Substation Design Exercises: Power Up Your Skills in Trinité-et-T

1. Given data — The transformer rating is 15 MVA stepping down from 66 kV to 11 kV.
2. Formula for three-phase current — For a three-phase system, the power is given by $S=\sqrt{3}{V}_L{I}_L$ where $V_L$ is line-to-line voltage and $I_L$ is line current.
   $$S=\sqrt{3}{V}_L{I}_L$$
3. Convert MVA to VA — Convert the transformer rating from MVA to VA: 15 MVA = 15 000 000 VA.
   $$S=15000000\text{VA}$$
4. Calculate HV side current — On the high voltage side, $V_{HV}=66\text{kV}=66000\text{V}$. Using $I_{HV}=S/(\sqrt{3}{V}_{HV})$:
   $$I_{HV}=\frac{15000000}{\sqrt{3}\times 66000}$$
5. Calculate LV side current — On the low voltage side, $V_{LV}=11\text{kV}=11000\text{V}$. Using $I_{LV}=S/(\sqrt{3}{V}_{LV})$:
   $$I_{LV}=\frac{15000000}{\sqrt{3}\times 11000}$$

$$I_{HV}=131.2\text{A},I_{LV}=787.3\text{A}$$

→ High voltage side current: 131.2 A, Low voltage side current: 787.3 A

1. Given data — Power transmitted is 50 MW at 132 kV over 30 km with conductor resistance 0.1 Ω/km and power factor 0.9.
2. Calculate line current — Using $P=\sqrt{3}VI\cos \varphi$, solve for current $I$ where $P=50\text{MW}=50000000\text{W}$ and $V=132000\text{V}$:
   $$I=\frac{P}{\sqrt{3}V\cos \varphi }=\frac{50000000}{\sqrt{3}\times 132000\times 0.9}$$
3. Calculate total line resistance — Total resistance for the 30 km line: $R_{total}=0.1\mathrm{\Omega }/km\times 30km=3\mathrm{\Omega }$.
   $$R_{total}=3\mathrm{\Omega }$$
4. Calculate power loss — For a three-phase system, power loss is $P_{loss}=3I^2{R}_{line}$:
   $$P_{loss}=3I^2{R}_{total}$$

$$P_{loss}=1296\text{kW}$$

→ Power loss: 1.3 MW (1 296 kW)

1. Calculate future load — With 20% growth, future load = 10 MVA × 1.20 = 12 MVA.
   $$S_{future}=10\text{MVA}\times 1.20=12\text{MVA}$$
2. Select transformer rating — Choose a standard transformer rating slightly above future load. Common ratings are 12.5 MVA or 15 MVA. We'll select 15 MVA for this design.
   $$S_{transformer}=15\text{MVA}$$
3. Calculate HV side current — High voltage side current for 15 MVA at 33 kV: $I_{HV}=S/(\sqrt{3}{V}_{HV})$
   $$I_{HV}=\frac{15000000}{\sqrt{3}\times 33000}$$
4. Calculate LV side current — Low voltage side current for 15 MVA at 11 kV: $I_{LV}=S/(\sqrt{3}{V}_{LV})$
   $$I_{LV}=\frac{15000000}{\sqrt{3}\times 11000}$$
5. Select circuit breaker ratings — Circuit breakers should have continuous rating above normal current and interrupting rating above maximum fault current. Typical practice is to select breakers with ratings 2-3 times the normal current for safety.
   $$I_{breaker\_HV}=2.5\times I_{HV}=2.5\times 262.4\text{A}=656\text{A}$$
6. Determine number of feeders — Each outgoing feeder typically carries 2-4 MVA. For 12 MVA future load, we need at least 3 feeders (12 MVA / 4 MVA per feeder = 3 feeders).
   $$n_{feeders}=\lceil \frac{12\text{MVA}}{4\text{MVA/feeder}}\rceil =3$$

$$S_{transformer}=15\text{MVA},I_{breaker\_HV}=656\text{A},I_{breaker\_LV}=1968\text{A},n_{feeders}=3$$

→ Transformer rating: 15 MVA, HV breaker rating: 656 A, LV breaker rating: 1 968 A, Number of outgoing feeders: 3

