Question 1 |

The fuel cost functions in rupees/hour for two 600 MW thermal power plants are
given by

Plant 1: \; C_1=350+6P_1+0.004P_1^2

Plant 2: \; C_2=450+aP_2+0.003P_2^2

where P_1 and P_2 are power generated by plant 1 and plant 2, respectively, in MW and a is constant. The incremental cost of power (\lambda ) is 8 rupees per MWh. The two thermal power plants together meet a total power demand of 550 MW. The optimal generation of plant 1 and plant 2 in MW, respectively, are

Plant 1: \; C_1=350+6P_1+0.004P_1^2

Plant 2: \; C_2=450+aP_2+0.003P_2^2

where P_1 and P_2 are power generated by plant 1 and plant 2, respectively, in MW and a is constant. The incremental cost of power (\lambda ) is 8 rupees per MWh. The two thermal power plants together meet a total power demand of 550 MW. The optimal generation of plant 1 and plant 2 in MW, respectively, are

200, 350 | |

250, 300 | |

325, 225 | |

350, 200 |

Question 1 Explanation:

\frac{dC_1}{dP_1}=IC_1=6+0.008P_1

\frac{dC_2}{dP_2}=IC_2=a+0.006P_2

For optimum generation,

IC_1=IC_2=\lambda

Therefore, 6+0.008P_1=8

P_1=250MW

Given, P_1+P_2=550MW

P_550-250=300MW

\frac{dC_2}{dP_2}=IC_2=a+0.006P_2

For optimum generation,

IC_1=IC_2=\lambda

Therefore, 6+0.008P_1=8

P_1=250MW

Given, P_1+P_2=550MW

P_550-250=300MW

Question 2 |

Two generators have cost functions F_{1} and F_{2}. Their incremental-cost characteristics are

\dfrac{dF_{1}}{dP_{1}}=40+0.2P_{1}

\dfrac{dF_{2}}{dP_{2}}=32+0.4P_{2}

They need to deliver a combined load of 260 \text{~MW}. Ignoring the network losses, for economic operation, the generations P_{1} and P_{2} (in \text{MW}) are

\dfrac{dF_{1}}{dP_{1}}=40+0.2P_{1}

\dfrac{dF_{2}}{dP_{2}}=32+0.4P_{2}

They need to deliver a combined load of 260 \text{~MW}. Ignoring the network losses, for economic operation, the generations P_{1} and P_{2} (in \text{MW}) are

P_{1}=P_{2}=130 | |

P_{1}=160, P_{2}=100 | |

P_{1}=140, P_{2}=120 | |

P_{1}=120, P_{2}=140 |

Question 2 Explanation:

\begin{aligned} I C_{1} &=I C_{2} \\ 40+0.2 P_{1} &=32+0.4 P_{2} \\ 0.4 P_{2}-0.2 P_{1} &=8 \\ P_{2}+P_{1} &=260 \end{aligned}

Solving equation (i) and (ii),

P_{1}=160 \mathrm{MW} ; P_{2}=100 \mathrm{MW}

Solving equation (i) and (ii),

P_{1}=160 \mathrm{MW} ; P_{2}=100 \mathrm{MW}

Question 3 |

Two generating units rated 300 MW and 400 MW have governor speed regulation of 6% and
4% respectively from no load to full load. Both the generating units are operating in parallel to
share a load of 600 MW. Assuming free governor action, the load shared by the larger unit is
_______ MW.

50 | |

100 | |

200 | |

400 |

Question 3 Explanation:

Given: speed regulation of generating unit is from no load to full load.

Hence, the rated frequency (50 Hz) will be common for both units at no load nad will drop to lower value as loaded to full load.

Let, P_1= Load shared by 400 MW machine

P_2= Load shred by 300 MW machine

For 400 MW machine drop in frequency from no load to full load =\frac{50 \times 4}{100}=2 Hz

New frequecy of operation of 400 MW machine, f_1=50-\frac{2}{400}P_1

For 300 MW machine drop in frequency from no load to full load =\frac{50 \times 6}{100}=3 Hz

New frequecy of operation of 300 MW machine, f_2=50-\frac{3}{300}P_2

we also know,

\begin{aligned} P_1+P_2&=600 \\\ (\text{Load shared }& \text{between shared units})\\ P_2&=600-P_1 \\ f_1&=f_2=f \\ (\text{Units operatio }& \text{ is in parallel}) \\ 50-\frac{2}{400}P_1 &=50-\frac{3}{300}(600-P_1) \\ \frac{P_1}{200}&=\frac{600-P_1}{100} \\ 100P_1 &=200(600-P_1) \\ P_1&=1200-2P_1\\ 3P_1&=1200\\ P_1&=400 MW \end{aligned}

Hence, the rated frequency (50 Hz) will be common for both units at no load nad will drop to lower value as loaded to full load.

