characterisation of electrical equipment in the neb
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Jonathan McDermott Raymond McNamara John Mitchell Peter Doyle. Characterisation of Electrical Equipment in the NEB. Introduction. The Engineering Building is instrumented throughout with electrical power sensors. - PowerPoint PPT PresentationTRANSCRIPT
Characterisation of Electrical Equipment in the NEB
Jonathan McDermottRaymond McNamaraJohn MitchellPeter Doyle
IntroductionThe Engineering Building is instrumented throughout with electrical power sensors. These sensors show the live performance of the building which help us describe the technical detail of the different electrical loads within the building.
The data is accessible through the website http://engbm.srv.nuigalway.ie/where a wide range of parameters can be selected, for calculations, graph creationand analytic purposes.
The Atrium Lighting
Our goal was to analyse the atrium
lighting electrical load.
Our aims :
• Describe the technical detail
of this load and relate it to the
functionality of the new
engineering building .
• Describe power supply
requirements
• Power consumption
• Power quality issues.
Topics Covered
Jonathan: Calculated power vs. measured power comparison for phase A , B, C.
Raymond: 3 phase voltage & current waveform when the atrium is at (i)full load (ii)light load. The corresponding, voltage, current, power & power factor.
John: Analysis of graphs and trends, relating them to the function of the building. Research into lighting technology used and investigation into possible improvements.
Peter: Cost of energy to supply the atrium over a typical week from different suppliers and sources.
Calculated Power vs. Measured Power
We aimed to produce plots of power consumption (KW), for a typical working week (7th Oct. 2011 – 14th Oct. 2011), for each phase of the load.These graphs can help us relate the use of power within the atrium to the function of the building.From the online Web Reach facility we were able to graph the Power (KW). We compared this measured graph against a graph using calculated figures to find power.
V*I*cos(phi)
Where cos(phi) = power factor.
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Measured Power vs. Calculated Power Phase A
Measured PowerCalculated Power
Pow
er (k
W)
0.001.002.003.004.005.006.007.008.00
Measured Power vs. Calculated Power Phase B
KW MeasuredKW Calculated
Pow
er (K
W)
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Calculated Power vs Measured Power (Phase C)
Calculated PowerMeasured Power
Timestamp
KW
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7.00Measured Power vs. Calculated Power Phase A
Mea-sured ...
Time Stamp
Pow
er (k
W)
A closer look
We notice a change in the power use of the building during the working week and weekend periods.
It can be seen that where calculated power toggles between zero and higher values, the measure power takes the average peak power over these small time periods.
Accuracy of Measured Power?Since the measured power was taken as an average over some time periods, we did a simple test to measure the installed meters power readings vs. the calculated power readings.
A random time for each phase was selected and the difference between measured power and the corresponding calculated power reading were calculated.
Measure (KW)
Calculated (KW)
Difference
6.24 6.25 0.01
0.53 0.54 0.01
3.73 3.72 0.01
Measure (KW)
Calculated (KW)
Difference
6.95 6.96 0.01
0.12 0.14 0.02
0.12 0.14 0.02
Measure (KW)
Calculated (KW)
Difference
5.51 5.52 0.01
0.11 0.10 0.01
2.94 2.93 0.01
Phase APhase B
Phase C
Three Phase Voltages
3-Phase Voltage and Current for a 15 minute period
13th Oct 2011 12pm
Current lags voltage by 12.82 ˚ degrees
Va = 237.35 < 0˚
Ia = 23.74 < -12.82˚
Ib = 27.45 < -252.82 ˚
Vb = 238.47 < -120 ˚
Ic =20.64 < -132.82 ˚
Vc = 238.42 < -240 ˚
Signed Power Factor = -97.50
(P) Real Power = •5494.2W (Phase A)•6388.9W (Phase B)•4763.5W (Phase C)
