Exam (elaborations) TEST BANK FOR Energy Management 5th Edition International Version By Klaus Dieter E. Pawlik (Solutions Manual)

Exam (elaborations) TEST BANK FOR Energy Management 5th Edition International Version By Klaus Dieter E. Pawlik (Solutions Manual) Solutions Manual for Guide to Energy Management, Fifth Edition Klaus-Dieter E. Pawlik v Table of Contents Chapter 1: Introduction to Energy Management .................................. 1 Chapter 2: The Energy Audit Process: An Overview ........................ 15 Chapter 3: Understanding Energy Bill .................................................. 21 Chapter 4: Economic Analysis and Life Cycle Costing ..................... 37 Chapter 5: Lighting ................................................................................... 53 Chapter 6: Heating, Ventilating, and Air Conditioning .................... 69 Chapter 7: Combustion Processes and the Use of Industrial Wastes ...................................................... 83 Chapter 8: Steam Generation and Distribution ................................. 103 Chapter 9: Control Systems and Computers ......................................111 Chapter 10: Maintenance ......................................................................... 119 Chapter 11: Insulation .............................................................................. 127 Chapter 12: Process Energy Management ............................................ 141 Chapter 13: Renewable Energy Sources and Water ........................... 149 Management Supplemental ......................................................................... 159 Introduction to Energy Management 1 1 Chapter 1 Introduction to Energy Management Problem: For your university or organization, list some energy management projects that might be good “fi rst ones,” or early selections. Solution: Early projects should have a rapid payback, a high probability of success, and few negative consequences (increasing/ decreasing the air-conditioning/heat, or reducing lighting levels). Examples: Switching to a more effi cient light source (especially in conditioned areas where one not only saves with the reduced power consumption of the lamps but also from reduced refrigeration or air-conditioning load). Repairing steam leaks. Small steam leaks become large leaks over time. Insulating hot fl uid pipes and tanks. Install high effi ciency motors. And many more 2 Solutions Manual for Guide to Energy Management Problem: Again for your university or organization, assume you are starting a program and are defi ning goals. What are some potential fi rst-year goals? Solution: Goals should be tough but achievable, measurable, and specifi c. Examples: Total energy per unit of production will drop by 10 percent for the fi rst and an additional 5 percent the second. Within 2 years all energy consumers of 5 million British thermal units per hour (Btuh) or larger will be separately metered for monitoring purposes. Each plant in the division will have an active energy management program by the end of the fi rst year. All plants will have contingency plans for gas curtailments of varying duration by the end of the fi rst year. All boilers of 50,000 lbm/hour or larger will be examined for waste heat recovery potential the fi rst year. Introduction to Energy Management 3 Problem: Perform the following energy conversions and calculations: a) A spherical balloon with a diameter of ten feet is fi lled with natural gas. How much energy is contained in that quantity of natural gas? b) How many Btu are in 200 therms of natural gas? How many Btu in 500 gallons of 92 fuel oil? c) An oil tanker is carrying 20,000 barrels of #2 fuel oil. If each gallon of fuel oil will generate 550 kWh of electric energy in a power plant, how many kWh can be generated from the oil in the tanker? d) How much coal is required at a power plant with a heat rate of 10,000 Btu/kWh to run a 6 kW electric resistance heater constantly for 1 week (16 8 hours)? e) A large city has a population which is served by a single electric utility which burns coal to generate electrical energy. If there are 500,000 utility customers using an average of 12,000 kWh per year, how many tons of coal must be burned in the power plants if the heat rate is 10,500 Btu/kWh? f) Consider an electric heater with a 4,500 watt heating element. Assuming that the water heater is 98% effi - cient, how long will it take to heat 50 gallons of water from 70 degree F to 140 degree F? 4 Solutions Manual for Guide to Energy Management Solution: a) V = 4/3 (PI) P = 4/3 × 3.14 × 53 523.33 ft3 E = V × 1,000 Btu/cubic foot of natural gas = 523.33 ft3 X 1,000 Btu/ft3 = 523,333 Btu b) E = 200 therms × 100, 000 Btu/therm of natural gas = 20,000,000 Btu E = 500 gallons × 140,000 Btu/gallon of #2 fuel oil 70,000,000 Btu c) E = 20,000 barrels × 42 gal./barrel × 550 kWh/gal. 