CHAPTER

SEVENVOLTAGE-DROP AND POWER LOSS CALCULATION

7-1 THREE-PHASE BALANCED PRIMARY LINES

As discussed in Chap. 5, a utility company strives to achieve a well-balanced distribution system in order to improve system voltage regulation by means of equally loading each phase. Figure 7-1 shows a primary system with either a three-phase three –wire or a three- phase four-wire main. The laterals can be either (1) three-phase three-wire, (2) three-phase four-wire, (3) single-phase with line-to-line voltage, undergrounded, (4) single-phase with line-to-neutral voltage, grounded, or (5) two-phase plus neutral, open-wye.

7-2 NON-THREE-PHASE PRIMARY LINES

Usually there are many laterals on a primary feeder which are not necessarily in three-phase, e.g., single-phase which causes the voltage drop and power loss due to load current not only in the phase conductor but also in the return path.

7-2-1 Single-Phase Two-Wire Laterals with Undergrounded Neutral

Assume that an overloaded single-phase lateral is to be changed to an equivalent three-phase three-wire and balanced lateral, holding the load constant. Since the power input to the lateral is the same as before,

where the subscripts 1ф and 3ф refer to the single-phase and three-phase circuits, respectively. Equation (7-1) can be rewritten as

(√ 3x Vs)I1ф = 3 VsI3ф (7-2)

where Vs is the line-to-neutral voltage. Therefore, from Eq. (7-2),

I1ф = √3 x I3ф (7-3)

Which means that the current in the single-phase lateral is 1.73 times larger than the one in the equivalent three-phase lateral.

The voltage drop in the three-phase lateral can be expressed as

VD3ф = I3ф(R cos θ + X sin θ) (7-4)

And in the single-phase lateral as

VD1ф(KR R cos θ + Kx X sin θ) (7-5)

Where KR and KX are conversion constants of R and X and are used to convert them from their three-phase values to the equivalent single-phase values.

KR = 2.0KX = 2.0 when underground cable is usedKx ≅ 2.0 when overhead line is used, with aproximately a ± 10% accuracy

Therefore Eq. (7-5) can be rewritten as

VD1ф = I1ф(2R cos θ + 2X sin θ) V (7-6)

Or substituting Eq. (7-3) into Eq. (7-6),

VD1ф = 2√3 x I3ф(R cos θ + X sin θ) V (7-7)

By divinding Eq. (7-7) by Eq. (7-4) side by side,

VD1фVD3ф

= 2√3 (7-8)

Which mean that the voltage drop n the single-phase undergrounded lateral is aproximately 3.46 times larger than the one in the equivalent three-phase lateral. Since base voltages for the single-phase and three-phase and three-phase lateral are

VB(1ф) = √3 x Vs.L-N V (7-9)

And VB(3ф) = √3 x Vs. L-N V (7-10)

Eq. (7-8) can be expressed in per units as

VDpu.1фVDpu.1ф

= 2.0 (7-11)

Which mean that the per unit voltage drop in the single-phase undergrounded lateral is two times larger yhan the one in the equivalent three-phase lateral. For example, if the per unit voltage drop in the single-phase lateral is 0.10, it would be 0.05 in the equivalent three-phase lateral.

The power losses due to the load currents in the conductors of the single-phase lateral and the equivalent three-phase lateral are

PL.S1ф = 2 x I21фR W (7-12)

And PLS. 3ф = 3 x I23фR W (7-13)

Respectively. Substituting Eq. (7-3) into Eq. (7-12),

PLS. 1ф = 2(√3 x I3ф)2R (7-14)

And dividing the resultant Eq. (7-14) by Eq. (7-13) side by side,

PLS .1фPLS .3ф

= 2.0 (7-15)

Which means that the power loss due the load currents in the conductors of the single-phases lateral is two times larger than the one in the equivalent three-phase lateral.

Therefore, one can conclude that by changing a single-phase lateral to an equivalent three-phase lateral both the per unit voltage drop and the power loss due to copper losses in the primary line are approximately halved.