1. Calculate system impedance at 33 kV — System impedance at 33 kV bus: $Z_{sys}=V^2/S_{sc}$ where $V=33\text{kV}$ and $S_{sc}=500\text{MVA}$:
   $$Z_{sys}=\frac{V^2}{S_{sc}}=\frac{{(33000)}^2}{500000000}=2.178\mathrm{\Omega }$$
2. Calculate transformer impedance — Transformer impedance on its own base: $Z_{transformer}=Z_{pu}\times (V_{LV}^2/S_{transformer})$ where $Z_{pu}=10\%=0.10$ and $S_{transformer}=15\text{MVA}$:
   $$Z_{transformer}=0.10\times \frac{{(11000)}^2}{15000000}=0.807\mathrm{\Omega }$$
3. Convert transformer impedance to 33 kV base — To combine impedances, convert transformer impedance to the 33 kV base: $Z_{transformer\_33kV}=Z_{transformer}\times (V_{HV}^2/V_{LV}^2)$:
   $$Z_{transformer\_33kV}=0.807\times \frac{{33}^2}{{11}^2}=7.263\mathrm{\Omega }$$
4. Combine impedances — Total impedance to fault at 11 kV bus: $Z_{total}=Z_{sys}+Z_{transformer\_33kV}=2.178+7.263=9.441\mathrm{\Omega }$
   $$Z_{total}=9.441\mathrm{\Omega }$$
5. Calculate short-circuit current — Short-circuit current at 11 kV bus: $I_{sc}=V_{phase}/Z_{total}$ where $V_{phase}=11000/\sqrt{3}$:
   $$I_{sc}=\frac{11000/\sqrt{3}}{9.441}=672.3\text{A}$$
6. Determine breaker rating — The circuit breaker should have an interrupting rating at least equal to the calculated short-circuit current. Standard ratings are 8 kA, 12.5 kA, 16 kA, 25 kA, etc. We select the next standard rating above 0.672 kA, which is 8 kA.
   $$I_{breaker\_rating}=8\text{kA}$$

$$I_{sc\_LV}=0.672\text{kA},I_{breaker\_rating}=8\text{kA}$$

→ Prospective short-circuit current: 0.672 kA, Required breaker interrupting rating: 8 kA

1. Convert energy savings to kWh — 12 GWh = 12 000 000 kWh.
   $$E_{saved}=12000000\text{kWh}$$
2. Calculate annual monetary savings — Annual savings = Energy saved × Tariff = 12 000 000 kWh × 0.45 TTD/kWh:
   $$Saving{s}_{annual}=12000000\times 0.45=5400000\text{TTD}$$
3. Calculate payback period — Simple payback period = Upgrade cost / Annual savings = 2 500 000 TTD / 5 400 000 TTD/year:
   $$T_{payback}=\frac{2500000}{5400000}$$

$$T_{payback}=0.46\text{years}$$

→ Simple payback period: 0.46 years (approximately 5.5 months)

1. Calculate protective level requirement — The protective level of arresters should be less than 80% of BIL for adequate protection: $V_{protective}<0.8\times BIL=0.8\times 95\text{kV}=76\text{kV}$.
   $$V_{protective}<76\text{kV}$$
2. Select arrester rating — Choose standard arrester rating below 76 kV. Common ratings are 72 kV or 69 kV. We'll select 72 kV arresters.
   $$V_{arrester}=72\text{kV}$$
3. Determine number of arresters using rolling sphere method — Using the rolling sphere method with sphere radius based on strike current (30 kA corresponds to ~20 m radius), place arresters at corners and along perimeter. For a 50 m × 40 m substation, 8 arresters (one at each corner and one mid-side on each long side) provide adequate coverage.
   $$n_{arresters}=8$$
4. Design grounding grid — For a rectangular grid, ground resistance can be approximated by $R=\rho /(4r)$ for a circular grid of equivalent area, where $r=\sqrt{A/\pi }$ and $A=L\times W$. For a rectangular grid, a more accurate formula is $R=\rho [\ln (2L/W)+1]/(2\pi L)$ for L >> W.
   $$R=\frac{\rho }{2\pi L}[\ln \left(\frac{2L}{W}\right)+1]$$
5. Calculate ground resistance — Substituting values: $L=50\text{m},W=40\text{m},\rho =100\mathrm{\Omega }\cdot m$:
   $$R=\frac{100}{2\pi \times 50}[\ln \left(\frac{2\times 50}{40}\right)+1]=0.318\times [\ln (2.5)+1]=0.318\times 1.916=0.61\mathrm{\Omega }$$
6. Specify grid parameters — Typical mesh size is 5-10 m. For this substation, use 6 m mesh size. Total conductor length for a 50 m × 40 m grid with 6 m spacing: approximately 2 × (50/6 + 40/6) × 6 m ≈ 180 m of conductor.
   $$L_{grounding}=180\text{m},mes{h}_{size}=6\text{m}$$

$$n_{arresters}=8,L_{grounding}=180\text{m},mes{h}_{size}=6\text{m},R_{ground}=0.61\mathrm{\Omega }$$

