Calculating Motor Speed:

A squirrel cage induction motor is a constant speed device. It cannot operate for any length of time at speeds below those shown on the nameplate without danger of burning out.

To Calculate the speed of a induction motor, apply this formula:

Srpm = 120 x F            P

Srpm = synchronous revolutions per minute.120   = constantF       = supply frequency (in cycles/sec)P       = number of motor winding poles

Example: What is the synchronous of a motor having 4 poles connected to a 60 hz power supply?

Srpm = 120 x F            P

Srpm = 120 x 60            4

Srpm = 7200             4

Srpm = 1800 rpm

Calculating Braking Torque:

Full-load motor torque is calculated to determine the required braking torque of a motor.To Determine braking torque of a motor, apply this formula:

T =  5252 x HP     rpm

T      = full-load motor torque (in lb-ft)5252 = constant (33,000 divided by 3.14 x 2 = 5252)HP    = motor horsepowerrpm = speed of motor shaft

Example: What is the braking torque of a 60 HP, 240V motor rotating at 1725 rpm?

T =  5252 x HP     rpm

T =  5252 x 60      1725

T =  315,120      1725

T = 182.7 lb-ft

Calculating Work:

Work is applying a force over a distance. Force is any cause that changes the position, motion, direction, or shape of an object. Work is done when a force overcomes a resistance. Resistance is any force that tends to hinder the movement of an object.If an applied force does not cause motion the no work is produced.

To calculate the amount of work produced, apply this formula:

W = F x D

W = work (in lb-ft)F  = force (in lb)D  = distance (in ft)

Example: How much work is required to carry a 25 lb bag of groceries vertically from street level to the 4th floor of a building 30' above street level?

W = F x DW = 25 x 30W = 750 -lb

Calculating Torque:

Torque is the force that produces rotation. It causes an object to rotate. Torque consist of a force acting on distance. Torque, like work, is measured is pound-feet (lb-ft). However, torque, unlike work, may exist even though no movement occurs.

To calculate torque, apply this formula:

T = F x D

T = torque (in lb-ft)F = force (in lb)D = distance (in ft)

Example: What is the torque produced by a 60 lb force pushing on a 3' lever arm?

T = F x DT = 60 x 3

T = 180 lb ft

Calculating Full-load Torque:

Full-load torque is the torque to produce the rated power at full speed of the motor. The amount of torque a motor produces at rated power and full speed can be found by using a horsepower-to-torque conversion chart. When using theconversion chart, place a straight edge along the two known quantities and read the unknown quantity on the third line.

To calculate motor full-load torque, apply this formula:

T = HP x 5252    rpm

T = torque (in lb-ft)HP = horsepower5252 = constantrpm = revolutions per minute

Example: What is the FLT (Full-load torque) of a 30HP motor operating at 1725 rpm?

T = HP x 5252    rpm

T = 30 x 5252     1725

T = 157,560     1725

T = 91.34 lb-ft

Calculating Horsepower:

Electrical power is rated in horsepower or watts. A horsepower is a unit of power equal to 746 watts or 33,0000 lb-ft per minute (550 lb-ft per second). A watt is a unit of measure equal to the power produced by a current of 1 amp across the potential difference of 1 volt. It is 1/746 of 1 horsepower. The watt is the base unit of electrical power. Motor power is rated in horsepower and watts.Horsepower is used to measure the energy produced by an electric motor while doing work.

To calculate the horsepower of a motor when current and efficiency, and voltage are known, apply this formula:

HP = V x I x Eff        746

HP = horsepowerV    = voltageI     = curent (amps)Eff. = efficiency

Example: What is the horsepower of a 230v motor pulling 4 amps and having 82% efficiency?

HP = V x I x Eff        746

HP = 230 x 4 x .82        746

HP = 754.4        746

HP = 1 Hp

Eff = efficiency / HP = horsepower / V = volts / A = amps / PF = power factor

Horsepower Formulas

To Find Use FormulaExample

Given Find Solution

HPHP = I X E X Eff.

       746240V, 20A, 85% Eff. HP

HP = 240V x 20A x 85%       746HP=5.5

II = HP x 746

     E X Eff x PF10HP, 240V, 

90% Eff., 88% PFI

I = 10HP x 746      240V x 90% x 88%

I = 39 A

To calculate the horsepower of a motor when the speed and torque are known, apply this formula:

HP = rpm x T(torque)         5252(constant)

Example: What is the horsepower of a 1725 rpm motor with a FLT 3.1 lb-ft?

