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EIN 432I
Spring 2006 Assignment #
Due: February 7,2006'
Reading assignmeni: Chapters 3 and5 in the Guide to Energt Management' Review sections 2,
and 4 in^ the class notes.
This home work considers TWO sets of Problems:
Problems Set Part - I:
- (^) Explain in your (^) own words the difference between a (^) "kilowafi" and a "kilowatt-hour." Why
do electric utilities sometimes charge customers for kilowatts as well as kilowaff-hours?
- On the following graph,^ what is the peak demand for the (^) month? What is the month's energy
consumption? The customer is on a ratchetless rate structure and is charged $6.50/kW and
$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
- A large manufacturing company's peak demand history (^) for the last year (^) is shown below. They have found (^) a way (^) to reduce their deinand in the oFpeak season by (^100) kW, but the peak season demand will be the same (ie, the demand in each month of November through May would be reduced by l00kW), Assuming (^) they are on the 650/o ratchet, what is their dollar savings? Assume the high month was July of the previous.year (^) at 1150 kW. If the demand rbduction (^) of 100 kW occurred in the peak (^) season, what would be the dollar savings (ie, the demand in June through October would (^) be reduced (^) by l00kw)?
Rate schedule infonhation :
On peak demand cha/ge: $l4.85lkWmonth (June-October)
Off peak^ demand chaige: (^) $5. l4lkWmonth (November-May)
200 150 100 lsot { I +
l
Ratchet clause: The
month is 65Yo of the the actual demand of
Jan Feb March April
May
demand for
Company's Peak Historl
lity has a ratchet (^) clause which (^) says that the billed demand for any
on-peak season morimum demand of the previous 12 months or
month, whichever is greater.
(^1 ) 1000 900 600 500 J
- Gator-One, a small turing (^) company, has 2 motors and some lights. They run one shift of 8 hours every of the week. The lights remain (^) "on" during the entire shift and contribute a steady 20 k to Gator-One's load. Assume that each of the (^2) motors require
100 kW when
JI I Gator-One is beirig charged (^) $I2lkWmonth and are on a70o/o ll-month ratchet. The demand-averhging t- (^) interval is 30 minutes. : If each of the mptors nrn a constant 5 hours during the 8-hour (^) shift, compute Gatorr One's
495 550 580 600 610
July
Aug
Sept Oct
Nov
a a
a) billed peak demand and annual demand cost. (Assume that (^) leak month of this year^ and the past^ year^ is sarne.)
What is the new peak demand?
What is 's annual demand costs for the first year after smart scheduling?
For the year
- A utility charges for based on a 30-minute averaging
Gator-One's manager suddenly discovers that they only need to run the motors
four hours a day, if consultant, he spaces
do some smart scheduling. Based on the advice of an energy
the usage of the motors; i.e. one of thb motors runs during the first
half only and the other runs during the second half (^) only. This way, at any given time during the shift,,there (^) i only one motor running.
b) c) d)
below, what is their bi{ing demand and how many^ kWh did.fhey use in that period (0-60 minutes)? (^) |
Load profile
600 5 q.
i 3oo 200 100 "o 0 20 40 60 Minutcs
45 l5 6
l0 50
25 30
80
EIN 4321 Spring 20A
Assignm eifi #3 Solutions
IF YOU HAVE ANY (^) QUESTIONS ABOUT ANY OF THESE PROBLEMS, PLEASE STOP
BY DURING OFFICE HOURS AND WE WILL BE GLAD TO DISCUSS THEM WITH YOU.
- A kilowatt is a rate of consumption of energy, or "power" at any (^) time, dt. It is known as the power of a system and is often referred to as "demand." Kilowatt-hour (^) is the time integral of
the kilowatt and is the most coillmon billing unit of electrical energy. Electric companies
charge larger customers both a consumption and demand charge (^) in order to regulate the
amount of power that thej'have to provide^ to the customers at one given time.
- The peak^ demand occurred onthe"l5th at$SS
Demand (^) charge = 425 kW x (^) $6.50/kW :^ $2,
Energy consumption =
lrat =^ ltrtw>a, 00
kWh = (175^ kW) x (72 hours) +^ Ill2 x (200 k* - tZS k\M) r 72tus
- (200 (^) kSl) * (72 hours) + (^) lll2 x (250 (^) kW - 200 kW) * 72 hrsl
- (^) (250 k\trI)' (72 hours) + (^) lll2 x (350 (^) kW - 250 kSD "
72 hrsl
(72 hours) +^ lll2 x (325 kW - 350 kW) '
72 hrsl
- (^) (325 (^) kVl) r (72 hours) + (^) |/2 x (425 (^) kW - 325 k\Ir) r 72 hrsl
+ (425 kW)
"
(72 hours) +^ Ill2 x (350 (^) kW - 425 k$l) * (^72) hrsl
+ (350 kW)
"
(72 hours) + (^) lll2 x (350 kW - 350 k\M) "
72tasl
- (350 k$l) r (72 hours) + (^) $/2 x (275 (^) kW - 350 kSl) "
7Zhrsl
- Q75 k'\M) , (72 hours) + pl2 (^) x (^) Q50 kW - 275 k\tr|) "
(^72) hrsl
- (^) (250 kSl) * (ZZ (^) hows) + (^) lll2 x (300 (^) kW - 250 k$l) r (^72) hrsl
kWh: (^) 13,
- (^) (14,400 + (^) 1,800)
- (19,000 + (^) 3,600)
- (25,200 (^) - 900)
- (^) (23,400) + (^) 3,600)
- (30,600 (^) -2,700)
- (25,200 (^) -2,700)
- (19,800 - (^) 900)
- (19,000 + (^) 1,900)
:216,900 kWh
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
t8-
2t-
Intenral 0- 3- 6- 9- t2-t 15-
t8-
2t-
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
Solution: "For^ the first 30 minutes: minutes:
average
Time (minutes)^ average kW Time (minutes) kW
l0 250 t0 200
Weighted Weighted
average: 216.67kW average: 275.00kW
Therefore, the mar<imum of the two,275.00 kW, is
the billed demand. See notes on synchronoirs
metering (^) on page 31 on the (^) notes
kVh (^) <216.67 kWX0:5 hours) +^ (275^ kWX0.5 hours) +^ (400^ kW)(0.05 (^) hours)
:re,m (o-65minutes)
kWh:( 2t6.67 kWX0.5 hours) +^ (275 kWX0.5 hours) = 245.85,
. (for^ 0-60 minutes)
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
There will probably be demand savings, since the lights will be
turned offduring the day. It is probable that their peak demand
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 +
12 fixtures x 12 Wfixture
= 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
