For consumers to benefit from electrification, electricity must be transmitted over distribution lines from the power supply to these consumers. And because the conductor used for these lines is one of the more expensive components of a mini-grid, there is an incentive to make electrification more affordable by using a smaller, cheaper conductor. However, in the process of transmitting electricity, resistance in the conductor leads to a drop in voltage along the line and to an associated loss of power. Reducing
conductor size can result in (1) poor quality of power at the consumer end of the line (low voltage and more pronounced voltage fluctuations) and (2) loss of power (due to resistive losses in the conductor). As is discussed in Chapter IV, low voltage can result in poor service (e.g., decreasing the light output of incandescent bulbs, making it difficult to ignite fluorescent tubes, or burning out electric motors). Loss of power along the line means extra power must be generated and paid for, if there is sufficient excess generating capacity in the first place; otherwise, fewer consumers could be served by that supply.
Assuming that the power supply is operating to specification, the size and type of conductor used for the mini-grid and, to a lesser extent, the power factor are the sole factors that determine whether an
acceptable voltage can be maintained. The purpose for this section is to show how to calculate the required conductor size. But before this issue can be addressed, the following data is required:
• The first requirement for ensuring properly sized conductor is to establish the expected load the mini-grid is to serve. This is described in the section on demand assessment (p. 44) in Chapter IV. The proper performance of this task is important. Underestimating load will mean that the
grid is undersized, leading to problems mentioned above. Overestimating the expected load will unnecessarily increase the cost of the system. An estimate should also be made for the power factor, ranging from 1.0 for only resistive loading like incandescent lighting to roughly 0.6 for cases where the principal load is uncorrected fluorescent lighting.
• The second requirement is to have mapped the community to be served and, on the map, to have laid out the distribution lines and indicated the locations and expected peak coincident demand of all the envisioned loads, as explained in Chapter V.
• The third requirement is that the desired option for line configuration explained in Chapter VI has been identified or the options at least reduced.
To introduce the various ways of calculating voltage drop along a conductor, the case shown in Fig. 24 will be used. A 2-wire single-phase ACSR line is envisioned, with a 0.30 m spacing between conductors and twelve 25-m spans. It is to be supplied by a 230-V, 50-Hz generator. A major use of the electricity in this example is a mixture of incandescent and fluorescent lighting with an average power factor of 0.9.
The peak coincident demand, which occurs during the early evening hours, expected over the life of the project, is indicated in watts in the figure. (Note that the assumed values for loads are higher than might be the case in a typical mini-grid. They were only selected for the purpose of illustrating the procedures described.)
For this example, it is assumed that the maximum acceptable voltage drop is 6 %. A 13 mm2 aluminum conductor has been tentatively selected and the first task is to determine whether this conductor is suitable.
The basic equations that will be used are reviewed in Appendix 6. The properties of available ACSR conductors under the conditions spacing noted previously are shown in Table 5.
Several approaches to determining voltage drop will be illustrated, beginning with what is usually the roughest estimate but the easiest to make and ending
200
Fig. 24. A portion of a village mini-grid for which the maximum voltage drop is to be calculated.
Conductor
* At 0.30 m conductor spacing served by 50 Hz.
Table 5. Resistance and reactance for ACSR conductor.
with a more involved procedure relying on the use of a spreadsheet and considerably more calculations, although it lends itself easily to computerization. With two of these approaches, it is also possible to estimate the power losses that will be incurred along the conductor.
Rough estimate of voltage drop
While considerable time may be required to obtain a precise estimate of the voltage drop at the end of a line, this first method permits a very good approximation with little effort. Detailed calculations might seem worth the effort, but it should be kept in mind that estimates of load used to calculate the voltage drop are very approximate in the first place, regardless of the care with which this load estimates are made. Performing unduly accurate calculations with estimated numbers is not worth the effort. If carefully applied, both of the first two approaches described here should be sufficient to make a
sufficiently accurate determination of the conductor size which would meet the conditions set for voltage drop.
This first method assumes that the load is more or less evenly distributed along the line. The more reality diverges from this assumption, the less accurate will this method be. For this approach, the maximum peak coincident loads for all consumers are simply summed and this value used in Eqn. (12) in Appendix 6. For the case illustrated in Fig. 24, the total load P = 5.8 kW. This equation leads to the following:
Because this is too high, the next larger conductor size (21 mm2) must be tried:
%
This maximum voltage drop of 5.2 % is acceptable.
The dependence of power loss on actual current in each span varies as the square of that current. If the load seems more or less uniformly distributed along the line, Eqn. 13 can be used of give a rough estimate of power loss in the line:
W
This second method can be used whether or not the distribution of customer load along a line is uniform.
In this case, rather than simply summing the loads along the line, the product of the power taken off at each point along the line and the distance from the beginning of the distribution line to that point must be summed. This is done in the Table 6. Once this total has been obtained, it is substituted in Eqn. (15) to derive the voltage drop:
This voltage drop remains within the acceptable range.
