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Performing an Audit and Assessing Data

Performing an Audit and Assessing Data. Perform an Audit. Types of Energy Audits. Desktop Use only nameplate data (horsepower or Kw) May use only our Inventory Table How are we operating our processes? Pitfall to this is it uses assumed efficiency & load Energy Survey and Analysis

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Performing an Audit and Assessing Data

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  1. Performing an Audit and Assessing Data

  2. Perform an Audit Types of Energy Audits • Desktop • Use only nameplate data (horsepower or Kw) • May use only our Inventory Table • How are we operating our processes? • Pitfall to this is it uses assumed efficiency & load • Energy Survey and Analysis • Use voltmeters, ammeters or, preferably, a Power meter, and flow measurement • How are we operating our processes? • Computer Modeling • Usually requires a consultant or more technical expertise • How are we operating our processes?

  3. Table of Equipment • Motor List - Area of Plant Equipment • Rated hp. • Number of units • Hours/day that each runs • Age • Efficiency • Power factor • Daily/monthly cost for group • All of the above can be obtained using nameplate data • Or can actually measure using volt, amps, or a Power meter. Suggest that we do this for all motors. Why?

  4. The ENERGY AUDIT • Identify priorities: • Look for Low Hanging Fruit • Get biggest bang for our buck • Operating Procedures • Older equipment • Outdated technology • Simple vs. complex Operator incentives will generally find ways to save. Must understand billing structure

  5. Motor Nameplate

  6. Desktop- Evaluation Equipment Evaluation Calculation: Use Nameplate Data • Reliance Electric, 75 hp., • 230/460 volt, amps 175/87.6, • 1780 rpm, 94.1 % efficiency • Power Factor 84.9% • Run time 6.25 hours per day 1.73 X Volts X Amps X Power Factor ÷ 1000 = Kilowatt (Kw) 1.73 X 460 volts X 87.6 Amps X 0.849 Power Factor ÷ 1000 = 60.2 Kw Demand Annual Kw = 60.2 (Kw) X 6.25 hr/day X 365 days/year = 137,331 Kw hr/year

  7. Energy Survey & Analysis Equipment Evaluation Calculation Use Field Gathered Data • Reliance Electric, • 75 hp. - 230/460 volt, amps 175/87.6 • 1780 rpm, 94.1 % efficiency • Measured Data: Volts : L1/L2 477, L2/L3 472, L1/L3 482 = avg. 477 • amps 86.9, 87.0, 86.2 = average 86.7 • Amperage imbalance 0.57 % • Power Factor 0.849 • Run time 6.25 hours per day 1.73 X Volts X Amps X Power Factor ÷ 1000 = Kilowatt (Kw) 1.73 X 477 volts X 86.7 Amps X 0.849 Power Factor ÷ 1000 = 60.7 Kw Demand Annual Kw = 60.7 (Kw) X 6.25 hr/day X 365 days/year = 138,472 Kw hr/year

  8. Monitor and Measure Results • Track performance • Set up a method to continuously gather the data • Pay attention to other changes made that may affect the data • Develop a plan to maintain the new equipment or processes • Remember that this is not a one-time event. It should be ongoing continuous improvement

  9. Correcting Power Factor • Installing capacitors (large investment) • Minimizing the operation of idling or lightly-loaded motors • Avoid operation of equipment above the rated voltage • Replace standard motors with high efficiency as the old one wear out or fails

  10. Example Cost Calculation • Data • 25 hp pump • Operates an average of 18 hours/day • Load factor = 65% • Motor efficiency = 90% • Power cost = $0.08/kwh • Monthly cost = 25 hp x 0.746 kwh/hp x 1.0/0.90 x $0.08/kwh x 0.65 x 18 hours/day x 30 days/month = $582.00 • What is the approximate Demand for this motor?

  11. Calculating The Cost • Assumptions • Not all motors run at full load • Will assume Power costs are ~ $0.08/kwh • Motor efficiency = 90% • Load factor on pump = 65% • {Hp X (0 .746)/motor efficiency (.90)} x monthly hours x $/kwhr x load factor (0.65) = Monthly costs

  12. Field Surveyand Analysis Better Than a Desktop Survey but requires more skills & reliable instruments

  13. WARNING • You need a qualified electrician to use the following equipment. • You will be going into live panels to take these measurements. • You will need very specific PPE depending on our activities. • DO NOT ATTEMPT TO DO THIS IF we HAVE NOT HAD THE REQUIRED SAFETY TRAINING