4.6E+08 kWh d) V = 10,000 Btu/kWh × 6 kW × 168 h/25,000,000 Btu/ton coal = 0.40 tons of coal e) V = 500,000 cus. × 12,000 kWh/cus. × 10,500 Btu/kWh × I ton/25,000,000 Btu = 2,520,000 tons of coal f) E = 50 gal. × 8.34 lbm/gal. × (140F - 70F) × 1 Btu/F/lbm = 29,190 Btu = 29,190 Btu/3,412 Btu/kWh = 8.56 kWh = 8.56 kWh/4.5 kW/0.98 = 1.94 h Introduction to Energy Management 5 Problem: If you were a member of the upper level management in charge of implementing an energy management program at your university or organization, what actions would you take to reward participating individuals and to reinforce commitment to energy management? Solution: The following actions should be taken to reward individuals and reinforce commitment to energy management: Develop goals and a way of tracking their progress. Develop an energy accounting system with a performance measure such as Btu/sq. ft or Btu/unit. Assign energy costs to a cost center, profi t center, an investment center or some other department that has an individual responsibility for cost or profi t. Reward (with a monetary bonus) all employees who control cost or profi t relative to the level of cost or profi t. At the risk of being repetitive, note that the level of cost or profi t should include energy costs. 6 Solutions Manual for Guide to Energy Management Problem: A person takes a shower for ten minutes. The water fl ow rate is three gallons per minute, the temperature of the shower water is 110 degrees E Assuming that cold water is at 65 degrees F, and that hot water from a 70% effi cient gas water heater is at 140 degrees F, how many cubic feet of natural gas does it take to provide the hot water for the shower? Solution: E = 10 min × 3 gal./min × 8.34 lbm/gal × (110 F - 65 F) × 1 Btu/lbm/F = 11,259 Btu V = 11,259 Btu × 1 cubic foot/1,000 Btu/0.70 = 16.08 cubic feet of natural gas Introduction to Energy Management 7 Problem: An offi ce building uses 1 Million kWh of electric energy and 3,000 gallons of #2 fuel oil per year. The building has 45,000 square feet of conditioned space. Determine the Energy Use Index (EUI) and compare it to the average EUI of an offi ce building. Solution: E(elect.) = 1,000,000 kWh/yr. × 3,412 Btu/kWh = 3,412,000,000 Btu/yr. E(#2 fuel) = 3,000 gal./yr. × 140,000 Btu/gal. = 420,000,000 Btu/yr. E = 3,832,000,000 Btu/yr. EUI = 3,832,000,000 Btu/yr./45,000 sq. ft = 85,156 Btu/sq. ft/yr. which is less than the average offi ce building 8 Solutions Manual for Guide to Energy Management Problem: The offi ce building in Problem 1.6 pays $65,000 a year for electric energy and $3,300 a year for fuel oil. Determine the Energy Cost Index (ECI) for the building and compare it to the ECI for an average building. Solution: ECI = ($65,000 + $3,300)/45,000 sq. ft = $1.52/sq. ft/yr. which is greater than the average building Introduction to Energy Management 9 Problem: As a new energy manager, you have been asked to predict the energy consumption for electricity for next month (February). Assuming consumption is dependent on units produced, that 1,000 units will be produced in February, and that the following data are representative, determine your estimate for February. ————————————————————— Units Consumption Average Given: Month produced (kWh) (kWh/unit) ————————————————————— January 600 600 1.00 February 1,500 1,200 0.80 March 1,000 800 0.80 April 800 1,000 1.25 May 2,000 1,100 0.55 June 100 700 7.00 Vacation month July 1,300 1,000 0.77 August 1,700 1,100 0.65 September 300 800 2.67 October 1,400 900 0.64 November 1,100 900 0.82 December 200 650 3.25 1-week shutdown January 1,900 1,200 0.63 Solution: First, since June and December have special circumstances, we ignore these months. We then run a regression to fi nd the slope and intercept of the process model. We assume that with the exception of the vacation and the shutdown that nothing other then the number of units produced affects the energy used. Another method of solving this problem may assume that the weather and temperature changes also affects the energy use. 10 Solutions Manual for Guide to Energy Management ————————————————————————— Units Consumption Average Month produced (kWh) (kWh/unit) ————————————————————————— January 600 600 1.00 February 1,500 1,200 0.80 March 1,000 800 0.80 April 800 1,000 1.25 May 2,000 1,100 0.55 July 1,300 1,000 0.77 August 1,700 1,100 0.65 September 300 800 2.67 October 1,400 900 0.64 November 1,100 900 0.82 January 1,900 1,200 0.63 From the ANOVA table, we see that if this process is modeled linearly the equation describing this is as follows: kWh (1,000 units) = 623 + 0.28 × kWh/unit produced = 899 kWh 2,500 2,000 1,500 1,000 500 January Febuary March April May June July August September October November December Units produced Comsumption (kWh) Introduction to Energy Management 11 —————————————————————————————————— Coeffi cients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95. 