This is a question about a wind turbine generator a) The blades of a 20kW,400V machine rotate at 2 revs per second at rated wind speed i) Calculate the speed of the generator if the gear box ratio is 1:10 ii) Calculate the current generated at rated power iii) This four pole machine has a lap wound armature consisting of 16 coils 20\times 10^(3) each with a resistance of 0.3 ohms. Calculate the resistance of the armature. 0.3\times 16 iv) Draw the circuit diagram for the generator with this armature winding resistance and the shunt resistance is 50 ohms Calculate the winding losses and hence the efficiency of the wind turbine at rated wind speed.
The Correct Answer and Explanation is:
Given Data:
- Rated power of the wind turbine generator: 20 kW
- Rated voltage: 400 V
- Speed of the blades: 2 revs per second (at rated wind speed)
- Gearbox ratio: 1:10
- Number of poles of the machine: 4
- Number of coils in the armature: 16
- Each coil has a resistance of 0.3 ohms
- Shunt resistance: 50 ohms
Part (i): Speed of the Generator
To find the speed of the generator, we need to consider the gearbox ratio. The gearbox ratio of 1:10 means that for every 1 revolution of the blades, the generator shaft will make 1/10th of a revolution.
The speed of the blades is 2 revolutions per second (2 rev/s), so the speed of the generator shaft is: Generator speed=Blade speedGearbox ratio=2 rev/s10=0.2 rev/s\text{Generator speed} = \frac{\text{Blade speed}}{\text{Gearbox ratio}} = \frac{2 \text{ rev/s}}{10} = 0.2 \text{ rev/s}Generator speed=Gearbox ratioBlade speed=102 rev/s=0.2 rev/s
Thus, the generator speed is 0.2 revolutions per second (rev/s).
Part (ii): Current Generated at Rated Power
The rated power of the generator is 20 kW, and the rated voltage is 400 V. To calculate the current generated at rated power, we use the formula: P=VIP = VIP=VI
Where:
- P=20,000 WP = 20,000 \text{ W}P=20,000 W (since 20 kW = 20,000 W)
- V=400 VV = 400 \text{ V}V=400 V
Rearranging to find the current III: I=PV=20,000400=50 AI = \frac{P}{V} = \frac{20,000}{400} = 50 \text{ A}I=VP=40020,000=50 A
Thus, the current generated at rated power is 50 A.
Part (iii): Resistance of the Armature
Each coil in the armature has a resistance of 0.3 ohms, and there are 16 coils. The total armature resistance is simply the resistance of one coil multiplied by the number of coils: Rarmature=0.3 Ω×16=4.8 ΩR_{\text{armature}} = 0.3 \, \Omega \times 16 = 4.8 \, \OmegaRarmature=0.3Ω×16=4.8Ω
Thus, the total resistance of the armature is 4.8 ohms.
Part (iv): Circuit Diagram and Efficiency Calculation
Circuit Diagram:
The circuit consists of the following components:
- The wind turbine generator (with its armature winding resistance)
- A shunt resistor of 50 ohms
- The armature resistance of 4.8 ohms
The generator’s output voltage is 400 V, and the armature resistance is in series with the current, leading to power losses.
Winding Losses:
Winding losses in the armature can be calculated using the formula for resistive power loss: Pwinding loss=I2RarmatureP_{\text{winding loss}} = I^2 R_{\text{armature}}Pwinding loss=I2Rarmature
Where:
- I=50 AI = 50 \text{ A}I=50 A
- Rarmature=4.8 ΩR_{\text{armature}} = 4.8 \, \OmegaRarmature=4.8Ω
Pwinding loss=(50)2×4.8=2500×4.8=12,000 WP_{\text{winding loss}} = (50)^2 \times 4.8 = 2500 \times 4.8 = 12,000 \, \text{W}Pwinding loss=(50)2×4.8=2500×4.8=12,000W
Efficiency:
The efficiency of the wind turbine generator can be calculated using the formula: η=Output powerInput power×100\eta = \frac{\text{Output power}}{\text{Input power}} \times 100η=Input powerOutput power×100
The input power is the total power supplied by the wind turbine, which is the rated power plus the losses (winding losses and shunt losses). For simplicity, we assume that the total power loss is the winding loss, as the shunt resistance is usually a separate consideration for regulation. Input power=Poutput+Pwinding loss=20,000+12,000=32,000 W\text{Input power} = P_{\text{output}} + P_{\text{winding loss}} = 20,000 + 12,000 = 32,000 \, \text{W}Input power=Poutput+Pwinding loss=20,000+12,000=32,000W η=20,00032,000×100=62.5%\eta = \frac{20,000}{32,000} \times 100 = 62.5\%η=32,00020,000×100=62.5%
Thus, the efficiency of the wind turbine generator at rated wind speed is 62.5%.
Summary:
- The speed of the generator is 0.2 rev/s.
- The current generated at rated power is 50 A.
- The total resistance of the armature is 4.8 ohms.
- The efficiency of the wind turbine at rated wind speed is 62.5%.