7-2-2 Single-Phase Two-Wire Unigrounded Laterals

In general, this system is presently not used due to the following disadvantages. There is no earth current in this system. It can be compared to a three-phase four-wire balanced lateral in the following manner. Since the power input to the lateral is the same as before,

S1ф = S3ф (7-16)

Or Vs x I1ф = 3 x Vs x I3ф (7-17)

From which I1ф = 3 x I3ф (7-18)

The voltage drop in the three-phase lateral can be expressed as

VD3ф = I3ф(R cos θ + X sin θ) V (7-19)

And in the single-phase lateral as

VD1ф = I1ф(KR R cos θ + Kx X sin θ) V (7-20)

Where KR = 2.0 when a full-capacity neutral is usedKR > 2.0 when a reduced-capacity neutral is usedKx ≅ 2.0 when overhead line is used

Therefore, if KR = 2.0 and Kx = 2.0, Eq. (7-20) can be rewritten as

VD1ф = I1ф(2R cos θ + 2X sin θ) V (7-21)

Or substituting Eq. (7-18) into Eq. (7-21),

VD1ф = 6 x I3ф(R cos θ + X sin θ) V (7-22)

Dividing Eq. (7-22) by Eq. (7-19) side by side,

VD1фVD3ф

= 6.0 (7-23)

Which means that the voltage drop in the single-phase two-wire unigrounded lateral with full-capacity neutral is six times larger than the one in the equivalent three-phase four-wire balanced lateral.

The power losses due to the load currents in the conductors of the single-phase two-wire unigrounded lateral with full-capacity neutral and the equivalent thre-phase four-wire balanced lateral are

PLS. 1ф = I21ф (2R) W (7-24)

And PLS. 3ф = 3 x I21фR W (7-25)

Respectively. Substituting Eq. (7-18) into Eq. (7-24),

PLS. 1ф = (3 x I3ф)2(2R) W (7-26)

And dividing Eq. (7-26) by Eq. (7-25) side by side,

PLS .1фPLS .3ф

= 6.0 (7-27)

Therefore, the power loss due to load currents in the conductors of the single-phase two-wire unigrounded lateral with full-capacity neutral is six times larger than the one in the equivalent three-phase four-wire lateral.

7-2-3 Single-Phase Two-Wire Laterals with Multigrounded Common Neutrals

Figure 7-2 shows a single-phase two-wire lateral with multigrounded common neutral. As shown in the figure, the neutral wire is connected in parallel (i.e., multigrounded) with the ground wire at various places through ground electrodes in order to reduce the current in the neutral wire. Ia is the current in the phase conductor, Iw is the return current in the neutral wire, and Id is the return current in the Carson’s equivalent ground conductor. According to Morrison [1], the return current in the neutral wire is

In = ζ 1 Ia where ζ1 = 0.25 to 0.33 (7-28)

And it is almost independent of size of the neutral conductor.

In Fig. 7-2, the constant KR is less than 2.0 and the constant Kx is more or less equal to 2.0 because of conflictingly large Dm (i.e., mutual geometric mean distance or geometric mean radius, GMR) of the Carson’s equivalent ground (neutral) conductor.

Therefore, Morrison’s data [1] (probably empirical) indicate taht

VDpu. 1ф = ζ2 x VDpu. 3ф V where ζ2 = 3.8 to 4.2 (7-29)

And PLS, 1ф = ζ3 x PLS. 3ф W where ζ3 = 3.5 to 3.75 (7-30)

Therefore, assuming that the data from Morrison [1] are accurate,

KR < 2.0 and Kx < 2.0

The per unit voltage drops and the power losses due to load currents can be approximated as

VDpu.1ф ≅ 4.0 x VDpu. 3ф V (7-31)

And PLS. 1ф ≅ 3.6 x PLS. 3ф W (7-32)

For the illustrative problems.

7-2-4 Two-Phase Plus Neutral (Open Wye) Laterals

Figure 7-3 shows an open-wye-connected lateral with two-phase and neutral. Of course, the neutral conductor can be unigrounded or multigrounded, but because of disadvantages the unigrounded neutral is generally not used. If the neutral is unigrounded, all neutral current is in the neutral conductor it self. Theoretically, it can be expressed that

V =Z I (7-33)

Where

V a = Za I a (7-34)

V b = Zb I b (7-35)

It is correct for equal load division between the two phases.

