Study Guide for the

Certified Energy Manager Exam July, 2014

The information on the following pages was personally compiled during my studies to prepare for the Certified Energy Manager exam. These notes and formulas served as a reference during classes and the exam itself, and I hope it can be of benefit to others. The initial section, Energy Units and Formulas by Topic Area, is organized in the same as order as the information that follows in Sections 1-16. Thus, formulas with examples are first given for Audits, followed by Energy Procurement and Electrical Systems, Energy Cost Analysis, Lighting, etc. on pages 3-58. On a personal note, I recommend taking the on-line CEM practice exam 2-3 weeks prior to the actual exam using this guide or similar reference information.

Larry S. Katz, CEM Long Branch, New Jersey

LSKatz10@gmail.com

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Contents page

Energy Units and Formulas by Topic Area 3 1. Codes and Standards and Indoor Air Quality 59 2. Energy Audits and Instrumentation 76 3. Energy Procurement - Energy Bills Electrical Systems 86 IV. ELECTRICAL SYSTEMS 4. Energy Cost Analysis and Life Cycle Cost 97 5. Lighting 99 6. HVAC Systems 114 7. Boilers 128 VII. INDUSTRIAL SYSTEMS 8. Steam Distribution Systems / Combustion 130 VII. INDUSTRIAL SYSTEMS 9. Control Systems and Computers 141 10. Energy Systems Maintenance 142 XVI. MAINTENANCE AND COMMISSIONING 11. Insulation 144 VIII. BUILDING ENVELOPE 12. Process Energy Management - Motors 147 13. Renewable Energy Systems and Water Management 162 IX. CHP SYSTEMS and RENEWABLE ENERGY XIII. THERMAL ENERGY STORAGE SYSTEMS 14. Distributed Generation CHP 164 IX. CHP SYSTEMS and RENEWABLE ENERGY 15. Building Automation and Control Systems / Energy Info Systems 178 XI. BUILDING AUTOMATION AND CONTROL SYSTEMS 16. Green Buildings 185 XII. GREEN BUILDINGS, LEED, AND ENERGY STAR

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Energy Units and Formulas by Topic Area

Electricity 1 kWh = 3412 Btu (IT) 1 kWh = .03412 therms 1 kWh = .003413 MCF Natural Gas 1 therm = 100,000 Btu 1 therm = 29.31 kWh 1 DTherm = 10 therms = 1,000,000 Btu = 1 MMBTU 1 therm = 100 ft3 natural gas 1 Dtherm = 1000 ft3 natural gas = 1 MCF = 10 CCF 1 ft3 natural gas = 1,000 BTU Weight 1 LB = .454 kg Capacitance C(farads) = KVAR/ [ 2(pi)(F in Hz)V2] F = 50 or 60 Hz, usually = KVAR / [ 377 x V2 ] Light 1 foot-candle = 1 lumen per sq.ft. = approx. 10 lux ENERGY USE INDEX (EUI) = BTUs/ft2 per year ENERGY COST INDEX (ECI) = $ /ft2 per year Solar Power Incident Solar power averages: 350 Wh/square foot per day in the U.S.

Power Measure of Energy/Time 1 BTU/hr is a measure of Power 1 ton = 12,000 Btu/hr 1 horsepower = 746 W = .746 KW 1 horsepower = 2545 Btu/h 1 watt (W) = 1 J/sec Water, Steam 1 Gallon water = 8.345 LBs water 1 LB water = .1198 Gallons water 1 LB water is heated 1 degF by 1 BTU 1 LB water at 212F liquid boils off as steam or steam condenses to water using about 970 BTU 1 LB water at 32F changes from liquid to solid or ice melts to liquid using about 144 BTU latent specific heat or reverse Water Flow: 500 LB/hr water = 1 gpm water Air 1 ft = 0.07788 lbs at 50degF 1 ft = 0.07640 lbs at 60degF 1 ft = 0.07495 lbs at 70degF Pressure 1 Bar = 14.50 psig 1 psia = 14.7 + psig a bicycle tire pumped up to 65 psi(g) above atmospheric pressure will have a pressure of 65 + 14.7 = 79.7 psia 1 psi = 144 pound force/square foot Geometry Area Tank (cylinder) = 2 x 2(pi)r2 (top and bottom) + 2(pi)r x length of tank Volume of Tank = (pi)r2 h Area of Sphere = 4(pi)r2 Volume of Sphere = 4/3(pi)r3

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Degree Days Heating Degree Days (HDD): when its cold and you need heating HDD = (65degF - Avg Temp of period) x (days of the given period) Cooling Degree Days (CDD) when its hot and you need cooling CDD = (Avg Temp of period - 65degF) x (days of the given period) ______________________________________________________________________

Approximate Heating Value of Common Fuels

Gas Natural Gas 1,000 Btu/cu ft 100,000 Btu/therm 23,600 Btu/lb Propane 2,500 Btu/cu ft 92,500 Btu/gal (LP) 21,000 Btu/lb LP Gas 95,000 Btu/gal 14,600 Btu/lb Methane 1,000 Btu/cu ft Landfill gas 500 Btu/cu ft Butane 3,200 Btu/cu ft 130,000 Btu/gal (liquid) Methanol 57,000 Btu/gal Ethanol 84,400 Btu/gal Crude Oil 5,100,000 Btu/barrel Gasoline 125,000 Btu/gal 20,000 Btu/lb Fuel Oil Light Heating Oil 140,000 Btu/gal also called #2 Fuel Oil Kerosene 135,000 Btu/gal #4 145,000 Btu/gal #6 153,000 Btu/gal Waste oil 125,000 Btu/gal Biodiesel 120,000 Btu/gal (Waste vegetable oil) Diesel/heating oil 19,700 Btu/lb Coal 12,500 Btu/lb 25,000,000 Btu/ton

Hard Coal (anthracite) 13,000 Btu/lb 26,000,000 Btu/ton Soft Coal (bituminous) 12,000 Btu/lb 24,000,000 Btu/ton

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Others Rubber pelletized 16,000 Btu/lb 32-34,000,000 Btu/ton Plastic 18-20,000 Btu/lb Nuclear fission 33,000,000,000 Btu/lb Hydrogen 61,000 Btu/lb Woods Softwood 2-3,000 lb/cord 1015,000,000 Btu/cord Hardwood 4-5,000 lb/cord 1824,000,000 Btu/cord Sawdust green 10-13 lb/cu ft 8-10,000,000 Btu/ton Sawdust kiln dry 8-10 lb/cu ft 14-18,000,000 Btu/ton Chips 45% moisture 10-30 lb/cu ft 7,600,000 Btu/ton Hogged 10-30 lb/cu ft 16-20,000,000 Btu/ton Bark 10-20 lb/cu ft 9-10,500,000 Btu/ton Wood pellets 40-50 lb/cu ft 16,000,000 Btu/ton 10% moisture Corn shelled 7,800-8,500 Btu/lb 15-17,000,000 Btu/ton cobs 8,000-8,300 Btu/lb 16-17,000,000 Btu/ton

conversions, and to more complex conversions and calculations in the future. The principle of performing unit conversions is simply to carry out algebraically correct multiplications and divisions using correct units at each step, starting with the given piece of information and transforming it into the desired units with the use of one or more conversion factors. If we ever perform one of these basic unit conversion calculations, and find that we have different units on the left and the right. We do not have the correct answer in terms of the desired units. This method is given the colloquial name Railroad Track Method, because the vertical separation lines remind us of railroad tracks. Before we start, here are the conversion factors that we will use in class and because youre in the energy management field Youll use these extensively in your work!

Conversion factors for Unit Conversions: 1 kWh3,412 Btu 1 ft3 natural gas..1,000 Btu 1 Ccf natural gas100 ft3 natural gas 1 Mcf natural gas1,000 ft3 natural gas 1 therm natural gas100,000 Btu 1 barrel crude oil5,100,000 Btu 1 ton coal25,000,000 Btu 1 gallon gasoline125,000 Btu 1 gallon ethanol84,400 Btu 1 gallon #2 fuel oil140,000 Btu 1 gallon LP gas.95,000 Btu 1 HP..746 Watts 1 Mbtu1,000 Btu 1 MMBtu.1,000,000 Btu 1 Decatherm..1,000,000 Btu 1 therm ..100,000 Btu 1 Quad.1x1015 Btu 1 MW1,000,000 Watts

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Temperature Range for Heat Transfer

Ammonia -95 and 140F Mercury 375 and 1,000F Methanol -50 to 240F Silver 2,700 to 3,600F. Water 40 and 425F.

Pour point The pour point is the lowest temperature at which an oil will continue to behave like a normal liquid. Most oils have a pour point of 5 degrees above the solidification point. Pour point is significantly influenced by the amount of wax in the oil. It is important to know the pour point of an oil so as to arrange the necessary heating during storage and use.

For air: 1 lb/hr = 4.5 Q 1 ton = Q h 2670 Small fans 0.40 0.50 efficiency Large fan 0.55 0.60 efficiency For water: 1 lb/hr = 500 gpm Small pumps 0.40 0.60 efficiency Large pumps 0.70 0.85 efficiency Air Weights by Temperature Temp Specific Weight (oF) (lb/ft3) x 10-2 -40 9.456 -20 9.026 0 8.633 10 8.449 20 8.273 30 8.104 40 7.942 50 7.786 60 7.636 70 oF 7.492(lb/ft3) x 10-2 = .07492 lb/ft3 80 7.353 90 7.219 100 7.09

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120 6.846 140 6.617 160 6.404 180 6.204 200 6.016 300 5.224 400 4.616 500 4.135 750 3.28 1,000 2.717 1,500 2.024

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AUDIT Energy Indexes Energy Use Index (EUI): BTUs/sq ft/yr energy in BTUs per conditioned square foot per year. Energy Cost Index (ECI): $/sq ft /yr energy cost in $s per conditioned square foot per year.

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Energy Procurement and Electrical Systems Contract

Fuel Availability Purchase price Contract terms, clauses Volume commitments RISK tolerance Environmental (emissions, etc.)

