Electrical Power Plant

LessonAbout 1 h

Electrical Power Plant

Power Plant is an industrial facility used for generating electric power from primary energy sources such as hydro, thermal, or wind energy.

1. Theoretical Power of a Water Turbine

  • Hydraulic Head: vertical distance water falls from reservoir to turbine (\(h\))
  • Discharge Rate: volume rate of water flow passing through turbine (\(Q\))
  • Specific Weight: weight of water per unit volume (\(w = 9810\ \text{N/m}^3\))

Theoretical Power of Water Turbine

\[ P = Q w h \]
\[ \text{Hp}_w = \dfrac{Q w h}{k} \]

Where: \(P\) = theoretical power (W), \(\text{Hp}_w\) = theoretical output horsepower (hp), \(Q\) = discharge flow rate (\(\text{m}^3/\text{s}\) or \(\text{ft}^3/\text{s}\)), \(w\) = specific weight of water (\(9810\ \text{N/m}^3\); \(62.4\ \text{lbs/ft}^3\)), \(h\) = head of water (\(\text{m}\); \(\text{ft}\)), \(k\) = unit conversion factor

Conversion Constants Table

Discharge (\(Q\)) Specific Weight (\(w\)) Head (\(h\)) Conversion Constant (\(k\))
\(\text{ft}^3/\text{sec}\) \(62.4\ \text{lbs/ft}^3\) \(\text{feet}\) \(550\)
\(\text{ft}^3/\text{min}\) \(62.4\ \text{lbs/ft}^3\) \(\text{feet}\) \(33000\)
\(\text{m}^3/\text{sec}\) \(9810\ \text{N/m}^3\) \(\text{meter}\) \(746\)
\(\text{m}^3/\text{min}\) \(9810\ \text{N/m}^3\) \(\text{meter}\) \(44760\)

EE Board Exam April 1994

Example. A waterfall is \(60\ \text{m}\) high and discharges at a rate of \(1.0\ \text{m}^3/\text{sec}\). A mini-hydroelectric plant below the waterfall has a turbine efficiency of \(80\%\) and a generator efficiency of \(95\%\). Calculate the kW output of the generator.

Solution.

\[ P_{\text{in}} = Q w h = (1.0)(9810)(60) = 588600\ \text{W} = 588.6\ \text{kW} \]
\[ P_{\text{out}} = P_{\text{in}} \eta_t \eta_g = 588.6(0.80)(0.95) = 447.34\ \text{kW} \]

Answer: \(447.34\ \text{kW}\)

2. Specific Speed of a Water Turbine

  • Definition: speed of a turbine generating one horsepower under a one-foot head
  • Design Parameter: dimensionless classification parameter used for turbine runner selection

Specific Speed and Actual Speed Equations

\[ N_s = \dfrac{5050}{h + 32} + 19 \]
\[ N = \dfrac{N_s h^{1.25}}{\sqrt{\text{Hp}_w}} \]

Where: \(N_s\) = specific speed (\(\text{rpm}\)), \(N\) = actual rotational speed (\(\text{rpm}\)), \(h\) = head of water (\(\text{foot}\)), \(\text{Hp}_w\) = theoretical horsepower of water turbine (\(\text{hp}\))

Example. A water turbine operating at \(85\%\) efficiency operates under a head of \(100\ \text{m}\) with a constant discharge of \(1000\ \text{gal/sec}\). Evaluate the operating speed of the turbine.

Solution.

\[ \text{Hp}_w = \dfrac{Q w h}{746} = \dfrac{(3.785)(9810)(100)}{746} = 4977.32\ \text{hp} \qquad\qquad h = 100(3.281) = 328.1\ \text{ft} \]
\[ N_s = \dfrac{5050}{328.1 + 32} + 19 = 33.024\ \text{rpm} \implies N = \dfrac{33.024(328.1)^{1.25}}{\sqrt{4977.32}} = 654\ \text{rpm} \]

Answer: \(654\ \text{rpm}\)

3. Turbine-Generator Setup

  • Mechanical Coupling: water turbine shaft directly drives electric generator rotor
  • Cascaded Efficiency: overall plant efficiency equals product of component efficiencies

