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For example a boiler cycling on off ie such that its

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Unformatted text preview: lding Energy Performance – Spring 2012 - Topic 17 Aspects of Building Energy Simulation (Part 2) The efficiency of some types of equipment operating at part-load can be significantly different than their fullload efficiency. — For example, a boiler cycling on-off (i.e. such that it’s average output matches) a load might operate at a lower efficiency than it’s full-load (steady-state) efficiency. boiler output (cycling on-off) load losses Energy Flowrate input boiler output time losses input output boiler heat distribution loop average load (e.g. over 1 hour) “capacity” , boiler output (cycling on-off) Energy Flowrate , load time Energy Balance on the Heating Loop losses input boiler boiler output heat distribution loop Energy Addition = average load (e.g. over 1 hour) Energy Withdrawal = ∙ = ∙ 1 hour 1 hour A common simulation technique for estimating the energy impact of operating at part-load is the use of a “Part-load Performance Adjustment Curve” e.g. In DOE-2 (hourly energy calculation software), the following is the form of a basic equation for calculating boiler input energy at part-load: , = , × × () hourly average energy input rate (boiler input) “heat input ratio” at full-load full-load output rate (boiler output capacity) heat input ratio correction factor (as a function of part-load ratio) = Heat Input Ratio at Full-Load Definition of “Heat Input Ratio”: Heat Input Ratio = INPUT OUTPUT inverse of “efficiency” losses boiler = = = ∙ ! = Heat Input Ratio at full-load losses , boiler , = , , full-load output rate (CAPACITY) full-load input rate “full-load efficiency” “rated efficiency” (“steady-state efficiency”) 1 = , = , () Heat Input Ratio Correction Factor (as a function of Part-Load-Ratio) • corrects for actual output required (part-load ratio) and operating efficiency impacts resulting from operating at part- load Recall: load , = = capacity , load this hour (i.e. required output) output capacity e.g. DOE-2 Default Boiler Performance Curve 1.0 0.8 ,-. /0. 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 /0. ,-. /0. = 1 + 3 ∙ /0. + 4 ∙ /0.5 1 = 6. 6859:; 3 = 6. ::<;<= 4 = −6. 6;:?<@ Note: The DOE-2 default curve is a generic estimate, and is not meant to be representative of any particular piece of equipment. It may provide a reasonable representation of performance for some types of boilers. However, there are numerous types of boilers and performance characteristics are highly dependent on the type of boiler and several other factors. The performance curve used in the example is provided for the purpose of discussing a typical simulation procedure. Note: When operating at full-load, i.e. = 1 losses input boiler output , = , × × () output capacity 1 1 Return to the example calculation… Boiler Characteristics losses input boiler rated capacity = , = 5000 E output rated input = , = 6250 E , rated efficiency = = = 0.8 (80%) , full load HIR = = ! = 0.8 ! = @. 59 physical meaning? We will perform the calculation assuming that boiler’s operational characteristics are such that the DOE-2 default performance curve will provide a reasonable estimate of the part-load performance. For the hour being analyzed: net heat removal from loop load = , = 2812 E Determine /0. for the current hour: load , 2812 W = = = capacity , 5000 W = 6. 9<5 (56.2%) Determine ,-.(/0.) for the current hour: = Q + R ∙ + S ∙ T where: and: Q = 0.082597 R = 0.996764 S = −0.079361 = 0.562 ,-. /0. = (0.082597) + (0.996764) ∙ (0.562) + (−0.079361) ∙ (0.562)T = 6. <@8 Operating point plotted on the default performance curve: 1.0 0.8 0.6 ,-. /0. 0.4 0.2 0.0 0.0 0.2 0.4 0.6 /0. 0.8 1.0 Determine input rate (, ) for the current hour: , = , × × () = 5000 W × 1.25 × 0.618 6250 W physical meaning? , = , × × () input rate when operating at full-load , , = , physical meaning of (): ratio of fuel consumption rate for this hour relative to fuel consumption rate when operating at capacity Boiler input rate (, ) for the current hour: , = , × × () = 5000 W × 1.25 × 0.618 = 6250 W × 0.618 = ?8<? Y energy flowrate of boiler fuel input (e.g. natural gas) losses input boiler output Overall System: 1302 W flue (boiler losses) boiler 1400 W Zone 1 98 W loss gain electricity heat distribution loop 930 W boiler output fan-coil 1000 W Zone 2 loss gain 70 W 2812 W electricity boiler input 930 W pump 3863 W natural gas fan-coil fan-coil 1000 W 70 W 350 W shaft power motor 500 W electricity motor losses 150 W electricity loss Zone 3 gain HVAC Energy Totals for Analyzed Hour: Electricity Natural Gas Hourly Average Rate Hourly Energy Use Hourly Average Rate Hourly Energy Use Fans 238 W 0.238 kWh - - Pumps 500 W 0.500 kWh - - Boiler - - 3863 W 0.373 m3 Total 738 W 0.738 kWh 3863 W 0.373 m3 Energy-Use (Repeat analysis 8759 more times to complete a one year simulation…) ASIDE: Approximate Energy Content of Natural Gas: 10.33 ekWh/m3 1 ^ Natural Gas Hourly Energy Use = 3.863 Z[E\ = 0.373 ^ 10.33 Z[E\ An Alternate View of Part-Load Performance Curve 1 , = , × × () 2 , = , Substitute , 2 + 3 into 3 1 1 = : , 1 = × × () , , 1 = × × () Rearrange: , () = , ∙ Define “Part Load Efficiency”: , = , losses , boiler , , () = , ∙ 1 1 () = ∙ Rearr...
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