0% Upper 95.0% —————————————————————————————————— Intercept 623. 93. 6. 9.19E-05 411. 834. 411. 834. X Variable 1 0, 0. 3. 0. 0, 0. 0. 0. —————————————————————————————————— SUMMARY OUTPUT ———————————— Regression Statistics ———————————— Multiple R 0. R Square 0. Adjusted R Square 0. Standard Effort 118. Observations 11 ———————————— ANOVA —————————————————————————— df SS MS F Signifi cance F —————————————————————————— Regression 1 .9788 .9 15.54545 0. Residual 9 ..07 Total 10 .5455 —————————————————————————— 500 1,000 1,500 2,000 2,500 • • • • • • • • • • • 1,400 1,200 1,000 800 600 400 200 Units produced Energy Used (kWh) 12 Solutions Manual for Guide to Energy Management Problem: For the same data as given in Problem 1.8, what is the fi xed energy consumption (at zero production, how much energy is consumed and for what is that energy used)? Solution: By looking at the regression run for problem 1.8 (see ANOVA table), we can see the intercept for the process in question. This intercept is probably the best estimate of the fi xed energy consumption: 623 kWh. This energy is probably used for space conditioning and security lights. Introduction to Energy Management 13 Problem: Determine the cost of fuel switching, assuming there were 2,000 cooling degree days (CDD) and 1,000 units produced in each year. Given: At the Gator Products Company, fuel switching caused an increase in electric consumption as follows: —————————————————————————— Actual energy Expected consumption energy after consumption switching fuel —————————————————————————— Electric/CDD 75 million Btu 80 million Btu —————————————————————————— Electric/units of production 100 million Btu 115 million Btu —————————————————————————— The base year cost of electricity is $15 per million Btu, while this year’s cost is $18 per million Btu. Solution: Cost variance = $18/million Btu - $15/million Btu = $3/million Btu Increase cost due to cost variance = Cost variance × Total Actual Energy Use = ($3/million Btu) × ((80 million Btu/CDD) × (2,000 CDDs) + (115 million Btu/unit) × (1,000 units)) = $825,000 CDD electric variance = 2,000 CDD × (80 - 75) million Btu/CDD = 10,000 million Btu Units electric variance = 1,000 units × (115 - 100) million Btu/unit = 15,000 million Btu 14 Solutions Manual for Guide to Energy Management Increase in energy use = CDD electric variance + Units electric variance = 10,000 million Btu + 15,000 million Btu = 25,000 million Btu Increase cost due to increased energy use = (Increase in energy use) × (Base cost of electricity) = 25,000 billion Btu × $15/million Btu = $375,000 Total cost of fuel switching = Increase cost due to increased energy use + Increased cost due to cost variance = $375,000 + $825,000 = $1,200,000 The Energy Audit Process: An Overview 15 15 Chapter 2 The Energy Audit Process: An Overview Problem: Compute the number of heating degree days (HDD) associated with the following weather data. Tempera- 65F -Temture Number perature Hours Given: Time Period (degrees F) of hours (degrees F) × dT Midnight - 4:00 AM 4:00 AM - 7:00 AM 7:00 AM - 10:00 AM 10:00 AM - Noon 22 2 43 86 Noon - 5:00 PM 5:00 PM - 8:00 PM 8:00 PM - Midnight ——— 1,028 Solution: From the added columns in the given table, we see that the number of hours times the temperature difference from 65 degrees F is 1,028 F-hours. Therefore, the number of HDD can be calculated as follows: HDD = 1,028 F-hours/24 h/day = 42.83 degree-days 16 Solutions Manual for Guide to Energy Management Problem: Select a specifi c type of manufacturing plant and describe the kinds of equipment that would likely be found in such a plant. List the audit data that would need to be collected for each piece of equipment. What particular safety aspects should be considered when touring the plant? Would any special safety equipment or protection be required? Solution: The following equipment could be found in a wide variety of manufacturing facilities: Equipment Audit data Heaters Power rating Use characteristics (annual use, used in conjunction with what other equipment, how is the equipment used?) Boilers Power rating Use characteristics Fuel used Air-to-fuel ratio Percent excess air Air-conditioners Power rating Chillers Effi ciency Refrigeration Cooling capacity Use characteristics Motors Power rating Effi ciency Use characteristics Lighting Power rating Use characteristics Air-compressors Power rating Use characteristics Effi ciency Various air pressures An assessment of leaks The Energy Audit Process: An Overview 17 Specifi c process equipment for example for a metal furniture plant one may fi nd some sort of electric arc welders for which one would collect its power rating and use characteristics. The following include a basic list of some of the safety precautions that may be required and any safety equipment needed: Safety precaution Safety equipment As a general