HP = rpm x T         5252

HP = 1725 x 3.1         5252

HP = 5347.5         5252HP = 1 hp

Calculating Synchronous Speed:

AC motors are considered constant speed motors. This is because the synchronous speed of an induction motor is based on the supply frequency and the number of poles in the motor winding. Motor are designed for 60 hz use have synchronous speeds of 3600, 1800, 1200, 900, 720, 600, 514, and 450 rpm.

To calculate synchronous speed of an induction motor, apply this formula:

rpmsyn =  120 x f               Np

rpmsyn = synchronous speed (in rpm)f           = supply frequency in (cycles/sec)Np       =  number of motor poles

Example: What is the synchronous speed of a four pole motor operating at 50 hz.?

rpmsyn =  120 x f               Np

rpmsyn =  120 x 50

              4rpmsyn =  6000

              4rpmsyn = 1500 rpm

Calculations Voltage Drop  Part ONE

The purpose of the National Electrical Code is the practical safeguarding of persons and property from hazards arising by the use of electricity. The NEC does not generally consider voltage drop to be a safety issue. As a result, the NEC contains six recommendations (Fine Print Notes) that circuit conductors be sized sufficiently large enough so that reasonable efficiency of equipment operation can be provided. In addition, the NEC has five rules that required conductors be sized to accommodate the voltage drop of the circuit conductors.

Fine Print Notes in the NEC are for informational purposes only and are not enforceable by the inspection authority [90-5(c)]. However, Section 110-3(b) requires equipment to be installed in accordance with the equipment instructions. Therefore, electrical equipment must be installed so that it operates within its voltage rating as specified by the manufacturer. Figure 1.

Author’s Comment: Figures are not posted on the internet.

Due to voltage drop within the circuit conductors, the operating voltage at electrical equipment will be less than the output voltage of the power supply. Inductive loads (i.e. motors, ballasts, etc.) that operate at voltage below its rating can overheat resulting in shorter equipment operating life and increased cost, as well as inconvenience for the customer. Under-voltage for sensitive electronic equipment such as computers, laser printers, copy machines, etc., can cause the equipment to lock up or suddenly power down resulting in data loss, increased cost and possible equipment failure. Resistive loads (heaters, incandescent lighting) that operate at under-voltages simply will not provide the expected rated power output, Figure 1.

Author’s Comment: Voltage drop on the conductors can cause incandescent lighting to flicker when other appliances, office equipment, or heating and cooling systems cycle on. Though this might be annoying for some, it’s not dangerous and it does not violate the NEC.

NEC RECOMMENDATIONS

The National Electrical Code contains six Fine Print Notes to alert the Code user that equipment can have improved efficiency of operation if conductor voltage drop is taken into consideration.

1. Branch Circuits – This FPN recommends that branch circuit conductors be sized to prevent a maximum voltage drop of 3%. The maximum total voltage drop for a combination of both branch circuit and feeder should not exceed 5%. [210-19(a) FPN No. 4], Figure 2.

2. Feeders – This FPN recommends that feeder conductors be sized to prevent a maximum voltage drop of 3%. The maximum total voltage drop for a combination of both branch circuit and feeder should not exceed 5%. [215-2(d) FPN No. 2], Figure 2.

Example: What is the minimum NEC recommended operating voltage for a 120 volt load that is connected to a 120/240-volt source, Figure 3 (8-11).

(a) 120 volts                 (b) 115 volt                  (c) 114 volts (d) 116 volts

Answer: (c) 114 volts The maximum conductor voltage drop recommended for both the feeder and branch circuit is 5 percent of the voltage source; 120 volts x 5% = 6 volts. The operating voltage at the load is determined by subtracting the conductor’s voltage drop from the voltage source, 120 volts – 6 volts drop = 114 volts.

3. Services – Interestingly there is no recommended voltage drop for service conductors, but this FPN reminds the Code user to consider voltage drop of the service conductors [230-

31(c) FPN].

Author’s Comment: Voltage drop on long service conductors can cause incandescent lighting in the building to flicker when appliances, heating or cooling systems cycle on. For information on how to solve or reduce incandescent lighting flicker, go to: www.mikeholt.com/Newsletters.

4. Conductor Ampacity – This FPN identifies the fact that the ampacities listed in Table 310-16 do not take voltage drop into consideration [310-15 FPN No. 1].

5. Phase Converters – Phase converters have their own recommendation that the voltage drop from the power supply to the phase converter should not exceed 3% [455-6(a) FPN].