Assuming equal load division among phases, the two-phase plus neutral lateral can be compared to an equivalent three-phase lateral, holding the total kilovoltampere load constant. Therefore

S2ф = S3ф (7-36)

Or 2VsI2ф = 3VsI3ф (7-37)

From which

I2ф = 32

I3ф (7-38)

The voltage-drop analysis can be performed depending upon whether the neutral is unigrounded or multigrounded. If the neutral is unigrounded and the neutral conductor impedance (Zn) is zero, the voltage drop in each phase is

VD2ф = I2ф(KR R cos θ + Kx X sin θ) V (7-39)

Where KR = 1.0

Kx = 1.0

Therefore

VD2ф = I2ф(R cos θ + X sin θ) V (7-40)

Or substituting Eq. (7-38) into Eq. (7-40),

VD2ф = 32

I3ф(R cos θ + X sin θ) V (7-41)

Dividing Eq. (7-41) by Eq. (7-19) side by side,

VD2фVD3ф

= 32

(7-42)

However, if the neutral is unigrounded and the neutral-conductor impedance (Zn) is larger than zero,

VD2фVD3ф

> 32

(7-43)

Therefore in this case some unbalanced voltages are inherent.

However, if the neutral is multigrounded and Zn > 0, the data from Morrison [1] indicate that the per unit voltage drop in each phase is

VDpu.2ф = 2.0 x VDpu.3ф (7-44)

When a full-capacity neutral is used and

VDpu. 2ф = 2.1 x VDpu. 3ф (7-45)

When a reduced-capacity neutral (i.e., when the neutral conductor employed is one or two sizes smaller than the phase conductors) is used.

The power loss analysis also depends upon whether the neutral is unigrounded or multigrounded. If the neutral is unigrounded, the power loss is

PLS. 2ф = I22ф(KR R) (7-46)

Where KR = 3.0 when a full-capacity neutral is usedKR > 3.0 when a reduced-capacity neutral is used

Therefore, if KR = 3.0,

PLS .2фPLS .3ф

= 3 I 22ф R3 I 2 3фR

(7-47)

OrPLS .2фPLS .3ф

= 2.25 (7-48)

On the other hand, if the neutral is multigrounded,

PLS .2фPLS .3ф

< 2.25 (7-49)

Based on the data from Morrison [1], the approximate value of this ratio is

PLS .2фPLS .3ф

≅ 1.64 (7-50)

Which means that the power loss due to load currents in the conductors of the two-phase three-wire lateral with multigrounded neutral is approximately 1.64 times larger than the one in the equivalent three-phase lateral.

7-3 FOUR-WIRE MULTIGROUNDED COMMON-NEUTRAL DISTRIBUTION SYSTEM

Figure 7-4 shows a typical four-wire multigrounded common-neutral distribution system. Because of the economic and operating advantages, this system is used extensively. The assorted secondaries can be, for example, either (1) 120/240 V

Single-phase three-wire, (2) 120/240 V three-phase four-wire connected in delta, (3) 120/240 V three-phase four-wire connected in open-delta, or (4) 120/208 V three-phase four-wire connected in grunded-wye. Where primary and secondary system are both existent, the same conductor is used as “common” neutral for both system. The neutral is grounded at each distribution transformer, at various places where no transformers are connected, and to water pipes or driven ground electrodes at each user’s service entrance. The secondary neutral is also grounded at the distribution transformer and the service drops. Typical values of the resistances of the ground electrodes are 5, 10, or 15 Ω. Under no circumstances should they be larger than 25 Ω. Usually, a typical metal water pipe system has a resistance value of less than 3 Ω. A part of the unalanced, or zero sequence, load current flows in the neutral wire, and the remaining part flows in the ground and/or the water system. Usually the same conductor size is used for both phase and neutral conductors.