rule of thumb the auditor should never touch anything: just collect data. If a measurement needs to be taken or equipment manipulated ask the operator. Beware of rotating machinery Beware of hot machinery/pipes Asbestos gloves Beware of live circuits Electrical gloves Have a trained electrician take any electrical measurements Avoid working on live circuits, if possible Securely lock and tag circuits and switches in the off/open position before working on a piece of equipment Always keep one hand in your pocket while making measurements on live circuits to help prevent accidental electrical shocks. When necessary, wear a full face respirator mask with adequate fi ltration particle size. Use activated carbon cartridges in the mask when working around low concentrations of noxious gases. Change cartridges on a regular basis. Use a self-contained breathing apparatus for work in toxic environments. Use foam insert plugs while working around loud machinery to reduce sound levels by nearly 30 decibels (in louder environments hearing protection rated at higher noise levels may be required) Always ask the facility contact about special safety precautions or equipment needed. Additional information can be found in OSHA literature. For our metal furniture plant: Avoid looking directly Tinted safety goggles at the arc of the welders 18 Solutions Manual for Guide to Energy Management Problem: Section 2.1.2 of the Guide to Energy Management provided a list of energy audit equipment that should be used. However, this list only specifi ed the major items that might be needed. In addition, there are a number of smaller items such as hand tools that should also be carried. Make a list of these other items, and give an example of the need for each item. How can these smaller items be conveniently carried to the audit? Will any of these items require periodic maintenance or repair? If so, how would you recommend that an audit team keep track of the need for this attention to the operating condition of the audit equipment? Solution: Smaller useful audit equipment may include: A fl ashlight Extra batteries A hand-held tachometer A clamp-on ammeter Recording devices These smaller items can be conveniently be carried in a tool box. As with most equipment, these items will require periodic maintenance. For example, the fl ashlight batteries and light bulbs will have to be changed. For these smaller items, one could probably just include the periodic maintenance as part of a pre-audit checklist. For items that require more than just cursory maintenance, one could include the item in their periodic maintenance system. The Energy Audit Process: An Overview 19 Problem: Section 2.2 of the Guide to Energy Management discussed the point of making an inspection visit to a facility at several different times to get information on when certain pieces of equipment need to be turned on and when they are unneeded. Using your school classroom or offi ce building as a specifi c example, list some of the unnecessary uses of lights, air conditioners, and other pieces of equipment. How would you recommend that some of these uses that are not necessary be avoided? Should a person be given the responsibility of checking for this unneeded use? What kind of automated equipment could be used to eliminate or reduce this unneeded use? Solution: Typically, one could visit a university at night and observe that the lights of classrooms are on even at midnight when no one is using the area. One idea would be to make the security force responsible for turning off non-security lights when they make their security tours at night. A better idea may be to install occupancy sensors so that the lights are on only when the area is in use. An additional benefi t of an occupancy sensors could be security; many thieves or vandals would be startled when lights come on. 20 Solutions Manual for Guide to Energy Management Problem: An outlying building has a 25 kW company-owned transformer that is connected all the time. A call to a local electrical contractor indicates that the core losses from comparable transformers are approximately 3% of rated capacity. Assume that the electrical costs are ten cents per kWh and $10/kW/month of peak demand, that the average building use is ten hours/month, and that the average month has 720 hours. Estimate the annual cost savings from installing a switch that would energize the transformer only when the building was being used. Given: Transformer power use 25 kW Core losses 3% Electrical energy cost $0.10 /kWh Demand charge $10/kW /month Building utilization 10 hrs /mo Hours in a month 720 hrs /mo Months in a year 12 mo /yr Solution: The energy savings (ES) from installing a switch that would energize the transformer only when the building was being used can be calculated as follows: ES = (Percentage of core losses) (Transformer power use)(Hours