initial commit / add all books

17 files changed, 389 insertions, 0 deletions

diff --git a/3733/CH2/EX2.1/Ex2_1.sce b/3733/CH2/EX2.1/Ex2_1.sce
new file mode 100644
index 000000000..efd2db49e
--- /dev/null
+++ b/3733/CH2/EX2.1/Ex2_1.sce
@@ -0,0 +1,13 @@
+// Example 2_1

+clc;funcprot(0);

+//Given data

+R=6.2;//Rainfall in  cm

+A=2346;// Area in km^2

+

+//Calculation

+Tr=A*10^6*(R/100);// Total rainfall in m^2

+V=(A*R*10^4)/86400;// Rainfall in day-sec-metre

+R_k=(A*R*10^4)/10^6;// Rainfall in km^2-m

+printf('\n Total rainfall=%0.4e m^3 \nVolume of rainfall=%0.0f day-sec-metre \nRainfall in km^2-m=%0.2f km^2-m',Tr,V,R_k);

+// The answer provided in the textbook is wrong

+

diff --git a/3733/CH2/EX2.10/Ex2_10.sce b/3733/CH2/EX2.10/Ex2_10.sce
new file mode 100644
index 000000000..22c85f9c9
--- /dev/null
+++ b/3733/CH2/EX2.10/Ex2_10.sce
@@ -0,0 +1,14 @@
+// Example 2_10

+clc;funcprot(0);

+//Given data

+H=40;// Head in m

+A=1.8;// Area of the reservoir in km^2

+P=24;// MW

+n_o=80/100;// The over all efficiency

+rho_w=1000;// kg/m^3

+g=9.81;// m/s^2

+

+//Calculation

+q=(P*1000*1000)/(rho_w*g*H*n_o);// m^3/sec

+x=(q*3600)/(A*10^6);// m/hr

+printf('\nThe rate of fall in height of reservoir=%0.3f m/hr',x);

diff --git a/3733/CH2/EX2.11/Ex2_11.sce b/3733/CH2/EX2.11/Ex2_11.sce
new file mode 100644
index 000000000..8eb2cd95d
--- /dev/null
+++ b/3733/CH2/EX2.11/Ex2_11.sce
@@ -0,0 +1,15 @@
+// Example 2_11

+clc;funcprot(0);

+//Given data

+V=6*10^6;// m^3

+H=75;// m

+F_l=0.6;// Load factor

+n_g=72/100;// The over all generation efficiency

+rho_w=1000;// kg/m^3

+g=9.81;// m/s^2

+

+//Calculation

+P=((V)/(365*24*3600))*(((rho_w)*g*H*n_g)/(1000));// The power capacity of the plant in kW

+E=P*F_l*365*24;// Energy produced in kWh

+printf('\nEnergy produced=%0.0f kW',E);

+// The answer provided in the textbook is wrong

diff --git a/3733/CH2/EX2.12/Ex2_12.sce b/3733/CH2/EX2.12/Ex2_12.sce
new file mode 100644
index 000000000..fd5de8ea3
--- /dev/null
+++ b/3733/CH2/EX2.12/Ex2_12.sce
@@ -0,0 +1,26 @@
+// Example 2_12

+clc;funcprot(0);

+//Given data

+H=30;// m

+A=250;// sq.km

+Ar=125;// Annual rainfall in cm

+Tr=70/100;// Total rainfall

+F_l=50/100;// Load factor

+h_l=8/100;// Head loss

+n_m=90/100;// Mechanical efficiency of the turbine

+n_g=95/100;// Generator efficiency

+rho_w=1000;// kg/m^3

+g=9.81;// m/s^2

+

+//Calculation

+V=A*10^6*(Ar/100)*Tr;//Water available during the year in m^3

+Q=(V)/(8760*3600);// Water flow per second in m^3/sec

+Q=Q*1000;// kg/sec

+n_h=(1-h_l);// Hydraulic efficiency

+n_o=n_h*n_m*n_g;//The over all efficiency

+P=(Q*9.81*H*n_o)/(1000);// kW

+//With 50% load factor

+Gc=P/F_l;// Generator capacity in kW

+printf('\nThe power=%0.0f kW \nGenerator capacity=%0.1f kW',P,Gc);

+// The answer provided in the textbook is wrong

+

diff --git a/3733/CH2/EX2.13/Ex2_13.sce b/3733/CH2/EX2.13/Ex2_13.sce
new file mode 100644
index 000000000..4300f4f70
--- /dev/null
+++ b/3733/CH2/EX2.13/Ex2_13.sce
@@ -0,0 +1,39 @@
+// Example 2_13

+clc;funcprot(0);