→ Number of lightning arresters: 8, Grounding conductor length: 180 m, Mesh size: 6 m, Expected ground resistance: 0.61 Ω

1. Calculate fundamental current — Fundamental current: $I_{fund}=S/(\sqrt{3}V)=20000000/(\sqrt{3}\times 11000)=1049.7\text{A}$.
   $$I_{fund}=\frac{20000000}{\sqrt{3}\times 11000}=1049.7\text{A}$$
2. Calculate 5th harmonic current — 5th harmonic current: $I_5=0.15\times I_{fund}=0.15\times 1049.7=157.5\text{A}$.
   $$I_5=0.15\times 1049.7=157.5\text{A}$$
3. Determine system impedance at 11 kV — System impedance: $Z_{sys}=V^2/S_{sc}={(11000)}^2/250000000=0.484\mathrm{\Omega }$. In per unit on 20 MVA base: $Z_{sys\_pu}=0.05\text{pu}$ (given).
   $$Z_{sys}=0.484\mathrm{\Omega }$$
4. Calculate required filter rating — To achieve THD < 5%, the filter should provide reactive power to reduce harmonic voltage distortion. A typical design targets harmonic current reduction by 70-80%. We'll design for 75% reduction.
   $$I_{5\_after}=0.25\times I_5=0.25\times 157.5=39.4\text{A}$$
5. Filter reactive power rating — Filter reactive power rating should match the fundamental current component: $Q_{filter}=\sqrt{3}V{I}_{fund}=1.732\times 11000\times 1049.7=20\text{MVAr}$ (approximately equal to load MVA rating for unity power factor correction).
   $$Q_{filter}=20\text{MVAr}$$
6. Filter tuning frequency — For 5th harmonic filter: $f_{tune}=h\times f_{fundamental}=5\times 50\text{Hz}=250\text{Hz}$.
   $$f_{tune}=250\text{Hz}$$
7. Calculate expected THD — With 75% harmonic current reduction, the harmonic voltage distortion is approximately $V_h=I_{5\_after}\times Z_{sys}=39.4\times 0.484=19.1\text{V}$. Fundamental voltage is $V_{fund}=11000/\sqrt{3}=6351\text{V}$. THD = $(V_h/V_{fund})\times 100\%=(19.1/6351)\times 100\%=0.30\%$.
   $$THD=\frac{19.1}{6351}\times 100\%=0.30\%$$

$$Q_{filter}=20\text{MVAr},f_{tune}=250\text{Hz},TH{D}_{after}=0.30\%$$

→ Filter reactive power rating: 20 MVAr, Tuning frequency: 250 Hz, Expected THD after filter: 0.30%

1. Calculate total failure rates — Total failure rate = ($n_{t}ransformer$ × lambd$a_T$) + ($n_{b}reaker$ × lambd$a_B$) = (2 × 0.02) + (8 × 0.01) = 0.04 + 0.08 = 0.12 failures/year.
   $${\lambda }_{total}=0.12\text{failures/year}$$
2. Calculate SAIFI — SAIFI = lambd$a_{t}otal$ = 0.12 interruptions/customer/year (assuming each failure affects all customers equally).
   $$SAIFI=0.12\text{interruptions/customer/year}$$
3. Calculate total repair time contribution — Total repair time contribution = ($n_{t}ransformer$ × lambd$a_T$ × $r_T$) + ($n_{b}reaker$ × lambd$a_B$ × $r_B$) = (2 × 0.02 × 168) + (8 × 0.01 × 24) = 6.72 + 1.92 = 8.64 hours/year.
   $$SAID{I}_{contribution}=8.64\text{hours/year}$$
4. Calculate SAIDI — SAIDI = 8.64 hours/customer/year.
   $$SAIDI=8.64\text{hours/customer/year}$$
5. Compare with targets — Current SAIFI (0.12) is below target (1.5), but current SAIDI (8.64 hours) exceeds target (2 hours). The main contributor is transformer repair time.
6. Propose upgrades — To reduce SAIDI, add a third transformer as backup (reducing transformer failure rate impact) and improve maintenance procedures to reduce repair time. Adding a third transformer reduces effective failure rate contribution from transformers by a factor of 3.
   Proposed: Add 1 transformer, reduce repair time to 72 hours
7. Calculate upgraded reliability — With 3 transformers and reduced repair time: New lambd$a_T$_contribution = (3 × 0.02 × 72) / 10 000 = 0.0432 hours/customer/year. Total SAIDI contribution = 0.0432 + 1.92 = 1.96 hours/year, which meets the target.
   $$SAID{I}_{upgraded}=1.96\text{hours/year}$$

$$SAIF{I}_{current}=0.12,SAID{I}_{current}=8.64\text{hours},SAID{I}_{upgraded}=1.96\text{hours}$$