in a month - Building utilization) (Months in a year) = 3% × 25 kW × 720 - 10) hrs/mo × 12 mo/yr = 6,390 kWh/yr Since we do not expect the monthly peak demand to be reduced by installing this switch, the only savings will come from energy savings. Therefore, annual savings (AS) can be calculated as follows: AS = ES × Electrical energy cost = 6,390 kWh/yr × $ 0.10/kWh = $ 639/yr Understanding Energy Bill 21 21 Chapter 3 Understanding Energy Bill Problem: By periodically turning off a fan, what is the total dollar savings per year to the company? Given: In working with Ajax Manufacturing Company, you fi nd six large exhaust fans are running constantly to exhaust general plant air (not localized heavy pollution). They are each powered by 30-hp electric motors with loads of 27 kW each. You fi nd they can be turned off periodically with no adverse effects. You place them on a central timer so that each one is turned off for 10 minutes each hour. At any time, one of the fans is off, and the other fi ve are running. The fans operate 10 h/day, 250 days/year. Assume the company is on the rate schedule given in Figure 3-10. Neglect any ratchet clauses. The company is on service level 3 (distribution service). (There may be signifi cant HVAC savings since conditioned air is being exhausted but ignore that for now.) Solution: Demand charge On-peak $12.22/kW/mo June-October 5 months/year Off-peak $4.45/kW/mo November-May 7 months/year Energy charge For fi rst two million kWh $0.03431/kWh All kWh over two million $0.03010/kWh Assumptions (and possible explanations) Assume the company uses well over two million kWh per month The fuel cost adjustment is zero, since the utility’s fuel cost is at the base rate. There is no sales tax since the energy can be assumed to be used for production 22 Solutions Manual for Guide to Energy Management The power factor is greater than 0.8 No franchise fees since the company is outside any municipality The demand savings (DS) can be calculated as follows: DS = [(DC on peak) × (N on peak) + (DC off peak) × (N off peak)] × DR where, DC = Demand charge for specifi ed period N = Number of months in a specifi ed period DR = Demand reduction, 27 kW since a motor using this amount is always turned off with the new policy Therefore, DS = [($12.22/kW/mo) × (5 mo/yr) + ($4.45/kW/mo) × (7 mo/yr)] × 27 kW = $2,491/yr The energy savings (ES) can be calculated as follows: ES = (EC >2 million) × (10 h/day) × (250 day/yr) × DR where EC = Marginal energy charge Therefore, ES = ($0.03010/kWh) × (10 h/day) × (250 day/yr) × 27 kW = $2,032/yr Finally, the total annual savings (TS) can be calculated as follows: TS = DS + ES = $4,523/yr Additional Considerations How much would these timers cost? How much would it cost to install these timers? Or an alternate control system? Does cycling these fans on and off cause the life of the fan motors to decrease? What would the simple payback period be? Net present value? Internal rate of return? Understanding Energy Bill 23 Problem: What is the dollar savings for reducing demand by 100 kW in the off-peak season? If the demand reduction of 100 kW occurred in the peak season, what would be the dollar savings (that is, the demand in June through October would be reduced by 100 kW)? Given: A large manufacturing company in southern Arizona is on the rate schedule shown in Figure 3-10 (service level 5, secondary service). Their peak demand history for last year is shown below. Assume they are on the 65% ratchet clause specifi ed in Figure 3-10. Assume the high month was July of the previous year at 1,150 kW. Month Demand (kW) Month Demand —————————————————————————— Jan 495 Jul 1100 Feb 550 Aug 1000 Mar 580 Sep 900 Apr 600 Oct 600 May 610 Nov 500 Jun 900 Dec 515 —————————————————————————— note italics indicates on-peak season Solution: Demand charge On-peak $13.27/kW/mo June-October 5 months/year Off-peak $4.82/kW/mo November-May 7 months/year Ratchet clause Dpeak = max (actual demand corrected for pf, 65% of the highest on-peak season demand corrected for pf) Assumptions (and possible explanations) Assume the company uses well over two million kWh per month. 24 Solutions Manual for Guide to Energy Management The fuel cost adjustment is zero, since the utility’s fuel cost is at the base rate. There is no sales tax since the energy can be assumed to be used for production. The power factor is greater than 0.8. No franchise fees since the company is outside any municipality. Estimated next year with a 100 kW decrease in the off-peak season Month Demand (kW) Ratchet Dollar savings ———————————————————————— Jan 395 747.5 0 Feb 450 747.5 0 Mar 480 747.5 0 Apr 500 747.5 0 May 510 747.5 0 Jun 900 747.5 0 Jul Aug Sep 900 715 0 Oct 600 715 0 Nov 400 715 0 Dec 415 715 0 ——— 0 Therefore, you would not save any money by reducing the peak demand in the off-season. This non-savings is due to the ratchet