Ratchet Clause: Demand is billed at a percent (Usually > 50%) of the largest kW demand over the past 11 months, or the current month demand, WHICHEVER IS GREATER. Load Factor Average Load Load Factor = ---------------------- for a specified period (month, hours, day(s), etc.) Peak Load Load Strategies Off Peak Air Conditioning (OPAC) OPERATING STRATEGIES Load Leveling Partial Load Shifting Partial shifting of AC load to off-peak hours Chiller runs at constant load or near constant load for 24 hours per day Very cost effective for new construction Less costly to purchase Less space needed But ~ less savings Full Shift Strategy operate at peak load hours only.

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Power Factor Reactive Power (KVAr)

Real Power (Actual work performed) (Total Power)2 = (Real Power)2 + (Reactive Power)2 Real Power Power Factor = ------------------- = cos (cos = adjacent / hypotenuse) Total Power

Single Phase: Real Power Power Factor = ------------------- Total Power Real Power (W) = Total Power (VxA) x PF W = V x A x pf Three Phase: Real Power = 3 x V x A x PF = 1.73 x V x A x PF

Synchronous motor having no load connected to its shaft is used for power factor improvement. BILLING: Some rate schedules will calculate the billed demand as follows: Base Power Factor (minimum acceptable) Billed Demand (kW) = Actual Demand x ------------------------- Actual Power Factor

Total Power (Apparent) KVA

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Power Voltage Current Calculations Y Circuit

V Line = V phase x 1.73 where V Line = Voltage from line to line (voltage measured between any two line conductors in a balanced three-phase system.) V phase = Voltage across component (can be Line to Neutral) (voltage measured across any one component (source winding or load impedance) in a balanced three-phase source or load.

Delta Circuit

I Line = I phase x 1.73 where I Line = Current from winding point to corresponding load point (current through any one line conductor) I phase = Current through each load resistor and source winding (current through any one component)

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Example: If electricity is selling for $0.06 per kilowatt-hour and is used for electric heating with an efficiency of 90%, what is the equivalent price of natural gas per therm if it can be burned with an efficiency of 80%? i. $1.33/therm ii. $1.47/therm iii. $1.56/therm iv. $1.65/therm v. $1.780/therm Solution: Same amount of heat generated with less efficient gas heater. $ .06/kWh Equiv. Rate --------------- = ------------------------ 90% 80% Equivalent Rate = $ .0533/kWh x 29.31 kWh/therm = $ 1.56/therm Example: An audit for one firm showed that the power factor is almost always 70% and that the demand is 1000 KW. What capacitor size is needed to correct power factor to 90%? Find KVAR at PF=.70 Find KVAR at PF = .90 Difference = Capacitor in KVARs required .7 = Real / Total .7 = 1000 KVA / Total Total = 1428.6 kVA (Total Power)2 = (Real Power)2 + (Reactive Power)2 (1428.6)2 = (1000)2 + (Reactive Power)2 Reactive Power = 1020 kVA .9 = Real / Total .9 = 1000 KVA / Total Total = 1111.1 (Total Power)2 = (Real Power)2 + (Reactive Power)2 (1111.1)2 = (1000)2 + (Reactive Power)2 Reactive Power = 484.3 kVA KVARs to add to get to 90% PF = 1020-484.3 = 535.7 KVARs

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Example: The amount of the active power that must be supplied by capacitors to correct a power factor of 84% to 95% in a 400HP motor at 75% load and 98% efficiency is: 400 HP x .746 kW/HP = 298.40kW 298.4 kW x 75% load = 223.80 kW 223.8 kW / 98% efficiency = 228.4 kW PF = .84 .84 = Real / Total .84 = 228.4 KVA / Total Total = 271.9 kVA (Total Power)2 = (Real Power)2 + (Reactive Power)2 (271.9)2 = (228.4)2 + (Reactive Power)2 Reactive Power = 147.46 kVA .95 = Real / Total .95 = 228.4 KVA / Total Total = 240.42 kVA (Total Power)2 = (Real Power)2 + (Reactive Power)2 (240.42)2 = (228.4)2 + (Reactive Power)2 Reactive Power = 75.07 kVA KVARs to add to get to 95% PF = 147.46 75.07 = 72.43 KVARs Example: A 3/4 HP electric motor has a power factor of .85. The nameplate current is 10 Amps at 115 Volts, or 1150 Volt Amps. What is Reactive Power? Based on Provided Current Draw and Volts: Total Power = 1150 VA PF = .85 = Real Power / Total Power Real power = 977.50 VA (Total Power)2 = (Real Power)2 + (Reactive Power)2 (1150)2 = (977.5)2 + (Reactive Power)2 Reactive Power = 605.8 VA

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Example: Given: 75% ratchet clause highest peak over last 11 months: 940kW Current month peak: 650kW What is billed demand for the current month? 75% x 940kW = 705kW 705kW > 650kW Therefore: Billed demand = 705kW

Example:

V line = V phase x 1.73 240V = V phase x 1.73 240 V / 1.73 = V phase V Phase = 138.7V

X

V Line = V phase x 1.73

V Line = Voltage from line to lineV Phase = Voltage from Line to Neutral

Thus:240 = V phase x 1.73V Phase = 138.7

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Energy Cost Analysis and Life Cycle Cost Discounted After-Tax Cash Flow (ATCF) ATCF will have a rate that is less than MARR. ATCF with straight-line depreciation: ATCF = Annual profit { (Annual profit Annual depreciation) x tax rate) }

Annual profit = Annual energy savings, e.g. Straight-line Depreciation = Total Cost of Project / life of project years

Discounted Benefit Cost Ratio PV (Benefits) BCR = ------------------------ PV (Costs)

NPV = PV Initial Project Cost Net Gain Total Benefit Total Cost ROI = ---------------- = ----------------------------------------- Total Cost Total Cost Example: A $1.4M project , 10 years, straight line depreciation, annual savings of $235,000, tax bracket of 34%, has an ATCF of: Depreciation = $1,400,000/10 years = $140,000 ATCF = 235,000 ((235,000-140,000) x .34) = $ 202,700

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Example: An Energy Saving device saves $25,000 for 8 years What should a company pay (PV) if MARR [minimum attractive rate of return] is 15%? SOLVE FOR PV N=8 PMT=$25,000 I/Y=15% Solve for PV = $112,183 or less

Example: A project costs $38,250, and will result in $30,500 savings per year for 15 years. What is the Rate of Return (ROR) ? SAME AS (IRR) SOLVE FOR i (I/Y) N=15 PV = - $38,250 PMT = $30,500 [CPT] I/Y = 79.3%

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Example: A project costs $38,250, and will result in $30,500 savings per year for 15 years. What is the Net Present Value (NPV) if interest rate is 10% ? SOLVE FOR PV N=15 I=10 Payment = $30,500

PV = $231,985 (calculated) NPV = $231,985 $38,250 = $193,735 Example: A project will cost $100,000, and save $23,400 of energy a year for 12 years. If the MARR is 12%, is this a worthwhile project? How much money (FV) will be available in 12 years if all energy savings are banked and earn 10%?

SOLVE FOR I/Y = MARR N=12 PV = - $100,000 PMT= $23,400 I/Y = 21.03% > 12% MARR

SOLVE FOR FV using I/Y N=12 PV = $100,000 PMT= -$23,400 I/Y = 10 FV = $186,549

Example: What is the BCR for the above where a project will cost $100,000, and save $23,400 of energy a year with 10% interest rate for 12 years? PV Benefits (23,400 per year; 10% IR) = $159,440 BCR = $159,440 / $100,000 = 1.59

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Lighting Typical lamp efficacies (lumens/watt) 100 W filament lamp 14 lumens/watt 58 W fluorescent tube 89 lumens/watt 400 W high-pressure sodium 125 lumens/watt 131 W low-pressure sodium 198 lumens/watt Light Units

1 foot-candle = 1 lumen per sq.ft Lighting Formulas:

1 Lux (lx) = 1 Footcandle (fc) x 10.76 Lux = Total Lumens Area

Ballast factor (BF) is a ratio: Lamps lumen output on commercial ballast BF = -------------------------------------------------------- Lamp's rated light output

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Calculating Average Light Level Throughout a Space (three formulas) Lighting Design Methods Average Maintained Illumination (Footcandles); LLF = light loss factor Fixtures x Lamps per Fixture x Lumens per Lamp x CU x LLF Foot Candles = ------------------------------------------------------------------------------------- (Area of the Room in ft2 ) To Calculate number of Fixtures: FC x (Area of the Room) # Fixtures = ----------------------------------------------------------------- Lamps per Fixture x Lumens per Lamp x CU x LLF or FC x (Area of the Room) # Fixtures = ----------------------------------------- Lumens in the Room x CU x LLF Lumen Method / Zonal Cavity Method (determining cavity ratios) FC x (Area of the Room) Total Lumens in Room/Fixture = ------------------------------------------------ # Fixtures x CU x LLF x Ballast Factor

Light Loss Factor (LLF) = Ballast Factor x Fixture Ambient Temperature Factor x Supply Voltage Variation Factor x Lamp Position Factor x Optical Factor x Fixture Surface Depreciation Factor x Lamp Burnouts Factor x Lamp Lumen Depreciation Factor x Fixture Dirt Depreciation Factor x Room Surface Dirt Depreciation Factor Lamp Burnout Factor = 1 - Percentage of Lamps Allowed to Fail Without Being Replaced

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Zonal Cavity Design Method N = F A / (Lu L Cu) where

N is number of lamps needed F is the required foot-candle level at the work area A is area of the room square feet Lu is lamp output in lumens L is the depreciation factor for the lamp and fixture Cu is coefficient of utilization

Inverse Square Law

E = I/D2 where E is luminance, or intensity of light at a specific point, in lumens per

square foot I is the luminous intensity, or the radiated energy density in Watts

per steradian D is the distance in feet from the source

This is the Point to point method to establish the required irradiated light to satisfy a specific luminance requirement. This relationship can only be used when surface is directly under the source and normal (perpendicular) to the light ray. For all other positions a more generalized formula is:

E = I x cos /D2 where: = the angle between the line joining the source to the point on the illuminated surface and a line normal (perpendicular) to the illuminated surface. To Find D: (its a triangle!)

