11.9 Applications

← Back to Fundamentals of Electric Circuits Overview

11.9 Applications

In this section, we consider two important application areas: how power is measured and how electric utility companies determine the cost of electricity consumption.

11.9.1 Power Measurement

The average power absorbed by a load is measured by an instrument called the wattmeter.

The wattmeter is the instrument used for measuring the average power.

Figure 11.30 sho ws a w attmeter that consists essentially of tw o coils: the current coil and the voltage coil. A current coil with very low impedance (ideally zero) is connected in series with the load (Fig. 11.31) and responds to the load current. The voltage coil with very high impedance (ideally infinite) is connected in parallel with the load as shown in Fig. 11.31 and responds to the load v oltage. The current coil acts lik e a short circuit because of its low impedance; the voltage coil behaves like

Reactive power is measured by an instrument called the varmeter. The varmeter is often connected to the load in the same way as the wattmeter.

Some wattmeters do not have coils; the wattmeter considered here is the electromagnetic type.

Practice Problem 11.15

A wattmeter.

Figure 11.31 The wattmeter connected to the load.

an open circuit because of its high impedance. As a result, the presence of the w attmeter does not disturb the circuit or ha ve an ef fect on the power measurement.

When the two coils are energized, the mechanical inertia of the moving system produces a deflection angle that is proportional to the average value of the product v(t)i(t). If the current and voltage of the load are v(t) = Vm cos(ωt + θv) and i(t) = Im cos(ωt + θi), their corresponding rms phasors are

$$\mathbf{V}_{\rm rms} = \frac{V_m}{\sqrt{2}} \underline{\theta_v} \quad \text{and} \quad \mathbf{I}_{\rm rms} = \frac{I_m}{\sqrt{2}} \underline{\theta_i} \quad (11.62)$$

and the wattmeter measures the average power given by

$$P = |\mathbf{V}_{\text{rms}}||\mathbf{I}_{\text{rms}}| \cos(\theta_{\nu} - \theta_{i}) = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}) \qquad (11.63)$$

As shown in Fig. 11.31, each wattmeter coil has two terminals with one marked ±. To ensure upscale deflection, the ± terminal of the current coil is toward the source, while the ± terminal of the voltage coil is connected to the same line as the current coil. Re versing both coil connections still results in upscale deflection. However, reversing one coil and not the other results in downscale deflection and no wattmeter reading.

Solution:

1. Define. The problem is clearly defined. Interestingly, this is a problem where the student could actually v alidate the results by doing the problem in the laboratory with a real wattmeter.

• 2. Present. This problem consists of finding the average power delivered to a load by an external source with a series impedance.
• 3. Alternative. This is a straightforward circuit problem where all we need to do is find the magnitude and phase of the current through the load and the magnitude and the phase of the voltage across the load. These quantities could also be found by using PSpice, which we will use as a check.
• 4. Attempt. In Fig. 11.32, the w attmeter reads the average power absorbed by the (8 − j6) Ω impedance because the current coil is in series with the impedance while the v oltage coil is in parallel with it. The current through the circuit is

The impedance while the voltage coil is in
int through the circuit is

$$I_{\rm rms} = \frac{150/0^{\circ}}{(12 + j10) + (8 - j6)} = \frac{150}{20 + j4} A$$

The voltage across the (8 − j6) Ω impedance is

$$\mathbf{V}_{\rm rms} = \mathbf{I}_{\rm rms}(8 - j6) = \frac{150(8 - j6)}{20 + j4} \text{ V}$$

The complex power is

$$\mathbf{S} = \mathbf{V}_{\text{rms}} \mathbf{I}_{\text{rms}}^* = \frac{150(8 - j6)}{20 + j4} \cdot \frac{150}{20 - j4} = \frac{150^2(8 - j6)}{20^2 + 4^2}$$ $$= 423.7 - j324.6 \text{ VA}$$

The wattmeter reads

$$P = \text{Re}(S) = 432.7 \text{ W}$$

5. Evaluate. We can check our results by using PSpice.

To check our answer , all we need is the magnitude of the current (7.354 A) flowing through the load resistor:

$$P = (I_L)^2 R = (7.354)^2 8 = 432.7 \text{ W}$$

As expected, the answer does check!

6. Satisfactory? We ha ve satisf actorily solv ed the problem and the results can now be presented as a solution to the problem.

For Practice Prob. 11.16.

Answer: 1.437 kW.