and the degree of unevenness of demand. Understanding Energy Bill 25 Estimated next year with a 100 kW decrease in the on-peak season Month Demand (kW) Ratchet Dollar savings ———————————————————————— Jan 495 747.5 0 Feb 550 747.5 0 Mar 580 747.5 0 Apr 600 747.5 0 May 610 747.5 0 Jun 800 747.5 $1,327 Jul 1000 650 $1,327 Aug 900 650 $1,327 Sep 800 650 $1,327 Oct 500 650 $863 Nov 500 650 $313 Dec 515 650 $313 ——— The fi rst year they would save: $6,797 ——— Every year after that they would save the following: Savings = 65% × 100 kW × 7 mo/yr × $4.82/kW/mo + 100 kW × 5 mo/yr × $13.27/kW/mo = $8,828/yr 26 Solutions Manual for Guide to Energy Management Problem: Use the data found in Problem 3.2. How many months would be ratcheted, and how much would the ratchet cost the company above the normal billing? Solution: Assuming that the 100 kW reduction is not made Month Demand (kW) Ratchet Ratchet Cost ———————————————————————— Jan 495 747.5 $1,217.05 Feb 550 747.5 $951.95 Mar 580 747.5 $807.35 Apr 600 747.5 $710.95 May 610 747.5 $662.75 Jun 900 747.5 $— Jul 1100 715 $— Aug 1000 715 $— Sep 900 715 $— Oct 600 715 $1,526.05 Nov 500 715 $1,036.30 Dec 515 715 $964.00 ———————————————————————— 8 months would be ratcheted at a cost of $7,876.40 ———— Understanding Energy Bill 27 Problem: Calculate the savings for correcting to 80% power factor? How much capacitance (in kVARs) would be necessary to obtain this correction? Given: In working with a company, you fi nd they have averaged 65% power factor over the past year. They are on the rate schedule shown in Figure 3-10 and have averaged 1,000 kW each month. Neglect any ratchet clause and assume their demand and power factor are constant each month. Assume they are on transmission service (level 1). Solution: Demand Charge On-peak $10.59/kW/mo June-October 5 months/year Off-peak $3.84/kW/mo November -May 7 months/year Billed Demand = Actual Demand × (base pf/actual pf) = 1000 kW × 0.8/0.65 = 1231 kW pf correction savings = 231 kW × (5 mo/yr × $10.59/kW/mo + 7 mo/yr × $3.82/kW/mo) = $18,422/yr pf = cos(theta) = 0.65 theta = 0. radians kVAR initial = 1000 kW × tan (0. 86) = 1169 kVAR pf = cos(theta) = 0.8 theta = 0. radians kVAR initial = 1000 kW × tan (0. 86) 750 kVAR capacitor size needed = 419 kVAR Also, using a pf correction table for 0.65 => 0.80: kVAR = (0.419) × (1000 kW) = 419 kVAR 28 Solutions Manual for Guide to Energy Management Problem: How much could they save by owning their own transformers and switching to service level 1? Given: A company has contacted you regarding their rate schedule. They are on the rate schedule shown in Figure 3-10, service level 5 (secondary service), but are near transmission lines and so can accept service at a higher level (service level 1) if they buy their own transformers. Assume they consume 300,000 kWh/month and are billed for 1,000 kW each month. Ignore any charges other than demand and energy. Solution: ———————————————————————————————— Service level 1 (proposed) Demand Charge On-peak $10.59 /kW/mo June-Oct. 5 months/year Off-peak $3.84 /kW/mo Nov.-May 7 months/year Energy Charge For fi rst two million kWh $0.03257 /kWh All kWh over two million $0.02915 /kWh ———————————————————————————————— Service level 5 (present) Demand Charge On-peak $13.27 /kW/mo June-Oct. 5 months/year Off-peak $4.82 /kW/mo Nov.-May 7 months/year Energy Charge For fi rst two million kWh $0.03528 /kWh All kWh over two million $0.03113 /kWh ———————————————————————————————— Rate savings: Demand Charge On-peak $2.68 /kW/mo June-Oct. 5 months/year Off-peak $0.98 /kW/mo Nov.-May 7 months/year Energy Charge For fi rst two million kWh $0.00271 /kWh All kWh over two million $0.00198 /kWh ———————————————————————————————— ES = 300,000 kWh/mo × 12 mo/yr × $0.00271/kWh = $9,756 /yr DS = 1,000 kW ($2.68/kW/mo×5 mo/yr+$0.98/kW/mo×7 mo/yr) = $20,260/yr TS = $30,016/yr Understanding Energy Bill 29 Problem: What is the savings from switching from priority 3 to priority 4 rate schedule? Given: In working with a brick manufacturer, you fi nd for gas billing that they were placed on an industrial (priority 3) schedule (see Figure 3-12) some time ago. Business and inventories are such that they could switch to a priority 4 schedule without many problems. They consume 7,000 Mcf of gas per month for process needs and essentially none for heating. Solution: Priority 3 (present) Monthly Cost Schedule Rate (for 7,000 Mcf/mo) First 1 ccf $19.04 $19.04 Next 2.9 Mcf/mo $5.490 /Mcf $15.92 Next 7 Mcf/mo $5.386 /Mcf $37.70 Next 90 Mcf/mo $4.372 /Mcf $393.48 Next 100 Mcf/mo $4.127 /Mcf $412.70 Next 7800 Mcf/mo $3.445 /Mcf $23,426.00 Over 8000 Mcf/mo $3.399 /Mcf $——— Total present monthly cost: $24,304.84 ——— Total present annual cost: $291,658.12 Priority 4 (proposed) Monthly Cost Schedule Rate (for 7,000 Mcf/mo) First 4,000 Mcf/mo or fraction thereof $12,814 $12,814 Next 4000 Mcf/mo $3.16 /Mcf $9,504 Over 8000 Mcf/mo $3.122 /Mcf $——— ——— Total present monthly cost: $22,318.00 ———— Total present annual cost: $267,816.00 Annual savings from switching: $23,842.12 Additional Considerations What-if there exists a 20% probability that switching to the proposed rate schedule will disrupt production one more time a year for an hour? 30 Solutions Manual for Guide to Energy Management Problem: Calculate the January electric bill for this customer. Given: A customer has a January consumption of 140,000 kWh, a peak 15-minute demand during January of 500 kW, and a power factor of 80%, under the electrical schedule of the example in Section 3.6. Assume that the fuel adjustment is: $0. 