Electrical and Energy

Welcome to the first in a series of articles focusing on electrical calculation basics. This month, we'll discuss the most fundamental of calculations — those for current (I) and kilowatts (kW). We'll also show you how you can do these calculations “in your head,” with very reasonable accuracy, through the use of constants.

You may ask, “What exactly is a constant?” An example of a constant with which you're very much familiar is pi (π), which is derived by dividing a circle's circumference by its diameter. No matter what the circumference and diameter of the respective circle, their ratio is always pi. You can use constants that apply to specific single- and 3-phase voltages to calculate current (I) and kilowatts (kW). Let's see how to do this.

Basic electrical theory tells us that for a single-phase system,

kW = (V × I × PF) ÷ 1,000.

For the sake of simplicity, let's assume the power factor (PF) is unity. Therefore, the above equation becomes

kW = (V × I) ÷ 1,000.

Solving for I, the equation becomes

I = 1,000kW ÷ V (Equation 1)

Now, if we look at the “1,000 ÷ V” portion of this equation, you can see that by inserting the respective single-phase voltage for “V” and dividing it into the “1,000,” you get a specific number (or constant) you can use to multiply “kW” to get the current draw of that load at the respective voltage.

For example, the constant for the 120V calculation is 8.33 (1,000 ÷ 120). Using this constant, Equation 1 becomes

I = 8.33kW.

So, if you have a 10kW load, you can calculate the current draw to be 83.3A (10 × 8.33). If you have a piece of equipment that draws 80A, then you can calculate the relative size of the required power source, which is 10kW (80 ÷ 8.33).

By using this same procedure but inserting the respective single-phase voltage, you get the following single-phase constants, as shown in Table 1.

For 3-phase systems, we use the following equation:

kW = (V × I × PF × 1.732) ÷ 1,000.

Again, assuming unity PF and solving this equation for “I,” you get:

I = 1,000kW ÷ 1.732V.

Now, if you look at the “1,000 4 1.732V” portion of this equation, you can see that by inserting the respective 3-phase voltage for “V” and multiplying it by 1.732, you can then divide that quantity into the “1,000” to get a specific number (or constant) you can use to multiply “kW” to get the current draw of that 3-phase load at the respective 3-phase voltage. Table 2 lists each 3-phase constant for the respective 3-phase voltage obtained from the above calculation.

Single-phase loads may be connected to a three-phase system in two ways. Either a load may be connected across two of the live conductors, or a load can be connected from a live phase conductor to the neutral conductor. 

Single-phase loads should be distributed evenly between the phases of the three-phase system for efficient use of the supply transformer and supply conductors. If the line-to-neutral voltage is a standard load voltage, for example 230 volt on a 400 volt three-phase system, single-phase loads can connect to a phase and the neutral. Loads can be distributed over three phases to balance the load. Where the line-to-neutral voltage is not the standard voltage for example 347 volts produced by a 600 V system, single-phase loads are connected through a step-down transformer.

In a symmetrical three-phase system, the system neutral has the same magnitude of voltage to each of the three-phase conductors. The voltage between line conductors (Vl) is √3 times the phase conductor to neutral voltage (Vp). That is: Vl = √3Vp.

In some multiple-unit residential buildings of North America, three-phase power is supplied to the building but individual units have only single-phase power formed from two of the three supply phases. Lighting and convenience receptacles are connected from either phase conductor to neutral, giving the usual 120 V required by typical North American appliances. In the split-phase system, high-power loads are connected between the opposite "hot" poles, giving a voltage of 240 V. In some cases, they may be connected between phases of a three-phase system, giving a voltage of 208 V. This practice is common enough that 208 V single-phase equipment is readily available in North America. Attempts to use the more common 120/240 V equipment intended for split-phase distribution may result in poor performance since 240 V heating and lighting equipment will only produce 75% of its rating when operated at 208 V. Motors rated at 240 V will draw higher current at 208 V; some motors are dual-labelled for both voltages.

Where three-phase at low voltage is otherwise in use, it may still be split out into single-phase service cables through joints in the supply network or it may be delivered to a master distribution board (breaker panel) at the customer's premises. Connecting an electrical circuit from one phase to the neutral generally supplies the country's standard single phase voltage (120 V AC or 230 V AC) to the circuit.

The currents returning from the customers' premises to the supply transformer all share the neutral wire. If the loads are evenly distributed on all three phases, the sum of the returning currents in the neutral wire is approximately zero. Any unbalanced phase loading on the secondary side of the transformer will use the transformer capacity inefficiently.

If the supply neutral of a three-phase system with line-to-neutral connected loads is broken, the voltage balance on the loads will no longer be maintained. The neutral point will tend to drift toward the most heavily loaded phase, causing undervoltage conditions on that phase and overvoltage on a lightly loaded phase; the lightly loaded phases may approach the line-to-line voltage, which exceeds the line-to-neutral voltage by a factor of √3, causing overheating and failure of many types of loads.

For example, if several houses are connected through a 240 V transformer, which is connected to one phase of the three-phase system, each house might be affected by the imbalance on the three-phase system. If the neutral connection is broken somewhere in the system, all equipment in a house might be damaged due to over-voltage. A similar phenomenon can exist if the house neutral (connected to the center tap of the 240 V pole transformer) is disconnected. This type of failure event can be difficult to troubleshoot if the drifting neutral effect is not understood. 

With inductive and/or capacitive loads, all phases can suffer damage as the reactive current moves across abnormal paths in the unbalanced system, especially if resonance conditions occur. For this reason, neutral connections are a critical part of a power distribution network and must be made as reliable as any of the phase connections.

Where a mixture of single-phase 120-volt lighting and three-phase, 240-volt motors are to be supplied, a system called high-leg delta is used.