distribution system design
TRANSCRIPT
DISTRIBUTION SYSTEM DESIGN
PRESENTED BY:
SNEHA CHERUVATTATH
MOHAMMED PITHAPUR
I. INTRODUCTION
Planning is the most important part of a power system. By definition, it means preparing for the future in an organized manner giving space for various possibilities of system growth over time keeping in mind possible factors like investment, system expansion and upgrade etc. It helps anticipate requirements and identifies ways to meet them. Reliability, flexibility, fixed cost, maintenance cost, load growth in the future etc. should be properly analyzed.
Site Description
The site in consideration is given below:
Fig 1: Planning Site
The building parameters are as mentioned below.
Name Dimensions Connected Load, kW
Demand Factor Power Factor
Length (m) Width (m)
Building 1 120 50 2000 0.6 0.73
Building 2 70 50 3500 0.7 0.75
Building 3 70 50 2500 0.72 0.7
Building 4 70 40 2100 0.7 0.8
Building 5 70 40 1700 0.72 0.78
Building6 70 40 1400 0.7 0.74
Building 7 100 60 * * *
Building 8 75 40 500 0.7 0.78
Table 1: Buildings description
Different loads for the building 7 are as given:
Load Type Number of devices, n
Real Power of individual device, kW Kuf Power Factor
Machines 28 5.5 0.33 0.56
Machines 24 4.0 0.3 0.5
Machines 8 1.5 0.34 0.5
Machines 10 0.5 0.35 0.55
Machines 15 0.5 0.35 0.55
Machines 7 0.7 0.35 0.5
Welding machines 12 25.0 0.33 0.5
Cranes 10 7.5 0.5 0.6
Cranes 4 3.5 0.5 0.6
Fans 12 5.5 0.8 0.85
Fans 6 15.0 0.8 0.85
Fans 7 1.25 0.8 0.85
Water pumps 7 2.5 0.8 0.85
Heaters 6 12.5 0.85 1.0
Furnaces 4 75.0 0.9 1.0
Table 2: Load description for building 7
The report is organized as follows: Section II describes the Load calculations for Building 7. Section III discusses the lighting loads and their calculations. Section IV explains the Power Factor Correction. Section V focusses on Transformer Selection. Section VI discusses the two proposed designs and their load Flow analysis. Economic evaluation is carried out in Section VII. The report is finally concluded in Section VIII.
II. LOAD CALCULATION FOR BUILDING 7
Consider the case of the welding machines from table 2. The given values are:
n = 12
P = 25kW
Kuf = 0.33
p.f = 0.5
Step 1. Find Total Power Consumed
Hence, Total Power Consumed = P*n*Kuf = 12*25*0.33 = 99kW
Step 2. Determine Apparent Power
And, Apparent Power S = Total Power/ p.f = 99/ 0.5= 198 kVA
Step 3. Find the reactive power
From the equation, P2+Q2=S2 or Q = S sinα
we obtain Q = 75.18 kVAR
Following this example for all the loads of building 7, we obtain the following results.
Name Number of devices, n
Individual device
Rating, kWKuf Power
FactorTotal power
Power Consumed =
P*KufS Q
Machines 28 5.5 0.33 0.56 154.00 50.82 90.75 75.19Machines 24 4 0.3 0.5 96.00 28.80 57.60 49.88Machines 8 1.5 0.34 0.5 12.00 4.08 8.16 7.07Machines 10 0.5 0.35 0.55 5.00 1.75 3.18 2.66Machines 15 0.5 0.35 0.55 7.50 2.63 4.77 3.99Machines 7 0.7 0.35 0.5 4.90 1.72 3.43 2.97Welding machines 12 25 0.33 0.5 300.00 99.00 198.00 171.47
Cranes 10 7.5 0.5 0.6 75.00 37.50 62.50 50.00Cranes 4 3.5 0.5 0.6 14.00 7.00 11.67 9.33Fans 12 5.5 0.8 0.85 66.00 52.80 62.12 32.72Fans 6 15 0.8 0.85 90.00 72.00 84.71 44.62Fans 7 1.25 0.8 0.85 8.75 7.00 8.24 4.34
Water 7 2.5 0.8 0.85 17.50 14.00 16.47 8.68
pumpsHeaters 6 12.5 0.85 1 75.00 63.75 63.75 0.00
Furnaces 4 75 0.9 1 300.00 270.00 270.00 0.00Total 160 712.84 945.34 462.91
Table 3: Load calculations for building 7
III. LIGHTNING LOADS
To ascertain the lighting loads for this system, we will be using the following procedure.
