Book "B": Power, Energy and Running Costs II - 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

A circuit is connected as shown.

Calculate the following:

1. the total power dissipation for the circuit;
2. the power dissipated in each individual resistor;
3. the cost to run the whole circuit for 10 hours with a tariff of 27 cents per unit.

Solution

This circuit needs to be analysed first, so that the power dissipation in each resistor can be calculated.

NOTE: Although the answers have been asked for in a certain order, it is not necessary to do the calculations in that order. You can present the answers later.

Power in \( R_4 \) and \( R_5 \)

\( R_4 \) and \( R_5 \) are connected directly across the 200 V supply. Their power dissipations can be calculated immediately.

• \( P_{R4} = \frac{V_{\mathrm{s}}^2}{R_4} = \frac{200^2}{33} = \frac{40000}{33} = \) 1212.12 W (6 sf).
• \( P_{R5} = \frac{V_{\mathrm{s}}^2}{R_5} = \frac{200^2}{88} = \frac{40000}{88} = \) 454.545 W (6 sf).

Power in \( R_1 \), \( R_2 \) and \( R_3 \)

The power in these resistors is a bit more complicated.

Two new resistances will be defined.

• \( R_6 \), the parallel combination of \( R_2 \) and \( R_3 \).
• \( R_7 \), the total resistance of \( R_1 \), \( R_2 \) and \( R_3 \).

These new resistances can be calculated immediately.

• \( R_6 = R_2 \parallel R_3 = 99 \parallel 22 = \) 18 Ω.
• \( R_7 = R_1 + R_2 \parallel R_3 = R_1 + R_6 = 40 + 18 = \) 58 Ω.

The voltage at the junction of \( R_1 \), \( R_2 \) and \( R_3 \), \( V_{\mathrm{A}} \) is given below. This voltage will be used to calculate power dissipations.

\( V_{\mathrm{A}} = V_{\mathrm{s}} \cdot \frac{R_6}{R_7} = 200 \cdot \frac{18}{58} = \) 62.0690 V (6 sf).

The power dissipations of \( R_1 \), \( R_2 \) and \( R_3 \) can be calculated now.

• \( P_{R1} = \frac{(V_{\mathrm{s} - V_{\mathrm{A}}})^2}{R_1} = \frac{(200-62.0690)^2}{40} = \frac{137.931^2}{40} = \) 475.624 W (6 sf).
• \( P_{R2} = \frac{V_{\mathrm{A}}^2}{R_2} = \frac{62.0690^2}{99} = \frac{3852.56}{99} = \) 38.9147 W (6 sf).
• \( P_{R3} = \frac{V_{\mathrm{A}}^2}{R_3} = \frac{62.0690^2}{22} = \frac{3852.56}{22} = \) 175.116 W (6 sf).

Power in \( R_1 \), \( R_2 \) and \( R_3 \): Kirchhoff's Voltage Law/Ohm's Law Method

The power dissipations of \( R_1 \), \( R_2 \) and \( R_3 \) using the current through \( R_1 \). The current through \( R_1 \) (\(I_{R1}\)) is the same as the current through the hypothetical \( R_7 \). This is because \( R_1 \) carries the whole current of \( R_2 \) and \( R_3 \).

\( I_{R1} = \frac{V_{\mathrm{s}}}{R_7} = \frac{200}{58} = \) 3.44828 A (6 sf).

The power dissipated in \( R_1 \) can be calculated using \( I^2 R \).

\( P_{R1} = I_{R1}^2 R_1 = 3.44828^2 \cdot 40 = 11.8906 \cdot 40 = \) 475.625 W (6 sf).

The voltage at the junction of \( R_1 \), \( R_2 \) and \( R_3 \), \( V_{\mathrm{A}} \) is given below. This voltage will be used to calculate power dissipations.

\( V_{\mathrm{A}} = V_{\mathrm{s}} - I_{R1} R_1 = 200 - 3.44828 \cdot 40 = 200 - 137.931 = \) 62.0688 V (6 sf).

The power dissipations of \( R_2 \) and \( R_3 \) can be calculated now.

• \( P_{R2} = \frac{V_{\mathrm{A}}^2}{R_2} = \frac{62.0688^2}{99} = \frac{3852.54}{99} = \) 38.9145 W (6 sf).
• \( P_{R3} = \frac{V_{\mathrm{A}}^2}{R_3} = \frac{62.0688^2}{22} = \frac{3852.54}{22} = \) 175.115 W (6 sf).

Total Power in Circuit (parts a) and b))

The resistors \( R_1 \) through \( R_6 \) are the only components in the circuit, so the power in the circuit as a whole is simply the sum of the power dissipation of each resistor. The resistors \( R_6 \) and \( R_7 \) are not included, as they representcombinations of resistors that already exist in the circuits.