D = (H2 + L2) or D2 = H2 + L2 I = Candlepower in candelas (cd) D = Direct distance between the lamp and the point where light level is calculated H = Distance between the lamp and the point direct below on the workplane L = Distance between that point and the point where light level is being calculated

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Room Cavity Ratio RCR = 5 MH (L + W) / (L x W) where:

Mounting height (MH): Distance in feet between the bottom of the fixture and the workplane. Room Area = length x width Rooms are rectangular

Ceiling Cavity Ratio = [5 x Ceiling Cavity Depth x (Room Length x Room Width)] (Room Length x Room Width) Floor Cavity Ratio = [5 x Floor Cavity Depth x (Room Length x Room Width)] Room Length x Room Width Room Cavity Ratio (for irregular-shaped rooms) = (2.5 x Room Cavity Depth x Perimeter) Area in Square Feet Reflected Reading Room Surface Reflectance (%) = -------------------------- where: Incident Reading

Reflected Reading = Measurement from a light meter holding it about 1.5 feet away from the surface with the sensor parallel and facing the surface. Incident Reading = Measurement from a light meter held flat against the surface and facing out into the room.

Calculating Number of Lamps And Fixtures And Spacing

Maximum Allowable Spacing Between Fixtures = Fixture Spacing Criteria x Mounting Height Mounting height (MH): Distance in feet between the bottom of the fixture and the workplane Spacing Between Fixtures = (Area in Square Feet Required # Fixtures) Number of Fixtures to be Placed in Each Row (Nrow) = Room Length Spacing Number of Fixtures to be Placed in Each Column (Ncolumn) = Room Width Spacing

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For above two formulas, round results to the nearest whole integer. Spacing row = Room Length (Number of Fixtures/Row - 1/3) Spacing column = Room Width (Number of Fixtures/Column -1/3) If the resulting number of fixtures does not equal the originally calculated number, calculate impact on the designed light level: % Design Light Level = Actual No. of Fixtures Originally Calculated No. of Fixtures To calculate fixtures mounted in continuous rows: Number of Luminaires in a Continuous Row = (Room Length Fixture Length) - 1 Number of Continuous Rows = Total Number of Fixtures Fixtures Per Row

Lamp Maintenance Calendar Lamp Life (Years) = Rated Lamp Life (Hours) Annual Hours of Operation (Hours/Year) Lamp Burnout Factor = [1 - Percentage of Lamps Allowed to Fail Without Being Replaced]

Group Relamping Cost Annualized Cost ($) = A x (B + C) A = Operating Hours/Year Operating Hours Between Relampings B = (Percentage of Lamps Failing Before Group Relamping x Number of Lamps) x (Lamp Cost + Labor Cost to Spot Replace 1 Lamp) C = (Lamp Cost, Group Relamping + Labor Cost to Group Relamp 1 Lamp) x Number of Lamps

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Spot Relamping Cost Average Annual Cost ($) = (Operating Hours/Year Rated Lamp Life) x (Lamp Cost + Labor Cost to Replace 1 Lamp) x Total Number of Lamps Cleaning Cost ($) = Time to Wash 1 Fixture (Hours) x Hourly Labor Rate ($) x Number of Fixtures in Lighted Space To Spot Replace or Group Relamp Spot replacement relamp = Lamp cost + Labor Cost Group Relamping = (Lamp cost + Labor Cost) / Relamping Interval which is the % of rated lamp life Determine Labor rates, Compare spot vs. group relamping.

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Calculate Annual Savings using Group Relamp Address 3 parts Watts saved x 8760 x $/kWh = USAGE energy savings kW saved x monthly demand chg x 12 months = DEMAND energy savings Annual savings for rated hours based on cost of lamp: Calculate total lamp hours = # lamps x 8760 Equiv. Lamps needed original = Lamp hours / Lamp Life Equiv. Lamps needed replacement = Lamp hours / Lamp Life Equiv. Lamps needed original x cost/lamp = Annualized cost/lamp orig Equiv. Lamps needed replacement x cost/lamp = Annualized cost/lamp replacement Annualized Cost replacement - Annualized Cost original = Annual cost savings Example: Given: 15 minutes to relamp one; 25 minutes to group relamp 8 Labor rate = $10 Replace Rate: 80% of rated life Which labor rate is better spot or group? Spot Relamp labor = $2.50 ($10/hr x hr) Lamp Cost = $ .85 One relamp = $3.35/lamp Group Relamp labor = 8 lamps in 25min is roughly 3 min per lamp 3/60 per hour for one lamp = $ .50 per lamp Lamp Cost = $ .85 G = (.50 + .85)/80% = $1.69/lamp

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Illuminance Number of Footcandles Facilities and Rooms HOUSE Room Foot Candles Needed Living Room 10-20 Kitchen General 30-40 Kitchen Stove 70-80 Kitchen Sink 70-80 Dining Room 30-40 Bedroom 10-20 Hall Way 5-10 Bathroom 70-80

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Table 1. ASHRAE/IES 90.1 - Lighting Power Allowances using the Building Area Method. Maximum Lighting Power Density (W/sq.ft.) Allowed per ASHRAE/IES 90.1 Standard - 2010

W/sq.ft. W/sq.ft. Automotive Facility 0.982 Manufacturing Facility 1.11 Convention Center 1.08 Motel 0.88 Court House 1.05 Movie Theater 0.83 Dining: Bar Lounge/Leisure 0.99 Multi-Family 0.60 Dining: Cafeteria/Fast Food 0.90 Museum 1.06 Dining: Family 0.89 Office 0.90 Dormitory 0.61 Parking Garage 0.25 Exercise Center 0.88 Penitentiary 0.97 Gymnasium 1.00 Performing Arts Theatre 1.39 Healthcare Clinic 0.87 Police/Fire Station 0.96 Hospital 1.21 Post Office 0.87 Hotel 1.00 Religious Building 1.05 Library 1.18 Retail 1.40 Manufacturing Facility 1.11 School/University 0.99 Motel 0.88 Sports Arena 0.78 Motion Picture Theater 0.83 Town Hall 0.92 Multi-Family 0.60 Transportation 0.77 Museum 1.06 Warehouse 0.66 Office 0.90 Workshop 1.20

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Lamp Types Characteristics - Overview Type (code) Common

ratings (watts) Colour rendering Colour temperature (K) Life (hours) Compact fluorescent lamps (FS) 555 good 2,7005,000 5,00010,000 High-pressure mercury lamps (QE) 80750 fair 3,3003,800 20,000 High-pressure sodium lamps (S-) 501,000 poor to good 2,0002,500 6,00024,000 Incandescent lamps (I) 5500 good 2,700 1,0003,000 Induction lamps (XF) 2385 good 3,0004,000 10,00060,000 LED lamps 17-36 75-80 3,500- 5,000 50,000 + Low-pressure sodium lamps (LS) 26180 monochromatic yellow colour 1,800 16,000 Low-voltage tungsten halogen lamps (HS) 12100 good 3,000 2,0005,000 Metal halide lamps (M-) 352,000 good to excellent 3,0005,000 6,00020,000 Tubular fluorescent lamps (FD) 4100 fair to good 2,7006,500 10,00015,000 Tungsten halogen lamps (HS) 1002,000 good 3,000 2,0004,000

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HVAC Efficiency Ratings, Formulas

COP = Coefficient of Performance heat absorbed by EVAPORATOR COP = --------------------------------------------------------------------------------------- (heat rejected by CONDENSER - heat absorbed by EVAPORATOR]

CAPACITY 3.517 x Tons COP = ------------------ = ----------------------- LOAD kW Higher COP means more efficiency. Air Conditioning EER or SEER 12 x Tons

EER (Btu/Wh) = ------------- kW 12 kW -------- = ------- EER Ton (= Btu/12,000 of course) 3.517 kW -------- = ------- COP Ton EER (Btu/Wh) = COP x 3.412 Btu/Wh

BTU/h BTU/h

LOAD (W) = -------------- = ---------- BTU/Wh EER Load (W) x Hrs of Operation Energy (Wh) = --------------------------------------- COP

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Affinity Laws (Fan/Pump Laws) Air Law 1. (the impeller diameter (D) is held constant): Law 1a. Flow is proportional (per ft, e.g.) to shaft speed:

Nx = speed in CFM for a fan Law 1b. Pressure or Head (per in2, eg) is proportional to the square of shaft speed:

Nx = speed in CFM for a fan Law 1c. Power is proportional to the cube of shaft speed (per ft3, e.g.): Nx = speed in CFM for a fan Px = energy, in kW, e.g.

Water/Liquid Law 2. (the shaft speed (N) held constant) Law 2a. Flow is proportional to the impeller diameter to the 3rd power: Law 2b. Pressure (Head) is proportional to the square of impeller diameter: Law 2c. Power is proportional to the fifth power of impeller diameter: Where:

is the volumetric flow rate (e.g. CFM, GPM or L/s), is the impeller diameter (e.g. in or mm), is the shaft rotational speed (e.g. rpm), is the pressure or head developed by the fan/pump (e.g. psi or Pascal), and is the shaft power (e.g. W). These laws assume that the pump/fan efficiency remains constant i.e. .