  14. Equipment Needed • Power meter (or ammeter and volt meter) • Tachometer (to measure pump speed) • Calibrated flow meter • Magnetic flow meters rarely need calibration, unless exposed to solids • Calibrated pressure gages • Always best to use a new calibrated pressure gage • Gages that have been installed for some time can be reading incorrectly, especially if subjected to solids • Usually a waste of money to install gages on all pumps • Best to install fittings and bring a good gage to take measurements when needed

  15. Measuring Power • It is best to measure with a Power meter (if we have a qualified electrician to set it up). This involves going into the electrical panel. • The Power meter will give us • Exact kw, kwh, and Power factor • A correlation with actual flowrate, etc., if a flowmeter is available

  16. System Measurements • For 460 volt systems measure all three legs of the Power feed. • The voltage to the motor should be within 10% of the nameplate. Example nameplate 460v : Range 414v – 506v • The voltage deviation in any one leg should be < 1% from the average. Example 1 % of 460v = 4.6v • The amperage Imbalance should be < 5% from the average • If it does not meet the above criteria, take steps to troubleshoot the system

  17. Data to Collect • Pump Speed (rpm) • Flowrate (gpm or cfm) • Pressure in and out (psig) • Volts and amps on all legs of power in • Power used (kw) • Power factor • VFD % of full scale

  18. Measuring Flowrate • Best if you have an installed flow meter, that is calibrated • Magnetic flow meters require little calibration, unless in a sludge line and subject to buildup • Venturi meters require calibration • Can install a temporary flow meter (strap-on) • If possible, pump in or out of a tank and measure levels

  19. Measure Pressure • You may have a pressure gage installed on the pump discharge. It probably does not read correctly • Best to install a calibrated gage (certified) on both suction and discharge side of the pump • TDH is the discharge pressure minus the suction pressure

  20. Impeller Size: 10” Total Head: 45 ft Capacity: 200 gpm Eff.: 58% bhp: 3.9 Impeller Size: 10” Total Head: 40 ft Capacity: 400 gpm Eff.: 73% bhp: 5.5

  21. Pump Curves • Provided by the manufacturer when delivered • Always good practice to verify the curve during commissioning (make it a requirement of the specification) • Good practice to check each pump’s performance on a schedule

  22. Pump Curves • Plots flow vs. TDH • Also shows Best Efficiency Point (BEP) – minimum stress on the pump system, max efficiency • Shows brake horsepower (bhp) vs. flow • If a VFD, will show several curves • If a tested pump is on the curve and operating at the BEP, we have done our job well.

  23. How to determine actual pump efficiency Ep= 3.14 X Head ft. X Flow MG Power kWh X Eff. Motor

  24. Calculate theoretical pump efficiency A raw sewage pump operating at a head of 30 feet draws down a 25 ft. by 25 ft. wet well by 3 ft. with the inlet valve closed in 3 minutes. During this period the current draw on the 50 hp. 460 volt motor was measured to be 63 amps. What is the efficiency of the pump? Flow Quantity MG = 25 ft. X 25 ft. X 3 ft. X 7.48 gal/ft3 = .o14 MG 1,000,000 Power Consumption KwHr = 1.73 X 460V X 63 Amps X .81 PF X 3 minutes / 60 min/hr. = 2 kwhr 1000 Ep= 3.14 X 30 Head ft. X .014 Flow MG 2.0 Power kWh X 0.90 Eff. Motor

  25. Calculate theoretical pump efficiency A raw sewage pump operating at a head of 30 feet draws down a 25 ft. by 25 ft. wet well by 3 ft. with the inlet valve closed in 3 minutes. During this period the current draw on the 50 hp. 460 volt motor was measured to be 63 amps. What is the efficiency of the pump? Flow Quantity MG = 25 ft. X 25 ft. X 3 ft. X 7.48 gal/ft3 = .o14 MG 1,000,000 Power Consumption KwHr = 1.73 X 460V X 63 Amps X .81 PF X 3 minutes / 60 min/hr. = 2 kwhr 1000 Ep = 3.14 X 30 Head ft. X .014 Flow MG = 0.73 2.0 Power kWh X 0.90 Eff. Motor

  26. Wire-To-Water Efficiency • A 100% efficient pump/motor combination would require one kWHr of electricity to raise .318 MG of water one foot. Ep = 3.14 X Head ft. X Flow MG = Power X Ep X Em = .318 Power kWh X Eff. Motor 3.14 X Head • So to determine the theoretical gallons/kWHr for any other head, simply divide 318,000 by the head. • Then actual efficiency of a particular pump/motor combination equals: Actual Gallons / kWHr Theoretical Gallons / kWHr

