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
partload can be significantly different than their fullload efficiency.
— For example, a boiler cycling onoff (i.e. such that it’s
average output matches) a load might operate at a lower
efficiency than it’s fullload (steadystate) efficiency.
boiler output (cycling onoff)
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 onoff) 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 partload is the use of a
“Partload Performance Adjustment Curve”
e.g. In DOE2 (hourly energy calculation software), the
following is the form of a basic equation for calculating
boiler input energy at partload: , = , × × ()
hourly average
energy input rate
(boiler input) “heat input ratio”
at fullload fullload output rate
(boiler output capacity) heat input ratio
correction factor
(as a function of
partload ratio) = Heat Input Ratio at FullLoad Definition of “Heat Input Ratio”:
Heat Input Ratio = INPUT
OUTPUT inverse of
“efficiency” losses boiler =
= = ∙ ! = Heat Input Ratio at fullload
losses , boiler ,
=
, ,
fullload
output rate
(CAPACITY) fullload
input rate “fullload efficiency”
“rated efficiency”
(“steadystate efficiency”) 1
= ,
=
, ()
Heat Input Ratio Correction Factor (as a function
of PartLoadRatio)
• corrects for actual output required (partload
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. DOE2 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 DOE2 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 fullload, 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 DOE2 default performance curve
will provide a reasonable estimate of the partload 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 fullload , , = , 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 fancoil 1000 W Zone
2 loss
gain 70 W 2812 W electricity boiler
input 930 W pump 3863 W
natural gas fancoil fancoil 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 EnergyUse (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 PartLoad 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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 Spring '12
 DavidMather

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