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Water Flow Rate in Heating Systems and Chiller Systems H = 500 x Q x T or Q = H/ (500 x T) H = 500 x Q x T x COP of system or Q = H/ (500 x T x COP)

Q = Water flow rate (GPM) H = Heat flow rate (Btu/hr) T = Temperature difference (deg F) (usually around 20F or sofyi ) COP = coeff. of performance [or system efficiency ] when applicable

[ 500 = 8.34 lb/gal. x 1 Btu/lb degF x 60 min/hr ] Evaporator Flow Rate The evaporator water flow rate can be expressed as Qe = Htons x 24 / T

Htons = Air conditioned cooling load (tons) Qe = evaporator water flow rate (gal/min) T = temperature difference (deg F) between inlet and outlet

Condenser Flow Rate The condenser water flow rate can be expressed as Qc = Htons x 30 / T

Qc = Condenser water flow rate (GPM) Htons = Air conditioning cooling load (tons) T = Temperature differential between inlet and outlet (F)

Flow Rate through a Valve Q = Cv (P/G) Where

Q = (Capacity) flow rate (GPM) Cv = Valve Sizing Coefficient (unique for each style and size of valve) P = Pressure differential (psi) G = Specific Gravity of Fluid (water at 60F = 1.000)

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Air Heat transfer equation sensible heat gain:

Q sensible (BTU/hr) = CFM x T x 1.085 Btuh Q sensible = Sensible heat gain (Btu/hr) CFM = ventilation air flow rate cu.ft./min T = (To Tc )

o To = outside dry bulb temp F o Tc = Dry bulb temperature of air leaving the cooling coil, F

Heat transfer equation change in ENTHALPY:

Q total (Btu/hr) = CFM x 4.5 x (h2 h1) Q total = enthalpy change (Btu/hr) CFM = ventilation air flow rate cu.ft./min h2 = Outside/Inside air enthalpy (Btu/lb) (dry air) h1 = Enthalpy of air leaving the cooling coil (Btu/lb) (dry air)

Heat transfer equation air flow through openings of various size: Impact change with weather stripping, caulking, etc. to limit loss of indoor air to the outdoor elements, and vice versa. BTU/Year = CFM x 25.92 x (HDD+CDD)

Q total = enthalpy change (Btu/hr) CFM = ventilation air flow rate cu.ft./min 25.92 = constant in [min lb BTU / day ft3 lb degF ]

Waste Heat Recovery Amount of potentially available waste heat, expressed in power (BTUs/hour) is calculated as follows: Q (Btus/h) = M Cp T where:

M is the mass flow rate in pounds/hour Cp is specific heat of the medium in BTU/lb.-degree F T is the starting temperature of the medium minus the ambient air temperature

32

Air Compressors Air Leak Rate (SCFM)

V x (P1 P2) Leak Rate (SCFM) = ------------------ T x 14.7 SCFM = standard cubic feet per minute = the volumetric flow rate of a gas corrected to "standardized" conditions of 14.73 psia

V = Volume Air (cu. ft.) P1 = Pressure P2 = Pressure T = temperature difference (deg F) Air Leaks size of hole

Typical compressor efficiency of 18kW/100cfm = .18 kW/cfm VOLUME IN cfm:

33

Heat and Cooling Load: People H sensible = N * (H sensible) * (CLF) H latent = N * (H latent) Where: H sensible = Total Sensible heat gain (Btu/hr) per table H latent = Total latent heat gain (Btu/hr) per table N = number of people in space. CLF = Cooling Load Factor, by hour of occupancy per table Note: CLF = 1.0, if operation is 24 hours or if cooling is off at night or during weekends.

32 TRG-TRC002-EN

notes

period twoCooling Load Estimation

Internal Heat Gains The next component of the space cooling load is the heat that originates within the space. Typical sources of internal heat gain are people, lights, cooking processes, and other heat-generating equipment, such as motors, appliances, and office equipment.

While all of these sources contribute sensible heat to the space, people, cooking processes, and some appliances (such as a coffee maker) also contribute latent heat to the space.

As mentioned in Period One, people generate more heat than is needed to maintain body temperature. This surplus heat is dissipated to the surrounding air in the form of sensible and latent heat. The amount of heat released by the body varies with age, physical size, gender, type of clothing, and level of physical activity. This table is an excerpt from the 1997 ASHRAE HandbookFundamentals. It includes typical sensible and latent heat gains per person,

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34

TRG-TRC002-EN 33

period twoCooling Load Estimation

notes based on the level of physical activity. The heat gains are adjusted to account for the normal percentages of men, women, and children in each type of space.The equations used to predict the sensible and latent heat gains from people in the space are:

QS = number of people sensible heat gain/person CLF

QL = number of people latent heat gain/person

where,

! QS = sensible heat gain from people, Btu/hr [W]

! QL = latent heat gain from people, Btu/hr [W]

! CLF = cooling load factor, dimensionless

Similar to the use of the CLTD for conduction heat gain and SCL for solar heat gain, the cooling load factor (CLF) is used to account for the capacity of the space to absorb and store heat. Some of the sensible heat generated by people is absorbed and stored by the walls, floor, ceiling, and furnishings of the space, and released at a later time. Similar to heat transfer by conduction through an external wall, the space can therefore experience a time lag between the time that the sensible heat is originally generated and the time that it actually contributes to the space cooling load. For heat gain from people, the value of CLF depends on 1) the construction of the interior partition walls in the space, 2) the type of floor covering, 3) the total number of hours that the space is occupied, and 4) the number of hours since the people entered the space.

Figure 37, CLF Factors for People, is an excerpt from the 1997 ASHRAE HandbookFundamentals. It shows that one hour after people enter the space, 35% (1 0.65) of the sensible heat gain from the people is absorbed by the surfaces and furnishings in the space, and 65% is the actual cooling load in the space. Following the table to the right, however, you see that, as the people are in the space for a longer period of time, the surfaces and furnishings of the space can no longer absorb as much heat, and they release the heat that was

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35

Heating and Cooling Load: Glass, Solar Radiation Sensible Heat Load from the Conduction Through Glass H = U * A * (CLTD)

H = Sensible heat gain (Btu/Hr) U = Thermal Transmittance for roof or wall or glass. (Btu/Hr Sq-ft F) A = area of roof, wall or glass calculated from building plans (sq-ft) CLTD = Cooling Load Temperature Difference (in F) for glass.

Radiant sensible loads - from the transparent/translucent elements such as window glass, skylights and plastic sheets: H = A*(SHGC)*(SC)*(CLF)

H = Sensible heat gain (Btu/Hr) A = area of roof, wall or glass calculated from building plans (sq-ft) SHGC = Solar Heat Gain Coefficient SC = Shading Coefficient CLF = Solar Cooling Load Factor.

Cooling Load: q (BTU/hr) = sum of all surfaces [ A x SC x MSHG x CLF ] Where:

q = cooling load (Btu/hr) the load to be cooled A = Window Area (Sq. ft) SC = Shading Coefficient MSHG = maximum solar heat gain (BTU / hr / ft2) SEE a TABLE CLF = cooling load factor SEE a TABLE To reduce cooling load: Reduce area A where solar radiation enters Improve shading coefficient SC (add shades, sun screens, reflective materials) Use lower MSHG based on direction for given location, month

36

Heat Load - Motors Single Phase:

BTU/Hr = kW x use factor x 3412 BTU/kWh or BTU/Hr = kW x use factor x PF x 3412 BTU/kWh

Three-Phase: BTU/Hr = [Voltage x Current per phase x PF x 1.732 W] x 3.412 BTU/Wh x use factor Heat Load - Lighting H (Btu/Hr) = K(kW) x 3412 BTU/kWh K is the lighting load in kW H = 3.41 * W * F UT * F BF * (CLF)

H = Sensible heat gain (Btu/hr) W = Installed lamp watts input from electrical lighting plan or lighting load data F UT = Lighting use factor, as appropriate F BF = Blast factor allowance, as appropriate CLF = Cooling Load Factor, by hour of occupancy Note: CLF = 1.0, if operation is 24 hours or if cooling is off at night or during weekends .

HVAC Control

Reheat Coil Reset Selects the zone/area with the greatest need for reheat, and establishes the minimum temperature of the heating hot water so that it is just hot enough to meet the reheat needs for that time period.

37

Example: What is kW load of a 100 ton system with COP of 3.5?

3.517 kW -------- = ------- COP Tons 3.517 x Tons kW 351.7 ----------------- = = ------------- = 100.49 kW COP 3.5

Example: Run time = 2000 hours/year 10.8 kW load reduction COP = 2.6

How much energy is saved? Energy = kWh (10.8kW x 2000 hrs per yr) Energy = ------------------------------------ 2.6 Energy Savings = 8,307.7 kWh/year

Example: A 500-ton absorption chiller operating at a COP of 0.70; what is kW load? 3.517 kW -------- = ------- COP Ton 3.517 kW -------- = ------- kW = 2,512 .70 500 Example: Given:

6000 HDD 900 Ft/min air flow;

38

8x10 panes of glass missing qty 6 What are BTUs lost /year? BTU/Year = CFM x 25.92 x (HDD+CDD) Find CFM and use formula: 8 x 10 = 8/12 ft x 10/12 ft = 80/144 ft2 x 6 windows = 3.33 ft2 total opening CFM = 900 ft/min x 3.33 ft2 = 3,000 CFM BTU/Year = 3,000 x 25.92 x 6,000 = 466,560,000 BTU/yr Example: 10. An absorption system with a COP of 0.8 is powered by hot water that enters at 200 F and exits at 180 F at a rate of 25 gpm. The chilled water operates on a 10 degF temperature difference and the condenser cooling water on a 22 F temperature difference. Calculate the chilled water flow. H = Enthalpy calculated by energy of hot water entering and exiting.