+//Given data

+m=[1 2 3 4 5 6 7 8 9 10 11 12];// Month

+D=[500 200 1500 2500 3000 2400 2000 1500 1500 1000 800 600];// Discharge in millions of m^3 per month

+H=80;// Available head in m

+n_o=80/100;// Overall efficiency of the generation 

+g=9.81;// The acceleration due to gravity in m/s^2

+

+// Calculation

+// (a)

+Q_a1=(D(1)+D(2)+D(3)+D(4)+D(5)+D(6)+D(7)+D(8)+D(9)+D(10)+D(11)+D(12))/12;// The average monthly flow in millions of m^3/month

+m_1=[0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12];

+D_1=[500 500 200 200 1500 1500 2500 2500 3000 3000 2400 2400 2000 2000 1500 1500 1500 1500 1000 1000 800 800 600 600 3200];

+Q_a=[Q_a1,Q_a1];

+m=[0,12];

+xlabel('Month');

+ylabel('Discharge in millions of m^3 per month');

+subplot(2,1,1);

+plot(m_1',D_1','b',m',Q_a','r-');

+a=gca();

+a.x_ticks.labels=["0","J","F","M","A","M","J","J","A","S","O","N","D"];

+a.x_ticks.locations=[0;1;2;3;4;5;6;7;8;9;10;11;12];

+legend('Hydrograph','Mean flow');

+D=[200 500 600 800 1000 1500 2000 2400 2500 3000];

+M=[12 11 10 9 8 7 4 3 2 1];// Total number of months during which flow is available

+for(i=1:10)

+    T(i)=(M(i)/12)*100;

+end

+subplot(2,1,2);

+xlabel('Percentage of time');

+ylabel('Discharge in millions of cu.m.month');

+plot(T,D','b');

+legend('Flow duration curve');

+

+m=(Q_a1*10^6/(30*24*3600));// The average flow available in m^3/sec

+P=(((m*1000*g*H)/1000)*(n_o/1000));// Average kW available in MW

+printf('\nAverage kW available at the site=%0.3f MW',P);

+// The answer provided in the textbook is wrong

diff --git a/3733/CH2/EX2.14/Ex2_14.sce b/3733/CH2/EX2.14/Ex2_14.sce
new file mode 100644
index 000000000..2b837c8b3
--- /dev/null
+++ b/3733/CH2/EX2.14/Ex2_14.sce
@@ -0,0 +1,38 @@
+// Example 2_14

+clc;funcprot(0);

+//Given data

+m=[1 2 3 4 5 6 7 8 9 10 11 12];// Month

+D=[80 50 40 20 0 100 150 200 250 120 100 80];// Discharge in millions of m^3 per month

+H=100;// Available head in m

+n_o=80/100;// Overall efficiency of the generation 

+g=9.81;// The acceleration due to gravity in m/s^2

+

+// Calculation

+// (a)

+Q_a1=(D(1)+D(2)+D(3)+D(4)+D(5)+D(6)+D(7)+D(8)+D(9)+D(10)+D(11)+D(12))/12;// The average monthly flow in millions of m^3/month

+m_1=[0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12];// Month for hydrograph

+D_1=[80 80 50 50 40 40 20 20 0 0 100 100 150 150 200 200 250 250 120 120 100 100 80 80 260];// Discharge in millions of m^3 per month

+Q_a=[Q_a1,Q_a1];// Mean flow

+m=[0,12];// month

+xlabel('Month');

+ylabel('Discharge in millions of m^3 per month');

+subplot(2,1,1);

+plot(m_1',D_1','b',m',Q_a','r-');

+a=gca();

+a.x_ticks.labels=["0","J","F","M","A","M","J","J","A","S","O","N","D"];

+a.x_ticks.locations=[0;1;2;3;4;5;6;7;8;9;10;11;12];

+legend('Hydrograph','Mean flow');

+D=[0 20 40 50 80 100 120 150 200 220];// Discharge in millions of m^3 per month

+M=[12 11 10 9 8 7 4 3 2 1];// Total number of months during which flow is available

+for(i=1:10)

+    T(i)=(M(i)/12)*100;

+end

+subplot(2,1,2);

+xlabel('Percentage of time');

+ylabel('Discharge in millions of cu.m.month');

+plot(T,D','b');

+legend('Flow duration curve');

+m=((Q_a1*10^6)/(30*24*3600));// The average flow available in m^3/sec

+P=(((Q_a1*10^6*1000*g*H)/(30*24*3600*1000))*(n_o/1000));// Average kW available in MW

+printf('\nAverage kW available at the site=%0.3f MW',P);

