ASABE PE Review Session Circuits,

Controls, and Sensors

Robert J. Gustafson, P.E. , PhDHonda Professor for Engineering Education

Director, Engineering Education Innovation CenterProfessor, Food, Agricultural and Biological

EngineeringOhio State University

Gustafson.4@osu.edu 614 292 0573

Topics to be Addressed:

• Review of Some Basic Terms and Concepts• Wire Sizing and Selection• Service Entrance• Sizing a Farm Service• Motor Control and Protection• Lighting Levels and Selection• Power Factor CorrectionProcess - I will introduce topic, do example, ask you to

do a similar example in some cases.

References Being Used:

NFEC, 2009, Agricultural Wiring Handbook, 15th Ed. (order at www.rerc.org, $22)

MWPS-28, 2005,Wiring Handbook for Rural Facilities, 3nd Ed. (order at www.mwps.org , $20)

Gustafson and Morgan. 2004. Fundamentals of Electricity for Agriculture ASABE, St. Joseph, MI (ASABE.org; Technical Library – Publications Included – Textbooks – Fund of Elec – Chaps 1 – 5, 8, and 19 (appendix))

Review of Some Basic Terms and Concepts

SUMMARY OF SYMBOLS, ABBREVIATION AND RELATIONS QUANTITY SYMBOL BASE UNIT ABBREV. ___________________________________________________ Current I Ampere A Voltage E Volt V Energy Ws Joule J (kWh 3.6 x 106 J) Resistance R Ohm ohm or Ω Power P Watt W Resistivity ρ Ohm-cm Ω cm

Ohm's Law

E = IR R = E/I I = E/R

Power Relations ac System P = EIcos ɸ = I2Rcos ɸ = (E2/R)cos ɸ cos ɸ = power factor

Power Factor =True Power/Apparent Power

= Watts /(Volts * Amperes) dc System P = EI = I2R = E2/R

3-phase systems

P = 3EPIP cos = 30.5EL-LIL

cos

where P = true power

EP = phase voltage EL-L = line-to-line voltage

IP = phase current IL = line current

cos = power factor

What you will most likely need for planning:

Assume cos power factor Power = 30.5EL-LIL

EnergyEnergy = Power * Time

or Power integrated over time

Water pump

Drain

p1 (psi) p2Gage Pressure

Direction of water flow

q = water flow (gpm)50

Water flow in a pipeWater flow in a pipe

0 p3

10

Voltage to Ground

Current flow in a wireCurrent flow in a wire

V1 (volts)V2

Voltage Source

0

Ground

Direction current flow

50I = current flow (amps) 10

V3

Fluid Circuit

Electrical CircuitLoad

Voltage to Ground

Current flow in a wireCurrent flow in a wire

V1 (volts)

Voltage Source

0

Ground

Direction current flow

50I = current flow (amps)

V3

For this circuit. If the current is measured at 10 amperes, what is the total resistance of the circuit?

a) 5 wattsb) 5 ohmsc) 10 ohmsd) 10 amperes

Load

Ohm's Law E = IR R = E/I I = E/R

Problem 1

Voltage to Ground

Current flow in a wireCurrent flow in a wire

V1 (volts)

Voltage Source

0

Ground

Direction current flow

50I = current flow (amps)

V3

For this circuit. If the current is measured at 10 amperes,

what is the total resistance of the circuit?

a) 5 wattsb) 5 ohmsc) 10 ohmsd) 10 amperes

Solution: R = E / I

R = 50 V / 10 A = 5 ohms

Load

Energy Management (Power and Energy Terms)

Recall:Power units Watts

If just resistive loads P = E x I (volts x amps)If with Power Factor P = E x I x power factor

Energy units kWh (integral of power over time)Energy = P x Time (Watts x Hr)

Cost = Energy X Rate (kWh x $/kWh)

Example ProblemsHurd Farms has a 1,200 sow farrowing facility with 4

farrowing rooms, each room having 50 heat lamps. If they switch from 250 W heat lamps to 175 W heat lamps, how much can they expect to save in a year? (They pay on average $0.09/kWh)

a) $11,800 per yearb) $3,500 per yearc) $350 per yeard) -$210.00 per year

Solution

Power4 rm x 50 lamps/rm x (250 -175 W/lamp)