H = 500 x Q x T x COP of system = 500 x 25 x 20 x .8 = 200,000 BTU/hr To solve for chilled water flow: Q = H/ (500 x T) = 200,000 BTU/hr / (500 x 10degF) Q = 40gpm

Example: A equal pct. valve has coeff of flow of .62, water temp is 182, diff pressure = 36 psi. What is flow rate? Q = .62 (36/.972 ) = 3.77 GPM

39

Example: 11. 10,000 cfm of air leaves an air handler at 50 F; it is delivered to a room at 65F. No air was lost in the duct. No water was added or taken away from the air in the duct. How many BTU/hr was lost in the ductwork due to conduction? i. 162,000 BTU/hr ii. 75,000 BTU/hr iii. 126,550 BTU/hr iv. 256,000 BTU/hr v. 10,000 BTU/hr Q Sensible (BTU/hr) = CFM x T x 1.085 Btuh = 10,000 x 15 x 1.085 = 162,750 BTU/hr Example: 34. A building has solar gray glass with a shading coefficient of 0.5 and Venetian blinds with a shading coefficient of 0.4. The combined shading coefficient for the building is:

SC must be measured in this example as there is no mathematical way to combine shading coefficients. The shading coefficient SC is the % heat gain that passes through the window. Example: window glazing has SC = .35. Therefore, 35% of heat gain passes through the window, and 65% of heat gain is stopped through the window.

Example: book ex. 6-4 200 fixtures, 4 lamps per fixture 40Watt bulbs replaced by 34 W bulbs no air conditioning operate 24x7

40

With Air Conditioning, what is additional savings? HVAC unit has COP = 2.8 200 x 6 x 4 = 4800 W load saved kWh/year saved = 4.8 x 8760 = 42,048 kWh/yr Savings from air conditioning with new lamps: kWh/year saved = 42,048 kWh/yr / 2.8 = 15,017 kWh/yr OR BTU/Hr = K (kW) x 3412 BTU/kWh BTU/Hr = 4.8 x 3412 BTU/kWh = 16377 BTU/hr BTU = 16377 x 8760 = 143,467,776 BTU/year Energy to remove heat of BTUs saved by A/C: Watt-hrs = 143,467,776 BTUs/[2.8 x 3.412 BTU/Wh] = 15,021 kWh Example: 6-6: A 5-hp fan motor is 84% eff; , replace with 1.5 hp motor with eff=85.2. What is load reduction? 5hp x .746kW/.84 = 4.44 kW used 1.5hp x .746kW/.852 = 1.31 kW used Energy saved = 4.44 1.31 = 3.13 kW Review: Maximum Solar Heat Gain - MSHG Cooling Load Factors - CLF

41

Boilers, Steam Formulas Boiler Blowdown Rate

The blow down rate of a boiler depends on

steam consumption (steam used in the process and not returned as condensate to the boiler)

concentration of impurities in the feed water maximum allowable Total Dissolved Solids (TDS) in the boiler

The blowdown rate can be calculated:

qBD = qS fc / (bc - fc) where qBD = blowdown rate (kg/h) qS = steam consumption (kg/h) [ the % of condensate NOT

returned as condensate to the boiler fc = Total Dissolved Solids - TDS - in the feed water (ppm) bc = maximum allowable Total Dissolved Solids(TDS) in the

boiler water (ppm)

42

Cost of Steam Loss Given:

steam pressure psi ambient outside temperature degF leak losses lb/hr boiler efficiency fuel cost Solve: Use Steam Tables for BTU/lb Find fuel amount lost per year in BTUs or LBS

Example: For a 600psia steam pressure and 75F ambient temperature with 3750 lb/hr of losses, with a boiler efficiency of 85% and fuel of $65/ton coal at 14,500 BTU/lb. Calculate cost of lost steam per YEAR: Use Steam Tables for BTU/lb 600psia -> 1203.7 BTU/lb STEAM 75F -> 43 BTU/lb SATURATED LIQUID -------------------- 1160.7 BTU/lb Find fuel amount lost per year (tons of coal): 1160.7 BTU/lb x 3750 lb/hr x 8760 hrs/yr x $65/ton Tons Coal = -------------------------------------------------------------------- = $100,543/year 14,500 BTU/ lb x 2000 lb/ton x .85 Steam Leak Through Orifice Lost Steam (Lb/h) = .70 x .0165 x 3600 x A x P.97 Lost Steam (Lb/h) = 41.58 x A x P.97 where

P = Pressure of steam line in psia A = Area of hole in square inches (TAKE NOTE IF YOU ARE GIVEN THE DIAMETER OR RADIUS OF THE HOLE!)

43

% Flash Steam Generated h(f1) h(f2) Flash % = ----------------------- H(fg2)

hf1 = Specific Enthalpy of Saturated Water at Inlet hf2 = Specific Enthalpy of Saturated Water at Outlet hfg2 = Latent Heat of Saturated Steam at Outlet

Example: Steam enters a heat exchanger at: 1200psia and 567degF Leaves as water at: 120psia and 300degF How much heat is exchanged per pound of entering steam?

Use Steam Tables to find saturated liquid enthalpies (BTU/lb) delta h = h1 h0 h1 = 571.9 BTU/lb enthalpy for 1200psia and 567F h2 = 269.7 BTU/lb enthalpy for 300 degF ------------------ h0 = 302.2 BTU/lb NOTE (I think this is right): For h2 1200 psi has temp of 567F; thus the enthalpy is 571.9 For h1 120 psi shows a steam temp of 341.27F; therefore, use enthalpy for 300F saturated liquid.

Example: Hole diameter: Pressure: 100 psig Steam Leak (lb/hr) = 41.58 x (.252 x 3.14159) x 114.7.97 = 812 lb/hr

44

Combustion Efficiency and Temperature: determine parameters with Chart: % excess air % flue gas Oxygen % flue gas CO2

45

Air - Excess Air Factor

Mass (kg) Air To Combust 1 kg Of Fuel EA = ------------------------------------------------------------ Stochiometric Air (AF)

Excess air factors found in practice As mentioned, the excess air factor of a burner furnace or boiler is a yardstick about its efficiency as well as the skill of the operator. Standard average figures are Gas burners, forced draft 1.1 - 1.3 Atmospheric gas burners 1.25 - 1.5 Oil burners 1.15 - 1.3 Coal dust burners 1.2 - 1.3 Coal firing (mechanical) 1.3 - 1.5 Coal firing (hand) 1.5 - 2.5

Table 1: Air-to-fuel ratio of various fuels Fuel Phase AF CO2 max wet

CO2 max dry Very light fuel oil liquid 14.27 13.56 Light fuel oil liquid 14.06 13.72 Medium heavy fuel oil liquid 13.79 14.00 Heavy fuel oil liquid 13.46 14.14 Bunker C liquid 12.63 16.23 Generic Biomass (maf) solid 5.88 17.91 Coal A solid 6.97 16.09 LPG (90 P : 10 B) gas 15.55 11.65 Carbon solid 11.44 21.00

46

Insulation Heat Transfer - Insulation Thermal Conductivity (K) ability of a material to conduct heat K is in (BTU)(in) / (ft2 )(hr)(degF) at room temperature; K varies by Temperature; charts give K values for varying temperature Thermal Resistance (R Value): R = d / K R is expressed (ft2 )(hr)(degF) / BTU d = thickness of material (inches) Total R value of materials = sum of individual R values of each material Conductance: 1 U = --- BTU / ft2 x h x degF R Total U value of multiple materials IS NOT THE sum of individual U values Heat Load / Transfer Conduction

Obtain K value from table Obtain Rinsulation values and Rsurface values from tables. Use d = thickness of material given Calculate U Calculate Heat Loss Q in BTU/hr

47

Heat transfer equation change in TEMPERATURE: Q total = U x A x T or Q total = U x A x (DD/time period) x 24h/day

Q total = Rate of heat transfer per surface area involved (BTU/hr) U = Conductance; [U = 1/R] A = Area of heat transfer surface T = temperature difference (deg F) between adjacent space and room

temp DD/year = total of HDD and CDD during the year or time period (deg F days)

Pipes For calculating heat transfer in pipes: a) Table 11-4 nominal pipe size (diam) to determine outside radius b) Determine R material using K and d thickness c) Table 11-3 to determine R surface d) R material + R surface = R total

U = 1 / R total Solve for Q = heat loss without insulation = U x A x T e) Determine d thickness including insulation: d = r2 ln (r2/r1) where: r1= outside radius of pipe r2= outside radius of pipe plus insulation f) calculate new R total g) calculate Q with insulation

48

Tanks a) Table 11-3 obtain R surface coefficient b) Use R tank R total = R tank + R surface coefficient d R tank = ---- where d = thickness K c) Solve for Q: 1. heat loss WITHOUT insulation 2. heat loss WITH insulation Q = U x A x T

Example: Insulation Problems: Identify a material with a certain R value Note that R may be expressed as: R = hr ft2 F/Btu inch per inch = d / K R = 1 / K when thickness is per inch (per 1 inch) Reference a books K table to find 1/R = K and the material.

Example: Problem 6-3: p267 Wall Area = 100 ft2 U = .25 BTU / ft2 hr degF DD = 3,000 heating season What is heat amount lost through wall? Q total = U x A x (DD/time period) x 24h/day = .25 BTU / ft2 hr degF x 100 sq ft x 3000 x 24 = 1,800,000 BTU / year

49

Example: Nominal 4 Steel Pipe Thickness of .25 500 ft long 180F hot water Ambient air = 80F Add 2 silicate insulation (K=.4) Heating efficiency = 80% What amount of heat is saved on BTUs? a) Table 11-4 nominal pipe size (diam) to determine outside radius r = 2.25 b) Determine R material using K and thickness d steel material K = 314.4 thickness = .25 R material = d/K = .25 / 314.4 = .0008 c) Table 11-3 to determine R surface R surface = 0.46 at Ts - Ta = 100F d) add R material + R surface = R total .0008 + .46 = .4608 U = 1 / R total = 2.17 Solve for Q = heat loss without insulation Q = U x A x T A = 2(pi)r x length = (2 x (3.1415) x 2.25)/12 in/ft x 500 ft = 589 ft2 Q = 2.17 x 589 x 100 = 127,821 BTU/hr e) Determine d thickness including insulation: d = r2 ln (r2/r1) where: r1= outside radius of pipe r2= outside radius of pipe plus insulation r2 = 2.25 + 2 = 4.25

