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. Single-Phase CalculationsBasic electrical theory tells us that for a single-phase system,
For the sake of simplicity, let's assume the power factor (PF) is unity. Therefore, the above equation becomes
Solving for I, the equation becomes
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
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. 3-Phase CalculationsFor 3-phase systems, we use the following equation:
Again, assuming unity PF and solving this equation for “I,” you get:
Now, if you look at the “1,000 ÷ 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 resulting quantity into “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. For most electrical loads like motors, the current I is lagging behind the voltage V by an angle φ. If currents and voltages are perfectly sinusoidal signals, a vector diagram can be used for representation. In this vector diagram, the current vector can be split into two components: one in phase with the voltage vector (component Ia), one in quadrature (lagging by 90 degrees) with the voltage vector (component Ir). See Fig. L1. Ia is called the active component of the current. Ir is called the reactive component of the current.  Fig. L1 – Current vector diagram The previous diagram drawn up for currents also applies to powers, by multiplying each current by the common voltage V. See Fig. L2. We thus define:
Fig. L2 – Power vector diagram In this diagram, we can see that:
This formula is applicable for sinusoidal voltage and current. This is why the Power Factor is then designated as "Displacement Power Factor". A simple formula is obtained, linking apparent, active and reactive power: S 2 = P 2 + Q 2 {\displaystyle S^{2}=P^{2}+Q^{2}}
A power factor close to unity means that the apparent power S is minimal. This means that the electrical equipment rating is minimal for the transmission of a given active power P to the load. The reactive power is then small compared with the active power. A low value of power factor indicates the opposite condition. Useful formulae (for balanced and near-balanced loads on 4-wire systems):
where: V = Voltage between phase and neutral An example of power calculations (see Fig. L3)Fig. L3 – Example in the calculation of active and reactive power
The calculations for the three-phase example above are as follows: Pn = delivered shaft power = 51 kW P = active power consumed P = P n ρ = 51 0.91 = 56 k W {\displaystyle P={\frac {Pn}{\rho }}={\frac {51}{0.91}}=56\,kW}
S = apparent power S = P c o s φ = 56 0.86 = 65 k V A {\displaystyle S={\frac {P}{cos\varphi }}={\frac {56}{0.86}}=65\,kVA}
So that, on referring to Figure L16 or using a pocket calculator, the value of tan φ corresponding to a cos φ of 0.86 is found to be 0.59 Q = P tan φ = 56 x 0.59 = 33 kvar (see Figure L4). Alternatively: Q = S 2 − P 2 = 65 2 − 56 2 = 33 k v a r {\displaystyle Q={\sqrt {S^{2}-P^{2}}}={\sqrt {65^{2}-56^{2}}}=33\,kvar}
Fig. L4 – Calculation power diagram |
Which of these formulas is used to calculate apparent power of a three-phase power circuit?
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