= 15,000 W = 15 kW

Solution

Power4 rm x 50 lamps/rm x (250 -175 W/lamp)

= 15,000 W = 15 kWEnergy = Power x Time (need to assume time)

Solution

Power4 rm x 50 lamps/rm x (250 -175 W/lamp)

= 15,000 W = 15 kWEnergy = Power x Time (need to assume time)

Assume on 30% of time15kW x .3 x 8760 hr/yr = 39,420 kWh/yr

Solution

Power4 rm x 50 lamps/rm x (250 -175 W/lamp)

= 15,000 W = 15 kWEnergy = Power x Time (need to assume time)

Assume on 30% of time15kW x .3 x 8760 hr/yr = 39,420 kWh/yrCost = Energy x Rate39,420 kWh/yr x $0.09/kWhr = $3,547/yr

Your Problem

How much should they expect to pay to run heat lamps for a year?Summary of data: 4 rooms; 50 lamps/rm; 175 W/lamp; $0.09/kwHra) $413b) $13,797c) $4,139d) $8,278

Could do a similar calculation for Energy and Cost savings for compact fluorescent for incandescent bulbs.

Another Example - Fact - Typical passenger car and light truck alternators are rated around 50-

70 amperesIf you assume 50% efficient, approx. how many

horsepower is going into running the alternator at rated load?

a. 0.2 hpb. 2 hpc. 20 hpd. 20 kW

Example Solution

Assume 12 V

Output P = E x I = 70 A x 12 V = 840 WInput P = Output P/Efficiency = 840/0.5 = 1680 W

1680 W x 1 hp/746W = 2.2 hp

b. 2 hp

Quick Summary

• Voltage – Electrical Pressure E Volts• Current – Electrical Flow I amperes • Resistance – OhmsOhms Law E = I x R

Power = E x I x power factor (W)

Energy = Power x Time (Wh or kWhr)

What is Coming Next 1) Look at selecting and sizing conductors to do

branch circuits or feeders2) Look at Sizing a building service entrance

panel 3) Look at Sizing the main service for a

farmstead

Wire Sizing and Selection1) Wire Size (cross-section of material)2) Material (copper or aluminum)3) Insulation (cable or conduit)• Keys

– Material Suitable to Environment– Adequate for “Allowable Ampacity”– Acceptable for “Voltage Drop”

Meeting Criteria• Allowable Ampacity - Limit to Not Overheat the Wire

and Insulation - From Tables

• Allowable Voltages Drop – Voltage drop due to resistance of conductors– Usually use 2% for Branch Circuits and 3% for Feeders

(Total should not exceed 5%)

• Must meet both Criteria– Short run Allowable Ampacity– Long Run Voltage Drop

Wire Size & MaterialSize• AWG – American Wire Gauge

– No. 14 No. 12 No. 10 …..No. 0…No. 0000 • Larger Number Smaller Size• No. 12 smallest for agr wiring

• kcmil – Thousands Circular MilsLarger Number Larger Area

Suitable for Environment• Material

– Copper– Aluminum

• Insulation (Used in Agr Wiring) – NM, nonmetallic sheathed cable Branch Circuits– UF, Underground feeder Branch Circuits– USE, Underground Service Entrance Underground Service

Tables for Wire Selection – MWPS Version

Tables for Wire Selection – Ag Wiring Handbook Version

Allowable Ampacity by Insulation Type

Voltage Drop at 2% - Distance Factor

Example Exercise• Select a UF Cable for a 20 Amp, 120 Volt load

at a distance of 190 feet (2% drop)

Assume a 30 Amp load at 240 V (2% drop allowable) with a one way distance of 90 ft. What copper wire size with UF insulation should I use?

a. #12 b. #10 c. #8 d. #6

Service Entrance Panel (SEP)

Circuit Breakers• 240 V (double)• 120 V (single)• Circuit breakers are rated in

amperes. Except for motor circuits, the circuit breaker must have a rating in amperes not greater than ampacity of wire

• Std. Ampacities – 15, 20, 30, 40• Correspond to Wire Sizes, like:

– 15 amp # 14 Cu, 20 amp #12 Cu

Sizing Building Panel (Ampacity at 240V) by Determining Demand Load

Load without diversity – largest combination of loads likely to operate at any time. (Based on judgment)

Example

Note: Use 1.5 A @ 120 V for each light and Duplex Convenience outlets (DCOs), unless you know the load.

