Economic Power Generation and Load Dispatch

 Question 1
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 1 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 2
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 2 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 3
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 3 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}
 Question 4
The fuel cost functions of two power plants are

Plant $P_1:C_1\equiv 0.05Pg^2_1+APg_1+B$
Plant $P_2:C_2\equiv 0.10Pg^2_2+3APg_2+2B$

where, $Pg_1 \; and \; Pg_2$ are the generated powers of two plants, and A and B are the constants. If the two plants optimally share 1000 MW load at incremental fuel cost of 100 Rs/MWh, the ratio of load shared by plants $P_1 \; and \; P_2$ is
 A 1:4 B 2:3 C 3:2 D 4:1
GATE EE 2014-SET-1   Power Systems
Question 4 Explanation:
Given, $P_{g1}+P_{g2}=1000 \;\;...(i)$
Also, incremental cost of production of
$g_1=\frac{dC_1}{dP_{g1}}=0.1P_{g1}+A \;\;\;...(ii)$
and incremental cost of production of
$g_2=\frac{dC_2}{dP_{g2}}=0.2P_{g2}+3A\;\;\;...(iii)$
\begin{aligned} \frac{dC_1}{dP_{g1}}&=\frac{dC_2}{dP_{g2}} \\ 0.1P_{g1}+A&=0.2P_{g2}+3A \\ 0.1P_{g1}-0.2P_{g2}&=2A \\ P_{g1}-2P_{g2}&=20A \;\;\;...(iv) \\ \text{Given, } \frac{dC_1}{dP_{g1}}&=\frac{dC_2}{dP_{g2}}=100\; Rs/MWh \end{aligned}
Putting this value in equation (ii), we have
\begin{aligned} 100&=0.1P_{g1}+A\\ A&=(100-0.1P_{g1})\;\;...(v) \end{aligned}
Using equation (iv) and (v), we have:
\begin{aligned} P_{g1}-2P_{g2}&=20(100-0.1P_{g1})\\ 3P_{g1}-2P_{g2}&=20004\;\;...(vi) \end{aligned} \begin{aligned} P_{g1}&=800MW \text{ and } P_{g2}=200MW\\ \therefore \;P_{g1}:P_{g2}&=800:200=4:1 \end{aligned}
 Question 5
The figure shows a two-generator system applying a load of $P_D$= 40MW, connected at bus 2.

The fuel cost of generators G1 and G2 are :
$C_1(P_{G1})$=10,000 Rs/MWh and $C_2(P_{G2})$=12,500 Rs/MWh and the loss in the line is $P_{loass(pu)}=0.5P_{G1(pu)}^2$, where the loss coefficient is specified in pu on a 100 MVA base. The most economic power generation schedule in MW is
 A $P_{G1}=20,P_{G2}=22$ B $P_{G1}=22,P_{G2}=20$ C $P_{G1}=20,P_{G2}=20$ D $P_{G1}=0,P_{G2}=40$
GATE EE 2012   Power Systems
Question 5 Explanation:
From cordination equation
\begin{aligned} \frac{dF_n}{dP_{Gn}}\cdot \frac{1}{1-\frac{\partial P_L}{\partial P_{Gn}}}&=\lambda \\ \text{Given: }P_L&=0.5P_{G1}^2\\ \frac{\partial P_L}{\partial P_{G1}}=P_{G1} \text{ and } \frac{\partial P_L}{\partial P_{G2}}&= 0\\ \Rightarrow \; \frac{dF_{1}}{dP_{G1}}\frac{1}{1-P_{G1}}&=\frac{dF_2}{dP_{G2}}\frac{1}{1-0}\\ \Rightarrow \; 10000 \times \frac{1}{1-P_{G1}}&=12500\\ 1-P_{G1}=\frac{100}{125}&=\frac{4}{5}\\ P_{G1}&=\frac{1}{5}p.u.\\ \end{aligned}
as the base value is 100 MVA
\begin{aligned} P_{G1}&=\frac{1}{5} \times 100=20MW\\ \Rightarrow \; P_L&=0.5P_{G1}^2\\ &=0.5\left ( \frac{1}{5} \right )^2=0.02p.u.\\ \Rightarrow \; P_L&=0.02 \times 100=2MW\\ P_D&=P_{G1}+P_{G2}-P_L\\ 40&=20+P_{G2}-2\\ \Rightarrow \;\;P_{G2}&=22 MV \end{aligned}
 Question 6
A load center of 120 MW derives power from two power stations connected by 220 kV transmission lines of 25 km and 75 km as shown in the figure below. The three generators G1, G2 and G3 are of 100 MW capacity each and have identical fuel cost characteristics. The minimum loss generation schedule for supplying the 120 MW load is
 A $P_{1}=80MW+losses$, $P_{2}=20MW$, $P_{3}=20MW$ B $P_{1}=60MW$, $P_{2}=30MW+losses$, $P_{3}=30MW$ C $P_{1}=40MW$, $P_{2}=40MW$, $P_{3}=40MW+losses$ D $P_{1}=30MW+losses$, $P_{2}=45MW$, $P_{3}=45MW$
GATE EE 2011   Power Systems
Question 6 Explanation:

Let load center is connected at point x
If r is resistance/km of line
Resistance of section between 1 nad x
$R_1=r \times 25=R$
Resistance of section between x nad 2
$R_2=r \times 75=3R$
Current (I) fed by generator $\propto$ power produced by the generator.
$I_1 \propto P_1, I_2 \propto P_2 \text{ and }I_3\propto P_3$
Transmission losses
\begin{aligned} P_L&=I_1^2 R+I_2^2 3R+ I_3^23R \\ P_L&\propto (P_1^2+3P_2^2+3P_3^2) \\ P_L&=K[P_1^2+3P_2^2+3P_3^2] \end{aligned}
\begin{aligned} &\text{Option (A)}\\ P_1&=80MW\\ P_2&=20MW\\ P_3&=20MW\\ P_L&=K[80^2+3 \times 20^2+3 \times 20^2]\\ &=8800KMW \end{aligned}
\begin{aligned} &\text{Option (B)}\\ P_1&=60MW\\ P_2&=30MW\\ P_3&=30MW\\ P_L&=K[60^2+3 \times 30^2+3 \times 30^2]\\ &=9000KMW \end{aligned}
\begin{aligned} &\text{Option (C)}\\ P_1&=40MW\\ P_2&=40MW\\ P_3&=40MW\\ P_L&=K[40^2+3 \times 40^2+3 \times 40^2]\\ &=11200KMW \end{aligned}
\begin{aligned} &\text{Option (D)}\\ P_1&=30MW\\ P_2&=45MW\\ P_3&=45MW\\ P_L&=K[30^2+3 \times 45^2+3 \times 45^2]\\ &=13050 KMW \end{aligned}
So, option (A) gives minimum losses.
 Question 7
Three generators are feeding a load of 100 MW. The details of the generators are

In the event of increased load power demand, which of the following will happen ?
 A All the generator will share equal power B Generator-3 will share more power compared to Generator-1 C Generator-1 will share more power compared to Generator-2 D Generator-2 will share more power compared to Generator-3
GATE EE 2009   Power Systems
Question 7 Explanation:
Let $x_1,x_2 \text{ and }x_3$ are reactance of generator-1, generator-2 and generator-3 respectively.
Neglecting armature resistance of all the three generators
\begin{aligned} VR_1&=0.02p.u. \\ VR_2&=0.04p.u \\ VR_3&=0.03p.u. \\ VR_1 & \lt VR_3 \lt VR_2 \end{aligned}
Since, voltage regulation (VR) $\propto$ reactance of generator .
$X_1 \lt X_3 \lt X_2$
Power shared by a generator $\propto \frac{1}{\text{reactance of generator}}$
Power shared by Generator 1 $\gt$ Generator 2 $\gt$ Generator 3
 Question 8
A loss less power system has to serve a load of 250 MW. There are tow generation (G1 and G2) in the system with cost curves $C_1 \; and \; C_2$ respectively defined as follows:

$C_{1}(P_{G1})=P_{G1}+0.055\times P^{2}_{G1}$
$C_{2}(P_{G1})=3 P_{G2}+0.03 \times P^{2}_{G2}$

where $P_{G1} \; and \; P_{G2}$ are the MW injections from generator G1 and G2 respectively. Thus, the minimum cost dispatch will be
 A $P_{G1}=250MW;P_{G2}=0MW$ B $P_{G1}=150MW;P_{G2}=100MW$ C $P_{G1}=100MW;P_{G2}=150MW$ D $P_{G1}=0MW;P_{G2}=250MW$
GATE EE 2008   Power Systems
Question 8 Explanation:
\begin{aligned} C_1(P_{G1})&=P_{G1}+0.055P_{G1}^2\\ \frac{dC_1}{P_{G1}}&=1+0.11P_{G1}\\ C_2(P_{G2})&=P_{G2}+0.03P_{G2}^2\\ \frac{dC_2}{P_{G2}}&=3+0.06P_{G2} \end{aligned}
For minimum cost analysis
\begin{aligned} \frac{dC_1}{P_{G1}}&=\frac{dC_2}{P_{G2}}\\ 1+0.11P_{G1}&=3+0.06P_{G2}\;\;...(i)\\ P_{G1}+P_{G2}&=250MW\;\;...(ii) \end{aligned}
Solving equation (i) and (ii),
we get, $P_{G1}=100MW$ and $P_{G2}=150MW$
 Question 9
The incremental cost curves in Rs/MWhr for two generators supplying a common load of 700 MW are shown in the figures. The maximum and minimum generation limits are also indicated. The optimum generation schedule is :
 A Generator A : 400 MW, Generator B : 300 MW B Generator A : 350 MW, Generator B : 350 MW C Generator A : 450 MW, Generator B : 250 MW D Generator A : 425 MW, Generator B : 275 MW
GATE EE 2007   Power Systems
Question 9 Explanation:
Maximum incremental cost in Rs/Mwhr for generator A =600 (at 450 MW)
Minimum incremental cost in Rs/Mwhr for generator B =650 (at 150 MW)
As maximum value of incremental cost of A is less than minimum value of B.
Therefore, generator 'A' will operate at its maximum (o/p) 450 MW and B at (700-450)=250 MW.
 Question 10
A load centre is at an equidistant from the two thermal generating stations $G_{1}$ and $G_{2}$ as shown in the figure. The fuel cost characteristic of the generating stations are given by

$F_{1}=a+bP_{1}+cP_{1}^{2} Rs/hour$
$F_{2}=a+bP_{2}+2cP_{2}^{2} Rs/hour$

Where $P_1 \; and \; P_2$ are the generation in MW of $G_{1}$ and $G_{2}$, respectively. For most economic generation to meet 300 MW of load $P_1 \; and \; P_2$ respectively, are
 A 150, 150 B 100, 200 C 200, 100 D 175, 125
GATE EE 2005   Power Systems
Question 10 Explanation:
Fuel-cost curve of generating station 1
$F_1=a+bP_1+cP_1^2 \; \text{Rs/hour}$
Fuel-cost curve of generating station 2
$F_2=a+bP_2+2cP_2^2 \; \text{Rs/hour}$
The slope of the fuel-cost curve i.e. $\frac{dF_i}{dP_i}$ is called the incremental fuel-cost (IC) and is expressed in Rs/MWh.
\begin{aligned} IC_1&=\frac{dF_1}{dP_1}=b+2cP_1 \\ IC_2&=\frac{dF_2}{dP_2}=b42cP_2 \end{aligned}
For economic generation
\begin{aligned} IC_1&=IC_2\\ b+2cP_1&=b+4cP_2\\ P_1&=2P_2\;\;...(i) \\ &\text{To meet 300 MW of load}\\ P_1+P_2&=300\;\;...(ii)\\ &\text{Solving eq. (i) and (ii)}\\ P_1&=200MW\\ P_2&=100MW \end{aligned}
There are 10 questions to complete.