6. Recreational Vehicle Parks – Recreational Vehicles have a recommendation that the maximum voltage drop for branch circuit conductors not exceed 3% and the combination of both the branch and feeder not to exceed 5% [210-19(a) FPN No. 4 and 551-73(d) FPN].

NEC REQUIREMENTS

The National Electrical Code also contains five rules requiring circuit conductors to be increased in size to accommodate voltage drop.

Grounding Conductors – This rule states that where circuit conductors are increased in size to compensate for voltage drop, the equipment grounding conductors must also be increased in size [250-122(b)].

Author’s Comment: If, however, the circuit conductors are not increased in size to accommodate voltage drop, then the equipment grounding conductor is not required to be larger than listed in Table 250-122.

Motion Picture/Television Studios – Branch circuit conductor for 60/120 volt systems used to reduce noise in audio/video production or other similar sensitive electronics for motion and television studios must not exceed 1.5%, and the combined

voltage drop of the feeder and branch circuit conductors must not exceed 2.5% [530-71(d)]. In addition, FPN No. 1 to Section 530-72(b) reminds the Code user to increase the size of the grounding conductor in accordance with Section 250-122(b).

Fire Pumps – The operating voltage at the terminals of a fire pump controller must not be less than 15% from the voltage rating of the controller while the motor is starting (lock-rotor current). In addition, the operating voltage at the terminals of the fire pump motor must not be less than 5% from the voltage rating of the motor when the motor is operating at 115 percent of it full-load current rating [695-7].

Author’s Comment: Next month in this article, I’ll give examples and graphics demonstrating the application of the NEC voltage drop rules.

DETERMINING CIRCUIT VOLTAGE DROP

When the circuit conductors have already been installed, the voltage drop of the conductors can be determined by one of two methods: Ohm’s law or the VD formula.

Ohm’s Law Method – Single-Phase Only

Voltage drop of the circuit conductors can be determined by multiplying the current of the circuit by the total resistance of the circuit conductors: VD = I x R. “I” is equal to the load in amperes and ”R” is equal to the resistance of the conductor as listed in Chapter 9, Table 8 for direct current circuit, or in Chapter 9, Table 9 for alternating current circuits. The Ohm’s law method cannot be used for three-phase circuits.

120 volt Example: What is the voltage drop of two No. 12 conductors that supply a 16 ampere, 120 volt load which is located 100 feet from the power supply (200 feet of wire), Figure 4.

(a) 3.2 volts                  (b) 6.4 volts                  (c) 9.6 volts                  (d) 12.8 volts

Answer: (b) 6.4 volts

Voltage Drop = I x R

“I” is equal to 16 amperes

“R” is equal to 0.4 ohms (Chapter 9, Table 9: (2 ohm/1,000 feet) x 200 feet

Voltage Drop = 16 amperes x 0.4 ohms

Voltage Drop = 6.4 volts, (6.4 volts/120 volts = 5.3% volts drop)

Operating Voltage = 120 volts – 6.4 volts

Operating Voltage = 113.6 volts

Author’s Comment: The 5.3% voltage drop for the above branch circuit exceeds the NEC’s recommendations of 3%, but it does not violate the NEC unless the 16 ampere load is rated less than 113.6 volts [110-3(b)].

240 volt Single-Phase Example: What is the operating voltage of a 44 ampere, 240 volt, single-phase load located 160 feet from the panelboard, if it is wired with No. 6 conductors, Figure 5?

(a) 233.1 volts              (b) 230.8 volts              (c) 228.4 volts              (d) 233.4 volts

Answer: (a) 233.1 volts

Voltage Drop = I x R

“I” is equal to 44 amperes

“R” is equal to 0.157 ohms (Chapter 9, Table 9: (.49 ohm/1,000 feet) x 320 feet

Voltage Drop = 44 amperes x 0.157 ohms

Voltage Drop = 6.9 volts, (6.9 volts/240 volts = 2.9% volts drop)

Operating Voltage = 240 volts – 6.9 volts

Operating Voltage = 233.1 volts

Voltage Drop Using the Formula Method

When the circuit conductors have already been installed, the voltage drop of the conductors can be determined by using one of the following formulas:

VD = 2 x K x Q x I x D/CM - Single Phase

VD = 1.732 x K x Q x I x D/CM - Three Phase

“VD” = Volts Dropped: The voltage drop of the circuit conductors as expressed in volts.

