# Economic Power Generation and Load Dispatch

 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
 A 200, 350 B 250, 300 C 325, 225 D 350, 200
GATE EE 2022   Power Systems
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, $P_1+P_2=550MW$
$P_550-250=300MW$
 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
 A $P_{1}=P_{2}=130$ B $P_{1}=160, P_{2}=100$ C $P_{1}=140, P_{2}=120$ D $P_{1}=120, P_{2}=140$
GATE EE 2021   Power Systems
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),
$P_{1}=160 \mathrm{MW} ; P_{2}=100 \mathrm{MW}$

 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.
 A 50 B 100 C 200 D 400
GATE EE 2017-SET-2   Power Systems
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}
 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 ________.
 A 76.78 B 120.24 C 136.36 D 156.84
GATE EE 2015-SET-2   Power Systems
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 ______ .
 A 10 B 20 C 30 D 40
GATE EE 2015-SET-1   Power Systems
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.