11.9.2 Electricity Consumption Cost

In Section 1.7, we considered a simplified model of the way the cost of electricity consumption is determined. But the concept of po wer factor was not included in the calculations. Now we consider the importance of power factor in electricity consumption cost.

Loads with low power factors are costly to serve because they require large currents, as explained in Section 11.8. The ideal situation would be to draw minimum current from a supply so that S = P, Q = 0, and pf = 1. A load with nonzero Q means that energy flows back and forth between the load and the source, gi ving rise to additional po wer losses. In vie w of this, power companies often encourage their customers to have power factors as close to unity as possible and penalize some customers who do not improve their load power factors.

Utility companies divide their customers into categories: as residential (domestic), commercial, and industrial, or as small po wer, medium power, and large power. They have different rate structures for each category. The amount of energy consumed in units of kilowatt-hours (kWh) is measured using a kilo watt-hour meter installed at the customer’ s premises.

Although utility companies use different methods for charging customers, the tarif f or char ge to a consumer is often tw o-part. The first part is fixed and corresponds to the cost of generation, transmission, and distribution of electricity to meet the load requirements of the con sumers. This part of the tarif f is generally e xpressed as a certain price

per kW of maximum demand. Or it may be based on kVA of maximum demand, to account for the power factor (pf) of the consumer. A pf penalty charge may be imposed on the consumer whereby a certain percentage of kW or kVA maximum demand is charged for every 0.01 fall in pf below a prescribed value, say 0.85 or 0.9. On the other hand, a pf credit may be given for every 0.01 that the pf exceeds the prescribed value.

The second part is proportional to the ener gy consumed in kWh; i t may be in graded form, for example, the first 100 kWh at 16 cents/kWh, the next 200 kWh at 10 cents/kWh and so forth. Thus, the bill is determined based on the following equation:

Total Cost = Fixed Cost + Cost of Energy (11.64)

A manufacturing industry consumes 200 MWh in one month. If the Example 11.17 maximum demand is 1,600 kW, calculate the electricity bill based on the following two-part rate:

Demand charge: $5.00 per month per kW of billing demand. Energy charge: 8 cents per kWh for the first 50,000 kWh, 5 cents per kWh for the remaining energy.

Solution:

The demand charge is

$5.00 × 1,600 = $8,000 (11.17.1)

The energy charge for the first 50,000 kWh is

$0.08 \times 50,000 = $4,000 \tag{11.17.2}

The remaining energy is 200,000 kWh− 50,000 kWh = 150,000 kWh, and the corresponding energy charge is

$0.05 × 150,000 = $7,500 (11.17.3)

Adding the results of Eqs. (11.17.1) to (11.17.3) gives

Total bill for the month = $8,000 + $4,000 + $7,500 = $19,500

It may appear that the cost of electricity is too high. But this is often a small fraction of the overall cost of production of the goods manufactured or the selling price of the finished product.

The monthly reading of a paper mill’s meter is as follows:

Maximum demand: 48,000 kW Energy consumed: 750 MWh

Using the two-part rate in Example 11.17, calculate the monthly bill for the paper mill.

Answer: $279,000.

Practice Problem 11.17

Example 11.18 A 300-kW load supplied at 13 kV (rms) operates 520 hours a month at 80 percent power factor. Calculate the average cost per month based on this simplified tariff:

Energy charge: 6 cents per kWh

Power-factor penalty: 0.1 percent of energy charge for every 0.01 that pf falls below 0.85.

Power-factor credit: 0.1 percent of energy charge for every 0.01 that pf exceeds 0.85.

Solution:

The energy consumed is

$$W = 300 \text{ kW} \times 520 \text{ h} = 156,000 \text{ kWh}$$

The operating power factor pf = 80% = 0.8 is 5 × 0.01 below the prescribed power factor of 0.85. Since there is 0.1 percent energy charge for every 0.01, there is a power-factor penalty charge of 0.5 percent. This amounts to an energy charge of

$$\Delta W = 156,000 \times \frac{5 \times 0.1}{100} = 780 \text{ kWh}$$

The total energy is

$$W_t = W + \Delta W = 156,000 + 780 = 156,780 \text{ kWh}$$

The cost per month is given by

Cost = 6 cents × Wt = $0.06 × 156,780 = $9,406.80

An 500-kW induction furnace at 0.88 power factor operates 20 hours per day for 26 days in a month. Determine the electricity bill per month based on the tariff in Example 11.18. Practice Problem 11.18

Answer: $15,553.20.