Example ContinuedEquipment operating without diversity Amperage at 240V

All lights 18.0 ADCOs with heat lamps 20.0 ADCOs at 1.5 A 4.5 ACold Weather fans 7.2 ATwo heaters 25.0 AAuger motor 15.5 A

Total load without diversity 90.2 A

Compute the demand load:L.W.D at 100% 90.2 ARemaining load at 50% (105.7 A- 90.2 A x 0.5) 8.0Total Demand Load 98.2 AUUse a service entrance main breaker rated greater than 98.2 A; 100 A is the

next larger size. Consider increasing to allow for future expansion.Standard Sizes: 100, 125, 150, 200, 225, 300, 400, and 600 Amp.

Sizing the Main Disconnect for a Farmstead

Capacity of Main Farmstead Service(source: Natl Elec Code Table 220.41)

Computed Demand Load Demand Factor ____________________________________ Residence 100% All other loads: Largest load 100% 2nd largest load 75% 3rd largest load 65% Sum of remaining loads 50%

Example for Main Disconnect Ampacity

Load Demand

Residence 150 A x 100% = 150 A

Largest load - Barn 130 A x 100% = 130 2nd largest - P. House 80 A x 75% = 60 3rd largest - Shop 75 A x 65% = 49 Remainder - Well 15 A x 50% = 8

______ 397 A

The total demand load = 397 A at 240 V for the farmstead.

Minimum Farmstead Service would be:

a) 300 Ampb) 367 Ampc) 400 Ampd) 100 Amp

Quick Summary 2

• Wire Selection Keys– Ampacity– Line Loss– Insulation and wire materials

• Service Entrance Ampacity– Based on Demand Load System

• Full Farmstead Ampacity– Based on Demand Load System

Special Consideration: Motor Circuits (NFEC Agr Wiring Handbook, Part IV)

Likely Questions:1. Wiring Sizing2. Overcurrent

Protection Rating

MOTOR NAMEPLATE INFORMATION

• http://www.elongo.com/pdfs/MotorNameplate990519.pdf

MOTOR NAMEPLATE INFORMATION Design and rating standards developed by the National Electrical Manufacturer's Association (NEMA) permit the comparison of motors from different manufacturers.

Information on the nameplate may include any or all of the following: VOLTS , the proper operating voltage, may be either a single value or, for dual-voltage motors, a dual value. AMPS is the full-load current draw in amperes with the proper voltage supply. When a dual number is listed,

the motor will draw the smaller amperage when connected to the higher voltage source. RPM is the rotor speed when the motor runs at the full-load point on the torque-speed curve (Figure 3.12). HZ is the design operating frequency of the electrical supply. In the United States, it is 60 cycles per second. A

standard frequency of 50 cycles per second is used in some counties. FR is one of the standard frame numbers used by manufacturers to insure interchangeability. For motors with

power ratings below 0.75 kW (1.0 hp), common frame numbers are 42, 48, and 56. The frame number divided by 6.3 (16) gives the height in cm (inches) from the bottom of the mounting to the shaft centerline. Letters may be added to specify the type of mounting, for example, T-frame or the heavier U-frame. A replacement motor with the same frame number as the original motor will fit on the same mounting.

DUTY indicates whether the motor is rated for continuous or intermittent; HOURS may be used to indicate the length of time the motor can be safely operated during intermittent duty.

TEMPERATURE RISE ( C) may be stated as the allowable temperature rise above a 40 C (104 F) ambient temperature while the motor is operating at full load. Often, a motor can be operated at 10% to 15% overload without damage, but the motor temperature should never exceed 55 C (131 F). If, while operating, a motor is not too hot to touch, it is not overheated.

As an alternative to temperature rise, the allowable AMBIENT TEMPERATURE may be listed. Then the motor can be operated at full load in environments with temperatures below the stated ambient temperature.