“K” = Direct Current Constant: This is a constant that represents the direct current resistance for a one thousand circular mils conductor that is one thousand feet long, at an operating temperature of 75º C. The direct current constant value to be used for copper is 12.9 ohms and 21.2 ohms is used for aluminum conductors. The “K” constant is suitable for alternating current circuits, where the conductors do not exceed No. 1/0.

“Q” = Alternating Current Adjustment Factor: Alternating current circuits No. 2/0 and larger must be adjusted for the effects of self-induction (skin effect). The "Q" adjustment factor is determined by dividing alternating current resistance as listed in NEC Chapter 9, Table 9, by the direct current resistance as listed in Chapter 9, Table 8.

“I” = Amperes: The load in amperes at 100 percent, not 125 percent for motors or continuous loads.

“D” = Distance: The distance the load is located from the power supply, not the total length of the circuit conductors.

“CM” = Circular-Mils: The circular mils of the circuit conductor as listed in Chapter 9, Table 8.

Single-Phase Example: What is the voltage drop for a No. 6 conductor that supplies a 44 ampere, 240 volt, single-phase load located 160 feet from the panelboard, Figure 6?

(a) 4.25 volts                (b) 6.9 volts                  (c) 3 percent (d) 5 percent

Answer: (b) 6.9 volts

VD = 2 x K x I x D/CM

K = 12.9 ohms, Copper

I = 24 amperes

D = 160 feet

CM = No. 6, 26,240 circular mils, Chapter 9, Table 8

VD = 2 wires x 12.9 ohms x 44 amperes x 160 feet/26,240 Circular Mils

VD = 6.9 volts, (6.9 volts/240 volts = 2.9% volts drop)

Operating Voltage = 240 volts – 6.9 volts

Operating Voltage = 233.1 volts

Three-Phase Example: A three-phase 208 volt, 36 kVA load is located 80 feet from the panelboard and it is wired with No. 1 aluminum conductors. What is the voltage drop of the conductors to the equipment disconnect, Figure 7?

(a) 3.5 volts                  (b) 7 volts                     (c) 3 percent (d) 5 percent

Answer: (a) 3.5 volts

VD = 1.732 x K x I x D/CM

K = 21.2 ohms, Aluminum

I = 100 amperes

D = 80 feet

CM = No. 1, 83,690 circular mils, Chapter 9, Table 8

VD = 1.732 x 21.2 ohms x 100 amperes x 80 feet/83,690 Circular Mils

VD = 3.5 volts (3.5 volts/208 volts = 1.7%)

Operating Voltage = 208 volts – 3.5 volts

Operating Voltage = 204.5 volts

I hope this short summary was helpfull. If you want to know more about this subject, please attend our seminar or order our home study video program today.

Voltage Drop Homestudy Program (4 Hour CEU Credit) Voltage drop calculations for branch circuits and feeders are explained in great detail. Subjects covered include: wire sizing, maximum distance, voltage drop, and the effects of, Harmonic currents, multi-wire branch circuits, copper versus aluminum, AC versus DC, metallic versus nonmetallic raceways, skin effect and eddy currents. Includes Articles 210, 215, 230, 250 and 310. Textbook with 2-hour video - $79 [CLV3], Book only $25 [CLW3].

ORDERING INFORMATIONMike Holt Enterprises, Inc. 7310 West McNab Road #201 Tamarac, Florida E-Mail SarinaOur office hours - 8:30 am (sometimes earlier) to 5:00 PM (sometimes later) Eastern standard time.

·         BY PHONE Toll Free, 1-888-NEC Code! When you call us, a real person will be there to answer all of your questions courteously and promptly (no voice mail). If you call before or after hours, simply place your order or leave a message on the answering machine.

·         PAYMENT, We accept VISA, Master Charge, American

Express, Discover, personal and business checks, money orders, and cash. Please make all checks and money orders payable to Mike Holt Enterprises, Inc.

·         SHIPMENT, Orders received by 1 PM, are shipped that day and orders received after 1 PM are shipped the next day (unless back-ordered). To receive your orders quickly we must have your correct address, don't forget the Zip Code. We cannot ship to a Post Office Box. Orders are shipped regular UPS, however Next Day, Second Day, Third Day, and C.O.D. shipments are available.

·         DON'T FORGET to include shipping and handling charges of $6 for orders less than $100 and 4% for orders over $100

 Calculations Voltage Drop

  Part TWO

Electrical equipment is designed to operate at within a given voltage range, typically no less than 10% and no more than 5% from it’s voltage rating.

