Book "B": Power, Energy and Running Costs I - Answers and Commentary

Introduction

The two basic formulas for energy consumption and cost are:

• \( E = P t \), where \( E \) is the energy, \( P \) is the power, and \( t \) is the time.
  
  
• \( C = r E \), where \( C \) is the cost, \( r \) is the rate or tariff, and \( E \) is the energy used.
  

When using these formulas, it is important to keep the units consistent with each other, and to accurately know what units your results are in.

A table of common \( E = P t \) unit sets is shown below. What this table means is that if you have a power consumption in kilowatts (kW), a time in hours , and you use the formula \( E = P t \), the output \( E \) will be in units of kilowatt-hours (kWh).

Power Unit	Time Unit	Energy Unit
watt (W)	second (s)	joule (J)
watt (W)	hour	watt-hour (Wh)
kilowatt (kW)	hour	kilowatt-hour (kWh)

A table of common \( C = r E \) unit sets is shown below. What this table means is that if you have a tariff in cents per kilowatt-hour (c.kWh^-1), an energy in kilowatt-hours (kWh), and you use the formula \( C = r E \), the output \( C \) will be in units of cents.

Most domestic energy rates are in cents per kilowatt-hour. It may be more convenient to work in dollars per kilowatt-hour if the costs are required in dollars. Note that a unit is a kilowatt-hour, so a tariff of 25 c per unit is the same as a tariff of 25 c per unit.

Some generators price in MWh, but only kWh tariff units are considered in this course.

When doing these problems, take care when mixing different power units in calculations.

For example, the total combined power consumption of a 1 kW heater and a 60 W light bulb is 1.06 kW or 1060 W, not any of the following:

• 1.06 W;
  
• 61 W;
• 61 kW;
• 60.001 W;
• 60.001 kW.
  

It is usually better to convert all power consumption values to the same unit.

Ideally, this unit should match the energy unit that the tariff is based on. For example, if the tariff is "per unit" (i.e. per kWh), it is easiest to work with power consumption figures in kW.

If the final cost is in dollars, it is easiest to work with tariffs in dollars, rather than cents.

The most common tariff unit is the kilowatt-hour (kWh). 1 kWh = 3600000 J (3.6 MJ). There are also other energy units like watt-seconds, joules and such. A conversion table is given below. The "36" comes into the conversions because 1 hour is 3600 seconds.

Conversions can also be "chained". For example, 1 kWs = 1000 J, since 1 kWs = 1000 J, and 1 kJ = 1000 J.

Question 1

Find the annual cost to run a 6kW heater if it is operated for 8 hours a day, 5 days a week, for all 52 weeks of the year. One unit of electricity cost 14 cents.

Solution

The unit cost is $0.14 kWh^-1, in other words, every kilowatt-hour of electricity consumed will cost $0.14.

There are various approaches of calculating the total cost, depending on how times are counted. The crucial thing to keep in mind is keep your units consistent! This is especially important if you have multiple loads, some in watts and some in kilowatts.

The heater operates for 8 hours a day, 5 days a week, and 52 weeks a year. The total operating time per year is calculated below.

\( t = 8 \cdot 5 \cdot 52 = \) 2080 hours per year.

The heater is 6 kW, which is equivalent to saying it consumes 6 kWh of energy per hour i.e. 6 kW = 6 kWh.h^-1.

Multiplying the heater power by the annual number of hours of operation will give the annual energy consumption.

\( E = P t = 6 \cdot 2080 = \) = 12480 kWh per year.

Finally, the cost of a unit (kWh) of electricity is $0.14.

\( C = r E = 0.14 \cdot 12480 = \) $1747.20 per year.

In other words, a 6 kW heater, running 8 hours a day, 5 days a week, 52 weeks a year will cost $1787.20 to run at 14c per unit.

Question 2

A street is lit by 27 lamps that are all 160 watts each. A time clock keeps these lights on for 9 hours a night all 365 nights of the year. Calculate the yearly cost of operating these lights if one unit costs 18.5 cents.

Solution

The unit cost is $0.185 kWh^-1, in other words, every kilowatt-hour of electricity consumed will cost $0.185.

The street lights operate for 9 hours a day (night), 365 days (nights) a year.

\( t = 9 \cdot 365 = \) 3285 hours per year.

The street lights operate for 3285 hours every year. The energy consumption of the street lights is now calculated. Since the cost is based on kilowatt-hours, it makes sense to convert the power consumption to kilowatts.

