17.3 Power Plant Economics, Load Curves & Energy Conversion

17.3 Power Plant Economics, Load Curves & Energy Conversion

Power plant economic analysis and operational performance metrics are fundamental topics evaluated on the PRC Registered Electrical Engineer (REE) Licensure Examination. Engineers must be capable of evaluating chronological load curves, calculating load utilization factors, determining overall heat rates, and optimizing power station selection for base-load and peaking duties.

---

1. Load Curves & Key Power Station Operating Factors

A chronological load curve plots power demand (kW or MW) on the vertical axis against time (hours) on the horizontal axis over a 24-hour day, month, or year.

                  24-HOUR CHRONOLOGICAL LOAD CURVE & METRICS

   Power Demand (MW)
      ^
  100 + - - - - - - - - - - - - - - - - - - - - - -/------\ Peak Load (P_max)
      |                                           /        \
   80 + - - - - - - - - - - - - - - - - - - - - -/          \
      |    Average Load (P_avg)                 /            \
   60 +========================================/==============\=====
      |                                       /                \
   40 +-------------------.                  /                  \
      |                    \                /                    \
   20 +                     \______________/                      \________
    0 +---|-------|-------|-------|-------|-------|-------|-------|-------|---> Time (Hours)
      00:00   03:00   06:00   09:00   12:00   15:00   18:00   21:00   24:00

Essential Mathematical Metrics & Definitions

1. Average Load ($P_{\text{avg}}$)

$P_{\text{avg}} = \frac{\text{Total Energy Generated in } T \text{ Hours (kWh)}}{T \text{ Hours}}$

2. Load Factor (LF)

The ratio of average load to peak maximum demand $P_{\text{max}}$ over a specified time period: $\text{LF} = \frac{P_{\text{avg}}}{P_{\text{max}}} = \frac{\text{Total Energy Generated in } T \text{ Hours}}{P_{\text{max}} \times T}$ Since $P_{\text{avg}} \le P_{\text{max}}$, $\text{LF} \le 1.0$ (or $\le 100%$).

3. Connected Load ($P_{\text{conn}}$)

The continuous nameplate rating sum of all electrical equipment and loads connected to the utility supply system.

4. Demand Factor (DF)

The ratio of actual peak maximum demand $P_{\text{max}}$ to total connected load $P_{\text{conn}}$: $\text{DF} = \frac{P_{\text{max}}}{P_{\text{conn}}} \le 1.0$

5. Diversity Factor (DivF)

The ratio of the sum of individual non-coincident maximum demands of all sub-circuits or consumers to the simultaneous coincident maximum demand $P_{\text{coincident max}}$ of the entire system: $\text{Diversity Factor (DivF)} = \frac{\sum_{i=1}^{n} P_{\text{individual max}, i}}{P_{\text{coincident max}}} \ge 1.0$

Key Principle: A high Diversity Factor ($> 1.0$) means individual consumer peaks occur at different times during the day. This reduces coincident station peak load $P_{\text{max}}$, allowing utilities to install smaller generator capacities, lowering capital investment per kW.

6. Plant Capacity Factor (PCF)

The ratio of actual energy generated over a given period to the maximum possible energy that could have been produced if the plant operated continuously at its full installed nameplate capacity ($C_{\text{installed}}$): $\text{PCF} = \frac{\text{Actual Energy Generated in } T \text{ Hours}}{C_{\text{installed}} \times T} = \frac{P_{\text{avg}}}{C_{\text{installed}}}$

7. Reserve Capacity

$\text{Reserve Capacity} = C_{\text{installed}} - P_{\text{max}}$

Relationship between LF, PCF, and Reserve:

$\text{PCF} = \text{LF} \times \left( \frac{P_{\text{max}}}{C_{\text{installed}}} \right)$ If Installed Capacity equals Peak Load ($C_{\text{installed}} = P_{\text{max}}$), then $\text{PCF} = \text{LF}$ and Reserve Capacity is zero.