SF , the service factor, is multiplied by the rated power to obtain the permissible loading. For example, a service factor of 1.10 means the motor could be operated at 10% overload without overheating. Service factors for farm-duty motors can be 1.35 or more.

INSULATION CLASS is a temperature-resistance rating of the insulation on the wires in the motor. Typical classes are A, B, F, or H, where class A is the lowest temperature rating. Class A or B insulation is used in most farm-duty motors.

The CODE LETTER is used to determine the maximum rating of the motor branch-circuit protection and is based on the locked-rotor current drawn by the motor. The following equation may be used to calculate the locked-rotor starting current from the code letter:

(3.4)

where amps = starting current in amperes kVA = rating from the National Electric Code (NEC) hp = rated power from nameplate, in hp volts = supply voltage in volts C ph = constant = 1.0 for single-phase motor or 1.73 for three-phase motor

A DESIGN letter may be given on the nameplate as an indication of their starting-to-rated currents and starting-to-rated torques. The five classes for squirrel-cage motors are A, B, C, D, and F, with A and B being the most common. Design A has starting current 6 to 7 times rated current and starting torque 150% of rated. Design B has starting current 5.5 to 6 times rated current and starting torque 150% of rated.

A THERMAL PROTECTION indication on the nameplate indicates the motor is equipped with such protection to prevent overheating the windings. Protection may be provided by sensing motor current or temperature in the windings and shutting off the motor when either becomes excessive. After shutdown, the motor must be reset manually unless it is equipped with an automatic reset.

Motor Branch Circuit Wire Sizing

• Key is what ampacity to use:– Individual Motor -- use 125% of full load motor

current (from nameplate or table)– Several Motors (or motors and other equipment)

-- use 125% of full load current of LARGEST motor and 100% of all other loads

• Example

A grain dryer has a 5 hp/ 240 V fan motor (nameplate 28 Amp) and 2 hp (nameplate 12 Amp) stirring device. What is ampacity of wire needed for branch circuit needed?

Motor 28 A x 1.25 = 35 AStirring 12 ATotal = 35 A + 12 A = 47 A• Use 47 A to size wire

47 Amp, Assuming UF Cable and distance of 50 Ft, what is wire size?

a) 10b) 8c) 6d) 4

Special Consideration: Motor Circuits (NFEC Agr Wiring Handbook, Part IV)

Overcurrent Protection Rating

MOTOR OVERLOAD PROTECTION

(A) For motors over one hp; see NEC 430-32(a)-- Select overload devices using motor nameplate amp rating. One of the following is required:

Manufacturer Selected:I. A thermal protective device integral with the motor that will prevent dangerous overheating of the motor due to overload or failure to start.

You Select:2. A separate overcurrent device that will trip at 125% of full-load current for motors with a marked temperature rise not over 40°C or service factor of 1.15 or more. For all other motors, use 115%.

In cases where motor size does not match the size of the protective device for the motor, use the next higher size. The rating should not exceed 140% of full-load current for motors with a temperature rise up to 40°C, or with a service factor of 1.15 or more;

Use 130% for all other motors (NEC 430-34).

Example The nameplate for a 5-horsepower motor reads

as follows:Volts: 240 V Cycle: 60 Phase: singleService Factor: 1.15 Code: C Amp: 24What would you pick for Overcurrent a) 24 b) 28 c) 30 d) 35The maximum allowable rating for motor over-

load protection (amps) would be?

Suggested Reference for Motor CircuitsBuilding Approach:Section 26 and 27 of NFEC Agr. Wiring HandbookSection 8.8 of Fundamentals of Electricity for Agriculture

Machinery Systems Approach:Chapter 3 Electrical Power for Agricultural Machines Ajit K. Srivastava, Carroll E. Goering, Roger P. Rohrbach, Dennis R. Buckmaster

Published in Engineering Principles of Agricultural Machines, 2nd Edition Chapter 3, pp. 45-64 ASABE

Chapter 7 Electrical Systems Published in Engine and Tractor Power Chapter 7, pp. 143-182 ( 2004 ASABE).