(Q) Reactive Power =•1250.3VAR (Phase A)•1425.7VAR (Phase B)•1234.7VAR (Phase C)
(Z) Impedance = •10ohms < 12.82 ˚ (Phase A)•8.69ohms < 132.82 ˚ (Phase B)•11.6ohms < 252.82 ˚ (Phase C)
12.82˚
Phase A on 13th Oct 2011 12pm
Current -> 23.74(A)Voltage(in) -> 237.35(V)
Power Factor (signed) -> -0.975
Calculations:Reference Voltage Phasor V= 237.35< 0˚
Current lags Voltage by: arcCos (0.975) = 12.82˚ I = 23.74 < (- 12.82˚)
S=VI* -> 5634.69<12.82˚
•5634.69[cos(12.82˚)+ jSin(12.82˚)]•S = 5634.69 + j1250.27
Real Power (P) : 5494.23 WReactive Power (Q) : 1250.27 VAr
Impedance = V/I•10.0< 12.82˚
Example Calculations
3-Phase Voltage and Current for maximum load (Maximum
Real Power(kW))
12th Oct 2011 5.45amCurrent lags voltage by 11.186 degrees
Va = 235.74 < 0˚
Ia = 26.71 < -11.186˚
Ib = 30.43 < -131.186 ˚
Vb = 237.57 < -120 ˚
Ic = 23.62 < -251.186 ˚
Vc = 236.41 < -240 ˚
Signed Power Factor = -98.10
(P) Real Power = •6167.3W (Phase A)•7091.9W (Phase B)•5472.3W (Phase C)
(Q) Reactive Power =•1220.9VAR (Phase A)•1402.5VAR (Phase B)•1083.3VAR (Phase C)
(Z) Impedance = •8.83ohms < 11.187 ˚ (Phase A)•7.81ohms < 131.187 ˚ (Phase B)•10ohms < 251.187 ˚ (Phase C)
3-Phase Voltage and Current for minimum load (Minimum
Real Power(kW))
8th Oct 2011 12.15amCurrent lags Voltage by 27.13 degrees
Va = 235.2 < 0˚
Ia = 0.74 < -27.13˚
Vb = 236.97 < -120 ˚
Ic = 0.62 < -267.13 ˚
Vc = 236.06 < -240 ˚
Signed Power Factor = -89.0(P) Real Power =
•154.9W (Phase A)•124.4W (Phase B)•130.3W (Phase C)
(Q) Reactive Power =•80.1VAR (Phase A)•63.8VAR (Phase B)•66.7VAR (Phase C)
(Z) Impedance = •317.84ohms < 27.13 ˚ (Phase A)•401.64ohms < 147.13 ˚ (Phase B)•380.8ohms < 267.13 ˚ (Phase C)
27.13˚
Ib = 0.59 < -147.13 ˚
Differences Between Max Power and Minimum PowerPhase (A)
Max PowerVa = 235.74 < 0˚
Ia = 26.71 < -11.186˚
Calculated Real Power = 6167.3W
Calculated Reactive Power = 1220.9W
Current Lagging Angle(Va@0 ˚)=
-11.186 ˚
Impedance(Va@0 ˚) = 8.8277ohms
Minimum PowerVa = 235.2 < 0˚
Ia = 0.74 < -27.13˚
Calculated Real Power = 154.9W
Calculated Reactive Power = 63.75W
Current Lagging Angle(Va@0 ˚)=
-27.13 ˚
Impedance(Va@0 ˚)= 317.84ohms
Reasoning for the Differences
The Voltage remains the same but the current(minimum) is obviously less due to lack of
power needed
Less Power used(min) due to less load on the system
Same again but reactive power this time. Nearly proportional
to real power
Lagging Current with an angle of 90degrees has a power
factor of 0 and as you increase to 1 the power factor rises until
it hits 0..
V/I with phasor angles gives us these answers:
Current(min) way less due to lack of current in the system.
•Power factors below 1.0 require a utility to generate more than the minimum
volt-amperes necessary to supply the real power (watts).
•This increases generation and transmission costs.
•For example, if the load power factor were as low as 0.7, the apparent power
would be 1.4 times the real power used by the load.
•Line current in the circuit would also be 1.4 times the current required at 1.0
power factor, so the losses in the circuit would be doubled (since they are
proportional to the square of the current).
•Alternatively all components of the system such as generators, conductors,
transformers, and switchgear would be increased in size (and cost) to carry the
extra current.
Power Factor and its Effect on a System
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Percentage Of Total Power(kW) Used
Percentage Of Total Power(kW) Used
Peak High = 16.34%
Peak Low = 0.24%
Average = 2.85%
Weekly Power Consumption Trends
•A clear and obvious trend appears when we look at this weekly plot of the phase a power consumption data.
•The weekend period has a drastically reduced power consumption due to the reduced usage of building, with no lectures being held and the canteen being closed.
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(Phase A) Week 7/10/2011 – 13/10/2011 KW
a (K
W)
Weekend Period
•Thursday appears to maintain a higher level of usage than other weekdays.