1. Find the area of the building
Say for building 1, the Area = 120*50 = 600 square meters
2. With load density as 0.15 kW/m2 and demand factor = 0.7 (given)
Total demand for lighting load = area * lighting load density * demand factor
= 6000* 0.15* 0.7 = 630kW
3. Assuming the lighting power factor as 0.8 we can S for lighting loadS = P/ p.f = 630/0.8 = 787.5
Following this example, we can determine apparent power for lighting for the other buildings as well.
Area P lighting Lighting pf S lighting
Building 1 6000 630 0.8 787.5
Building 2 3500 367.5 0.8 459.375
Building 3 3500 367.5 0.8 459.375
Building 4 2800 294 0.8 367.5
Building 5 2800 294 0.8 367.5
Building 6 2800 294 0.8 367.5
Building 7 6000 630 0.8 787.5
Building 8 3000 315 0.8 393.75Table 4: Lighting load calculation
IV. POWER FACTOR CORRECTION
1. Find the total Active PowerTotal P = P + P lighting
2. Determine the total Apparent PowerTotal S = S + S lighting
3. From S and P determine the reactive powerP2 + Q2 = S2
4. Determine the value for power factor correction fromP = S cosα
5. Assume new power factor to be 0.95So tan(arc cos(0.95)) = 0.3286
6. Determine the new reactive power Q0.95Q0.95 = tan(arc cos(0.95)) * P
7. Find power factor correction Correction = Q-Q0.95
8. Determine the new apparent power
This leads us to the following results for the buildings 1 to 7,
Building Total P Total S Total Q New
pftan(arc
cos(0.95)) Q0.95 Q Correction
New S (kVA)
1 1830 2385.01 1529.50 0.95 0.3286 601.338 -928.17 1926.27
2 2817.5 3699.02 2396.76 0.95 0.3286 925.8305 -1470.93 2965.72
3 2167.5 3003.78 2079.58 0.95 0.3286 712.2405 -1367.34 2281.52
4 1764 2183.38 1286.65 0.95 0.3286 579.6504 -707.00 1856.80
5 1518 1915.11 1167.62 0.95 0.3286 498.8148 -668.80 1597.85
6 1274 1670.21 1080.05 0.95 0.3286 418.6364 -661.42 1341.02
7 1342.8 1686.52 1020.35 0.95 0.3286 441.257224 -579.09 1413.48
8 665 819.31 478.58 0.95 0.3286 218.519 -260.06 699.98
V. DESIGN CASES
CASE-1
The design case-1 is proposed by Mohammed and Sneha. The layout of the proposed design is depicted in the following figure:
The main highlights of the design are as follows:
1. One main substation (34.5/13.8kV)2. Four line substations (13.8/0.48kV)3. Tie feeders of 13.8kV4. Capacitive Banks for Power Factor Correction
TRANSFORMER SELECTION
Let’s consider the case, building 1 and 2. Both these buildings are taken as one load and their power requirement is around 4.481 MVA. If we are employing two transformers, one as a backup of another in case of an outage. So the power is divided equally and each transformer would require a rating of about 2.44 MVA. In his scenario we are considering an overloading of 1.3 times the actual rating. So,
(2.44 + x)* 1.3 = 4.891
1.3*x = 4.891-3.172
x = 1.322 MVA
Hence the actual required rating is, 2.44 + 1.322 = 3.76 MVA
In which case, we will be using transformers of rating 3.8 MVA
Voltage levels (High Tension Line):
In this model, high tension power lines will be used. To send high amount of power, high current or voltage are required. High voltage is preferred because lot of power is lost due to resistance in wires. But with high voltage this doesn’t happen as currents are smaller. Also problems like voltage sags can be eliminated.
Tie feeder:
A normal system feeder will have a fixed flow from sending end (source) to receiving end (load). This system has been upgraded with the help of a tie feeder, so that power flow will take place in both directions since both the sources will be synchronized.
Simulation was carried out in ETAP to find out the reliability and voltage levels of the proposed design.
Location: Buffalo, New York
Engineer: Mohammed Pithapur Study Case: LF
12.6.0EPage: 1
SN: RSW1QSJ
Filename: DesignProject2
Project: Distribution System Design ETAP
Contract:
Date: 05-05-2016
Revision: Base
Config.: Normal
LOAD FLOW REPORT
Bus
ID kV
Voltage
Ang.% Mag.
