ASABE PE Review
Session Circuits,
Controls, and Sensors
Robert J. Gustafson, P.E. , PhD
Honda Professor for Engineering Education
Director, Engineering Education Innovation Center
Professor, Food, Agricultural and Biological
Engineering
Ohio 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 Correction
Process - 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
(kWh
Joule
3.6 x 106 J)
Resistance
R
Ohm
Power
P
Watt
Resistivity
ρ
Ohm-cm
J
ohm or Ω
W
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 f = 30.5EL-LIL cos f
where P = true power
EP = phase voltage
EL-L = line-to-line voltage
IP = phase current
IL = line current
cos f = power factor
What you will most likely need for planning:
Assume cos f = power factor = 1
Power = 30.5EL-LIL
Energy
Energy = Power * Time
or Power integrated over time
Gage Pressure
p2
p1 (psi)
q = water flow (gpm)
Water
pump
Direction of water flow
p3
Fluid Circuit
Drain
Water flow in a pipe
V1 (volts)
V2
Voltage to Ground
50
I = current flow (amps)
Load
Direction current flow
Voltage Source
V3
0
Electrical Circuit
Ground
Current flow in a wire
10
Problem 1
V1 (volts)
Voltage to Ground
50
Direction current flow
Voltage Source
Load
0
V3
I = current flow (amps)
Ground
Current flow in a wire
For this circuit. If the current is measured at 10 amperes,
what is the total resistance of the circuit?
a)
b)
c)
d)
5 watts
5 ohms
10 ohms
10 amperes
Ohm's Law
E = IR
R = E/I I = E/R
V1 (volts)
Voltage to Ground
50
Direction current flow
Voltage Source
Load
0
V3
I = current flow (amps)
Ground
Current flow in a wire
For this circuit. If the current is
measured at 10 amperes,
what is the total resistance of the
circuit?
a)
b)
c)
d)
5 watts
5 ohms
10 ohms
10 amperes
Solution: R = E / I
R = 50 V / 10 A = 5 ohms
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 Problems
Hurd 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)
b)
c)
d)
$11,800 per year
$3,500 per year
$350 per year
-$210.00 per year
Solution
Power
4 rm x 50 lamps/rm x (250 -175 W/lamp)
= 15,000 W = 15 kW
Solution
Power
4 rm x 50 lamps/rm x (250 -175 W/lamp)
= 15,000 W = 15 kW
Energy = Power x Time (need to assume time)
Solution
Power
4 rm x 50 lamps/rm x (250 -175 W/lamp)
= 15,000 W = 15 kW
Energy = Power x Time (need to assume time)
Assume on 30% of time
15kW x .3 x 8760 hr/yr = 39,420 kWh/yr
Solution
Power
4 rm x 50 lamps/rm x (250 -175 W/lamp)
= 15,000 W = 15 kW
Energy = Power x Time (need to assume time)
Assume on 30% of time
15kW x .3 x 8760 hr/yr = 39,420 kWh/yr
Cost = Energy x Rate
39,420 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/kwHr
a) $413
b) $13,797
c) $4,139
d) $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 5070 amperes
If you assume 50% efficient, approx. how many
horsepower is going into running the
alternator at rated load?
a. 0.2 hp
b. 2 hp
c. 20 hp
d. 20 kW
Example Solution
Assume 12 V
Output P = E x I = 70 A x 12 V = 840 W
Input 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 – Ohms
Ohms 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 feeders
2) Look at Sizing a building service entrance
panel
3) Look at Sizing the main service for a
farmstead
Wire Sizing and Selection
1) 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 & Material
Size
• 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 Mils
Larger Number Larger Area
Suitable for Environment
• Material
– Copper
– Aluminum
• Insulation (Used in Agr Wiring)
– NM, nonmetallic sheathed cable
– UF, Underground feeder
– USE, Underground Service Entrance
Branch Circuits
Branch Circuits
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 Continued
Equipment operating without diversity
All lights
DCOs with heat lamps
DCOs at 1.5 A
Cold Weather fans
Two heaters
Auger motor
Total load without diversity
Amperage at 240V
18.0 A
20.0 A
4.5 A
7.2 A
25.0 A
15.5 A
90.2 A
Compute the demand load:
L.W.D at 100%
90.2 A
Remaining load at 50% (105.7 A- 90.2 A x 0.5)
8.0
Total Demand Load
98.2 AU
Use 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)
b)
c)
d)
300 Amp
367 Amp
400 Amp
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 Sizing
2. Overcurrent
Protection
Rating
MOTOR NAMEPLATE INFORMATION
• http://www.elongo.com/pdfs/MotorNameplate9905
19.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 Uframe. 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 lockedrotor 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 A
Stirring 12 A
Total = 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)
b)
c)
d)
10
8
6
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: single
Service Factor: 1.15 Code: C Amp: 24
What would you pick for Overcurrent
a) 24
b) 28
c) 30
d) 35
The maximum allowable rating for motor overload protection (amps) would be?
Suggested Reference for Motor Circuits
Building Approach:
Section 26 and 27 of NFEC Agr. Wiring Handbook
Section 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
Light Out
Efficiency
Light on Surface
Watts
Lumens
Lumens/Watt
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 lx
Convert to fc Conversion (Recall 1 fc = 10.76 lx)
4.6 to 9.2 fc
Using 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 voltage
Phase Shift caused by inductance or capacitance
True Power = Apparent Power x Power Factor
Watts = (Volts x Amperes) x p.f.
Why Should I care?
True Power = Apparent Power x
Power Factor
Watts
= (Volts x Amperes) x p.f
True Power (Watts)
does useful work
what meter measures
Reactive Power
does 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 singlephase 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 A
XC = E / IC Inductive Reactance (Ohms Law)
= 220 V/ 4.68 A = 47 ohm
Capacitance
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?