Example 7-1 Assume that the circuit shown in Fig. 7-5 represents a single-phase circuit if dimensional variables are used; it represents a balanced three-phase circuit if per unit variables are used. The R + jX represents the total impedance of lines and/or transformers. The power factor of the load is cos θ = cos (θV R - θI). Find the load power factor for which the voltage drop is maximum.

SOLUTION The line voltage drop is

VD = I(R cos θ + X sin θ)

By taking its partial derivative with respect to the θ angle and equating the result to zero,

∂(VD )∂θ

= - IR sin θ + IX cos θ = 0

OrXR

= sin θcosθ

= tan θ

Therefore

Θmax = tan-1 XR

And from the impedance triangle shown in Fig. 7-6, the load power factor for which the voltage drop is maximum is

PF = cos θmax = R

R2+X 2¿1/2¿ (7-51)

Also cos θmax = cos (tan-1 XR

) (7-52)

Example7-2 Consider the three-phase three-wire 240 V secondary system with balanced loads at A, B, and C as shown in Fig. 7-7. Determine the following:

(a) Calculate the total voltage drop, or as it is sometimes called, voltage regulation, in one phase of the lateral by using the approximate method.

(b) Calculate the real power per phase for each load.(c) Calculate the reactive power per phase for each load.(d) Calculate the kilovoltampere output and load power factor of the distribution trnasformer.

SOLUTION

(a) Using the approximate voltage-drop equation, that is,VD = I(R cos θ + X sin θ)

The voltage drop for each load can be calculated asVDA = 30(0.05 x 1.0 + 0.01 x 0) = 1.5 VVDB = 20(0.15 x 0.5 + 0.03 x 0.866) = 2.02 VVDC = 50(0.20 x 0.9 + 0.08 x 0.436) = 10.744 V

Therefore the total voltage drop is

∑VD = VDA

+ VDB + VDC

= 1.5 + 2.02 + 10.744= 14.264 V

Or14.264V

240V = 0.0594 pu V

(b) The real power per phase for each load can be calculated from

P = V I cos θ

Or

PA = 240 x 30 x 1.0 = 7.2 kW

PB = 240 x 20 x 0.5 = 2.4 kW

PC = 240 x 50 x 0.9 = 10.8 kW

Therefore the total real power per phase is

∑ P = PA + PB + PC

= 7.2 + 2.4 + 10.8

= 20.4 kW

(c) The reactive power per phase for each load can be calculated from

Q = VI sin θ

Or

QA = 240 x 30 x 0 = 0 kvar

QB = 240 x 20 x 0.866 = 4.156 kvar

QC = 240 x 50 x 0.436 = 5.232 kvar

Therefore the total reactive power per phase is

∑Q = QA + QB + QC

= 0 + 4.156 + 5.232

= 9.389 kvar

(d) Therefore the kilovoltampere output of the distribution transformer is

S = (P2 + Q2)1/2

= (20.42 + 9.3892)1/2

≅ 22.457 kVA/phase

This the total kilovoltampere output of the distribution transformer is

3 x 22.457 ≅ 67.37 kVA

Hence, the load power factor of the distribution transformer is

Cos θ = ∑ P

S

= 20.4 kW

22.457kVA

= 0.908 lagging

Example 7-3 This example is a continuation of Example 6-1. It deals with voltage drops in the secondary distribution system. In this and the following examples, a single-phase three-wire 120/240 V directly buried underground residential distribution (URD) secondary system will be analyzed, and calculations will be made for motor-starting voltage dip and for steady-state voltage drops of the time of annual peak load. Assume that the cable impedances given in Table 7-2 are correct for a typical URD secondary cable.

Transformer data The data given in Table 7-1 are for modern single-phase 65oC OISC distribution transformers of the 7200-120/240 V class. The data were taken from a recent catalog of a manufacturer. All given per unit values are based on the transformer-rated kilovoltamperes and voltages.

The 2400 V class transformers of the sizes being considered have about 15 percent less R and about 7 percent less X than the 7200 V transformers. Ignore the small variation of impedance with rated voltage and assume that voltage drop calculated with the given data will suffice for whichever primary voltage is used.