This approach assumes that the voltage drop along this distribution line is small and that each consumer actually received close to the supply voltage. In reality, the voltage decreases slightly as one proceeds away from the power supply. Slight changes in consumer voltage in turn imply slightly increased or decreased current demand by each consumer, depending on the nature of the end-use, and this modified current would mean increased or decreased voltage drop due to these current changes. However, with a properly designed system with limited voltage drop, the error introduced is negligible for purposes of line design.
While this approach will calculate the voltage drop for a given conductor size, it is also possible to derive an equation to calculate the (aluminum) conductor size for a given voltage drop:
If copper conductor were used, the required area would be A as calculated above divided by 1.6.*
While the precise conductor size depends on conductor reactance that in turn depends on the frequency of the supply and equivalent separation of the conductors, the above equation represents an estimate with an assumed frequency of 55 Hz and equivalent separation of 0.30 m. Errors introduced for sizing mini-grids are minimal. Applying this equation to the previous case gives a minimum conductor size of 18 mm2, verifying once more that 21-mm2 conductor is the proper size to use.
Unlike the initial approach which assumed a uniform distribution of loading along the line, there is not an easy way of estimating more accurately the power loss using the above approach. Use of a spreadsheet as is described below would have to be used for this purpose.
Spreadsheet estimate
This approach is an alternative to that just covered and gives the same results for voltage drop. It also permits the calculation of the power loss along a line. It relies on dividing up the entire line into separate spans and calculating the line current, voltage drop, and power loss in each span.† For a single-phase line,
* These equations are obtained by using Eqn. (10) in Appendix 6, substituting Eqn. (1) for r, substituting a linear approximation for x at f = 55 Hz as x = 0.363-0.00075A (see Fig. 137) and solving the resulting quadratic equation for A.
† If the causal relationship between voltage and the current demand of each consumer is known, this can also be included in this model. However, one should not blindly use Eqn. (4) for this purpose. A decrease in the
Node Pn(kW) Ln(km) Pn x Ln Table 6. A good estimate can be obtained by deriving the weighted loading of the distribution line.
Eqns. (4), (9), and (11) in Appendix 6 would be used, respectively. (For other configurations, the
equations in Table 8 would be used.) Then the total voltage drop and loss would be obtained by summing the individual drops and losses. While basically a simple technique, it requires the use of a spreadsheet either completed manually or, preferably, by computer. However, while reliance on a computer can greatly facilitate determination of the voltage drop and power loss at any point in the system under a number of different scenarios, extreme care must be taken to ensure that all embedded equations in the spreadsheet are correct. Results obtained by the computer should always be checked against the estimates obtained using the very simple methods noted above, as there should always be close agreement if all equations have been properly applied.
The contents of each column included in the spreadsheet in Table 7 are explained below:
• Column (1): The number of each span, starting with the remotest pole, in this case span #12 that ends on pole #12. Spans are numbered consecutively outward from the supply, but with the furthest pole listed first.
• Column (2): The power taken off by spurs (i.e., service connections) originating from the pole at the end of that span.
• Column (3): The total load served by that span. This is the sum of the total load served by the spur(s) shown in the previous column plus the total load served by the next span(s) in the outward direction (the value found in the position located immediately above it, or “0” in the case of the first row).
• Column (4): The voltage at the end of that span. As an approximation, this is always set to the generation voltage.
• Column (5): The current flowing in that span equals the current in the next span(s) in the outward direction (found in the previous row) plus the current needed to serve the load in the spurs at the end of that span as calculated using Eqn. (4) in Appendix 6.
• Column (6): The length of that span.
• Column (7): The voltage drop in that span, calculated using the first part of Eqn. (9) in
Appendix 6. The values inserted into that equation are those found in the same row, columns (5) and (6), along with the average power factor for the loads on that line and the values of r and x for the conductor being used along that span (which are shown at the end of Table 7).
• Column (8): The power loss in that span, calculated using the first part of Eqn. (11) in Appendix 6. The values inserted into that equation are the same as those used in column (7), except for reactance that does not affect power loss.
consumer voltage (due to voltage drop in the line from the supply) can cause either an increased or decreased demand for current, depending on the nature of the end-use. For example, at a lower voltage, a light bulb would draw less current while a motor might draw more current. Furthermore, the size and direction of the current change may not be as straightforward to calculate as may first appear. If the voltage supplied to a light bulb decreases 10 %, Eqn. (4) implies the current would increase 10 %. However, in real life, because a bulb is considered a resistive load, the current demand would be proportional to the voltage, e.g., a 10 % lower voltage would imply a current decrease of 10 %. To complicate matters further, at a lower voltage, the filament temperature is lower and its resistance increases, further decreasing the current. As a bottom line, no general approach exists that can be used to calculate actual current drawn by the consumer based on the nominal power drawn by the consumer.