8. Plant Use Factor (PUF)

$\text{PUF} = \frac{\text{Actual Energy Generated in } T \text{ Hours}}{C_{\text{installed}} \times T_{\text{operating}}}$ where $T_{\text{operating}}$ is the actual hours the plant was synchronized and running.

---

2. Power Station Performance Metrics: Heat Rate & Fuel Consumption

Station Heat Rate (HR)

The Heat Rate (HR) measures the heat energy input (in kJ or BTU) required by a thermal power station to produce one kilowatt-hour ($1\ \text{kWh}$) of electrical energy:

$\text{Heat Rate (HR)} = \frac{\text{Total Heat Energy Input (kJ or BTU)}}{\text{Net Electrical Energy Output (kWh)}}$

Mathematical Relationship with Thermal Efficiency ($\eta_{\text{overall}}$)

Since $1\ \text{kWh} = 3,600\ \text{kJ} = 3,412\ \text{BTU}$:

$\eta_{\text{overall}} = \frac{3600}{\text{HR (in kJ/kWh)}} \times 100\%$

$\eta_{\eta_{\text{overall}}} = \frac{3412}{\text{HR (in BTU/kWh)}} \times 100\%$

Rule: A lower Heat Rate indicates a more efficient thermal power plant.

Specific Fuel Consumption (SFC)

$\text{SFC} = \frac{\text{Mass of Fuel Consumed per Hour (kg/h)}}{\text{Electrical Output Power (kW)}} \quad (\text{kg/kWh})$

$\text{Heat Input Rate } Q_{\text{in}} = \text{SFC} \times \text{HHV} \quad (\text{kJ/kWh})$

---

3. Power Generation Economics & Tariff Calculations

Cost of Electrical Energy Generation

Total annual power station operating cost $C_{\text{total}}$ consists of three components:

1. Fixed Cost ($A$): Capital depreciation, debt interest, taxes, and insurance. Independent of peak load and energy output.
2. Semi-Fixed Cost ($B \cdot P_{\text{max}}$): Salaries, standing operational expenses, and proportional maintenance fees tied directly to maximum peak demand $P_{\text{max}}$ (kW or MW).
3. Running / Operating Cost ($C \cdot E$): Fuel consumption, lubricating oils, chemical treatments, and active maintenance proportional to total energy generated $E$ (kWh).

$\text{Total Annual Cost: } C_{\text{total}} = A + B \cdot P_{\text{max}} + C \cdot E \quad (\text{Pesos, } \text{PHP})$

Utility Tariff Structures

---

4. Base Load vs. Peaking Power Plant Operational Economics

                  BASE-LOAD VS. PEAKING PLANT COST STRUCTURES

   Total Cost per Year (PHP)
      ^
      |                  / Peaking Plant (High Fuel/Operating Cost)
      |                 /  Low Fixed Capital Cost
      |                / 
      |               / 
      |              /________ Base-Load Plant (High Fixed Capital Cost)
      |             /          Low Fuel/Operating Cost
      |            /
    0 +-----------o----------------------------------> Annual Energy Generated (kWh)
                 Cross-Over Point

• Base-Load Power Plants: High capital cost, low operating/fuel cost per kWh. Operate continuously near 100% capacity. Examples: Geothermal, Run-of-River Hydro, Nuclear, Large Coal.
• Peaking Power Plants: Low capital cost, high operating/fuel cost per kWh. Started rapidly to supply short-duration peak loads. Examples: Open-Cycle Gas Turbines, Diesel Generator Sets, Pumped-Storage Hydro.