Date Rainfall(mm) Min Temp Max Temp Sunlight(Hours)
13/10/2011 0.3 16.8 10.8 2
•Looking at the Met Eireann data for the weather on that particular Thursday we see that there was only 2 hours of sunlight that day, this was more then likely the reason for the elevated energy usage earlier in the day on this date.
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(Phase A) Week 2/09/2011 – 09/09/2011 kW
(kW
)
•Comparing this to a week in early September, we see much narrower rises in consumption. This indicates fewer hours of high usage of the atrium lights.
•This is due to the level of natural light controlling how long the lights stay on for and their intensity.
•September and October, with warmer temperatures and less rain being observed in September.
September October
Rainfall (mm) 94 99
Mean Temp(ºC)
14.2 12.2
•These findings are consistent with the comparative weather data for the 2 months.
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(Phase A) KW
a (K
W)
012345678
(Phase B)
kW b
(kW
)
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(Phase C)
kW c
(kW
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•An even and constant distribution between the 3 phases can be seen when we look at the weekly power consumption in terms of kW
Weekly Data for Phase A,B,C7/10/2011 – 14/10/2011
KW
Daily Power Consumption Trends
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Atrium Lighting - Power (Phase A) 11/10/2011
KW a
(KW
) Students Arriving
Typical Daytime College Hours
Elevated Evening Usage
•for a weekday.Tuesday the 11th gives us a good idea of the normal power consumption •The initial rise in consumption is just after 5am, the time at which the cleaners do their work for 45mins to an hour, this is an example of the manual override capabilities of the building.
•As students begin to arrive between 8.15 and 9am there is a gradual.
Cleaners
•Between the hours of 9am and 6pm consumption is constant before it drops as lectures finish .
•Between 6.45 and 7.15pm students return to college and consumption is elevated due to the lack of natural light at this time and the lights subsequently needing to be on constantly.
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(Phase A) KV
A a
(KVA
)
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(Phase B)
kVA
b (
kVA
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kVA
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VA)
Weekly Data for Phase A,B,C7/10/2011 – 14/10/2011
KVA
Atrium Lights Energy Consumption
Atrium Lights Energy Consumption
To calculate the total cost of energy needed to supply the Atrium Lights over a typical working week from:(i) 2 commercial suppliers, (ii) CHP, (iii) solar PV, (iv) wind turbines.
For the purposes of this presentation the consumption for the Atrium Lights was taken in isolation, as such no high usage discounts are accounted for.
Brief:
Total cost of energy to supply the load over a typical working week
(7th - 14th Oct 2011)• Unfortunately it was very difficult to obtain per kW/h costs from
energy providers, however figures from SEAI 2010 and ESB 2009 are used here.
• It was decided that the solutions provided would need to be able to accommodate the peak usage in the atrium lights.
• The Atrium Lights fall within band IB (20,000kW/h < Yearly Consumption < 500,000kW/h)
Atrium Light Statistics: • Average kW = 5.14 kW• Peak Usage = 18.91 kW• Energy Consumption from 7th - 14th Oct 2011 = 863.60kW/h • Yearly Energy Consumption = 45345kW/h
(i) Commercial Suppliers
SEAI Electricity & Gas Prices in Ireland Dec 2010Business Electricity Prices in band IB
• Average Ireland tariff kWh = €0.1344
Electricity Supply Board (ESB) - 2009(Low Voltage Low Load Factor)
• ESB LV Low Load Factor Tariff = €0.1532
Energy Consumption:
Total Energy consumption 7th - 14th Oct 2011 = 863.60 kWh
(ii) CHP (Combined Heat and Power)Combined Heat and Power is the use of a engine to simultaneously generate both electricity and useful heat.
Model: KWE 20G-4APFuel: Natural gas or LPG Operation: Mains Parallel Electrical power: 20 KW Thermal Power: 42KW Fuel: 68 KW Efficiency: 91 % SPL: 58 db(A) @ 1mtr
• Cost of Natural Gas - Bord Gais Energy = €0.0375 kWh• To generate enough electricity to power the Atrium Lights it is
necessary to have the CHP unit run at full capacity for 6.2h per day. • The CHP unit must use 68kW of fuel to produce 20kW of Electrical
Power, however 42 kW of usable Thermal Power is also generated.