Fortunately, all these variation in current with changes in voltage are minimal, and assuming a constant voltage in the analysis does not introduce significant errors in conclusions drawn when sizing mini-grids.
Each row is completed down to the first span. This is very quickly done using a computerized
spreadsheet after the equations have been inputted. This spreadsheet indicates a voltage drop along the entire line of 12.5 V or 5.4 %, confirming the results of the previous approaches. The power loss during times of peak demand is more precisely calculated to be 240 W, confirming that the estimate made earlier was fairly close.
Effect of conductor size on power loss
While limiting the voltage drop is important in order for electrical appliances to operate as they are intended to, power losses can be important because they represent lost revenues to the system owner. In the example above, energy loss at peak times is about 250 W. If the system operates for only four hours each night and the load remains constant, this will, for example, consume 1.0 kWh nightly or about 360 kWh annually. If the system is supplied by a small diesel genset, this will mean that about 180 liters of fuel costing perhaps $50 is wasted annually.
If one is trying to minimize cost, the question that must next be asked is whether spending more for a large conductor with fewer losses generates savings that are more than the cost of the larger conductor required to achieve this. For example, what would be the effect of selecting the next larger conductor size (34 mm2)?
Placing the new values for resistance and reactance for this new conductor into the spreadsheet in Table 7 will show a power loss of 150 W and a reduction in the voltage drop to 4 %. Supplying this power loss would require about 220 kWh, a reduction of about 100 W or 140 kWh annually. This would cost about
Table 7. This spreadsheet layout is another approach for calculating voltage drop and power loss along a length of distribution line. The span is identified by the same number as the pole at the end of that span.
(1) (2) (3) (4) (5) (6) (7) (8)
Pole (or Demand (kW) Voltage Current Length Volt drop Power loss
span) no. Spurs Main (V) (A) (km) (V) (kW)
12 0.600 0.600 230 2.9 0.025 0.2 0.001
11 0.200 0.800 230 3.9 0.025 0.3 0.001
10 0.600 1.400 230 6.8 0.025 0.5 0.003
9 0.400 1.800 230 8.7 0.025 0.6 0.005
8 0.600 2.400 230 11.6 0.025 0.8 0.009
7 0.600 3.000 230 14.5 0.025 1.0 0.015
6 0.200 3.200 230 15.5 0.025 1.1 0.017
5 0.800 4.000 230 19.3 0.025 1.4 0.026
4 0.000 4.000 230 19.3 0.025 1.4 0.026
3 0.400 4.400 230 21.3 0.025 1.5 0.032
2 1.000 5.400 230 26.1 0.025 1.8 0.048
1 0.400 5.800 230 28.0 0.025 2.0 0.055
TOTALS: 5.80 28.0 0.300 12.5 0.239
Assumptions:
Power factor = 0.9
Resistance = 1.41 ohm/km Reactance = 0.32 ohm/km
$30, resulting in an annual savings of $20 from the previous scenario. However, increasing conductor size would increase the cost of that 300 m of conductor by perhaps $40. Therefore, if the plant operates for more than two years under the assumed conditions, it would be cheaper to use the larger conductor.
In this case, using the larger conductor would not only be cheaper over a couple of years. It would also permit an increase in the size of the load in the future while still maintaining the voltage drop within acceptable limits (although with somewhat increased losses).
Generalized equations
The equations used in the previous example are for a 2-wire, single-phase line. But as was mentioned in Chapter VI, several other configurations are possible and phases will likely not be balanced. The approaches for calculating voltage drop and power loss above can be applied in the same fashion to the other common configurations but with the slight modifications to the governing equation as shown in Table 8.
To use Table 8, the variables Y and Z must first be calculated as follows:
( )
2 105P = total loading on line (kW), either located entirely at end or uniformly distributed along line L = length of the line (km)
r, x = resistance and reactance (ohm/km, see paragraphs beginning on p. 224) Balanced load of P
Three-wire Y/4 Z/4 Y/8 Z/12 1.8
Three-phase
Three-wire, delta Y/2 Z/2 Y/4 Z/6 1.1
Four-wire, wye Y/6 Z/6 Y/12 Z/18 1.5
Table 8. Equations for maximum percent voltage drop (%VD) and total power loss (Pl) along a conductor for different line configurations and load distributions.
E = nominal consumer voltage, i.e., 120 or 230 V
(for single-phase, three-wire: phase-neutral voltage) (for delta configuration: phase-phase voltage) (for wye configuration: phase-neutral voltage)
The percentage voltage drop and power loss in the first four column of data in Table 8 are for a balanced 3-wire, single-phase configuration and for balanced three-phase configurations. As an example of how voltage drop changes if loads are not balanced, the last column in Table 8 includes factors that must multiply the voltage drops for balanced loads (the first or third column of data) to obtain the maximum voltage drop for a 50 % unbalance in loading (for the meaning of a 50 % unbalance, see the footnote on p. 58).*
Using these equations, Box 5 provides a graphical solution for conductor size under the conditions noted.