How are voltage drop calculations used in sizing submersible pump power cables?

What happens when the power source for a submersible pump is not located near the pump's control box? For one thing, the pump's power cable must be sized to limit the resultant voltage drop to less than 5%. After all, the pump must be supplied with adequate voltage to ensure efficient operation and to prevent premature motor failure.

The first solution that comes to mind is the use of cable charts. Unfortunately, some cable charts are based on the current to be carried by the respective cable while other charts list only standard horsepowers and are based on the average current required for these horsepower rated motors. Also, some charts are based on a particular make of electric motor, specifically on its current and power factor (PF). The motor current and PF of another make of motor will differ because efficiencies and designs differ. A low efficiency motor with a low PF will draw more current than one with having high efficiency and high PF.

There is another approach, however, that is tailored to the specific situation at hand: voltage drop calculations. Let's see how this approach works.

Basic voltage drop equations

There are different equations and different impedance values used in determining the allowable conductor length for a specific situation. However, four equations can be used regardless of the situation to determine voltage drop.

[V.sub.D] = (L x R x I) [divided by] 1000 (eq. 1)

[V.sub.D] = (L x R x PF x I) [divided by] 1000 (eq. 2)

[V.sub.D] = [(L x I) x (Rcos[Phi] + Xsin[Phi])] [divided by] 1000 (eq. 3)

[V.sub.D] = (L x Z x I) [divided by] 1000 (eq. 4)

where [V.sub.D] = voltage drop (volts)

L = one way length of conductor (ft)

R = resistance of conductor per 1000 ft (ohms)

I = current (amperes)

PF = power factor of motor

X = reactance of conductor per 1000 ft (ohms)

Z = impedance of conductor per 1000 ft (ohms)

[Phi] = angle that voltage leads current (electrical degrees)

You can use Equation 1 when dealing with DC circuits, or AC circuits having a PF of 1. This equation, although frequently used, will not provide the correct voltage drop unless the PF is 1.

Equation 2, although sometimes used when the PF is less than 1, will not provide the correct result when this is the case. The voltage drop across the cable will always be equal to or greater than the current (I) times the resistance (R). Equation 2 will give a result equal to or less than I times R.

Equation 3, which uses both resistance and reactance, is adequate for determining voltage drop in cables providing power to motors with a PF of less than 1. This widely used equation will also work with a PF of 1.

Equation 4 is the most accurate but is not widely used. It turns out to be the same as Equation 3 if the impedance angle is equal to the PF angle. The impedance angle can be found by dividing the cable reactance (X) by the cable resistance (R), and then, using trigonometric tables, finding the inverse tangent (arctan) of this result. The cable's impedance (Z) is equal to the square root of the quantity resistance squared ([R.sup.2]) plus reactance squared ([X.sup.2]). In mathematical terms, this is shown by the following equation.

Z = [square root of([R.sup.2] + [X.sup.2])] (eq. 5.)

Impedance can also be found by using the following equation:

Z = Rcos[Phi] + Xsin[Phi] (eq. 6.)

If this equation is used to find impedance, then Equations 3 and 4 will provide the same answer.

Finding resistance

There are many ways in which you can obtain resistance values for Equations 1, 2, and 3. Conductors have different DC resistance values based on their temperature, length, and diameter. They also have different AC resistance values based on skin effect, proximity effect, the AC frequency, and the DC resistance. Skin effect is a phenomenon whereby the current in a conductor travels along its outer edge instead of uniformily through the conductor's entire cross section. Proximity effect is a phenomenon caused by other conductors close by that are also carrying current.

Determining maximum allowable voltage drop

Cable charts that provide allowable cable lengths are usually compiled through the use of the equations previously discussed; in other words, they are based on calculations that verify a maximum allowable voltage drop for a specific size of cable and make of motor. Regardless of motor make, however, the total voltage drop in a conductor should not exceed 5 % of the circuit voltage.

For 3-phase motors, the voltage reading is line-to-line, but the voltage drop is considered line-to-neutral. Thus, the cable voltage drop for 3-phase motors should not exceed 5% of the line-to-neutral voltage. To find this Voltage, you divide the line-to-line voltage by the square root of 3.

For single-phase motors, the voltage reading and the voltage drop are both considered line-to-line. The previously discussed voltage drop equations should be multiplied by two for single-phase motors because voltage is dropped in the cable both going to and returning from the motor.

Finding the maximum length of a conductor requires knowledge of the cable's impedance, current, and maximum allowable voltage drop. The maximum allowable voltage drop ([V.sub.Dmax]) for a 3-phase motor can be found by using the following equation.

[V.sub.Dmax] = (V [divided by] [square root of 3]) x 5% (eq. 7)

The maximum impedance ([Z.sub.max]) allowed can be found by using the following equation.

[Z.sub.max] = (V [divided by] I) x (0.05 [divided by] [square root of 3]) (eq. 8)

The maximum length allowed ([L.sub.max]) for a 3-phase motor's power cable can be found by using the following equation.

[L.sub.max] = (V [divided by] I) x (0.05 [divided by] [square root of 3]) x (1000 [divided by] Z) (eq. 9)

Looking at Equation 9, we can determine that 0.05 times 1000 divided the square root of 3 approximately equals 28.87. As such, Equation 9 can be simplified as follows.

[L.sub.max] = [V [divided by] (I x Z)] x 28.87 (eq. 10)

For power feeders to single-phase motors, the equation for finding [L.sub.max] is as follows.

[L.sub.max] = [V [divided by] (I x Z)] x [(0.05 x 1000) [divided by] 2] (eq. 11)

The impedance value (Z) of a specific size conductor can be found in the NEC, Chapter 9, Table 9. Various values are provided for both copper and aluminum wire, relative to the type of raceway used (PVC duct, aluminum conduit, steel conduit).

Sample calculation

Suppose we have a 150 hp, 460V motor having a PF of .85 and drawing 191 A. If we decide to use a 4/0 AWG copper cable having an impedance 0.078 ohms, what will the maximum length be so that the 5% maximum allowable voltage drop is not exceeded?

Using Equation 10, we have the following.

[L.sub.max] = [V [divided by] (I x Z)] x 28.87 (eq. 10) = [460V [divided by] (191A x 0.078)] x 28.87 = 891 ft

This is the maximum length allowed for the conductor to maintain a 5% or less voltage drop.

Jeff Duhan is assistant electrical engineer, Sun-Star Electric, Inc., Lubbock, Tex.