Example: A typical 230 volt load is designed to operate at not less than 207 volts (-10%) and not more than 242 volts (+5%), Figure 1.

Author’s Comment: Figures are not posted on the internet.

The actual operating voltage dependents on the output voltage from the electric utility and the voltage drop of the circuit conductors. Keep in mind that the voltage from the electric utility is not constant; its lower during peak utility loading and higher during off-peak load periods.

Generally, overvoltage in an electrical system is not a problem, unless there is a wiring error in the electrical system1, however reduced or under voltage can caused inconvenience by flickering lights2, erratic performance of electro-mechanical devices such as relays and contactors, fires, and equipment failures. In particular, sensitive electronic equipment operating at reduced voltage will not have sufficient “ride-through” capability for voltage sags, and fire pump equipment possibly could fail at inadequate voltage.

1See  http://www.mikeholt.com/Newsletters/campus.htm for case studies on overvoltage.

2 See  http://www.mikeholt.com/Newsletters/10-7-99.htm for case studies on flickering lights.

Electric utilities are required by public service commissions to supply electrical power with sufficient voltage and capacity for the loads to be served and for most installations, this is not a problem. Reduced or under voltage is often caused by excessive long service, feeder, and or branch circuit conductors. The sizing of these conductors is important to insure proper operating voltage for a safe and efficient electrical systems.

The actual equipment operating voltage is dependent on the originating voltage, the conductor size (actually its resistance), and the magnitude of the current flowing through the circuit conductors. The originating voltage at times can be increased by adjusting the taps on the transformer and the circuit voltage drop can be reduced by decreasing the load or increasing the conductor circular mil area.

Last month I explained that the Fine Print Notes (FPN) in the NEC about voltage drop is not enforceable as a Code rule. However, The National Electrical Code does require conductors to be sized to accommodate voltage drop for the following purposes:

·        Grounding Conductors – Section 250-122(b)

·        Motion Picture/Television Studios – Section 530-71(d)

·        Fire Pumps – Section 695-7

The following formulas can be used to properly size conductors to prevent excessive voltage drop:

CM (single-phase) = (2 x K x I x D)/VD

CM (three-phase) = (1.732 x K x I x D)/VD

Author's Comment: Download a free Windows 95 Voltage Drop Calculator from www.mikeholt.com.

“CM” = Circular-Mils: The circular mils of the circuit conductor as listed in Chapter 9, Table 8.

“K” = Direct Current Constant: The direct current constant value to be used for copper is 12.9 ohms and 21.2 ohms is used for aluminum conductors.

“Q” = Alternating Current Adjustment: Alternating current circuits No. 2/0 and larger must be adjusted for the effects of self-induction (skin effect). The "Q" adjustment factor is determined by dividing alternating current resistance as listed in NEC Chapter 9, Table 9, by the direct current resistance as listed in Chapter 9, Table 8.

“I” = Amperes: The load in amperes at 100 percent, not 125 percent for motors or continuous loads.

“D” = Distance: The distance the load is located from the power supply, not the total length of the circuit conductors.

“VD” = Volts Dropped: The voltage drop of the circuit conductors as expressed in volts.

Example – Single-Phase

A 5 horsepower motor is located 100 feet from a 120/240 volt panelboard. What size conductor should be used if the motor nameplate indicates the voltage range is between 208-230 volts. Limit the voltage drop to 7.2 volts (3% of the voltage source) and the terminals are rated 75ºC, Figure 2.

(a) No. 10 THHN        (b) No. 8 THHN          (c) No. 6 THHN          (d) No. 4 THHN

• Answer: (a) No. 10 THHN

Section 430-22(a) requires motor conductors to be sized not less than 125 percent of the motor full-load current (28 amperes) as listed in Table 430-148. A No. 10 is rated 35 amperes at 75ºC [Table 310-16 and Section 110-14(c)] and it is suitable to meet the NEC requirements (28 ampere x 1.25 = 35 ampere). In addition, a No. 10 conductor limits the voltage drop to meet the manufacture’s voltage limitation rating [110-3(b)].

Conductor required to limit voltage drop to 3%

CM = (2 x K x I x D)/VD

CM = Wire size, Chapter 9, Table 8

K = 12.9 ohm, copper 

I = 28 ampere

D = 100 feet    

VD = 240 volts x 3% = 7.2 volts

CM = (2 x 12.9 ohms x 28 amperes x 100 feet)/7.2 volts

CM = 10,033, No. 10, Chapter 9, Table 8

Example – Three-Phase

Example: A 25 horsepower, 208 volt three-phase fire pump motor is located 175 feet the service. The fire pump motor controller is located 150 feet from the service (motor 25 feet from controller). What size conductor must be installed to the fire pump motor? Note: Terminals are rated 75ºC, Figure 3.

