The main objective of the governor system in power systems for all types of turbines is to control the:
The main objective of the governor system in power systems for all types of turbines is to control the frequency. A governor protects the prime mover from over speed and keeps the prime mover speed at or near the desired revolutions per minute. When a prime mover drives an alternator supplying electrical power at a given frequency, a governor must be used to hold the prime mover at a speed that will yield this frequency.
In a hydroelectric plant, the available head is 80 m for a discharge 12m^{3}/sec, and overall efficiency of the PowerStation of 80%. The specific weight of water is 9.81 k/m2. Then the power developed is – (in MW)
P = WQHη KW
= 9.81 × 12 × 80 × 0.8
= 7534.08 KW = 7.53 MW
An industrial consumer has a daily load pattern of 2500 kW, 0.8 lag for 8 hours, 2000 kW, 0.6 lag for 8 hours and 1500 kW, unity power factor for 8 hours. What is the load factor?
= 48000/60000
= 0.8
Consider a power system with three identical generators. The transmission losses are negligible. One generator(G1) has a speed governor which maintains its speed constant at the rated value, while the other generators (G2 and G3) have governors with a drop of 5%. If the load of the system is increased, then in steady state.
Given power system with these identical generators G_{1} has Speed governor G_{2} and G_{3 }has drop of 5% When load increased, in steady state generation of G_{1} is only increased while generation of G_{2} and G_{3 }are unchanged.
consider the two bus power system network with given loads as shown in the figure. All the value is shown in the figure in per unit. The reactive power supplied by generator G_{1} and G_{2} is Q_{G1}and Q_{G2} respectively. The per unit values of Q_{G1}, Q_{G2} and line reactive power loss Q_{loss} respectively are
Loss less line
= 1.34 PU
∴ Q_{G1} = 5 + 1.34 = 6.24 PU
Q_{L2} = Q_{G2} + Q_{R}
Q_{G2} = 10 – (1.34) = 11.34 PU
Q_{loss} = Q_{S}  Q_{P} = 1.34 – (1.34) = 2.68
A generating station has the following data:
Installed capacity = 420 MW; capacity factor = 60%
Annual load factor = 70%.
The minimum reserve capacity of the station is – (in MW)
By above two equations,
Maximum demad = Installed capacity x
Reserve capacity = Installed capacity − Maximum demand
= 420 – 360 = 60 MW
Determine the incremental cost of received power of the plant shown in the figure below, if the incremental cost of production is ________
Penalty Factor = 20/15 = 4/3
Cost of received power =
Assume that the fuel input in Btu per hour for units 1 and 2 are given by
F_{1} = (4P_{1} + 0.036P^{2}_{1} + 60)10^{6}
F_{2} = (3P_{2} + 0.02P^{2}_{2} + 100)10^{6}
The maximum and minimum loads on the units are 100 MW and 10 MW respectively. The minimum cost of a generation when the following load is supplied is _______ (in Rs.) Where the cost of fuel is Rs. 1 per million Btu. (Approximately)
From the fuel input characteristics
When the load is 50 MW, for economic loading.
⇒ 0.072P_{1} + 4 = 0.04 P_{2} + 3
⇒ 0.072P_{1} – 0.04P_{2} + 1 = 0
P_{1} + P_{2} = 50
⇒ P_{1} = 8.93 MW P_{2} = 41.07 MW.
Minimum load on P_{1} = 10 MW
⇒ (i) P_{1} = 10 MW, P_{2} = 40 MW (or)
(ii) P_{1} = 0 MW, P_{2} = 50 MW
(i) P_{1} = 10 MW, P_{2} = 40 MW
F_{1} = 103.6 million Btu / hr
F_{2} = 252 million Btu / hr
F_{1} + F_{2} = 355.6 million Btu / hr
(ii) P_{1} = 0 MW, P_{2} = 50 MW
F_{1} = 60 million Btu / hr
F_{2} = 300 million Btu / hr
F_{1} + F_{2} = 360 million Btu / hr
So for P_{1} = 10 MW, P_{2} = 40 MW, the fuel cost is less.
When the load is 100 MW.
⇒ 0.072 P_{1 }− 0.04 P_{2 }+ 1 = 0
⇒ P_{1} + P_{2} = 100
⇒ P_{1} = 26.79 P_{2} = 73.21
F_{1} = 193 million Btu / hr
F_{2} = 512.58 million Btu / hr
Now, total fuel cost
= (103.60 + 252 + 193 + 512.58) (12) (1)
= Rs. 12734.16
A steam power station spends Rs. 30 lakhs per annum for coal used in the station. The coal has a calorific value of 5000 kcal/kg and costs Rs. 300 per ton. If the station has thermal efficiency of 33% and electrical efficiency of 90%, find the average load on the station. (in kW)
Overall efficiency, η_{overall} = 0.33 × 0.9 = 0.297
Coal used/annum =
Heat of combustion = coal used/annum × calorific value
= 10^{7} × 5000 = 5 × 10^{10} kcal
Heat output = η_{overall} × heat of combustion
= (0.297) × (5 × 10^{10}) = 1485 × 10^{7} kcal
Units generated per annum =
Average load on station =
In the above system shown, a transmission loss of 2.5 MW is incurred if 50 MW is transmitted from. Plant – 1 to the load LaGrange’s multiplier λ is Rs. 20 / MWH.
The incremental fuel cost is given.
I_{C1} = 0.01P_{1 }+ 12Rs/MWH,I_{C2 }= 0.04P_{2} + 16Rs/MWH
For optimal load scheduling.
The powers of plants 1 and plant 2 and losses are respectively.
Loss P_{L}= B_{11}P^{2}_{1}
Load Power P_{D} = P_{1} + P_{2 }− PL
For P_{1} = 50MW,P_{L} = 2.5
So, 2.5 = B_{11} × (50)^{2}
⇒ B_{11 }= 0.001 MW^{−1
}
For optimal load scheduling.
And
1 × (0.04P_{2} + 16) = 20
P_{2} = 100MW
0.01P_{1} + 12 = 20(1 − 0.002P_{1})
P_{1} = 160MW
P_{L} = 0.001 × (160)^{2} = 25.6MW
P_{P} = 100 + 160 − 25.6 = 234.4MW
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