Generation
MW Mvar
Load
MW Mvar MW Mvar AmpID %PF %Tap
Bus1* 34.500 0.0 100.000 2.524 7.069 Bus2 2.090 0.752 37.2 94.10 0
Bus3 4.980 1.771 88.4 94.2
Bus2 13.800 -0.499.590 Bus6 2.086 0.735 92.9 94.30 0 0 0
Bus1 -2.086 -0.735 92.9 94.3
Bus3 13.800 -1.099.039 Bus8 4.961 1.669 221.1 94.80 0 0 0
Bus1 -4.961 -1.669 221.1 94.8
Bus6 13.800 -0.499.525 Bus2 -2.085 -0.734 92.9 94.30 0 0 0
Bus7 1.043 0.367 46.5 94.3
Bus7 1.043 0.367 46.5 94.3
Bus7 0.480 -2.297.943 0.6562.074 Bus6 -1.037 -0.328 1335.8 95.30 0
Bus6 -1.037 -0.328 1335.8 95.3
Bus8 13.800 -1.098.732 Bus3 -4.945 -1.666 221.1 94.80 0 0 0
Bus11 0.627 0.222 28.2 94.2
Bus10 2.626 0.934 118.1 94.2
Bus9 1.693 0.509 74.9 95.8
Bus9 0.480 -2.597.708 0.4581.688 Bus8 -1.688 -0.458 2153.4 96.50 0
Bus10 0.480 -3.496.942 0.8082.615 Bus8 -2.615 -0.808 3395.2 95.50 0
Bus11 13.800 -1.098.713 Bus8 -0.627 -0.222 28.2 94.20 0 0 0
Bus13 0.627 0.222 28.2 94.2
Bus13 0.480 -2.297.616 0.2060.624 Bus11 -0.624 -0.206 810.2 95.00 0
# Indicates
Indicates a voltage regulated bus (voltage controlled or swing type machine connected to it)
XFMRLoad Flow
*
The Load Flow Analysis clearly indicates that the voltages at all buses are within permissible limits.
Location: Buffalo, New York
Engineer: Mohammed Pithapur Study Case: LF
12.6.0EPage: 1
SN: RSW1QSJ
Filename: BuffStateProject2
Project: Distribution System Design ETAP
Contract:
Date: 05-05-2016
Revision: Base
Config.: Normal
Alert Summary Report
Cable 95.0 100.0
Bus
MarginalCritical
95.0 100.0Loading
% Alert Settings
95.0 100.0
95.0 100.0
95.0 100.0
Line
Transformer
Reactor
Panel
95.0 100.0
95.0 98.0
102.0 105.0
95.0 100.0
95.0 100.0
95.0 100.0
100.0
Generator Excitation
Bus Voltage
UnderExcited (Q Min.)
OverExcited (Q Max.)
UnderVoltage
OverVoltage
Protective Device
Generator
Inverter/Charger 100.0 95.0
Marginal Alerts Report
Device ID Type Rating/LimitCondition Unit Operating % Operating Phase Type
97.1 3-PhaseUnder VoltageBus10 Bus 0.480 kV 0.466
Bus13 97.8 3-PhaseUnder VoltageBus 0.480 kV 0.470
Bus7 97.9 3-PhaseUnder VoltageBus 0.480 kV 0.470
Bus9 97.9 3-PhaseUnder VoltageBus 0.480 kV 0.470
Location: Buffalo, New York
Engineer: Mohammed Pithapur Study Case: LF
12.6.0EPage: 1
SN: RSW1QSJ
Filename: BuffStateProject2
Project: Distribution System Design ETAP
Contract:
Date: 05-05-2016
Revision: Base
Config.: Normal
Bus Loading Summary Report
ID
BuskV
Directly Connected Load
MW Mvar MVA % PF AmpRated Amp LoadingMW Mvar MW Mvar
Constant kVA Constant Z Constant IPercent
Generic
MW Mvar
Total Bus Load
Bus1 34.500 7.517 125.8 94.2 0 0 0 0 0 0 0 0
Bus2 13.800 2.212 92.9 94.3 0 0 0 0 0 0 0 0
Bus3 13.800 5.245 221.6 94.8 0 0 0 0 0 0 0 0
Bus6 13.800 2.211 92.9 94.3 0 0 0 0 0 0 0 0
Bus7 0.480 2.881 3537.7 72.0 0 0 2.074 0.656 0 0 0 0
Bus8 13.800 5.240 221.6 94.8 0 0 0 0 0 0 0 0
Bus9 0.480 2.093 2570.8 81.0 0 0 1.695 0.460 0 0 0 0
Bus10 0.480 3.382 4187.7 77.6 0 0 2.625 0.811 0 0 0 0
Bus11 13.800 0.668 28.2 94.2 0 0 0 0 0 0 0 0
Bus13 0.480 0.784 963.6 80.0 0 0 0.627 0.207 0 0 0 0
* Indicates operating load of a bus exceeds the bus critical limit ( 100.0% of the Continuous Ampere rating).