Each street light consumes 160 W, and there are 27 of them.

\( P = 160 \cdot 27 = \) 4320 W or 4.32 kW.

Multiplying the street light power by the annual number of hours of operation will give the annual energy consumption.

\( E = P t = 4.32 \cdot 3285 = \) = 14191.2 kWh per year.

Finally, the cost of a unit (kWh) of electricity is $0.185.

\( C = r E = 0.185 \cdot 14191.2 = \) $2625.37 per year.

Question 3

A car wash business runs for an average of 9 hours a day and only 330 days of the year, to allow for maintenance of the equipment.

Installed are:

• Three 2 kW motors;
• 18 LED floodlights each 35 watts;
• Two 2.5kW heater blower units;
• Three 500w water pumps;
• Seven 350 W vacuum units;
• Seven 200 W controller kiosks.

Calculate the annual electricity running costs if the unit rate is negotiated at 14.25 cents per kWh.

NOTE: The car wash is described as running "only" 330 days a year. It is important not to make "value judgements" about how long a business "should" operate.

Solution

The unit cost is $0.1425 kWh^-1, in other words, every kilowatt-hour of electricity consumed will cost $0.1425.

The car wash operates for 9 hours a day, 330 days a year.

\( t = 9 \cdot 330 = \) 2970 hours per year.

The total power is given below.

\( P = \) 16.98 kW.

Multiplying the power by the annual number of hours of operation will give the annual energy consumption.

\( E = P t = 16.98 \cdot 2970 = \) = 50430.6 kWh per year.

Finally, the cost of a unit (kWh) of electricity is $0.1425.

\( C = r E = 0.1425 \cdot 50430.6 = \) $7186.36 per year.

Question 4

A 230 V electric blanket has 3 heat positions: low, medium or high. Two 1000 Ω elements are used inside the electric blanket.

What will be the total cost to run the blanket on low for 10 hours, then medium for 20 hours and then high for 2 hours?

The cost of electricity is 22 cents per unit.

Solution

The unit cost is $0.22 kWh^-1, in other words, every kilowatt-hour of electricity consumed will cost $0.22.

First, the power levels of the electric blanket are calculated. A three-heat setup has two identical resistors. In terms of the resistances, the connections are as follows. The power dissipations are calculated using \( P = \frac{V^2}{R} \).

• Low: Both 1000 Ω resistors in series. \( P = \) 26.45 W or 0.02645 kW.
  
• Medium: One 1000 Ω resistor only. \( P = \) 52.9 W or 0.0529 kW.
  
• High: Both 1000 Ω resistors in parallel. \( P = \) 105.8 W or 0.1058 kW.

For this question, it is easier to calculate the total number of units consumed. A table of calculations is shown below. The power figures are in kilowatts, as the price is per kilowatt-hour.

The total energy is given below.

\( E = \) 1.5341 kWh.

Finally, the cost of a unit (kWh) of electricity is $0.22.

\( C = r E = 0.22 \cdot 1.5341 = \) $0.3375.

The cost could also be thought of $0.34.

Question 5

An industrial welding machine electrically welds together mesh gates.

This machine operates at 50 V and draws 200 A when it welds.

Each mesh gate takes 15 seconds to complete the welding and over a 24 hour period 4000 gates are welded.

Calculate the cost to supply the electricity to run the welder over 5 days of production if electricity is supplied at 25 cents per unit.

Solution

The unit cost is $0.25 kWh^-1, in other words, every kilowatt-hour of electricity consumed will cost $0.25.

In this case, it is easier to calculate the energy use per welding operation, then convert it through to cost.

The energy use per gate is calculated below.

\( E = V I t = 50 \cdot 200 \cdot 15 = \) 150000 J or 15 kJ.

The energy use per welding operation is kept in joule units, to keep repeating decimals from converting to kWh to a minimum.

The welding machine is operated 4000 times a day for five days, i.e. there are 20000 operations. The total energy use over five days is calculated below.

\( E_{\mathrm{tot}} = N E = 20000 \cdot 15 = \) 3000000 kJ or 3000 MJ.

The gate welding machine consumes 3000 MJ of energy over five days. This corresponds to 833.3333 kWh to the nearest \( \frac{1}{10000} \) kWh (1 kWh = 3.6 MJ).

The cost of a unit (kWh) of electricity is $0.25.

\( C = r E = 0.25 \cdot 833.3333 = \) $208.33.

Question 6

This question is based on filling in a table. The answers are given in bold.