Let, P_1= Load shared by 400 MW machine

P_2= Load shred by 300 MW machine

For 400 MW machine drop in frequency from no load to full load =\frac{50 \times 4}{100}=2 Hz

New frequecy of operation of 400 MW machine, f_1=50-\frac{2}{400}P_1

For 300 MW machine drop in frequency from no load to full load =\frac{50 \times 6}{100}=3 Hz

New frequecy of operation of 300 MW machine, f_2=50-\frac{3}{300}P_2

we also know,

\begin{aligned} P_1+P_2&=600 \\\ (\text{Load shared }& \text{between shared units})\\ P_2&=600-P_1 \\ f_1&=f_2=f \\ (\text{Units operatio }& \text{ is in parallel}) \\ 50-\frac{2}{400}P_1 &=50-\frac{3}{300}(600-P_1) \\ \frac{P_1}{200}&=\frac{600-P_1}{100} \\ 100P_1 &=200(600-P_1) \\ P_1&=1200-2P_1\\ 3P_1&=1200\\ P_1&=400 MW \end{aligned}

Question 4 |

The incremental costs (in Rupees/MWh) of operating two generating units are functions of their
respective powers P_1 \; and \; P_2 in MW, and are given by

\frac{dC_{1}}{dP_{1}}=0.2 P_{1}+50

\frac{dC_{2}}{dP_{2}}=0.24P_{2}+40

where

20MW\leq P_{1}\leq 150MW

20 MW \leq P_{2} \leq 150MW.

For a certain load demand, P_1 \; and \; P_2 have been chosen such that dC_1/d P_1=76Rs/MWh and dC_2/d P_2=68.8 Rs/MWh. If the generations are rescheduled to minimize the total cost, then P_2 is ________.

\frac{dC_{1}}{dP_{1}}=0.2 P_{1}+50

\frac{dC_{2}}{dP_{2}}=0.24P_{2}+40

where

20MW\leq P_{1}\leq 150MW

20 MW \leq P_{2} \leq 150MW.

For a certain load demand, P_1 \; and \; P_2 have been chosen such that dC_1/d P_1=76Rs/MWh and dC_2/d P_2=68.8 Rs/MWh. If the generations are rescheduled to minimize the total cost, then P_2 is ________.

76.78 | |

120.24 | |

136.36 | |

156.84 |

Question 4 Explanation:

\begin{aligned} 0.2P_1+50&=76\\ P_1&=130mW\\ 0.24P-2+40&=68.8\\ P_2&=120mW\\ P_1+P_2&=250\;\;...(i)\\ \text{For minimized}& \text{ schedule,}\\ 0.2P_1+50&=0.24P_2+40\\ 0.2P_1-0.24P_2&=-10\;\;\;..(ii)\\ &\text{From eq. (i) and (ii).}\\ P_1&=113.63MW\\ P_2&=136.36MW \end{aligned}

Question 5 |

Consider the economic dispatch problem for a power plant having two generating units. The fuel costs in Rs/MWh along with the generation limits for the two units are given below:

C_{1}(P_{1})=0.01{P_{1}}^2+30P_{1}+10; 100MW\leq P_{1} \leq 150MW

C_{2}(P_{2})=0.05{P_{2}}^2+10P_{2}+10; 100MW\leq P_{2} \leq 180MW

The incremental cost (in Rs/MWh) of the power plant when it supplies 200 MW is ______ .

C_{1}(P_{1})=0.01{P_{1}}^2+30P_{1}+10; 100MW\leq P_{1} \leq 150MW

C_{2}(P_{2})=0.05{P_{2}}^2+10P_{2}+10; 100MW\leq P_{2} \leq 180MW

The incremental cost (in Rs/MWh) of the power plant when it supplies 200 MW is ______ .

10 | |

20 | |

30 | |

40 |

Question 5 Explanation:

\begin{aligned} C_1&=0.01P_1^2+30P_1+10\\ \frac{dC_1}{dP_1}&=2 \times 0.01P_1+30\\ &=0.02P_1+30\\ C_2&=0.05P_2^2+10P_2+10\\ \frac{dC-2}{dP_2}&=0.1P_2+10\\ \text{For optimum } &\text{incremental cost,}\\ \frac{dC_1}{dP_1}&=\frac{dC_2}{dP_2}\\ 0.02P_1+30&=0.1P_2+10\\ 0.02P_1-0.1P_2&=-20 \;\;...(i)\\ \text{Given, }P_1+P_2&=200 \;\;...(ii)\\ \text{Solving } & \text{equation (i) and (ii),}\\ P_2&=200, P_1=0\\ \text{But the }& \text{range of }P_1 \text{ and }P_2 \text{ are specified as}\\ 100MW \leq & P_1 \leq 150MW\\ 100MW \leq & P_2 \leq 180MW\\ \text{hence, }& P_2=200 MW \text{ is not valid}\\ \text{so, both }& P_1 \text{ and } P_2 \text{ are operating at}\\ P_1=100,P_2=100\;& \Rightarrow \; P_1+P_2=200\\ IC_1=\frac{dC_1}{dP_1}&=0.02 \times P_1+30=32 Rs/MW\\ IC_2=\frac{dC_2}{dP_2}&=0.1 \times P_2+10=20 Rs/MW \end{aligned}

There are 5 questions to complete.