# Indicates operating load of a bus exceeds the bus marginal limit ( 95.0% of the Continuous Ampere rating).
Location: Buffalo, New York
Engineer: Mohammed Pithapur Study Case: LF
12.6.0EPage: 2
SN: RSW1QSJ
Filename: BuffStateProject2
Project: Distribution System Design ETAP
Contract:
Date: 05-05-2016
Revision: Base
Config.: Normal
Branch Loading Summary Report
ID Type
LoadingAmp %
Capability(MVA) MVA %
Loading (output) CKT / Branch Cable & Reactor
Transformer
Loading (input)
%MVA
Ampacity(Amp)
Cable2 Cable 253.11 221.57 87.54
Cable3 Cable 132.62 28.24 21.29
T1 Transformer 10.000 2.221 22.2 2.212 22.1
T2 Transformer 10.000 5.296 53.0 5.245 52.4
T12 Transformer 2.000 0.668 33.4 0.660 33.0
T14 Transformer 3.600 2.798 77.7 2.748 76.3
T15 Transformer 3.600 1.775 49.3 1.756 48.8
T16 Transformer 1.600 1.105 69.1 1.088 68.0
T17 Transformer 1.600 1.105 69.1 1.088 68.0
* Indicates a branch with operating load exceeding the branch capability.
Location: Buffalo, New York
Engineer: Mohammed Pithapur Study Case: LF
12.6.0EPage: 3
SN: RSW1QSJ
Filename: BuffStateProject2
Project: Distribution System Design ETAP
Contract:
Date: 05-05-2016
Revision: Base
Config.: Normal
Branch Losses Summary Report
ID MW Mvar MW Mvar kW kvar From To
CKT / Branch From-To Bus Flow To-From Bus Flow Losses % Bus Voltage% Drop
Vd
in Vmag
3.3 18.0 100.0 99.6 0.41T1 2.090 0.752 -2.086 -0.735
18.6 102.2 100.0 99.0 0.96T2 4.989 1.778 -4.970 -1.676
1.4 0.3 99.6 99.5 0.06Cable4 2.086 0.735 -2.085 -0.734
4.7 2.7 99.0 98.9 0.10Cable2 4.970 1.676 -4.965 -1.673
5.5 39.0 99.5 97.9 1.58T16 1.043 0.367 -1.037 -0.328
5.5 39.0 99.5 97.9 1.58T17 1.043 0.367 -1.037 -0.328
0.1 0.0 98.9 98.9 0.02Cable3 0.629 0.223 -0.629 -0.223
11.2 127.3 98.9 97.1 1.79T14 2.636 0.938 -2.625 -0.811
4.5 51.2 98.9 97.9 1.03T15 1.700 0.512 -1.695 -0.460
2.3 16.2 98.9 97.8 1.10T12 0.629 0.223 -0.627 -0.207
57.0 396.0
Location: Buffalo, New York
Engineer: Mohammed Pithapur Study Case: LF
12.6.0EPage: 5
SN: RSW1QSJ
Filename: BuffStateProject2
Project: Distribution System Design ETAP
Contract:
Date: 05-05-2016
Revision: Base
Config.: Normal
0.000 0.000 0.000
0.000 0.000 0.000
Lagging 95.68 7.339
Total Generic Load:
Total Constant I Load:
0.000 0.000
0.396 0.057
Number of Iterations: 3
System Mismatch:
Apparent Losses:
SUMMARY OF TOTAL GENERATION , LOADING & DEMAND
0.000
Lagging
Lagging
2.135 7.021
0.000 0.000
94.16 7.517 2.531 7.078
0.000 0.000 0.000
94.16 7.517 2.531 7.078
% PFMVAMvarMW
Total Static Load:
Total Motor Load:
Total Demand:
Source (Non-Swing Buses):
Source (Swing Buses):