Turbine-Generator Set Overall Efficiency

\[ \eta_{\text{overall}} = \eta_t \times \eta_g \]
\[ \eta_{\text{overall}} = \dfrac{P_{\text{out}}}{\text{Hp}_w \times 746} \times 100\% \]

Where: \(\eta_{\text{overall}}\) = combined efficiency, \(\eta_t\) = turbine efficiency, \(\eta_g\) = generator efficiency, \(P_{\text{out}}\) = generator output power (W), \(\text{Hp}_w\) = theoretical water input horsepower (hp)

EE Board Exam October 1983 / EE Board Exam October 1990

Example. A waterfall \(50\ \text{m}\) high discharges \(1.3\ \text{m}^3/\text{sec}\). Calculate the annual energy in kW-hr generated by a plant at the bottom if turbine efficiency is \(70\%\) and generator efficiency is \(90\%\).

Solution.

\[ P_{\text{out}} = Q w h \eta_t \eta_g = (1.3)(9810)(50)(0.70)(0.90) = 401.72\ \text{kW} \]
\[ E_{\text{annual}} = P_{\text{out}} \times 8760 = 401.72(8760) = 3519067\ \text{kW-hr} \]

Answer: \(3519067\ \text{kW-hr}\)

4. Heat Energy Conversion Factors

  • Calorie to Joule: \(1\ \text{calorie} = 4.186\ \text{joules}\)
  • BTU to Calorie: \(1\ \text{BTU} = 252\ \text{calories}\)
  • Kilowatt-Hour to Joule: \(1\ \text{kW-hr} = 3.6 \times 10^6\ \text{joules}\)
  • Heating Value: heat liberated per unit mass of burned fuel (\(\text{BTU/lb}\) or \(\text{kcal/kg}\))

Heat-to-Electrical Energy Conversion

\[ 1\ \text{kW-hr} = 3.6 \times 10^6\ \text{J} = 860\ \text{kcal} = 3412\ \text{BTU} \]

Where: \(\text{J}\) = joules (SI unit), \(\text{cal}\) = calories, \(\text{BTU}\) = British Thermal Units (FPS unit), \(\text{kW-hr}\) = kilowatt-hours

EE Board Exam October 1997

Example. A power plant consumes \(100000\ \text{lbs}\) of coal per hour. The heating value of coal is \(12000\ \text{BTU/lb}\) and overall plant efficiency is \(30\%\). What is the kW output of the plant?

Solution.

\[ P_{\text{in}} = \dfrac{100000(12000)(252)(4.186)}{3600} = 351624000\ \text{W} = 351624\ \text{kW} \]
\[ P_{\text{out}} = \eta P_{\text{in}} = 0.30(351624) = 105487.2\ \text{kW} \approx 105500\ \text{kW} \]

Answer: \(105500\ \text{kW}\)

5. Terms Used in Prediction of Load

  • Average Load (\(\text{AL}\)): mean load during a specified period (24 hours, month, or year)
  • Peak Load (\(\text{PL}\)): maximum load demand registered during the period
  • Connected Load (\(\text{CL}\)): sum of continuous ratings of all equipment connected
  • Installed Capacity: total rated capacity of generating units in plant

Average Load Equations

\[ \text{AL}_{\text{day}} = \dfrac{\sum \text{demand of day}}{24} \]
\[ \text{AL}_{\text{year}} = \dfrac{\text{Annual energy in kW-hr}}{8760} \]

Where: \(\text{AL}\) = average load (\(\text{kW}\)), \(\text{Annual energy}\) = total yearly energy output (\(\text{kW-hr}\))

Load, Demand, and Diversity Factors

\[ \text{LF} = \dfrac{\text{AL}}{\text{PL}} \qquad\qquad \text{DF} = \dfrac{\text{PL}}{\text{CL}} \qquad\qquad \text{dF} = \dfrac{\sum \text{PL}_{\text{individual}}}{\text{PL}_{\text{station}}} \]

Where: \(\text{LF}\) = load factor (\(\le 1.0\)), \(\text{DF}\) = demand factor (\(\le 1.0\)), \(\text{dF}\) = diversity factor (\(\ge 1.0\)), \(\text{AL}\) = average load (\(\text{kW}\)), \(\text{PL}\) = peak load (\(\text{kW}\)), \(\text{CL}\) = connected load (\(\text{kW}\))