(a) No. 4 THHN          (b) No. 3 THHN          (c) No. 2 THHN          (d) No. 1 THHN

• Answer: (b) No. 3 THHN

When sizing conductor’s for fire pump motors the following rules must be considered.

Calculation 1.

Section 695-6(c)(2) – No. 3. Branch circuit conductors must be sized no less than 125 percent of the fire pump motor full-load current as listed in Table 430-148 or 430-150, based on 75°C terminal rating [110-14(c)(1)] as listed in Table 310-16.

74.8 ampere x 1.25 = 93.4 ampere, No. 3 THHN at 75°C is rated 100 ampere

Calculation 2.

Section 695-7 – No. 3. The operating voltage at the motor controller terminals shall not drop more than 15 percent below the controller-rated voltage when the motor starts (lock-rotor current).

CM = (1.732 x K x I x D)/VD

CM = Wire size, Chapter 9, Table 8

K = 12.9 ohms, copper

I = 404 ampere (locked-rotor, Table 430-151B)

D = 150 feet    

VD = 31.2 volts (208 volts x 15%)

CM = (1.732 x 12.9 ohms x 404 ampere x 150 feet)/31.2 volts

CM = 43,396, Chapter 9, Table 8 = No. 3

Calculation 3.

Section 695-7 – No. 4. The operating voltage at the terminals of the motor shall not drop more than 5 percent below the voltage rating of the motor while the motor is operating at 115 percent of the full-load current rating of the motor.

CM = (1.732 x K x I x D)/VD                  

CM = Wire size, Chapter 9, Table 8

K = 12.9 ohms, copper

I = 86 ampere (74.8 amperes @115%), Table 430-150

D = 175 feet

VD 5% = 10.4 volts (208 volts x 5%)

CM = (1.732 x 12.9 ohms x 86 ampere x 175 feet)/10.4 volts

CM = 32,332, Chapter 9, Table 8 = No. 4

Caution: For voltage drop, the No. 4 wire is okay from the controller to the motor, but Section 695-6(c)(2) requires the branch circuit conductors to be sized no less than No. 3.

I hope this short summary was helpfull. If you want to know more about this subject, please attend our seminar or order our home study video program today.

Voltage Drop Homestudy Program (4 Hour CEU Credit) Voltage drop calculations for branch circuits and feeders are explained in great detail. Subjects covered include: wire sizing, maximum distance, voltage drop, and the effects of, Harmonic currents, multi-wire branch circuits, copper versus aluminum, AC versus DC, metallic versus nonmetallic raceways, skin effect and eddy currents. Includes Articles 210, 215, 230, 250 and 310. Textbook with 2-hour video - $79 [CLV3], Book only $25 [CLW3].

ORDERING INFORMATIONMike Holt Enterprises, Inc. 

7310 West McNab Road #201 Tamarac, Florida E-Mail Sarina

Our office hours - 8:30 am (sometimes earlier) to 5:00 PM (sometimes later) Eastern standard time.

BY PHONE Toll Free, 1-888-NEC Code! When you call us, a real

person will be there to answer all of your questions courteously and

promptly (no voice mail). If you call before or after hours, simply place

your order or leave a message on the answering machine.

PAYMENT, We accept VISA, Master Charge, American Express,

Discover, personal and business checks, money orders, and cash.

Please make all checks and money orders payable to Mike Holt

Enterprises, Inc.

SHIPMENT, Orders received by 1 PM, are shipped that day and orders

received after 1 PM are shipped the next day (unless back-ordered). To

receive your orders quickly we must have your correct address, don't

forget the Zip Code. We cannot ship to a Post Office Box. Orders are

shipped regular UPS, however Next Day, Second Day, Third Day, and

C.O.D. shipments are available.