Lighting Levels and Selection

Some Useful Terms to recall:

Energy In WattsLight Out LumensEfficiency Lumens/WattLight on Surface lx (lux is SI unit)

fc (footcandle is English)Conversion 1 fc = 10.76 lx

Suggested References:

ASAE EP344.3 JAN2005 Lighting Systems for Agricultural Facilities- gives requirements for various tasks- give some data on sources

Sec 12 and 13 of Agr. Wiring Handbook- gives some general examples as guidance

Agr Wiring Handbook – Rules of Tumb

Assuming luminaires (light source) hangs 7 to 10 feet from floor (true for many livestock facilities) – Guideline 3 lumens of lamp output per square foot of floor area yields 1 footcandle at work level.

So if we want to light a 12 ft by 20 ft farrowing room with 26W Compact Fluorescent Bulbs (1,655 lumens per bulb), how many do we need?

Light level required: ASABE Std Farrowing 50 – 100 lxConvert to fc Conversion (Recall 1 fc = 10.76 lx)

4.6 to 9.2 fcUsing Rule of Tumb; 3 lumens for 1 fc/ft2

[4.6 to 9. 2 fc] x 3lumens/1 fc/ft2

(13.8 to 27.6) lu/ft2 x 12 ft x 20 ft /(1655 lu/bulb)

Result - 2 to 4 bulbs

Rules of Thumb on Spacing

• Spaced 1 to 1.5 times mounting height if desired illumination 5 to 10 fc.

• Greater than 10 fc limit to 1 times height

Power Factor Correction

Main Reference – Chapter 3 (3.5& 3.6) of Fundamental of Electricity for Agr.

Recall: Power Factor = cos ɸ = Cosine of phase shift angle

between AC current and voltagePhase Shift caused by inductance or capacitanceTrue Power = Apparent Power x Power Factor Watts = (Volts x Amperes) x p.f.

Why Should I care?True Power = Apparent Power x Power FactorWatts = (Volts x Amperes) x p.f

True Power (Watts)does useful workwhat meter measures

Reactive Powerdoes no “useful work”for inductor related to magnetic field

For a fixed voltage, if current “in phase” ( ɸ = 0 & p.f. = 1) with voltage, we have a minimum current to supply true power

Answer

Reduces Current between source and load-This reduces line loss for distribution system-This may mean smaller wire size between source and load

Example Power Factor Correction – Motor Given a 220 V single-phase 60 Hz induction motor that draws 7.6 amperes with a power factor of 0.787, calculate the size of a parallel connected capacitor required to return the power factor to unity (1.0).

If a capacitive current IC equal in magnitude to the inductive current of the motor, I L , is added, the circuit is balanced and the source current is now equal to just the resistive component of the motor current.

•

IR = IS = IM x (.787) = 6.0 A

No add capacitance

IS = IM = 7.6 A

Corrected

IS = IR = 6.0 A

Converting IC needed to Capacitance

IC = IL = IM * sin f = 4.68 AXC = E / IC Inductive Reactance (Ohms Law)

= 220 V/ 4.68 A = 47 ohmCapacitance

C = 106/(2π f XC) = 106/(2π (60Hz) 47 ohm)

= 56.4 μf

Table Method for Correction Calculation

Table A.8 of Fund. of Elec (Appendix)

Table Factor x kilowatt input = kVAr of capacitance required

Example

Assume a 500 kVA load with a power factor of 0.6. What size capacitor would be required to raise the power factor to 0.9?

Note: May not want to correct back to 1.0!

The capacitor rating in kVAr is found by multiplying the true power of the load (kW of load) by the factor taken from the table.

Table Method for Correction Calculation

First find the true power of the load as

True Power = Apparent Power x Power Factor = 500 kVA x 0.6 = 300 kW

kVAr Required = T.P. x factor= 300 x .85 = 255 kVAr

Need Capacitor Bank of 255 kVAr

Topics to be Addressed:• Review of Some Basic Terms and Concepts

– Voltage, current, resistance, power and energy• Wire Sizing and Selection

– Ampacity, allowable voltage drop, environment • Service Entrance

– Sizing using demand system• Sizing a Farm Service

– Sizing using demand system• Motor Control and Protection

– Size wire and overload protection (125%)• Lighting Levels and Selection

– Lighting level and number of sources• Power Factor Correction

ASABE PE Review Session

What Questions do you have?