Power Systems Crash Course: 5 Exercises to Master the Grid in the

1. Energy input calculation — Calculate the total energy input from diesel consumption per hour. Multiply fuel consumption by calorific value.
   $$E_{in}=Q_{\dot{}}\times CV=8\text{ L/h}\times 38\text{ MJ/L}=304\text{ MJ/h}$$
2. Electrical energy output — Apply the generator efficiency to find the electrical energy output per hour.
   $$E_{out}=E_{in}\times \eta =304\text{ MJ/h}\times 0.35=106.4\text{ MJ/h}$$
3. Power conversion — Convert the hourly energy output to power in kilowatts. Note that 1 kWh = 3.6 MJ.
   $$P_{out}=\frac{106.4\text{ MJ/h}}{3.6\text{ MJ/kWh}}=29.56\text{ kW}$$

$$P_{\text{out}}=29.6\text{kW}$$

→ The electrical power output of the diesel generator is approximately 29.6 kW.

1. Total line resistance — Calculate the total resistance of the 35 km transmission line.
   $$R_{\text{total}}=L\times R_{\text{per km}}=35\text{ km}\times 0.12\mathrm{\Omega }/\text{km}=4.2\mathrm{\Omega }$$
2. Power loss calculation — Calculate the power loss using the formula for resistive losses.
   $$P_{\text{loss}}=I^2\times R_{\text{total}}={(250A)}^2\times 4.2\mathrm{\Omega }=262500W=0.2625MW$$
3. Transmitted power — Calculate the total power transmitted using three-phase power formula.
   $$P_{\text{trans}}=\sqrt{3}\times V_{\text{line}}\times I=\sqrt{3}\times 132000V\times 250A=57157641W\approx 57.16MW$$
4. Percentage loss — Calculate the percentage of power lost during transmission.
   $$\text{Loss \%}=\frac{P_{\text{loss}}}{P_{\text{trans}}}\times 100=\frac{0.2625MW}{57.16MW}\times 100\approx 0.46\%$$

$$P_{\text{loss}}=0.26\text{MW},\text{Loss \%}=0.46\%$$

→ The power loss in the transmission line is 0.26 MW, representing approximately 0.46% of the transmitted power.

1. Line current calculation — Rearrange the three-phase power formula to solve for line current.
   $$I_{\text{line}}=\frac{P}{\sqrt{3}\times V_{\text{LL}}\times \text{pf}}=\frac{150000W}{\sqrt{3}\times 415V\times 0.85}=244.8A$$
2. Apparent power calculation — Calculate the apparent power using the power factor relationship.
   $$S=\frac{P}{\text{pf}}=\frac{150kW}{0.85}=176.47kVA$$

$$I_{\text{line}}=245\text{A},S=176\text{kVA}$$

→ The motor draws a line current of 245 A and has an apparent power of 176 kVA.

1. Load increase percentage — Calculate what percentage the 120 MW load increase represents of the total 2000 MVA capacity.
   $$\text{Load \%}=\frac{120MW}{2000MVA}\times 100=6\%$$
2. Governor response — Calculate the power change required from generators to compensate for the load increase using the droop characteristic.
   $$P_{\text{gov}}=\text{Load \%}\times \frac{100}{\text{droop \%}}=6\%\times \frac{100}{4}=150\%$$
3. Frequency deviation — Calculate the frequency deviation caused by the load increase and governor response.
   $$\mathrm{\Delta }f=f_{\text{initial}}\times \frac{\text{Load \%}}{\text{droop \%}}=50Hz\times \frac{6}{4}=75Hz$$
4. New frequency — Calculate the new steady-state frequency after the change.
   $$f_{\text{new}}=f_{\text{initial}}-\mathrm{\Delta }f=50Hz-0.75Hz=49.25Hz$$

$$f_{\text{new}}=49.25\text{Hz}$$

→ The new steady-state frequency after the Carnival load surge is 49.25 Hz.

1. Initial apparent power — Calculate the initial apparent power from real power and power factor.
   $$S_{\text{initial}}=\frac{P}{{\text{pf}}_{\text{initial}}}=\frac{1200kW}{0.75}=1600kVA$$
2. Initial reactive power — Calculate the initial reactive power using Pythagorean theorem.
   $$Q_{\text{initial}}=\sqrt{S_{\text{initial}}^2-P^2}=\sqrt{{1600}^2-{1200}^2}=1058.3kVAR$$
3. Target reactive power — Calculate the reactive power needed for the target power factor.
   $$S_{\text{target}}=\frac{P}{{\text{pf}}_{\text{target}}}=\frac{1200kW}{0.95}=1263.2kVA$$
4. Capacitor sizing — Calculate the required capacitor reactive power to achieve the target power factor.
   $$Q_c=Q_{\text{initial}}-Q_{\text{target}}=1058.3kVAR-392.1kVAR=666.2kVAR$$
5. Initial power loss — Calculate the initial power loss in the feeder using the initial current.
   $$I_{\text{initial}}=\frac{S_{\text{initial}}}{\sqrt{3}\times V_{\text{LL}}}[Assume{V}_{LL}=11kVforcalculation]I_{\text{initial}}=\frac{1600kVA}{\sqrt{3}\times 11kV}=84.2A{P}_{\text{loss, initial}}=3\times I_{\text{initial}}^2\times R=3\times {(84.2A)}^2\times 0.5\mathrm{\Omega }=10.6kW$$
6. Final power loss — Calculate the final power loss after capacitor installation.
   $$I_{\text{final}}=\frac{S_{\text{target}}}{\sqrt{3}\times V_{\text{LL}}}=\frac{1263.2kVA}{\sqrt{3}\times 11kV}=66.7A{P}_{\text{loss, final}}=3\times I_{\text{final}}^2\times R=3\times {(66.7A)}^2\times 0.5\mathrm{\Omega }=6.7kW$$
7. Loss reduction — Calculate the reduction in power loss due to capacitor installation.
   $$\mathrm{\Delta }{P}_{\text{loss}}=P_{\text{loss, initial}}-P_{\text{loss, final}}=10.6kW-6.7kW=3.9kW$$

$$Q_c=666\text{kVAR},\mathrm{\Delta }{P}_{\text{loss}}=3.9\text{kW}$$

→ The required capacitor size is 666 kVAR, which reduces power losses by 3.9 kW.