DON'T FORGET to include shipping and handling charges of $6 for

orders less than $100 and 4% for orders over $100

Determination of voltage dropScope and content of Wiki EIG

General rules of electrical installation design

Connection to the MV utility distribution network

Connection to the LV utility distribution network

MV & LV architecture selection guide

LV Distribution

Protection against electric shocks

Sizing and protection of conductors

Conductor sizing and protection

Methodology and definition   

Overcurrent protection principles   

Practical values for a protective scheme   

Location of protective devices   

Conductors in parallel   

Practical method for determining the smallest    

allowable cross-sectional area of circuit 

conductors

Determination of voltage drop

Short-circuit current   

Particular cases of short-circuit current   

Protective earthing conductor   

The neutral conductor   

Worked example of cable calculation   

LV switchgear: functions & selection

Protection against voltage surges in LV

Energy Efficiency in electrical distribution

Power factor correction and harmonic filtering

Power harmonics management

Characteristics of particular sources and loads

PhotoVoltaic (PV) installation

Residential electrical installations

ElectroMagnetic Compatibility (EMC)

The impedance of circuit conductors is low but not negligible: when carrying load

current there is a voltage drop between the origin of the circuit and the load terminals.

The correct operation of a load (a motor, lighting circuit, etc.) depends on the voltage at

its terminals being maintained at a value close to its rated value. It is necessary

therefore to determine the circuit conductors such that at full-load current, the load

terminal voltage is maintained within the limits required for correct performance.

This section deals with methods of determining voltage drops, in order to check that:

They comply with the particular standards and regulations in force

They can be tolerated by the load

They satisfy the essential operational requirements

Contents

 [hide]

1     Maximum voltage drop   

2     Calculation of voltage drop in steady load conditions   

o 2.1      Use of formulae   

o 2.2      Simplified table   

o 2.3      Examples   

[edit]

Maximum voltage drop

Maximum allowable voltage-drop vary from one country to another. Typical values for

LV installations are given below in Figure G25.

Type of installations Lighting circuits

Other uses (heating and power)

A low-voltage service connection from a LV public power distribution network

3% 5%

Consumers MV/LV substation supplied from a public distribution MV system

6% 8%

Fig. G25: Maximum voltage-drop between the service-connection point and the point of

utilization

These voltage-drop limits refer to normal steady-state operating conditions and do not

apply at times of motor starting, simultaneous switching (by chance) of several loads,

etc. as mentioned in Chapter A (factor of simultaneity, etc.).

When voltage drops exceed the values shown in Figure G25, larger cables (wires) must

be used to correct the condition.

The value of 8%, while permitted, can lead to problems for motor loads; for example:

In general, satisfactory motor performance requires a voltage within ± 5% of its rated

nominal value in steady-state operation,

Starting current of a motor can be 5 to 7 times its full-load value (or even higher). If an 8%

voltage drop occurs at full-load current, then a drop of 40% or more will occur during start-

up. In such conditions the motor will either:

- Stall (i.e. remain stationary due to insufficient torque to overcome the load torque) with

consequent over-heating and eventual trip-out

- Or accelerate very slowly, so that the heavy current loading (with possibly undesirable

low-voltage effects on other equipment) will continue beyond the normal start-up period

Finally an 8% voltage drop represents a continuous power loss, which, for continuous loads

will be a significant waste of (metered) energy. For these reasons it is recommended that

the maximum value of 8% in steady operating conditions should not be reached on circuits

which are sensitive to under-voltage problems (see Fig. G26).

Fig. G26: Maximum voltage drop

[edit]Calculation of voltage drop in steady load conditions

[edit]Use of formulae

Figure G27 below gives formulae commonly used to calculate voltage drop in a given

circuit per kilometre of length.

If:

IB: The full load current in amps

L: Length of the cable in kilometres

R: Resistance of the cable conductor in Ω/km

 for copper 

 for aluminium

Note: R is negligible above a c.s.a. of 500 mm2

X: inductive reactance of a conductor in Ω/km

Note: X is negligible for conductors of c.s.a. less than 50 mm2. In the absence of any

other information, take X as being equal to 0.08 Ω/km.

ϕ: phase angle between voltage and current in the circuit considered, generally:

- Incandescent lighting: cosφ = 1

- Motor power:

   - At start-up: cosφ = 0.35

   - In normal service: cosφ = 0.8

Un: phase-to-phase voltage

Vn: phase-to-neutral voltage

For prefabricated pre-wired ducts and bustrunking, resistance and inductive reactance

values are given by the manufacturer.

Circuit Voltage drop (ΔU)

in volts

Single phase: phase/phase

Single phase: phase/neutral

Balanced 3-phase: 3 phases (with or without neutral)

Fig. G27: Voltage-drop formulae

[edit]Simplified table

Calculations may be avoided by using Figure G28, which gives, with an adequate

approximation, the phase-to-phase voltage drop per km of cable per ampere, in terms

of:

Kinds of circuit use: motor circuits with cosφclose to 0.8, or lighting with a cosφclose to 1.