50

d = r2 ln (r2/r1) = 4.25 x ln (4.25/2.25) = 2.70 f) calculate new R total R material + R surface = R total K material = .4 from chart R material = 2.70 / .4 = 6.75 R TOTAL = R surface .46 + R material 6.75 = 7.21 U = 1 / R total = .14 g) calculate Q with insulation Q = U x A x T A = 2(pi) 4.25 x length = 1112 ft2 Q = .14 x 1112 x 100 = 127,821 BTU/hr = 15,576 BTU/hr Annual savings = (127,821 - 15,576 BTU/hr) x 8760 = 983 MMBTU/yr 983 x $6/MMBTU /.80 efficiency = $7,374/yr

51

Building Envelope Dry Bulb Dew Point Temp Enthalpy Wet Bulb Temp Humidity REFER TO PSYCHOMETRIC CHART Ventilation Rate Procedure Building Zone Outdoor Air Flow Vbz = RpPz + RaAz zone (ft2)

Vbz = breathing zone outdoor airflow Pz = zone population largest number of people occupying zone / typical usage (DESIGN OCCUPANCY) Rp = People outdoor airflow rate required per person FROM TABLE Ra = Area outdoor airflow rate per unit area FROM TABLE Az = zone floor

SEE TABLES for VENTILATION RATES by ROOM TYPE Example: For a 4000 sq ft science lab and calculated minimum outdoor air flow of 3000 cfm, what is max occupancy? V(bz) = 3000 A = 4000 Vbz = RpPz + RaAz zone (ft2) 3000 = 10 x Pz + .18 x (4,000 ft2) solve for Pz = 228 persons

52

Motors - Process Energy Management Motor Load Factors:

NPHP x ( .746 kW/HP) x (Load Factor) kW = ---------------------------------------------------- Efficiency where NPHP is nameplate horsepower

Example: 100 hp rated motor with 95% efficiency and 60% load (factor). Find the kW Load: kW = 100 x .746 x .6 / .95 = 47.12 kW Determining Motor Loads From Measured Data Note: Methods are listed from LEAST to MOST Accurate

At zero load, the motor operates at (or very near) the synchronous speed. At full load, the motor operates at its rated speed = nameplate speed.

S synchronous = (120 x F) / P S = speed in RPM F = frequency in hertz P = # of poles in motor Calcs in chart:

# of Poles

Synchronous Speed

2 3600 4 1800 6 1200 8 900

NOTE: for a POLE PAIR: DOUBLE the number of poles.

EXAMPLE: four 2-pole pairs = 8 POLES TOTAL

53

Slip Method A motor's speed and slip is proportional to its load.

(S synchronous - S ) Time Slip % Motor Load = ----------------------------------- = ------------------------------- ( S synchronous S Full Load) Design Slip

S = measured motor speed, RPM Ssynchronous = Motor's synchronous speed (zero or no load) = NLRPM SFull Load = Motor's full load (rated load) speed = FLRPM Note: Slip = (S synchronous - S)

Example: Given: Motor nameplate data: Ssynchronous = 1800 RPM Full Load (Rated) Speed: S Full Load = 1750 RPM Measured motor speed S = 1770 RPM

What is the motor load? Motor Load = (1800 - 1770)/(1800 - 1750) = 60%

Example: Given: Motor Load = 85% Full Load (Rated) Speed: S Full Load = 860 RPM Twin 4-pole motors Freq: 60 Hz (assumed) What is the measured motor speed?

Motor Load = (Ssynchronous - S)/( Ssynchronous - SFull Load) 85% = (Ssynchronous - S)/( Ssynchronous - 850) Ssynchronous (8 poles) = (120 x 60) / 8 = 900 85% = (900-S) / (900-860) S= 900-85% x (900-860) = 866 RPM

54

Voltage Compensated Slip Method Measurements required: S = measured motor speed, RPM

V = average RMS line-line current Inputs required: S synchronous = Motor's synchronous speed S Full Load = Motor's full load (rated) speed V rated = Motor's rated voltage Motor Load = (S synchronous - S)/ [ (S synchronous S FullLoad) x (Vrated/V)2 ] Slip = (S synchronous - S) Example: Motor nameplate data:

S synchronous = 1800 RPM S Full Load = 1750 RPM V rated = 240 V Measured motor speed = S = 1770 RPM V = average RMS line-line voltage = 224 V What is the motor load? Load = (1800 - 1770)/[(1800 - 1750)*(240/224)^2] = 52%

Voltage Compensated Current Ratio Measurements required: I = RMS motor current, average of 3 phases. V = average RMS line-line current Inputs required: Irated = nameplate rated current at full load Vrated = nameplate rated voltage at full load Motor Load = (I/Irated) x (V/Vrated) Example: Motor nameplate data: Irated = 20 Amps; Vrated = 240 V Measured RMS line current: Ia = 16.2 Amps, Ib = 15.5 Amps, Ic = 16.8 Amps Measured RMS line-line voltage: Vab = 232 V, Vbc = 228 , Vac = 236 V What is the Motor Load? I = (Ia + Ib + Ic)/3 = (16.2 + 15.5 + 16.8)/3 = 16.167 Amps V = (Vab + Vbc + Vac)/3 = 232 V Motor Load = (16.167/20) x (232/240) = 78.1%

55

Direct KWh Calculation Measurements required: P measured = measured motor load, kW or I = RMS motor current, average of 3 phases. V = average RMS line-line VOLTAGE PF = measured power factor Inputs required: HP = Motor's rated power output, HP (or kW) h Full Load = Motor's full load rated efficiency Formulas:

Motor Load = Pmeasured / Prated Pmeasured = 3 * V * I * PF / 1000 [kW] (if not measured directly from the meter) Prated = HP * (0.746 kW/HP) / hrated = motor's power input at rated full load

56

Example: Given: Motor nameplate data: HP = 40hp, hFull Load(rated) = 91.2% Measured RMS line current: Ia = 36 Amps, Ib = 38 Amps, Ic = 37 Amps Measured RMS line-line voltage: Vab = 469 V, Vbc = 473, Vac = 467 V Measured PF: PFa = 0.75, PFb = 0.78, PFc = 0.76 What is the Motor Load? I = (Ia + Ib + Ic)/3 = 37 Amps V = (Vab + Vbc + Vac)/3 = 469.67 V PF = (PFa + PFb + PFc)/3 = 0.763 Pmeasured = 3 * 469.67 * 37 * 0.763 = 22.9 kW Prated = 40 * (0.7457 kW/HP) / 91.2% = 32.7 kW Motor Load = 22.9 / 32.7 = 70% Computerized Modeling Techniques ORMEL 96 that uses an equivalent circuit technique, based on IEEE's standard 112, to determine motor load. It requires: Being implemented in the Motor Master+ Software

Measurements required: S = measured motor speed, RPM Inputs required: Motor nameplate data, including kVA code. Also:

Specialized Motor Testing Equipment Laboratory Methods

57

Air Compressors Air Leak Rate (SCFM)

V x (P1 P2) Leak Rate (SCFM) = ------------------ T x 14.7 SCFM = standard cubic feet per minute = the volumetric flow rate of a gas corrected to "standardized" conditions of 14.73 psia

V = Volume Air (cu. ft.) P1 = Pressure P2 = Pressure T = temperature difference (deg F) Air Leaks size of hole

Typical compressor efficiency of 18kW/100cfm = .18 kW/cfm VOLUME IN cfm:

58

Renewable Energy Systems and Water Management Thermal Storage Formula: C (BTUs) = m Cp T where:

C is the stored BTUs m is the mass of the substance (lb) Cp is the specific heat capacity of the substance (Btu/lboF) T is the temperature change (oF)

Thermal efficiency is defined as:

= Thermal efficiency = Total work output by all systems = Total heat input into the system

59

1. Codes and Standards and Indoor Air Quality ASHRAE ANSI/ASHRAE/IES Standard 90.1-2013 ASHRAE Energy Standard for Buildings Except Low-Rise Residential Buildings

US standard that provides minimum requirements for energy efficient designs for buildings except for low-rise residential buildings.

There are many states that apply the ASHRAE 90.1 standard to different buildings that are being constructed or under renovation. Most states apply the standard or equivalent standards for all commercial buildings while others apply the standard or equivalent standards for all government buildings.

Illuminating Engineering Society of North America IESNA and ANSI take part ASHRAE 90.1 covers:

Buildings building envelope majority of mechanical and lighting systems New buildings being constructed Additions / alterations to existing buildings and their systems

Not Covered: Single family homes multifamily of three stories or less homes, manufactured or modular homes, buildings that do not use electricity or fossil fuels equipment and building systems that are used for industrial, manufacturing, or commercial purposes

Energy Elements Covered: Envelope

o insulation o fenestration (window design) o doors o air leakage o type of building each has different requirements to meet

nonresidential conditioned space residential conditioned space semi-heated space.

o Roof o Walls o Floor

60

HVAC Hot Water Lighting lighting power density ASHRAE/IES 90.1 Lighting Power Allowances using the Building Area Method.

ASHRAE 90.1 Industrial processes (added in 2013) Economizers for data centers (2011) Building Envelope include skylights, solar reflectance, thermal emittance, air barriers, and solar orientation Revisions affect the maximum:

o fan power limits o pump head calculation o chilled water pipe sizing o radiant panel insulation o single-zone V A V o supply air temperature reset.

HVAC o Energy recovery is required for many more HVAC systems. o Several reheat exceptions were eliminated or modified. o Restrictions were placed on overhead air heating. o Economizer requirements were added for more climate zones and smaller systems.

