P)\), increasing \(V\) (voltage) means \(I\) (current) must decrease.

Higher voltage levels are desirable in power transmission for several reasons:

• Reduced Power Loss: As demonstrated in the exercise, higher voltage resulted in lower current, which minimized heat loss (\(I^2R\)). This is why power companies use high-voltage lines for long-distance transmission.
• Improved Efficiency: Lower current means that thinner wires can be used, which reduces costs and decreases the amount of energy lost as heat.

In our scenario, when power was transmitted at 22 kV, the efficiency was remarkable with almost negligible power loss compared to a much higher loss at 220 V.