Outcome 6: Electrical Power

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51 Outcome 6: Electrical Power

Outcome/Competency: You will be able to apply the laws of voltage, current, and resistance to solve electrical power problems.

Rationale:
Why is it important for you to learn this skill?

Electrical power is the commodity that is transported by power lines. This power is consumed to do work by anything that gets plugged in. You will understand the relationship between the three properties of electricity (voltage, resistance, and current) and the power it delivers. People pay for power; understanding units of electrical power allows you to understand how it is charged for.

Objectives:

To be competent in this area, the individual must be able to:

Define power in an electrical circuit
Solve problems of electrical power

Learning Goals

Draw comparison between mechanical and electrical energy and power
Calculate electric power in a circuit by identifying the correct formula
Solve problems using the energy cost formula

Introduction:

In this module, we will review comparisons between mechanical and electrical energy and power, how to calculate electrical power within a circuit, and how to apply the energy cost formula. This chapter will include class discussions, articles to be reviewed, review questions, and a cumulative test.

Topic 1: Electric Power

Prerequisite Skills:

Solving equations and combining two equations by substitution will be discussed. It may be a good opportunity to review points of algebra.

We know that horsepower is a rate of doing work. Electrical power is also referred to as the rate of doing work. A more correct definition of power is the rate at which electrical energy can be converted to heat, light, acoustic, or mechanical energy. The energy is this case must be thought of as work not as output. The metric unit for energy is newtons – meter.

One Newton – metre was called a “joule” for ease of use. Therefore, if power is a rate doing work we can see that:

Watt’s Law

$\text{Power}=\frac{\text{Joules}}{\text{Second}}=\text{Watts}$

One watt of power is produced when one volt forces one ampere of current through a resistance of one ohm.

Topic 2: Watt’s Law

Since electrical power is measured in watts and it has a direct relationship with voltage and amperage, we should be able to use it in a mathematical formula. The formula is:

Power equals Voltage times Amperage

P equals E times I

This formula is similar in format to Ohm’s law ( E = I X R ) so we can use the same tradesperson’s triangle method of transposing the equation:

we also know ohms law as:

Voltage equals Amperage times Resistance

E equals I times R

we can combine the two to include resistance in a formula:

P equals E times I and E equals I times R

P equals open paren I times R close paren times I

P equals I squared times R

another formula is:

P equals E times I and I equals E over R

P equals E times open paren E over R close paren

$P equals the fraction with numerator E squared and denominator R$

We can now solve for the power in a circuit using any two variables.

Example 1

Use all three power equations to find power in this circuit and see if they confirm each other.

E sub a. p p l i e d equals 5 A times 6 Omega equals 30 V

P equals E times I

P equals 30 v o l t s times 5 a. m p s equals 150 w a. t t s

-or-

P equals I squared times R

P equals 5 squared times 6 equals 150 w a. t t s

-or-

$P equals the fraction with numerator E squared and denominator R$

$P equals the fraction with numerator 30 squared and denominator 6 equals 150 w a. t t s$

Example 2

A radio is rated 36 watts. How much force is needed to push 3 amps through it?

E equals P over I

$E equals the fraction with numerator 36 cap W and denominator 3 cap A equals 12 V$

Example 3

Power ratings for heating and lighting (including toasters ovens, hair dryers, etc.) are usually rated or measured in watts. If this power rating was in mechanical terms, what would the appliance be rated in?

Answer: Horsepower.

The relationship between horsepower and watts is:

1 horsepower = 746 watts

Example 4

What would a 7 horsepower motor be rated at in watts?

2 lines Line 1: 7 H of P times 746 cap W over H of P equals 5222 cap W Line 2: o r 5.222 k cap W

Example 5

What horsepower is a 200 kW motor?

$H of P equals 200 comma 000 cap W times the fraction with numerator 1 H of P and denominator 746 cap W equals 268 H of P$

Topic 3: Electrical Energy

Electrical energy is converted to heat, light, or mechanical energy to create power.

P o w e r equals W o r k over T i m e

Electrical energy in the above equation is work.

Electrical utilities have another term also called electrical energy but with a different meaning. This can be somewhat confusing but can be explained by the fact that this electrical energy is a unit of measurement.

First, although we say we purchase and sell power what we really are selling is electrical energy.