Lighting o Power densities (LPD) dropped slightly on average. o Daylighting and associated lighting control requirements were added. o Many lighting control requirements were added, including independent functional testing of lighting controls, occupancy and vacancy controls, exterior lighting controls, and whole-building shutoff. o Offices and computer classrooms now require 50 percent of 120V

receptacles to be automatically switched. Requirements were added for service water booster pumps and elevators. Revised, stricter opaque element and fenestration requirements at a reasonable level of cost-effectiveness Revised equipment efficiencies for heat pumps, packaged terminal air conditioners (PTACs), single package vertical heat pumps and air conditioners (SPVHP and SPVAC), and evaporative condensers New provisions for commercial refrigeration equipment and improved

controls for heat rejection and boiler equipment Improved requirements for expanded use of energy recovery, small-motor efficiencies, and fan power control and credits Improved equipment efficiencies for chillers

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A new alternate compliance path to Section 6, "Heating, Ventilating, and Air-Conditioning," for computer room systems, developed with ASHRAE Technical Committee (TC) 9.9. The Federal Energy Policy Act of 2005 established a tax deduction for energy-efficient commercial buildings applicable to qualifying systems and buildings placed in service from January 1, 2006, through December 31, 2007. A tax deduction of $1.80 per square foot is available to owners of new or existing buildings who install: (1) interior lighting or (2) building envelope or (3) heating, cooling, ventilation, or hot water systems that reduce the buildings total energy and power cost by 50% or more in comparison to a building meeting minimum requirements set by ASHRAE Standard 90.1-2001.

ASHRAE 90.2 is a Residential Energy Std- low-rise residential buildings (single family to multi-family). ANSI/ASHRAE Standard 90.2-2007 - Published standard. (Supersedes ANSI/ASHRAE Standard 90.2-2004. Superseded 90A-1980 & 90B-1975 for all requirements for low-rise residential buildings) This standard provides minimum energy efficiency requirements for the design and construction of:

1) new residential dwelling units and their systems 2) where explicitly specified: o new portions of residential dwelling units and their systems o new systems and equipment in existing dwelling units. o Note: There are no requirements in this standard that apply to new portions of residential dwelling units and their systems, nor to new systems and equipment in existing dwelling units. For the purpose of this standard, 'residential dwelling units' include single-family

houses, multi-family structures (of three stories or fewer above grade), and modular houses.

This standard does not include 'transient' housing such as hotels, motels, nursing homes, jails, and barracks, or manufactured housing.

This standard applies to the building envelope, heating equipment and systems, air-conditioning equipment and systems, domestic water-heating equipment and systems, and provisions for overall building design alternatives and trade-offs.

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This standard does not apply to: o specific procedures for the operation, maintenance and use of residential buildings, o portable products such as appliances and heaters; and o residential electric service or lighting requirements. The Standard shall not be used to abridge any safety, health or environmental requirements.

ASHRAE Standard 62.1-2010 Ventilation for Acceptable Indoor Air Quality Specify minimum ventilation rates and other measures intended to provide IAQ that is acceptable to human occupants and that minimizes adverse health effects. 1. Intended for regulatory application to new buildings and additions 2. Guide the improvement of IAQ in existing buildings

All spaces intended for human occupancy excluding low-rise residential (62.2) Defines requirements for ventilation, air-cleaning design, commissioning,

installation and O&M Additional requirements & other standards may apply (labs, healthcare, industrial, etc.) May be applied to both new and existing buildings, not intended to be used retroactively Does not prescribe specific ventilation rates for smoking spaces Ventilation requirements based on chemical, physical, & biological contaminants Consideration or control of thermal comfort is not included In addition to ventilation, the standard contains requirements related

to certain sources ASHRAE Standard 62 prescribes a ventilation standard of 15 cubic feet of

outside air per building occupant. o This level may be ensured by controlling the indoor CO2 content. o The ventilation demand in each zone can be determined by remote CO2 sensors in a similar manner as a thermostat that regulates the degree of cooling or heating supplied.

Acceptable IAQ may not be achieved in all buildings meeting these requirements because of:

Diversity of sources and contaminants Air temperature, humidity, noise, lighting, and psychological/social factors Varied susceptibility in the occupants Introduction of outdoor contaminants

In an average HOME, the standard number of air changes per hour (ACH) = 0.35. In other words, it will take a little less than three hours for the air in the home to recycle entirely.

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ANSI/ASHRAE/ISO Standard 135-2008 - BACnet BACnet originated & supported by ASHRAE BACnet makes it possible to integrate a facilitys various control systems to a single workstation application for ease of operation.

allows users to expand and upgrade controls using technology from multiple vendors common communication infrastructure and front-end building automation systems

The protocol defines a model for building automation systems:

Data and control functions structured in an object oriented fashion Services that describe data requests and responses Network datalink types A scalable and flexible internetwork and network architecture.

ASHRAE/USGBC/IES Standard 189.1 The Green Standard Standard 189.1 provides a total building sustainability package for those who strive to design, build and operate green buildings. Site location and sustainability energy use recycling water use efficiency energy efficiency indoor environmental quality Standard 189.1 serves as a compliance option in the 2012 International Green Construction CodeTM (IgCC) published by the International Code Council. The IgCC regulates construction of new and remodeled commercial buildings.

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INTERNATIONAL IEC Codes International Electrotechnical Commission is a non-profit, non-governmental international standards organization that prepares and publishes International Standards for all electrical, electronic and related technologies collectively known as "electrotechnology". IEC standards cover a vast range of technologies including:

power generation transmission and distribution home appliances office equipment semiconductors fibre optics batteries solar energy nanotechnology marine energy PLC Programmable Logic Controller programming via the IEC 61131-3 standard others The IEC charter embraces all electrotechnologies including energy production and distribution electronics magnetics and electromagnetics electroacoustics multimedia telecommunication medical technology general disciplines such as terminology and symbols, electromagnetic compatibility (by its Advisory Committee on Electromagnetic Compatibility, ACEC), measurement and performance, dependability, design and development, safety and the environment.

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IECC - International Energy Conservation Code from the ICC Model Energy Code (MEC) was predecessor Published and maintained by the International Code Council (ICC) as the International Energy Conservation Code (IECC) as of 1998, contains energy efficiency criteria for new residential and commercial buildings and additions to existing buildings. It covers the buildings ceilings, walls, and floors/foundations; and the mechanical, lighting, and power systems.

IECC - International Energy Conservation Code is a building code created by the International Code Council in 2000. It is a model code adopted by many states and municipal governments in the United States for the establishment of minimum design and construction requirements for energy efficiency.

Building exterior Mechanical systems Lighting systems Internal power systems

Designed to address the design of energy-efficient building envelopes and installation of energy efficient mechanical, lighting and power systems through requirements emphasizing performance that will result in the optimal utilization of fossil fuel and nondepletable resources in all communities, large and small.

commercial buildings low-rise residential buildings (3 stories or less in height above grade.) establishes minimum regulations for energy efficient buildings using

prescriptive and performance-related provisions. Fully compatible with all of the International Codes (I-Codes) published by the International Code Council (ICC),

o International Building Code, o International Existing Building Code, o International Fire Code, o International Fuel Gas Code, o International Green Construction Code o International Mechanical Code, ICC Performance Code, International Plumbing Code, o International Private Sewage Disposal Code, o International Property Maintenance Code, o International Residential Code, o International Swimming Pool and Spa Code TM o International Wildland-Urban Interface Code o International Zoning Code.

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International Performance Measurement and Verification Protocol (IMVP) The International Performance Measurement and Verification Protocol (IMVP) provides an overview of current best practice techniques available for verifying results of: energy efficiency water efficiency renewable energy projects Indoor Air Quality May also be used by facility operators to assess and improve facility performance.

Energy conservation measures covered herein include: fuel saving measures water efficiency measures load shifting energy reductions through installation or retrofit of equipment modification of operating procedures.

ISO 50001: Energy management systems Requirements for establishing, implementing, maintaining and improving an energy management system, whose purpose is to enable an organization to follow a systematic approach in achieving continual improvement of energy performance, including energy efficiency, energy security, energy use and consumption. ISO 50001 requires an organization to demonstrate that they have improved their energy performance. Structure: 1.: General Requirements 2.: Management Responsibility 3.: Energy Policy 4.: Energy Action Plan 5.: Implementation and Operation 6.: Performance Audits 7.: Management Review Method: ISO 50001 provides a framework of requirements that help organizations

develop a policy for more efficient use of energy fix targets and objectives to meet the policy use data to better understand and make decisions concerning energy use and consumption measure the results review the effectiveness of the policy continually improve energy management

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FEDERAL REGULATION Federal Power Act

1930 Act created the Federal Power Commission Originally had authority for hydro-electric plants FPC is now FERC The Federal Energy Regulatory Commission (FERC) is the United States federal agency with jurisdiction over: interstate electricity sales wholesale electric rates hydroelectric licensing natural gas pricing oil pipeline rates

The responsibilities of FERC now include the following:

regulating the transmission and sale of natural gas for resale in interstate commerce regulating the transmission of oil by pipelines in interstate commerce regulating the transmission and wholesale sales of electricity in interstate commerce licensing and inspecting private, municipal, and state hydroelectric projects approving the siting of and abandonment of interstate natural gas facilities,

including pipelines, storage and liquefied natural gas ensuring the reliability of high voltage interstate transmission system monitoring and investigating energy markets Also: using civil penalties and other means against energy organizations and individuals who violate FERC rules in the energy markets overseeing environmental matters related to natural gas and hydroelectricity projects and major electricity policy initiatives administering accounting and financial reporting regulations regulating businesses of regulated companies In 1938, the Natural Gas Act gave FPC jurisdiction over interstate natural gas pipelines and wholesale sales. In 1942, this jurisdiction was expanded to cover the licensing of more natural gas

facilities.

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1948 - Federal Water Pollution Control Act (Clean Water Act expanded this in 1972) In 1954, the Supreme Court decision in Phillips Petroleum Co. v. Wisconsin extended FPC (pre-FERC) jurisdiction over all wellhead sales of natural gas in interstate commerce.

