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}
Question 6
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 6 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)
For optimum load sharing
\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 7
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 7 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 8
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 8 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 9
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 9 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 10
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 10 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
There are 10 questions to complete.