Type of cable; single-phase or 3-phase

Voltage drop in a cable is then given by:

K x IB x L

K is given by the table,

IB is the full-load current in amps,

L is the length of cable in km.

The column motor power “cosφ = 0.35” of Figure G28 may be used to compute the

voltage drop occurring during the start-up period of a motor (see example no. 1 after the

Figure G28).

c.s.a. in mm2

Single-phase circuit Balanced three-phase circuit

Motor power Lighting Motor power lighting

Normal service Start-up Normal service Start-up

Cu AI cos φ = 0.8 cos φ = 0.35 cos φ = 1 cos φ = 0.8 cos φ = 0.35 Cos o=1

1.5   24 10.6 30 20 9.4 25

2.5   14.4 6.4 18 12 5.7 15

4   9.1 4.1 11.2 8 3.6 9.5

6 10 6.1 2.9 7.5 5.3 2.5 6.2

10 16 3.7 1.7 4.5 3.2 1.5 3.6

16 25 2.36 1.15 2.8 2.05 1 2.4

25 35 1.5 0.75 1.8 1.3 0.65 1.5

35 50 1.15 0.6 1.29 1 0.52 1.1

50 70 0.86 0.47 0.95 0.75 0.41 0.77

70 120 0.64 0.37 0.64 0.56 0.32 0.55

95 150 0.48 0.30 0.47 0.42 0.26 0.4

120 185 0.39 0.26 0.37 0.34 0.23 0.31

150 240 0.33 0.24 0.30 0.29 0.21 0.27

185 300 0.29 0.22 0.24 0.25 0.19 0.2

240 400 0.24 0.2 0.19 0.21 0.17 0.16

300 500 0.21 0.19 0.15 0.18 0.16 0.13

Fig. G28: Phase-to-phase voltage drop ΔU for a circuit, in volts per ampere per km

[edit]Examples

Example 1 (see Fig. G29)

A three-phase 35 mm2 copper cable 50 metres long supplies a 400 V motor taking:

100 A at a cos φ = 0.8 on normal permanent load

500 A (5 In) at a cos φ = 0.35 during start-up

The voltage drop at the origin of the motor cable in normal circumstances (i.e. with the

distribution board of Figure G29 distributing a total of 1,000 A) is 10 V phase-to-phase.

What is the voltage drop at the motor terminals:

In normal service?

During start-up?

Solution:

Voltage drop in normal service conditions: 

Table G28 shows 1 V/A/km so that:

ΔU for the cable = 1 x 100 x 0.05 = 5 V

ΔU total = 10 + 5 = 15 V = i.e. 

This value is less than that authorized (8%) and is satisfactory.

Voltage drop during motor start-up:

Δ Ucable = 0.52 x 500 x 0.05 = 13 V

Owing to the additional current taken by the motor when starting, the voltage drop at the

distribution board will exceed 10 Volts.

Supposing that the infeed to the distribution board during motor starting is 900 + 500 =

1,400 A then the voltage drop at the distribution board will increase approximately pro

rata, i.e. 

ΔU distribution board = 14 V

ΔU for the motor cable = 13 V

ΔU total = 13 + 14 = 27 V i.e. 

a value which is satisfactory during motor starting.

http://www.electrical-installation.org/wiki/File:FigG29.jpg

Fig. G29: Example 1

Example 2 (see Fig. G30)

A 3-phase 4-wire copper line of 70 mm2 c.s.a. and a length of 50 m passes a current of

150 A. The line supplies, among other loads, 3 single-phase lighting circuits, each of 2.5

mm2 c.s.a. copper 20 m long, and each passing 20 A.

It is assumed that the currents in the 70 mm2 line are balanced and that the three

lighting circuits are all connected to it at the same point.

What is the voltage drop at the end of the lighting circuits?

Solution:

Voltage drop in the 4-wire line:  

Figure G28 shows 0.55 V/A/km

ΔU line = 0.55 x 150 x 0.05 = 4.125 V phase-to-phase

which gives:  phase to neutral.

Voltage drop in any one of the lighting single-phase circuits:

ΔU for a single-phase circuit = 18 x 20 x 0.02 = 7.2 V

The total voltage drop is therefore

7.2 + 2.38 = 9.6 V 

 

This value is satisfactory, being less than the maximum permitted voltage drop of 6%.

Fig. G30: Exampl