URD secondary cable data Cable insulations and manufacture are constantly being improved, especially for high-voltage cables. Therefore, any cable data soon become obselete. The following information and data have been abstracted from recent cable catalogs.

Much of the 600 V class cable now commonly used for secondary lines and services has Al conductor and cross-linked PE insulation which can stand 90oC conductor temperature. The trplexed cable assembly shown in Fig. 7-8 (quadruplexed for three-phase four-wire service) has three or four insulated conductors when aluminium is used. When copper is used, the one grounded neutral conductor is bare. The neutral conductor typically is two AWG sizes smaller than the phase conductors.

The twin concentric cable assembly shown in Fig. 7-9 has two insulated copper or aluminium phase conductors plus several spirally served small bare copper binding conductors which act as the current-carrying grounded neutral. The number and size of the spiral neutral wires vary so that the ampacity of the neutral circuit is equivalent to two AWG wire sizes smaller aluminium/copper XLPE 600 V class cable.

The triplex and twin concentric assemblies obviously have the same resistance for a given size of phase conductors. The triplex assembly has very slightly higher reactance than the concentric assembly. The difference in rectances is too small to be noted unless precise computations are undertaken for some special purpose. The reactances of those cables should be

increased by about 25 percent if they are installed in iron conduit. The reactances given below are valid only for balanced loading (where the neutral current is zero).

The triplex assembly has about 15 percent smaller ampacity than the concentric assembly, but the exact amount of reduction varies with wire size. The ampacities given are for 90oC conductor temperature, 20oC ambient earth temperature, direct burial in earth, and 10 percent daily load factor. When installed in burried duct, the ampacities are about 70 percent of those listed below. For load factors less than 100 percent, consult current literature or cable standards. The increased ampacities are significantly large.

Arbitrary criteria

1. Use the approximate voltage-drop equation, that is,VD = I(R cos θ + X sin θ)

And adapt it to per unit data when computing transformer voltage drops and adapt it to ampere and ohm data when computing service-drop (SD) and secondary-line (SL) voltage drops. Obtain all voltage-drop answers in per unit based on 240 V.

2. Maximum allowable motor-starting voltage dip (VDIP) = 3 percent = 0.03 pu = 3.6 V based on 120 V. This figure is arbitrary; utility parctices vary.

3. Maximum allowable steady-state voltage drop in the secondary system (transformer + SL + SD) = 3.50 percent = 0.035 pu = 4.2 V based on 120 V. This figure also is quite abritrary; regulatory commission rules and utility practices vary. More information about favorable and tolerable amounts of voltage drop will be discussed in connection with subsequent examples, which will involve voltage drops in the primary lines.

4. The loading data for computation of steady-state voltage drop is given in Table 7-3.5. As loadingndata for transient motor-starting VDIP, assume an air-conditioning

compressor motor located most unfavorably. It has a 3-hp single-phase 240 V 80 A locked rotor current, with a 50 percent PF locked rotor.

Assumptions1. Assume perfectly balanced loading in all three-wire single-phase circuits.2. Assume nominal operating voltage of 240 V when computing currents from

kilovoltampere loads.3. Assume 90 percent lagging power factor for all loads.

Using the given data and assumptions, calculate the K constant for any one of the secondary cable sizes, hoping to verify one of the given values in Table 7-2.

SOLUTION Let the secondary cable size be #2 AWG, arbitrarily. Also let the I current be 100 A and the length of the secondary line be 100 ft. Using the values from Table 7-2, the resistance and reactance values for 100 ft of cable can be found as

R = 0.334 Ω/1000 ft x 100 ft

1000 ft

= 0.0334 Ω

And

X = 0.0299 Ω/1000 ft x 100 ft

1000 ft

= 0.00299 Ω

Therefore, using the approximate voltage drop equation,

VD = I(R cos θ + X sin θ)

= 100(0.0334 x 0.9 + 0.00299 x 0.435)

= 3.136 V

Or, in per unit volts,

3.136V120V

= 0.0261 pu V

Which is very close to the value given in Table 7-2 for the K constant, that is, 0.02613 pu V/(104 A . ft) of cable.