Plant Capacity Factor and Plant Use Factor

\[ \text{PCF}_{\text{year}} = \dfrac{\text{Annual energy in kW-hr}}{\text{Installed capacity in kW} \times 8760} \]
\[ \text{PUF}_{\text{year}} = \dfrac{\text{Annual energy in kW-hr}}{\text{Installed capacity in kW} \times N} \]

Where: \(\text{PCF}\) = plant capacity factor, \(\text{PUF}\) = plant use factor, \(N\) = actual operating hours in one year

EE Board Exam October 1990

Example. A \(55000\ \text{kW}\) thermal power plant delivers an annual output of \(238000000\ \text{kW-hr}\) with a peak load of \(44\ \text{MW}\). Calculate the plant capacity factor.

Solution.

\[ \text{PCF} = \dfrac{238000000}{55000 \times 8760} \times 100\% = 49.4\% \]

Answer: \(49.4\%\)

6. Formula Summary

Formula Name Formula Variables
Theoretical Water Turbine Power \(P = Q w h\) \(P\) = power (\(\text{W}\))
\(Q\) = flow rate (\(\text{m}^3/\text{s}\))
\(w = 9810\ \text{N/m}^3\)
\(h\) = head (\(\text{m}\))
Theoretical Water Horsepower \(\text{Hp}_w = \dfrac{Q w h}{k}\) \(\text{Hp}_w\) = horsepower (\(\text{hp}\))
\(k\) = conversion constant
Specific Speed of Water Turbine \(N_s = \dfrac{5050}{h + 32} + 19\) \(N_s\) = specific speed (\(\text{rpm}\))
\(h\) = head (\(\text{ft}\))
Actual Turbine Operating Speed \(N = \dfrac{N_s h^{1.25}}{\sqrt{\text{Hp}_w}}\) \(N\) = actual speed (\(\text{rpm}\))
\(N_s\) = specific speed (\(\text{rpm}\))
\(h\) = head (\(\text{ft}\))
\(\text{Hp}_w\) = water hp
Overall Hydroelectric Efficiency \(\eta_{\text{overall}} = \eta_t \times \eta_g\) \(\eta_{\text{overall}}\) = plant efficiency
\(\eta_t\) = turbine efficiency
\(\eta_g\) = generator efficiency
Daily Average Load \(\text{AL}_{\text{day}} = \dfrac{\sum \text{demand of day}}{24}\) \(\text{AL}_{\text{day}}\) = average daily load (\(\text{kW}\))
Annual Average Load \(\text{AL}_{\text{year}} = \dfrac{\text{Annual energy in kW-hr}}{8760}\) \(\text{AL}_{\text{year}}\) = average yearly load (\(\text{kW}\))
Load Factor \(\text{LF} = \dfrac{\text{AL}}{\text{PL}}\) \(\text{LF}\) = load factor
\(\text{AL}\) = average load (\(\text{kW}\))
\(\text{PL}\) = peak load (\(\text{kW}\))
Demand Factor \(\text{DF} = \dfrac{\text{PL}}{\text{CL}}\) \(\text{DF}\) = demand factor
\(\text{PL}\) = peak load (\(\text{kW}\))
\(\text{CL}\) = connected load (\(\text{kW}\))
Diversity Factor \(\text{dF} = \dfrac{\sum \text{PL}_{\text{individual}}}{\text{PL}_{\text{station}}}\) \(\text{dF}\) = diversity factor
\(\text{PL}\) = peak load (\(\text{kW}\))
Utilization Factor \(\text{UF} = \dfrac{\text{PL}_{\text{station}}}{\text{Installed capacity}}\) \(\text{UF}\) = utilization factor
\(\text{PL}_{\text{station}}\) = station peak load (\(\text{kW}\))
Plant Capacity Factor \(\text{PCF} = \dfrac{\text{Annual energy in kW-hr}}{\text{Installed capacity in kW} \times 8760}\) \(\text{PCF}\) = plant capacity factor
Plant Use Factor \(\text{PUF} = \dfrac{\text{Annual energy in kW-hr}}{\text{Installed capacity in kW} \times N}\) \(\text{PUF}\) = plant use factor
\(N\) = annual operating hours