The power dissipations come from the voltage divider method. While the differences are practically negligible, the voltage divider has fewer "steps" between the inputs and answer. It is therefore considered more numerically stable.

Although the difference is nearly negligible, I will use the rounded "high-precision" total power (2356 W) for calculating the cost.

Total Cost (part c))

The unit cost is $0.27 kWh^-1, in other words, every kilowatt-hour of electricity consumed will cost $0.27.

The total power is 2.356 kW.

The time is 10 hours.

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

\( E = P t = 2.356 \cdot 10 = \) = 23.56 kWh.

The cost of a unit (kWh) of electricity is $0.27.

\( C = r E = 0.27 \cdot 23.56 = \) $6.36.

NOTE: There is a small difference between the figure calculated here and the figure in the book ($6.34). This appears to be due to a power consumption of 23.48 kWh being used to calculate the total energy cost.

Question 2

A 230 volt TPS cable supplies a heater that has a 15 ohm element.

The TPS cable has 0.5 ohms resistance in the phase conductor and has 0.5 ohms resistance in the neutral conductor.

The cable was damaged during installation and was joined in a connection box in the neutral wire. The terminal wasn't tightened sufficiently at this join and resulted in an extra cable resistance of 1 ohm.

Draw a circuit diagram of the above information and use that to calculate the following:

1. The voltage across the heater element;
2. The power dissipated in the cable (including the repair joint);
3. The cost to run the heater for 40 hours (including the power dissipated in the cables) at a tariff of 22 cents kWh.

Solution

The energy supplied (and paid for) is the total energy energy from the 230 V supply (\( V_{\mathrm{s}} \)). The power dissipated in the cable, and the defective joint contributes to this.

A snippet of some working showing the circuit diagram is shown below.

The cable resistance \( R_{\mathrm{c}} \) is calculated below. Both the phase and neutral conductor resistances are required, because any current supplied by the phase must have a return path, in this case through the neutral with its associated resistance.

\( R_{\mathrm{c}} = 0.5 + 0.5 + 1 = \) 2 Ω.

The load resistance \( R_{\mathrm{L}} \) is 15 Ω.

Heater Voltage

The heater (or load) voltage \( V_{\mathrm{L}} \) can be calculated using simple voltage division.

\( V_{\mathrm{L}} = V_{\mathrm{s}} \cdot \frac{R_{\mathrm{L}}}{R_{\mathrm{c}} + R_{\mathrm{L}}} = 230 \cdot \frac{15}{15 + 2} = 230 \cdot \frac{15}{17} = \) 202.941 V (6 sf) or 202.9 V (4 sf) #.

The heater (or load) current \( I_{\mathrm{L}} \) can be calculated using Ohm's Law.

\( I_{\mathrm{L}} = \frac{V_{\mathrm{s}}}{R_{\mathrm{L}}} = \frac{202.941}{15} = \) 13.5294 A (6 sf).

Cable Power Dissipation

The cable power dissipated (\( P_{\mathrm{c}} \)) can be calculated in three main ways:

• Voltage drop across the cable, and the cable resistance: \( P_{\mathrm{c}} = \frac{(V_{\mathrm{s}} - V_{\mathrm{L}})^2}{R_{\mathrm{c}}} = \frac{(230 - 202.941)^2}{2} = \frac{27.259^2}{2} = \) 366.095 W (6 sf).


• Current flow through the cable, and the cable resistance: \( P_{\mathrm{c}} = I_{\mathrm{L}}^2 R_{\mathrm{c}} = 13.5294^2 \cdot 2 = \) 366.089 W (6 sf).
  
  
• Voltage drop across the cable, and Current flow through the cable: \( P_{\mathrm{c}} = (V_{\mathrm{s}} - V_{\mathrm{L}}) I_{\mathrm{L}} = (230 - 202.941) \cdot 13.5294 = 27.059 \cdot 13.5294 = \) 366.092 W (6 sf).
  
  

All of these methods are close to each other. The simplest way is probably the \( I^2 R \) method.

Cost of Running

The heater is run for 10 hours, at a tariff of $0.22 per kWh.

The total power dissipation may be calculated using the supply voltage and load current.

\( P_{\mathrm{TOT}} = V_{\mathrm{s}} I_{\mathrm{L}} = 230 \cdot 13.5294 = \) 3111.76 W (6 sf) or 3.11176 kW.

\( E = P t = 3.11176 \cdot 40 = \) 124.470 kWh (6 sf).

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

\( C = r E = 0.22 \cdot 124.470 = \) $27.38 #.