“Power” is a rating of a motor or load. An example of this is a two kilowatt motor. This indicates how fast the energy is consumed (a 2 kW motor uses energy faster than a 1 kW motor.)

Electrical energy is measured in kilowatt hours. This is how we commercially sell energy.

Electric Energy

The amount of power used during a period of time.

E n e r g of y equals P o w e r open paren k cap W close paren times T i m e h

Energy is measured in kWh or “killowatt hours”

Example 6

If a 2 kW motor ran for six hours how much energy is used?

E n e of r of g of y open paren k cap W h close paren equals P o w e of r of open paren k cap W close paren times T i m e of h

E n e r g of y equals 2 k cap W times 6 h equals 12 k cap W h

Topic 4: Energy Cost Calculations

Since utilities are in the business of selling electrical energy, they must have some method of knowing how much energy is used by each customer. We do this by using a meter that measures watts and hours. In other words, the power consumed by a customer over a time frame.

Since the water – hour is a very small unit, we charged by the kWh instead. The formula for cost is:

$the fraction with numerator P o w e r sub w a. t t s times T i m e sub h of o u r s times C o s t sub dollar sign p e r k W h and denominator 1000$

There are three things to remember when doing cost calculations:

If you get your power (E X I) in watts, you must divide the equation by 1000 to get kilowatt hours.
Energy is charged in cents per kWh to get dollars. In our equation, we must convert cents to dollars. For example, 7 cents, or $0.07 = 0.07 dollars.
Since energy is in kilowatt hours, we must convert days, weeks, etc. into hours.

Example 7

What is the cost to operate for three hours at 5.6 cents per kWh

1 lines Line 1: P equals 100 cap A times 120 V equals 12 comma 000 cap W open paren o r 12 k cap W close paren

E n e r g of y equals P o w e r times T i m e

E n e r g of y equals 12 k cap W times 3 h of o u r s equals 36 k cap W h

$C o s t equals the fraction with numerator dollar sign 0.056 and denominator k cap W h times 36 k cap W h equals dollar sign 2.02$

Notice, in this case, we kept the energy in kWh to simplify the cost calculation. Therefore, there was no need to divide by 1000.

Example 8

What is the cost to operate for six weeks 12 hours per day at 5.6 cents per kWh

$E sub 3 equals the fraction with numerator open paren 960 cap W close paren and denominator open paren 4 cap A close paren equals 240 V$

E3=960W4A=240V

$I sub 2 equals the fraction with numerator 240 V and denominator open paren 10 ohm close paren equals 24 cap A$

I sub t o t a. l equals 24 cap A positive 12 cap A positive 4 cap A

I sub t o t a. l equals 40 cap A

P sub t o t a. l equals 240 V times 40 cap A

P sub t o t a. l equals 9600 cap W

To operate for 6 weeks, 12 hours per day, at 5.6 cents per kWh:

$E n e r g of y C o s t equals 9.600 k cap W times 12 h of r s over d a. y times 7 d a. y s over w e e k times 6 w e e k s times the fraction with numerator dollar sign 0.0569 and denominator k cap W h equals dollar sign 275$

Topic 5: Line Voltage and Power Loss

We have learned in the previous modules that current needs a push (volts) to overcome the resistance in conductors. Since all conductors possess resistance, some voltage and power is lost in the conductors that feed a load.

The power and voltage losses in the conductor itself will be covered in this topic. You will see how the resistance of a conductor effects the voltage and power available to a load.

Voltage Loss in the Line Wires

A two-wire circuit is a circuit that provides a path to and from a load.

Examples:

A Hi-con line to a farm transformer with the earth return (ground grid)
A duplex service to a streetlight.

In both examples, resistance in the conductors will affect the circuit.

We will represent the line wires as being in series with the source and show the affect a voltage drop across a resistor (the lines) has on the circuit.

Note: when we solve two-wire circuits, both wires have the same amount of resistance unless otherwise stated.

Practice Exercises

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Find the line current in the following circuits

a)

b)

c)

d)

e)

f)

g)

h)

Answer Key

a 2, b 1, c 4, d 2, e 5, f 10, g 10, h 4

License

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Powerline Tech Prep Program Manual Copyright © by Saskatchewan Indian Institute of Technologies-Trades and Industrial is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.