  27. Calculation of Wire-To-Water Efficiency A pump raises 10 million gallons of wastewater an average of 30 feet during a month, and consumes 1340 kWHr in the process. What is the approximate pump/motor efficiency? • Theoretical Pumped at 30 ft of head 318,000 = 10,607 gals / kWHr 30 • Actual Pumped = 10,000,000 = 7463 gals/ kWHr 1340 kWHr • Pump/motor combination equals: Actual Gallons / kWHr = 7463 gals/ kWHr = 70% Theoretical Gals / kWHr 10,607 gals/ kWHr

  28. Causes of Pump Inefficiency • Wrong type of application • Oversized • Poor system design • Cavitation – can lead to pump failure • Wear ring clearance excessive • Internal recirculation • Poor flow control maybe pumping more than process needs • Bearings worn • Mechanical seal leakage – or improper packing adjustment

  29. Possible Fixes to Pump Problems • If a pump is oversized (common) • Replace the impeller with a smaller one • Trim the impeller • Use a Variable Frequency Drive • Install a smaller pump for lower flow periods

  30. Energy Conservation Measures For Motors • Motor Rewind / Overhaul • Match Motor to Load • Improving power factor with capacitors • Replacement with energy efficient motors • You need to include cost of installation in your analysis.

  31. Energy Conservation Measures For Motors Motor Rewind / Overhaul • Usually remains the best option: • Especially when it’s a special order motor (non-stock). • Its already a premium efficiency motor • Its needed ASAP • Should evaluate this option before failure occurs • Will the overhaul reduce or improve efficiency – Talk to your motor shop • 97% of a motors Lifetime real cost is Energy Consumption. Less than 1% is purchase cost

  32. Energy Conservation Measures For Motors Motor Rewind / Overhaul Let’s Look at an example: You have a 10 hp. Return Sludge Motor that has failed. The efficiency of the motor is 88.5% and rewind will cost $700. A new premium efficiency motor 91.7% will cost $1100. The return sludge pump operates 24 hours / day. Rewound Motor Energy Consumption 10 hp X 0.746 Kw/hp= 8.4 kW 0.885 eff Annual energy cost = $7,358 8.4 kW X 24 hrs./day X 365 day/yr. X .10 / kWh Premium Motor Energy Consumption 10 hp X 0.746 Kw/hp= 8.1 kW 0.917 eff Annual energy cost = $7,096 8.1 kW X 24 hrs./day X 365 day/yr. X .10 / kWh New Premium vs Rewind $1,100 - $700 = $400 Annual Energy savings $7,358 - $7,096 = $262 / year Payback Period $400 / $262 = 1.5 years Note: There will also be a savings related to the lower demand charge • You need to include cost of installation in your analysis.

  33. Energy Conservation Measures For Motors Match Motor to Load • Usually not a good option: • Unless motor is loaded at less then 50% or OVERLOADED • Motors are generally most efficient at about 75 - 80% load. • Power Factor is a consideration when under loaded. • Designers sometimes oversize motors thinking about future upgrade • Smaller motors (< 10 hp) are more affected by underload • Load is difficult to determine. Suggest use amperage draw to estimate

  34. Energy Conservation Measures For Motors Match Motor to Load • Usually a good option when: • Motors are significantly oversized and under loaded. • Motors are moderately oversized and under loaded. Replace when Fail • Motors that are properly sized but standard efficiency. Do analysis • Target the replacement motor load at 75%

  35. Energy Conservation Measures For Motors Match Motor to Load Let’s Look at an example: You have a 10 hp. Motor. The motor is operating at ½ load and is 88.5% efficient. A new premium efficiency 7.5 hp. motor is 91.7% efficient and will cost $900. The motor operates 24 hours / day. 7.5 hp Motor 5 hp load 5 hp X 0.746 Kw/hp= 4.0 kW 0.917 eff Annual energy cost = $3,504 4.0 kW X 24 hrs./day X 365 day/yr. X .10 / kWh 10 hp Motor 5 hp. load 5 hp X 0.746 Kw/hp= 4.2 kW 0.885 eff Annual energy cost = $3,679 4.2 kW X 24 hrs./day X 365 day/yr. X .10 / kWh New 7.5 hp Premium = $900 Annual Energy savings $3,679 - $3,504 = $175 / year Payback Period $900 / $175 per year = 5 years Note: There will also be a savings related to the lower demand charge • You need to include cost of installation in your analysis.

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