+// The answer provided in the textbook is wrong

diff --git a/3733/CH2/EX2.15/Ex2_15.sce b/3733/CH2/EX2.15/Ex2_15.sce
new file mode 100644
index 000000000..7a9c67028
--- /dev/null
+++ b/3733/CH2/EX2.15/Ex2_15.sce
@@ -0,0 +1,34 @@
+// Example 2_15

+clc;funcprot(0);

+//Given data

+w=[1 2 3 4 5 6 7 8 9 10 11 12];// Week

+b=[6000 4000 5400 2000 1500 1000 1200 4500 8000 4000 3000 2000];// Weekly flow in m^3/sec

+

+//Calculation

+for(i=1:12)

+    c(i)=b(i)*7;

+end

+Cv(1)=c(1);// day-sec-metres

+Cv(2)=Cv(1)+c(2);// day-sec-metres

+Cv(3)=Cv(2)+c(3);// day-sec-metres

+Cv(4)=Cv(3)+c(4);// day-sec-metres

+Cv(5)=Cv(4)+c(5);// day-sec-metres

+Cv(6)=Cv(5)+c(6);// day-sec-metres

+Cv(7)=Cv(6)+c(7);// day-sec-metres

+Cv(8)=Cv(7)+c(8);// day-sec-metres

+Cv(9)=Cv(8)+c(9);// day-sec-metres

+Cv(10)=Cv(9)+c(10);// day-sec-metres

+Cv(11)=Cv(10)+c(11);// day-sec-metres

+Cv(12)=Cv(11)+c(12);// day-sec-metres

+w=[0 1 2 3 4 5 6 7 8 9 10 11 12];// Week for plot

+CV=[0 Cv(1) Cv(2) Cv(3) Cv(4) Cv(5) Cv(6) Cv(7) Cv(8) Cv(9) Cv(10) Cv(11) Cv(12)];// Cumulative volume in day-sec-metres for plot

+ylabel('Flow in thousands & day-sec-meter');

+plot(w,CV/1000)

+// The total flow in the week,Q=7*day-sec-metres.

+// From fig.prob.2.15

+C=42*10^3;// The capacity of the reservoir in day-sec-metre

+bc=5.7*20*10^3;// day-sec-metre

+ac=5.5;// day

+Q=bc/(ac*7);// Flow rate available in m^3/sec

+printf('\n The capacity of the reservoir=%0.1e day-sec-metre \nFlow rate available=%0.0f m^3/sec',C,Q);

+// The answer vary due to round off error

diff --git a/3733/CH2/EX2.16/Ex2_16.sce b/3733/CH2/EX2.16/Ex2_16.sce
new file mode 100644
index 000000000..cc4f17291
--- /dev/null
+++ b/3733/CH2/EX2.16/Ex2_16.sce
@@ -0,0 +1,33 @@
+// Example 2_16

+clc;funcprot(0);

+//Given data

+m=[1 2 3 4 5 6 7 8 9 10 11 12];// Month

+F=[100 50 20 80 10 10 190 40 30 200 170 80];// Flow in millions of cu-m-per month

+

+// Calculation

+Cv(1)=F(1);

+Cv(2)=Cv(1)+F(2);// Millions of cu-m

+Cv(3)=Cv(2)+F(3);// Millions of cu-m

+Cv(4)=Cv(3)+F(4);// Millions of cu-m

+Cv(5)=Cv(4)+F(5);// Millions of cu-m

+Cv(6)=Cv(5)+F(6);// Millions of cu-m

+Cv(7)=Cv(6)+F(7);// Millions of cu-m

+Cv(8)=Cv(7)+F(8);// Millions of cu-m

+Cv(9)=Cv(8)+F(9);// Millions of cu-m

+Cv(10)=Cv(9)+F(10);// Millions of cu-m

+Cv(11)=Cv(10)+F(11);// Millions of cu-m

+Cv(12)=Cv(11)+F(12);// Millions of cu-m

+m=[0 1 2 3 4 5 6 7 8 9 10 11 12];// Month

+CV=[0 Cv(1) Cv(2) Cv(3) Cv(4) Cv(5) Cv(6) Cv(7) Cv(8) Cv(9) Cv(10) Cv(11) Cv(12)];// Cumulative volume in millions of cu-m

+xlabel('Month');

+ylabel('Millions of cu.m')

+plot(m,CV,'b');

+// From Fig.Prob(2.16),from the mass curve

+Sc=80*10^6;// Storage capacity in m^3

+sc=85*10^6;// Spill way capacity required in m^3

+i=13;

+j=1;

+Q=((CV(i)-CV(j))/(m(i)-m(j)))*10^6;// The uniform discharge in m^3/month

+// The required storage capacity for the uniform supply Q,

+SC_u=233*10^6;// cu-m.