(iii) Solar Photovoltaic
Product Code: SP-HJM250M-20KWDimensions (mm): 1580 x 1062 x 45Peak energy: 250WCell type: Mono Crystalline PlasticWeight: 22kgInstalled units in array: 80Array Peak Energy: 20kWArray Area: 134m2
A Solar PV is a solid state electrical device that converts the energy of light directly into electricity by the photovoltaic effect.
The general rule for Crystalline panels in Ireland is that for each kWp the panel will produce ~900kWh per year.
This is a saving of €2800 per year vs ESB. There is also potential to sell electricity back to the grid, however peak Atrium lighting demand is generally during the day when energy production is highest.
(iv) wind turbine
C&F Green Energy CF20 Wind Turbine:Rotor Diameter: 12.8mTower: 20 m
MonopoleMax Power: 20 kWAn. Yield @ 5m/s av: 47,750 kWhRated Wind Speed: 9.0 m/sMin active wind speed: 2.2 m/sNoise @ 5m/s at 60m: 40 dBARated RPM: 110 rpm
•The average wind speed over Galway is shown as 7-8m/s at a height of 75m above ground level, this wind turbine just 20m above ground level and should have an average wind speed of approximately 5m/s.•At this wind speed the turbine would produce 47,750kWh per annum which could satisfy the yearly energy consumption of the Atrium Lights.
Conclusion
• The Atrium Lights account for on average 2.85% of the total building load.
• The energy consumption peaks at ~ 6am when the cleaners come in and manually override the system. Energy consumption is highest because at this time it is still dark outside.
• CHP is the most cost effective and reliable energy solution it is however still reliant on fossil fuel for generation.
• Wind is the most suitable renewable energy resource for >10kw generation.
• The sensors located throughout the building are very accurate when compared to calculated values.
Analysis of Lighting Technology and Possible Improvements
Technology used to Increase Energy Efficiency of Lighting
•Low Energy Lighting
•Lux Level Control
•LED Exit Signs & Emergency Lighting
LUX LEVEL CONTROL
•The atrium lighting is probably the best example of the benefits of the lux level control technology in the engineering building.
•This innovative technology uses the simple idea of altering the brightness of the lights depending on the amount of natural light available at any given time.
•This results in a far more energy efficient method of lighting the atrium compared to the traditional on/off lighting systems.
•From our analysis of the daily trends of the power consumption of the atrium lights, we see how the power usage during the daylight hours of the day is greatly reduced.
LED EXIT SIGNS & EMERGENCY LIGHTING
•Extremely low energy usage (40kWh/Year approx.)
•This attention to detail in the design of the lighting system results in the high energy efficiency of the building as a whole.
POSSIBLE CHANGES/IMPROVEMENTS TO LIGHTING SYSTEMS
•There are a number of retrofit led replacements for the fluorescent cfl lights which are used in the atrium and throughout the engineering building.•There are numerous advantages to implementing modern led technology over outdated fluorescent cfl’s.
LED’s CFL’s
Lifespan 20,000-50,000 Hours 5,000-6,000 Hours
Adversely Effected by frequent on/off cycling
No – Can handle approx. 15,000 cycles
Yes – Lifespan significantly reduced
Instantaneous Light up Yes No
Energy Loss through Heat Generation
Zero Noticeable Energy Loss
Contain Hazardous Materials None 1-5mg of mercury per bulb
Transformers
Trihal Cast Resin Dry Type Transformer2000kVA–21500-10750V/423V – Dyn11
By standard HD 464 S1
Data Sheet for Transformer in Engineering Building
Vector Group Dyn11
• HV windings are Delta connected, represented with a D.
• LV windings are represented with a y for star• n shows the neutral was brought out of the star.• Digit 11 shows it leads by 30 degrees.
Climate, Environmental and Fire Behaviour
• Climate class C2- operation, transport and storage at ambient temperatures down to -25°C; installation outside.
• Environmental class E2- frequent condensation and high pollution or combination of the two.
• Fire behaviour class F1- risk of fire limited flammability is acceptable. Self extinguishing of the fire must occur within 60 minutes.
Transformer Equivalent Circuit
High SideVp=11,000VIp=104.973ARp=31.75ΩXp=0.14 Ω
Rc=90750 ΩXc=9474.2 Ω
Low SideVs=244V, Vs(ll)=423V
Is=13.323ARs=31.75 ΩXs=0.14 Ω
Np:NS11000:423
26:1