01/kWh Solution: Quantity Cost Customer charge $21.00 /mo 1 mo $21 Energy charge $0.04 /kWh 140,000 kWh $5,600 Demand charge $6.50 /kW/mo 500 kW $3,250 Taxes 8% Fuel Adjustment $0.01 /kWh 140,000 kWh $1,400 ——— sub-total $10,271 ——— tax 822 ——— total $11,093 Understanding Energy Bill 31 Problem: Compare the following residential time-of-use electric rate with the rate shown in Figure 3-6. Given: Customer charge $8.22 /mo Energy charge $0.1230 /kWh on-peak $0.0489 /kWh off-peak This rate charges less for electricity used during off-peak hours—about 80% of the hours in a year—than it does for electricity used during on-peak hours. Solution: Each of the rates have a different on-peak period. However, if we assume that no matter which rate schedule is used that 80% of the energy is used off-peak, then average cost per kWh can be calculated as follows: AC = (Off-peak percentage of energy use)(Off-peak energy cost) + (1 - off-peak percentage of energy use)(On-peak energy cost) Therefore, the average cost per kWh with the above schedule is: AC = (80%)($0.0489/kWh) + (1-80%)($0,123/kWh) = $0.06372/kWh And the average cost per kWh with the schedule in fi gure 3-6 is: AC = (80%)($0.0058/kWh) + (1-80%)($0,10857/kWh) = $0.02635/kWh 32 Solutions Manual for Guide to Energy Management Problem: What is the power factor of the combined load? If they added a second motor that was identical to the one they are presently using, what would their power factor be? Given: A small facility has 20 kW of incandescent lights and a 25 kW motor that has a power factor of 80%. Solution: The lamp: ———————— 20 kW The motor: 18.75 pf = cos (theta) kVAR theta = 0. 25 kW kVAR = 25 tan (theta) = 18.75 kVAR Combined: 18.75 kVA = square root(kW2 + kVAR2) kVARF = square root(452+18.752) 45 kW = 48.75 kVA pf = kW/kVA = 45/48.75 = 0.92 Combined: 37.5 kVA = square root(kW2 + kVAR2) kVAR = square root(702 +37.52) 70 kW = 79.41 kVA pf = kW/kVA = 70/79.41 = 0.88 Understanding Energy Bill 33 Problem: For the load curve shown below for Jones Industries, what is their billing demand and how many kWh did they use in that period? Given: A utility charges for demand based on a 30-minute synchronous averaging period. Solution: For the fi rst 30 minutes: For the second 30 minutes: Time (minutes) average kW Time (minutes) average kW 5 100 Weighted average: 216.67 kW Weighted average: 275.00 kW Therefore, 275 kW is the billed demand. kWh = (216.67 kW)(0.5 hours) + (275 kW)(0.5 hours) + (400 kW)(5 minutes x 1 hour/60 minutes)) = 279.17 kWh Minutes kW 0 200 10 200 15 400 25 100 30 100 45 500 50 200 60 200 60 400 65 400 34 Solutions Manual for Guide to Energy Management Problem: Based on the hypothetical steam rate in Figure 3-13, determine their steam consumption cost for the month? Given: The Al Best Company has a steam demand of 6,500 lb/hr and a consumption of 350,000 lbs during the month of January. Solution: Steam consumption charge $3.50 /1000 lb. for the fi rst 100, 000 lb of steam per month $3.00 /1000 lb. for the next 400,000 lb of steam per month $2.75 /1000 lb. for the next 500,000 lb of steam per month $2.00 /1000 lb. for the next 1,000,000 lb of steam per month Consumption cost = $3.50/1,000 lb × 100,000 lb + $3.00/1,000 lb × 250,000 lb = $350 + $750 = $1,100 Understanding Energy Bill 35 Problem: What is Al’s cost for chilled water in July? What was their Btuh (Btu/hour) equivalent for the average chilled water demand? Given: Al Best also purchases chilled water with the rate schedule of fi gure 3-13. During the month of July, their chilled water demand was 485 tons and their consumption was 250,000 ton-hours Solution: Chilled water demand charge: $2,500 /mo for the fi rst 100 tons or any portion thereof $15 /mo/ton for the next 400 tons $12 /mo/ton for the next 500 tons $10 /mo/ton for the next 500 tons $9 /mo/ton for over 1500 tons Chilled water consumption charge: $0.069 /tonh for the fi rst 10,000 tonh/mo $0.060 /tonh for the next 40,000 tonh/mo $0.055 /tonh for the next 50,000 tonh/mo $0.053 /tonh for the next 100,000 tonh/mo $0.051 /tonh for the next 100,000 tonh/mo $0,049 /tonh for the next 200,000 tonh/mo $0.046 /tonh for the next 500,000 tonh/mo Demand cost = $2,500 + ($15/ton)(385 