→ Current SAIFI: 0.12 interruptions/year, Current SAIDI: 8.64 hours/year, Proposed upgrades: Add one transformer, reduce repair time to 72 hours, Resulting SAIDI: 1.96 hours/year

1. Calculate inverter output power — Inverter output power = Solar capacity × efficiency = 10 MW × 0.97 = 9.7 MW.
   $$P_{inverter}=10\text{MW}\times 0.97=9.7\text{MW}$$
2. Select transformer rating — Choose a standard transformer rating slightly above inverter output. A 12 MVA transformer provides adequate capacity and future expansion margin.
   $$S_{transformer}=12\text{MVA}$$
3. Determine transformer voltages — Standard configuration: Primary 33 kV (grid connection), secondary 400 V (inverter connection).
   $$V_{primary}=33\text{kV},V_{secondary}=400\text{V}$$
4. Calculate required reactive power capability — Grid codes typically require solar farms to operate at power factor between 0.95 lagging and 0.95 leading. For 9.7 MW output, reactive power capability: $Q=P\times \tan ({\cos }^{-1}(0.95))=9.7\times 0.329=3.2\text{MVAr}$.
   $$Q_{required}=9.7\times \tan ({\cos }^{-1}(0.95))=3.2\text{MVAr}$$
5. Protection coordination — Protection scheme must include: directional overcurrent relays (for bidirectional power flow), under/over voltage protection, frequency protection, anti-islanding protection, and reactive power control. Use numerical relays with communication to substation SCADA system.
6. Reactive power control strategy — Implement Volt/VAR control strategy: Inverter should absorb reactive power when voltage is high and supply reactive power when voltage is low, within the ±3.2 MVAr capability. Use droop control with 1% droop characteristic.
   $$Droop=1\%(V_{ref}=33\text{kV},Q_{max}=\pm 3.2\text{MVAr})$$

$$S_{transformer}=12\text{MVA},V_{primary}=33\text{kV},V_{secondary}=400\text{V},Q_{range}=\pm 3.2\text{MVAr}$$

→ Transformer rating: 12 MVA, Primary voltage: 33 kV, Secondary voltage: 400 V, Reactive power capability: ±3.2 MVAr, Protection: Directional overcurrent, Volt/VAR control with 1% droop

1. Select SCADA architecture — For a remote substation, use a hierarchical architecture: Local RTU at Tobago substation, Remote HMI in Trinidad control center, connected via microwave link. This provides local control capability and remote monitoring.
   Architecture: Hierarchical SCADA with local RTU and remote HMI
2. Specify communication protocols — Use IEC 61850 for substation automation (GOOSE messaging for fast protection signals, MMS for monitoring). Use IEC 60870-5-104 for remote telemetry over the microwave link. Use DNP3 for integration with existing systems if needed.
   $$Protocols:IEC61850(substation),IEC60870-5-104(remote),DNP3(legacyintegration)$$
3. RTU selection and configuration — Select an RTU with capacity for 120 analog and 250 digital inputs. Typical RTU can handle 256 digital and 128 analog points. Configure the RTU for local control and monitoring.
   RTU capacity: 256 digital, 128 analog points
4. Communication infrastructure — Use existing microwave link for data transmission. Add VPN encryption for security. Bandwidth requirement: approximately 2-5 Mbps for this I/O count.
   $$Bandwidth:2-5Mbps,Security:VPNencryption$$
5. HMI software and integration — Select SCADA software with IEC 61850 support and remote access capability. Configure HMI screens for substation one-line diagram, alarm management, trending, and reporting.
   HMI: IEC 61850 compatible, remote access enabled
6. Cost breakdown — Estimated costs: RTU (120 000 TTD), Communication equipment (50 000 TTD), HMI software license (150 000 TTD), Engineering and configuration (200 000 TTD), Installation and commissioning (100 000 TTD), Contingency (30 000 TTD). Total: 650 000 TTD.
   $$Totalcost=120000+50000+150000+200000+100000+30000=650000\text{TTD}$$

$$SCAD{A}_{arch}=\text{Hierarchical RTU+HMI},Protocols=\text{IEC 61850/104/DNP3},Cost=650000\text{TTD}$$

→ SCADA architecture: Hierarchical with local RTU and remote HMI, Communication protocols: IEC 61850/IEC 60870-5-104/DNP3, Estimated cost: 650 000 TTD