1963 / 1970 / 1990 The Clean Air Act gives the EPA the power to create and enforce standards related to air quality, designed to control air pollution on a national level. Requires the EPA to develop and enforce regulations to protect the public

from airborne contaminants known to be hazardous to human health. 1963 version of the legislation established a research program, expanded in 1967. 1970 amendments greatly expanded the federal mandate, requiring comprehensive federal and state regulations for both stationary

(industrial) pollution sources and mobile sources. It also significantly expanded federal enforcement. 1990 amendments addressed acid rain, ozone depletion and toxic air

pollution, established a national permits program for stationary sources, and increased enforcement authority. The amendments also established new auto gasoline reformulation requirements, set Reid vapor pressure (RVP) standards to control evaporative emissions from gasoline, and mandated new gasoline formulations sold from May to September in many states.

includes a provision for citizen suits. 1972 1977 1987 Clean Water Act (Federal Water Pollution Control Act) is the primary federal law governing water pollution.

restore and maintain integrity of the nation's waters (chemical, physical, and biological)

prevent point and nonpoint pollution sources provide assistance to publicly owned wastewater treatment works maintain the integrity of wetlands. does not directly address groundwater contamination

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Major amendments were enacted in the Clean Water Act of 1977 and the Water Quality Act of 1987. Groundwater protection provisions are included in:

Safe Drinking Water Act 1974 Resource Conservation and Recovery Act 1976 Superfund Act 1980

1976 The Resource Conservation and Recovery Act (RCRA) principal federal law governing the disposal of solid waste and hazardous waste.

1977 Department of Energy (DOE) Organization Act passed by Congress in response to an energy crisis in 1973. Consolidated various energy-related agencies into a Department of Energy. FPC was renamed the Federal Energy Regulatory Commission (FERC), a

separate independent regulatory body. As a further protection, when the Department of Energy proposes a rule, it must refer the proposal to FERC, and FERC can take over the proceeding if FERC determines that the rulemaking "may significantly affect" matters in its jurisdiction.[4] The DOE Act transferred regulation of interstate oil pipelines from the

Interstate Commerce Commission (ICC) to FERC. However, the FERC lost some jurisdiction over the imports and exports of gas and electricity. In 1978, FERC was given additional responsibilities for harmonizing the regulation of wellhead gas sales in both the intrastate and interstate markets. National Energy Act of 1978 included:

Public Utilities Regulatory Policy Act (PURPA) of 1978. o FERC administered program for new cogeneration and small power

production. o To encourage improved efficiency at central power plants: buy power from other energy producers.

The Natural Gas Policy Act, which reduced the scope of federal price regulation, to bring greater competition to both the natural gas and electric industries. Also protects consumers from sharp upticks in heating oil pricing.

FOR FERC/PURPA co-gen qualifying facility (QF) criteria SEE CHP SECTION 14.

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Fuel Use Act of 1978 declared that major fuel-burning installations could not use natural gas as a primary source of energy. The intention of this law was to force industrial concerns to use alternative fuels. 1979 to present (10 surveys conducted) - Commercial Buildings Energy Consumption Survey (CBECS) conducted by the Department of Energy provides data on commercial energy use.

covers primarily non-industrial usage concentrates on energy use for air conditioning and lighting.

1980 Superfund Act [Comprehensive Environmental Response, Compensation, and Liability Act (CERCLA) ] federal law designed to clean up sites contaminated with hazardous substances as well as broadly defined "pollutants or contaminants". 1982 IEEE-519 PQ (Power Quality) Standard - To minimize the impact of facility harmonic distortion on the utility power system and on neighboring facilities IEEE standard 519 provides recommended limits for total harmonic voltage and current distortion. THD total harmonic distortion In 1989, Congress ended federal regulation of wellhead natural gas prices, with the passage of the Natural Gas Wellhead Decontrol Act of 1989. 2005 The Energy Policy Act of 2005 expanded FERC's authority to impose:

mandatory reliability standards on the bulk transmission system impose penalties on entities that manipulate the electricity and natural

gas markets. FERC ORDERS

1985 FERC Order No. 436 required that natural gas pipelines provide open access to transportation services, enabling consumers to negotiate prices directly with producers and contract separately for transportation

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1989 FERC Order 500, issued in late 1989, was an addendum to FERC Order 436 and provided mechanisms for settling certain contract liabilities incurred by pipelines that could not take all of the gas they had ordered from producers. 1992 FERC issued Order No. 636 - Restructuring of Gas Pipeline Services (The Restructuring Rule). This mandated unbundling of sales services from transportation services, providing customers with full choice of providers and opening these markets to competition. FERC Order No. 636A clarified 636

determine rates for "small customers" capacity releases for any period of less than one calendar month without bidding for the released capacity. distribute revenue responsibility among customers requiring pipelines to recover 10 percent of their gas supply realignment costs from their Part 284 interruptible transportation service. true-up mechanisms that provide a reasonable opportunity for pipelines to recover (but not over-recover) costs.

FERC Order No. 637 further addressed inefficiencies in the capacity release market. 1998 FERC Orders 888 and 889 - opening the US energy market to competition.

FERC Order No. 888 Promoting Wholesale Competition Through Open Access Non-discriminatory Transmission Services by Public Utilities Unbundling of electrical services and the separation of marketing functions utilities to provide open access to their energy rate schedules (tariffs),

FERC Order 889

set standards making pricing and tariff information available to the marketplace established OASIS, a bulletin board system that allows energy customers on

the wholesale market to schedule and reserve capacity on the US regional energy grids to insure that energy can be delivered to customers without competitive interference. FERC is self-funding, in that the it pays for its own operations by imposing annual charges and fees on the industries it regulates.

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The FERC has recent activity: Regional Transmission Organizations (RTOs) Independent System Operators (ISOs) new law dealing with LNG terminals, electric reliability new merger regulations and new anti-market manipulation regulations. Also: FERC regulates approximately 1,600 hydroelectric projects in the U.S. Largely responsible for permitting construction of a large network of

interstate natural gas pipelines. works closely with the United States Coast Guard to review the safety, security, and environmental impacts of proposed LNG terminals and associated shipping.

Executive Order 13423 requires Federal agencies to lead by example in advancing the nations energy security and environmental performance by achieving the following goals:

Energy Efficiency: Reduce energy intensity 30 percent by 2015, compared to an FY 2003 baseline. Greenhouse Gases: Reduce greenhouse gas emissions through reduction of energy intensity 30 percent by 2015, compared to an FY 2003 baseline. Renewable Power: At least 50 percent of current renewable energy purchases must come from new renewable sources (in service after January 1, 1999). Building Performance: Construct or renovate buildings in accordance with sustainability strategies, including resource conservation, reduction, and use; siting; and indoor environmental quality. Water Conservation: Reduce water consumption intensity 16 percent by 2015, compared to an FY 2007 baseline. Vehicles: Increase purchase of alternative fuel, hybrid, and plug-in hybrid vehicles when commercially available. Petroleum Conservation: Reduce petroleum consumption in fleet vehicles by 2 percent annually through 2015, compared to an FY 2005 baseline. Alternative Fuel: Increase use of alternative fuel consumption by at least 10 percent annually, compared to an FY 2005 baseline. Pollution Prevention: Reduce use of chemicals and toxic materials and purchase lower risk chemicals and toxic materials. Procurement: Expand purchases of environmentally sound goods and services, including biobased products. Electronics Management: Annually, 95 percent of electronic products purchased must meet Electronic Product Environmental Assessment Tool standards where applicable; enable Energy Star features on 100 percent of computers and monitors; and reuse, donate, sell, or recycle 100 percent of electronic products using environmentally sound management practices.

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Energy Act 1992 restricted production of incandescent bulbs, T12s; established lighting efficacy and color index standards The Energy Policy Act of 2005 EPACT The act, described by proponents as an attempt to combat growing energy problems, changed US energy policy by providing tax incentives and loan guarantees for energy production of various types. General provisions

Federal buildings must be 30% more efficient than ASHRAE or IECC where life cycle cost effective Under an amendment in the American Recovery and Reinvestment Act of

2009, Section 406, the Energy Policy Act of 2005 authorizes loan guarantees for innovative technologies that avoid greenhouse gases

the Act increases the amount of biofuel (usually ethanol) that must be mixed with gasoline sold in the United States to 4 billion US gallons

$200 million annually for clean coal initiatives, repealing the current 160-acre (0.65 km2) cap on coal leases, allowing the advanced payment of royalties from coal mines and requiring an assessment of coal resources on federal lands that are not national parks; it authorizes subsidies for wind and other alternative energy producers; it adds ocean energy sources, including wave and tidal power for the first time as separately identified, renewable technologies; it authorizes $50 million annually over the life of the law for biomass grants; it includes provisions aimed at making geothermal energy more competitive with fossil fuels in generating electricity; o The federal Energy Policy Act of 2005 established a tax deduction for

energy-efficient commercial buildings applicable to qualifying systems and buildings placed in service from January 1, 2006, through December 31, 2007. This deduction was subsequently extended through 2008, and then again through 2013 by Section 303 of the federal [http://thomas.loc.gov/cgi-bin/query/z?c110:H.R.1424.enr: Energy Improvement and Extension Act of 2008] (H.R. 1424, Division B), enacted in October 2008. A tax deduction of $1.80 per square foot is available to owners of new or existing buildings who install (1) interior lighting; (2) building envelope, or (3) heating, cooling, ventilation, or hot water systems that reduce the buildings total energy and power cost by 50% or more in comparison to a building meeting minimum requirements set by ASHRAE Standard 90.1-2001. Energy savings must be calculated using qualified computer software approved by the IRS.

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The Energy Policy Act of 2005 EPACT Continued requires the Department of Energy to:

study and report on existing natural energy resources including wind, solar, waves and tides; study and report on national benefits of demand response and make a recommendation on achieving specific levels of benefits and encourages

time-based pricing and other forms of demand response as a policy decision;

designate National Interest Electric Transmission Corridors where there are significant transmission limitations adversely affecting the public (the Federal Energy Regulatory Commission may authorize federal permits for transmission projects in these regions); report in one year on how to dispose of high-level nuclear waste; it authorizes the Department of the Interior to grant leases for activity that involves the production, transportation or transmission of energy on the

Outer Continental Shelf lands from sources other than gas and oil it requires all public electric utilities to offer net metering on request to their customers; it prohibits the manufacture and importation of mercury-vap