tons) = $8,275 Consumption cost = $0.069 × 10,000 tonh/mo + $0.060 × 40,000 tonh/mo + $0.055 × 50,000 tonh/mo + $0.053 × 100,000 tonh/mo + $0.051 × 50,000 tonh/mo = $13,690 ———— Total bill = $21,965 ———— Average demand = consumption × 12,000 Btuh/ton 744 hr/July = 250,000 tonh × 12,000 Btuh/ton 744 hr/July = 4,032,258 Btuh in July 36 Solutions Manual for Guide to Energy Management This page intentionally left blank Economic Analysis and Life Cycle Costing 37 37 Chapter 4 Economic Analysis and Life Cycle Costing Problem: How much can they spend on the purchase price for this project and still have a Simple Payback Period (SPP) of two years? Using this fi gure as a cost, what is the return on investment (ROI), and the Benefi t-Cost Ratio (BCR)? Given: The Orange and Blue Plastics Company is considering an energy management investment which will save 2,500 kWh of electric energy at $0.08/kWh. Maintenance will cost $50 per year, and the company’s discount rate is 12%. Solution: Annual savings = annual kWh saved × electric energy cost-maintenance cost = 2,500 kWh/yr × $0.08/kWh - $50/yr = $150/yr Implementation cost = SPP × Annual savings = 2 yrs × $150/yr = $300 Since no life is given, assume the project continues forever. Therefore, use the highest n in the TMV tables: n = 360 P = A[P|A, i, N] 300 = 150 [P|A, i, 360] 2 = [P|A, i, 360] From the TMV tables, we see that i ~ 50%. Therefore, ROI = 50% 38 Solutions Manual for Guide to Energy Management BCR = PV(benefi ts)/PV (costs) PV (benefi ts) = A[P|A, i, N] = $150 × [P|A, 12%, 360 yr] = $150 × 8.3333 = $1,250 BCR = $1,250/$300 = 4.17 If N = 5 years: we read from the TMV tables that the factor 2 falls between 40% and 50% tables with the factors 2.0352 and 1.7366 respectively. Therefore, to fi nd a more precise percentage we linearly interpolate: (50% - 40%)/(1.7366 - 2.0352) (× - 40%)/(2 - 2.0352) Solving for X: X = 41.2 % = ROI BCR = PV (benefi ts)/PV (costs) PV (benefi ts) = A[P|A, i, N] = $150 [P|A, 12%, 5 yr] = $150 × 3.6048 = $541 BCR = $541/$300 = 1.80 Economic Analysis and Life Cycle Costing 39 Problem: Which model should she buy to have the lowest total monthly payment including the loan and the utility bill? Given: A new employee has just started to work for Orange and Blue Plastics, and she is debating whether to purchase a manufactured home or rent an apartment. After looking at apartments and manufactured homes, she decides to buy one of the manufactured homes. The Standard Model is the basic model that costs $20,000 and has insulation and appliances that have an expected utility cost of $150 per month. The Deluxe Model is the energy effi cient model that has more insulation and better appliances, and it costs $22,000. However, the Deluxe Model has expected utility costs of only $120/month. She can get a 10 year loan for 10% for the entire amount of either home. Solution: Assume the 10% is the compounded annual percentage rate. P = A [P|A, i, N] = A [P|A, 10%, 10 years] = A (6.1446) A (Standard) = $20,000/6.1446 yrs = $3,254.89/yr = $271.24/mo Monthly (Standard) = $271.24/mo + $150/mo —————— = $421.24/mo —————— A (Standard) = $22,000/6.1446 yrs = $3,580.38/yr = $298.36/mo Monthly (Deluxe) = $298.36/mo +$120/mo —————— = $418.36/mo Therefore, if she buys the deluxe, she will have a slightly lower monthly cost 40 Solutions Manual for Guide to Energy Management Problem: Determine the SPP, ROI, and BCR for this project: Given: The Al Best Company uses a 10-hp motor for 16 hours per day, 5 days per week, 50 weeks per year in its fl exible work cell. This motor is 85% effi cient, and it is near the end of its useful life. The company is considering buying a new high effi ciency motor (91% effi cient) to replace the old one instead of buying a standard effi ciency motor (86.4% effi cient). The high effi ciency motor cost $70 more than the standard model, and should have a 15-year life. The company pays $7 per kW per month and $0.06 per kWh. The company has set a discount rate of 10% for their use in comparing projects. Solution: Assume the load factor (If) is 60%. DR = If × Pm × 0.746 kW/hp × ((1/effs) - (1/effh)) where, DR = Demand reduction Pm = Power rating of the motor, 10-hp effs = Effi ciency of the standard effi ciency motor, 86.4% effh = Effi ciency of the high effi ciency motor, 91% Therefore, DR = 0.6 × 10 hp × 0.746 kW/hp × ((1/0.864) - (1/0.91)) = 0.26 kW DCR = DR × DC × 12 mo/yr where, DCR = Demand cost reduction DC = Demand cost, $7/kW/mo Therefore, DCR = 0.26 kW × $7/kW/mo × 12 mo/yr = $22.00/yr ES DR × 16 hr/day × 5 days/wk × 5 0 wk/yr Therefore, ES = 0.26 kW × 16 hr/day × 5 days/wk × 50 wk/yr = 1,047.5 KWh/yr ECS = ES × EC Economic Analysis and Life Cycle Costing 41 where, ECS = Energy cost savings EC = Energy cost, $0.06/kWh Therefore, ECS = 1,047.5 kWh/yr × $0.06/kWh = $62.85/yr Therefore, the annual cost savings (ACS) can be calculated as follows