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Material Type: Assignment; Class: INDUS ENERGY MANAGE; Subject: INDUSTRIAL ENGINEERING; University: University of Florida; Term: Spring 2006;
Typology: Assignments
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Spring 2006 Assignment #
and 4 in^ the class notes.
This home work considers TWO sets of Problems:
Problems Set Part - I:
$0.06/kWh, and^ is^ charged^ a^ "customer charge"^ of $50/month, all of which is subject to a ISo tax rate.^ What is this month's utilitv bill? 500 450, 400 350
-.s
. 250
15 18- 21 Day
Off peak^ demand chaige: (^) $5. l4lkWmonth (November-May)
200 150 100 lsot { I +
l
month is 65Yo of the the actual demand of
Jan Feb March April
Company's Peak Historl
lity has a ratchet (^) clause which (^) says that the billed demand for any
month, whichever is greater.
(^1 ) 1000 900 600 500 J
JI I Gator-One is beirig charged (^) $I2lkWmonth and are on a70o/o ll-month ratchet. The demand-averhging t- (^) interval is 30 minutes. : If each of the mptors nrn a constant 5 hours during the 8-hour (^) shift, compute Gatorr One's
495 550 580 600 610
July
Sept Oct
a a
a) billed peak demand and annual demand cost. (Assume that (^) leak month of this year^ and the past^ year^ is sarne.)
What is the new peak demand?
For the year
four hours a day, if consultant, he spaces
do some smart scheduling. Based on the advice of an energy
half only and the other runs during the second half (^) only. This way, at any given time during the shift,,there (^) i only one motor running.
b) c) d)
below, what is their bi{ing demand and how many^ kWh did.fhey use in that period (0-60 minutes)? (^) |
Load profile
600 5 q.
i 3oo 200 100 "o 0 20 40 60 Minutcs
45 l5 6
25 30
80
EIN 4321 Spring 20A
Assignm eifi #3 Solutions
IF YOU HAVE ANY (^) QUESTIONS ABOUT ANY OF THESE PROBLEMS, PLEASE STOP
charge larger customers both a consumption and demand charge (^) in order to regulate the
Demand (^) charge = 425 kW x (^) $6.50/kW :^ $2,
Energy consumption =
lrat =^ ltrtw>a, 00
kWh = (175^ kW) x (72 hours) +^ Ill2 x (200 k* - tZS k\M) r 72tus
(72 hours) +^ lll2 x (325 kW - 350 kW) '
72 hrsl
"
(72 hours) +^ Ill2 x (350 (^) kW - 425 k$l) * (^72) hrsl
"
(72 hours) + (^) lll2 x (350 kW - 350 k\M) "
72tasl
7Zhrsl
(^72) hrsl
kWh: (^) 13,
Bill :^ ($50^ +^ 52,763 +^ 1216,900 kWh x (^) $0.06/kWh)l) x 1. = ($15,827)^ x^ 1. : (^) $1.8,2qq
Interval 0- 3- 6- 9- tz-t 15-18 r
Intenral 0- 3- 6- 9- t2-t 15-
Problem:
Given:
What is the dollar savings forreducing (^) 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 $yings (that is, the demand in June through October would be (^) reduced by (^100) kWX
A large (^) manufacturing company in southern Arizona is on the rate schedule shown inFigure 3 - l0 (service level 5, secondary service). Their peak demand history for last year is ihown below. Assume they are on the 657o ratchet clause (^) specified in Figure 3 -10 (page (^) 105-109). Assume the higb month was July ofthe previous year at 1,150 kW.
Month Demand (kW)^ Month (^) Demand Jan Feb Mar Apr May Jun
495 550 580 600 610 900
JUI Aag sep Oct Nov Dec
I 100 1000 900 .' 600 500 515 note italics indicates on-peak season
Solution: Demand chmge On-peak$ Ofrpeak $
14.85 /kWmo 5.14 /kWmo
June - October Noveniber - May
5months/year Tmonths/year
Ratchet clause Dpeak: max(actual demand corrected for pf,65Yo ofthe highest on-peak (^) season demand corrected for pf)
Estimated next year with a 100 kW (^) decrehse in (^) the off-peak (^) season Month Demand (kW)^ Ratchet Dollar savings Jan 395 747.5 (^0) Feb (^450) Mar 480 Apr 500 May 510 Jan 900 JUI (^) II Aug 1000 Sep 900 Oct 600 Nov 400 Dec 415
747.5 0 747.5 (^0) 747.5 0 747.5 0 747.5 0 715 0 715 0 ,7r5 0 715 0 7t5 0 (^715 )
Therefore, you would not save (^) any money (^) by reducing (^) the peak
1
s)
For the second 30
average
metering (^) on page 31 on the (^) notes
kVh (^) <216.67 kWX0:5 hours) +^ (275^ kWX0.5 hours) +^ (400^ kW)(0.05 (^) hours)
Problem 5.1: How much canyou (^) soye by installing a photocell?
What is the payback period of this investment?
Given: When performing^ an en€rgy survey, you find twelve two.lamp F40T12 swurity lighting fixtures turned on during dayligtrt hours (averaging (^12) hours/day). (^) The lamps &aw 40 Watts each, the ballasts draw 12 Watts each, and the lights are currently left on 24 hours peT day. (^) ' Energy cost: (^) $ 0.0i5 /kWh Powercost: $ 7 /t(W Lamps: (^) $ 1 llamp Photocell (installbd):^ $ 85 /cell
Solution: Assuming one photocell^ can control all 12 fixtures
occurs during the day. Therefore, the demand reduction (^) @R) can be calculated as follows: DR-=Nfx Nlx Pl+Nfx Pb
'
where' Nf (^) = Number of fixtures, 12 fxtures Nl: Number of lamps per fixturg (^) 2 lamps/fixture Pl: Power use of lamps,40 Wlamp Pb = Power use of ballasts, 12 Wfxture Therefore, DR = 12 fixturesx^2 lamps/fixturex^40 Wlamp +
= 1,104 W = 1.104^ kW
Therefore, the energy savings (ES) (^) can be calculated as follows: ES = DR x l2hrlday x 365 days/yr = 1.104 kW x 12 tv/day x 365 days/yr = 4,835'.52 kWr/yr
Therefore, the cost savings (CS)^ can be calculated as follows: C.S (^) = ES^ x $ 0.055 lkWh +
" DRx^ $^7 /kW^ xt2mo/yr -. S 358.69 /yr
SPP= IClCS =- $85 /^ $" 358.69^ lyr = 0.24 years = 2.84^ months