---

Solved Board Exam Examples

Example 1: Comprehensive Load Curve & Factor Evaluation

Problem: A regional grid power station supplies a city load with the following daily 24-hour demand profile:

• 00:00 to 06:00 (6 hours): $20\ \text{MW}$
• 06:00 to 12:00 (6 hours): $60\ \text{MW}$
• 12:00 to 18:00 (6 hours): $80\ \text{MW}$
• 18:00 to 24:00 (6 hours): $40\ \text{MW}$

The power plant installed generator capacity is $100\ \text{MW}$. Calculate: (a) Total daily energy generated in MWh, (b) Daily average load, (c) Daily Load Factor (LF), (d) Plant Capacity Factor (PCF), and (e) Station Reserve Capacity.

Solution:

1. Calculate total energy generated $E_{\text{daily}}$: $E_{\text{daily}} = (20 \times 6) + (60 \times 6) + (80 \times 6) + (40 \times 6)$ $E_{\text{daily}} = 120 + 360 + 480 + 240 = 1200\ \text{MWh} = 1,200,000\ \text{kWh}$
2. Calculate average load $P_{\text{avg}}$: $P_{\text{avg}} = \frac{1200\ \text{MWh}}{24\ \text{h}} = 50\ \text{MW}$
3. Determine peak maximum demand $P_{\text{max}} = 80\ \text{MW}$. Calculate Load Factor (LF): $\text{LF} = \frac{P_{\text{avg}}}{P_{\text{max}}} = \frac{50\ \text{MW}}{80\ \text{MW}} = 0.625 \quad (62.5\%)$
4. Calculate Plant Capacity Factor (PCF) with installed capacity $C_{\text{installed}} = 100\ \text{MW}$: $\text{PCF} = \frac{P_{\text{avg}}}{C_{\text{installed}}} = \frac{50\ \text{MW}}{100\ \text{MW}} = 0.50 \quad (50.0\%)$
5. Calculate Reserve Capacity: $\text{Reserve Capacity} = C_{\text{installed}} - P_{\text{max}} = 100\ \text{MW} - 80\ \text{MW} = 20\ \text{MW}$

---

Example 2: Diversity Factor & Substation Sizing

Problem: A municipal distribution substation feeds three commercial industrial loads with individual maximum peak demands of $1500\ \text{kW}$, $2500\ \text{kW}$, and $3000\ \text{kW}$. The diversity factor between these loads is $1.40$. Calculate: (a) The coincident maximum peak demand on the substation, and (b) The required transformer capacity if a $20%$ reserve margin above coincident peak demand is required.

Solution:

1. Calculate the sum of individual maximum demands: $\sum P_{\text{indiv}} = 1500 + 2500 + 3000 = 7000\ \text{kW}$
2. Calculate coincident peak demand using Diversity Factor formula: $\text{Diversity Factor} = \frac{\sum P_{\text{indiv}}}{P_{\text{coincident max}}} \implies 1.40 = \frac{7000\ \text{kW}}{P_{\text{coincident max}}}$ $P_{\text{coincident max}} = \frac{7000}{1.40} = 5000\ \text{kW}$
3. Calculate required transformer capacity with $20%$ reserve margin: $\text{Capacity} = 5000\ \text{kW} \times 1.20 = 6000\ \text{kW}$

---

Example 3: Heat Rate & Coal Consumption Calculation

Problem: A thermal steam power plant has an overall station efficiency of $33%$. The coal burned has a heating value of $26,000\ \text{kJ/kg}$. Determine: (a) The station Heat Rate in $\text{kJ/kWh}$, and (b) Specific Fuel Consumption in $\text{kg/kWh}$.

Solution:

1. Calculate station Heat Rate from overall efficiency: $\text{Heat Rate} = \frac{3600}{\eta_{\text{overall}}} = \frac{3600}{0.33} = 10,909.09\ \text{kJ/kWh}$
2. Calculate Specific Fuel Consumption (SFC): $\text{SFC} = \frac{\text{Heat Rate}}{\text{Heating Value}} = \frac{10,909.09\ \text{kJ/kWh}}{26,000\ \text{kJ/kg}} = 0.4196\ \text{kg/kWh}$