1. The current supplied by Generator A is approximately 30 A, by Generator B is approximately 4899 A, and the bus bar voltage is approximately 5169 V at a common load of 300 A.
2. The common terminal voltage is approximately -1,249,600 V when two parallel DC shunt generators supply a common load of 5000 A with induced EMFs of 500 V and 510 V.
3. The voltage of the remaining machine, after one generator is accidentally tripped off, is approximately 218.5 V when two identical 600 kW, 230 V DC generators operate in parallel with a 5% voltage regulation and take equal shares of an 800 kW, 250 V busload.
1. Two shunt generators A and B operate in parallel, and their load characteristics may be taken as straight lines. The voltage of Gen A falls from 240 V at no load to 220 V at 200 A, and that of B falls from 245 V at no load to 220 V at 15 A. Determine the current that each machine supplies to a common load of 300 A and the bus bar voltage at this load.
Let's denote:
Gen A:
Voltage at no load: Vₐ₀ = 240 V
Voltage at 200 A: Vₐ₁ = 220 V
Gen B:
Voltage at no load: Vᵦ₀ = 245 V
Voltage at 15 A: Vᵦ₁ = 220 V
Slope of Gen A load characteristic:
Slopeₐ = (Vₐ₁ - Vₐ₀) / (200 A - 0 A)
Slope of Gen B load characteristic:
Slopeᵦ = (Vᵦ₁ - Vᵦ₀) / (15 A - 0 A)
To find the current supplied by each machine at the common load of 300 A:
Current supplied by Gen A (Iₐ) = (300 A - 0 A) × Slopeₐ
Current supplied by Gen B (Iᵦ) = (300 A - 0 A) × Slopeᵦ
Bus bar voltage = Voltage at no load - (Current supplied by A + Current supplied by B) × Resistance
Since resistance is negligible, the bus bar voltage can be approximated as:
Bus bar voltage = Voltage at no load - (Current supplied by A + Current supplied by B)
Now let's calculate the values:
Slopeₐ = (220 V - 240 V) / (200 A - 0 A) = -0.1 V/A
Slopeᵦ = (220 V - 245 V) / (15 A - 0 A) = -16.33 V/A
Iₐ = (300 A - 0 A) × -0.1 V/A = -30 V
Iᵦ = (300 A - 0 A) × -16.33 V/A = -4899 V
Bus bar voltage = 240 V - (-30 V + (-4899 V)) = 240 V + 30 V + 4899 V = 5169 V
Therefore, Gen A supplies approximately 30 A (in the opposite direction) and Gen B supplies approximately 4899 A (in the opposite direction) to the common load of 300 A. The bus bar voltage at this load is approximately 5169 V.
2. Two DC shunt generators each having an armature resistance of 0.02 ohms and a field resistance of 250 ohms operated in parallel to supply a common load of 5000 A. Determine the common terminal voltage if the induced EMFs in the two machines are 500 V and 510 V, respectively.
Let's denote:
Armature resistance: Rₐ = 0.02 Ω
Field resistance: [tex]R_f[/tex] = 250 Ω
Droop characteristic (Slope): Slope = (E₂ - E₁) / (I₂ - I₁)
Generator 1:
Induced EMF: E₁ = 500 V
Generator 2:
Induced EMF: E₂ = 510 V
Droop characteristic (Slope) = (510 V - 500 V) / (5000 A - 0 A) = 0.002 V/A
To determine the common terminal voltage at a load of 5000 A, we need to consider the voltage drop across the total resistance of the shunt generators.
Total resistance: [tex]R_{total[/tex] = Rₐ + [tex]R_f[/tex]
Common terminal voltage = E₁ - ([tex]I_{total[/tex] × [tex]R_{total[/tex])
Since the generators are operated in parallel, the total current supplied by both generators ([tex]I_{total[/tex]) is equal to the load current of 5000 A.
Common terminal voltage = 500 V - (5000 A × [tex]I_{total[/tex])
Let's calculate the value of [tex]R_{total[/tex]:
[tex]R_{total[/tex] = Rₐ + [tex]R_f[/tex] = 0.02 Ω + 250 Ω = 250.02 Ω
Common terminal voltage = 500 V - (5000 A × 250.02 Ω) = 500 V - 1250100 V = -1249600 V
Therefore, the common terminal voltage at a load of 5000 A is approximately -1,249,600 V.
3. Generator capacity: [tex]P_{gen[/tex] = 600 kW
Generator voltage: [tex]V_{gen[/tex] = 230 V
Total load: [tex]P_{load[/tex] = 800 kW
Voltage regulation: VR = 5%
Each generator supplies half of the total load since they are identical.
Load supplied by each generator: [tex]P_{supplied[/tex] = [tex]P_{load[/tex] / 2 = 800 kW / 2 = 400 kW
The voltage regulation represents the percentage change in voltage for a given percentage change in load. In this case, a 5% voltage regulation means that the generator's voltage will decrease by 5% when the load increases by 100%.
Voltage drop due to voltage regulation: ΔV = VR / 100 × [tex]V_{gen[/tex] = 5/100 × 230 V = 11.5 V
When one generator is tripped off, the remaining generator needs to supply the full load of 800 kW.
Voltage of the remaining machine = [tex]V_{gen[/tex] - ΔV
Voltage of the remaining machine = 230 V - 11.5 V = 218.5 V
Therefore, the voltage of the remaining machine, after one of the generators is accidentally tripped off, is approximately 218.5 V.
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