Presentation on theme: "ASABE PE Review Session Circuits, Controls, and Sensors Robert J. Gustafson, P.E., PhD Honda Professor for Engineering Education Director, Engineering."— Presentation transcript:
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.firstname.lastname@example.orgGustafson.email@example.com 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 Three-Phase Power 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, 15 th Ed. (order at www.rerc.org, $22)www.rerc.org MWPS-28, 2005,Wiring Handbook for Rural Facilities, 3 nd 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 10 6 J) Resistance R Ohm ohm or Power P Watt W Resistivity ρ Ohm-cm cm Ohm's Law E = IRR = E/I I = E/R Power Relations ac System P = EIcos ɸ = I 2 Rcos ɸ = (E 2 /R)cos ɸ cos ɸ = power factor Power Factor =True Power/Apparent Power = Watts /(Volts * Amperes) dc System P = EI = I 2 R = E 2 /R 3-phase systems P = 3E P I P cos = 3 0.5 E L-L I L cos where P = true power E P = phase voltageE L-L = line-to-line voltage I P = phase currentI L = line current cos = power factor What you will most likely need for planning: Assume cos power factor Power = 3 0.5 E L-L I L Energy Energy = Power * Time or Power integrated over time
Water pump Drain p 1 (psi) p2p2 Gage Pressure Direction of water flow q = water flow (gpm) 50 Water flow in a pipe 0 p3p3 10 Voltage to Ground Current flow in a wire V 1 (volts) V2V2 Voltage Source 0 Ground Direction current flow 50 I = current flow (amps) 10 V3V3 Fluid Circuit Electrical Circuit Load
Voltage to Ground Current flow in a wire V 1 (volts) Voltage Source 0 Ground Direction current flow 50 I = current flow (amps) V3V3 For this circuit. If the current is measured at 10 amperes, what is the total resistance of the circuit? a)5 watts b)5 ohms c)10 ohms d)10 amperes Load Ohm's Law E = IRR = E/II = E/R Problem 1
Voltage to Ground Current flow in a wire V 1 (volts) Voltage Source 0 Ground Direction current flow 50 I = current flow (amps) V3V3 For this circuit. If the current is measured at 10 amperes, what is the total resistance of the circuit? a)5 watts b)5 ohms c)10 ohms d)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 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) $11,800 per year b) $3,500 per year c) $350 per year d) -$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
Could do a similar calculation for Energy and Cost savings for compact fluorescent for incandescent bulbs.
Fact - Typical passenger car and light truck alternators are rated around 50-70 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
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 cableBranch Circuits – UF, Underground feederBranch Circuits – USE, Underground Service EntranceUnderground Service
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 diversityAmperage at 240V All lights18.0 A DCOs with heat lamps20.0 A DCOs at 1.5 A4.5 A Cold Weather fans7.2 A Two heaters25.0 A Auger motor15.5 A Total load without diversity90.2 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 Load98.2 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.
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 Amp b) 367 Amp c) 400 Amp d) 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 Amapacity – 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 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 5 kW electric heater. What is ampacity of Branch Circuit needed for this unit? Motor 28 A x 1.25 = 35 A Heater 5 kW / 240 V = 21 A Total = 35 A + 21 A = 56 A Use 56 A to size wire
Assuming UF Cable and distance of 50 Ft, what is wire size? a) 10 b) 8 c) 6 d) 4
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:
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: I. 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%.
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: I. 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%. 2. A thermal protective device integral with the motor that will prevent dangerous overheating of the motor due to overload or failure to start.
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: I. 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%. 2. A thermal protective device integral with the motor that will prevent dangerous overheating of the motor due to overload or failure to start. 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 VCycle: 60 Phase: single Service Factor: 1.15 Code: C Amp: 24 The maximum allowable rating for motor over- load protection (amps) is most nearly: a) 24b) 28c) 30d) 35
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 Watts Light OutLumens EfficiencyLumens/Watt Light 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 lx Convert to fc 4.6 to 9.2 fc Using Rule of Tumb; 3 lumens for 1 fc/ft 2 4.6 to 9. 2 fc x 3lumens/1 fc/ft 2 (13.8 to 27.6) lu/ft 2 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 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 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 of 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 I C 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. I R = I S = I M x (.787) = 6.0 A No add capacitance I S = I M = 7.6 A Corrected I S = I R = 6.0 A
Converting I C needed to Capacitance X C = E / I C Inductive Reactance (Ohms Law) = 220 V/ 4.68 A = 47 ohm Capacitance C = 10 6 /(2π f X C ) = 10 6 /(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
Three-Phase Power Main References: Gustafson and Morgan 4.5, 4.6 and 5.2 Why 3-Phase -Efficient distribution -Lower initial cost for motors -Smaller currents Means Smaller wires Fig 4.26
2 Basic Configurations Delta Line Current = Vector Sum of 2 phase currents E line-to-line = E phase I line = 1.73 x I phase Wye Line Voltage = Vector Sum of 2 Voltage s I P (Phase Current) = I L (Line Current E P = Phase Voltage E L-L = Line-to-Line Voltage E L-L = 1.73 * EP
Power Relationships P = 3E P I P cos = 3 0.5 E L-L I L cos where P = true power E P = phase voltageE L-L = line-to-line voltage I P = phase currentI L = line current cos = power factor What you will most likely need for planning: Assume cos power factor Power = 3 0.5 E L-L I L
Planning for 3-phase Systems Section 5.2 of Textbook Summary on Sizing for Services Same basic process as for single-phase Add loads up in Volts * Amperes (VA) BE SURE TO MULITPLE 3-PHASE LOADS TIMES SQUARE ROOT OF 3 Current Rating =
Sizing Brach Circuits and Conductors Branch circuits serving equipment like motors, resistance heat, or receptacles Two main categories: Motor loads- Assume circuit breakers Size breaker at 175% of full-load - Conductors 125% of full-load Non-Motor loads - Note- for continuous load on circuit breaker must not exceed 80% of rating Therefore use 125% of circuit load for breaker and wire sizing
Example 1 Motor – sizing overcurrent protection (breaker) and wire 3-Phase Irrigation Pump, 7.5 hp, 480 V located 20 feet from service panel Find the full load current for motor Text Table A.4 Full-load Current for 3-phase motors Induction-Type Squirrel Cage and Wound Rotor (Amperes) HpkW115V230V460V ½0.3722.214.171.124 ¾0.5606.43.21.6 10.74126.96.36.199 1 ½1.1212.06.03.0 21.49188.8.131.52 184.108.40.206 53.7315.27.6 7 ½5.602211 107.462814 1511.24221 2014.95427 2516.46834 3022.48040 29.810452 5037.313065
Overcurrent Protection 1.75 * 11 amp = 19.25 Amp So use 20 AMP breaker Wire Size 1.25 * 11 Amp = 13.75 Amp Use Allowable Ampacity Table A.6 Use 3 # 12 Copper Should we check line loss (voltage drop)?
Example 2 Heating Unit sizing circuit breaker and wire size 5 kW, 240 V, 3-phase Water Heater Phase Current = No more than 80% Rule for Breaker Load Use 15 Amp breaker Table for Wire Ampacity -
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 Three-Phase Power
ASABE PE Review Session What Questions do you have?
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