+printf('\nThe required reservoir capacity=%0.0e m^3 \nSpill way capacity=%0.1e m^3 \nAverage flow capacity=%0.2e m^3/month \nRequired capacity of the reservoir fo the uniform supply=%0.2e cu-m',Sc,sc,Q,SC_u);

diff --git a/3733/CH2/EX2.17/Ex2_17.sce b/3733/CH2/EX2.17/Ex2_17.sce
new file mode 100644
index 000000000..8f686b2f9
--- /dev/null
+++ b/3733/CH2/EX2.17/Ex2_17.sce
@@ -0,0 +1,28 @@
+// Example 2_17

+clc;funcprot(0);

+//Given data

+m=[1 2 3 4 5 6 7 8 9 10];// Month

+D=[200 100 20 20 260 180 40 280 60 120];// Discharge in millions of cu-m-per month

+Q=100;// millions of cu.m

+

+// Calculation

+Cv(1)=D(1);

+Cv(2)=Cv(1)+D(2);// Millions of cu-m

+Cv(3)=Cv(2)+D(3);// Millions of cu-m

+Cv(4)=Cv(3)+D(4);// Millions of cu-m

+Cv(5)=Cv(4)+D(5);// Millions of cu-m

+Cv(6)=Cv(5)+D(6);// Millions of cu-m

+Cv(7)=Cv(6)+D(7);// Millions of cu-m

+Cv(8)=Cv(7)+D(8);// Millions of cu-m

+Cv(9)=Cv(8)+D(9);// Millions of cu-m

+Cv(10)=Cv(9)+D(10);// Millions of cu-m

+m=[0 1 2 3 4 5 6 7 8 9 10];// Month

+CV=[0 Cv(1) Cv(2) Cv(3) Cv(4) Cv(5) Cv(6) Cv(7) Cv(8) Cv(9) Cv(10)];// Cumulative volume in millions of cu-m

+xlabel('Discharge in millions of cu-m  month');

+ylabel('Millions of cu.m');

+plot(m,CV);

+// From the mass curve

+Q_a=72.6;// Flow rate at point a in millions of cu-m/month

+Q_b=166.4;// Flow rate at point b in millions of cu-m/month

+Q_c=137.6;// Flow rate at point c in millions of cu-m/month

+printf('\nThe maximum flow available=%0.1f millions of cu-m/month \nThe minimum flow available=%0.1f millions of cu-m/month',Q_b,Q_a);

diff --git a/3733/CH2/EX2.2/Ex2_2.sce b/3733/CH2/EX2.2/Ex2_2.sce
new file mode 100644
index 000000000..8acf8118b
--- /dev/null
+++ b/3733/CH2/EX2.2/Ex2_2.sce
@@ -0,0 +1,10 @@
+//Example 2_2

+clc;funcprot(0);

+// Given data

+Pdr=400*10^6; // Per day requirement in L

+Pdr=Pdr/10^3;// convert L to m^3

+Aw=30000*10^6;// Available water in the dam in m^3

+

+//Calculation

+n=(Aw)/(Pdr);// days

+printf('No.of days water supplied,N=%0.0f days\n',n);

diff --git a/3733/CH2/EX2.3/Ex2_3.sce b/3733/CH2/EX2.3/Ex2_3.sce
new file mode 100644
index 000000000..f91a64469
--- /dev/null
+++ b/3733/CH2/EX2.3/Ex2_3.sce
@@ -0,0 +1,15 @@
+// Example 2_3

+clc;funcprot(0);

+//Given data

+D=[1 2 3 4 5 6 7];// Days

+F=[100 320 210 120 50 30 25];//Mean daily flow in m^3/sec

+

+//Calculation

+Tf=F(1)+F(2)+F(3)+F(4)+F(5)+F(6)+F(7);

+Tfv=24*3600*(Tf);// Total flow volume in m^3

+Tfv_1=Tfv/(10^6);// million-m^3

+Tfv_2=Tfv/86400;// day-sec-metre

+Tfv_3=Tfv/(3350*10^4);// cm

+Tfv_4=Tfv_1;// km^2-m as 1 km^2-m =1 million of cu-m.

+printf('\nTotal flow volume=%0.1f million-m^3 \nTotal flow volume =%0.1f day-sec-metre \nTotal flow volume=%0.1f cm \nTotal flow volum=%0.1f km^2-m',Tfv_1,Tfv_2,Tfv_3,Tfv_4);

+// The answer provided in the textbook is wrong

diff --git a/3733/CH2/EX2.4/Ex2_4.sce b/3733/CH2/EX2.4/Ex2_4.sce
new file mode 100644
index 000000000..64726eaa2
--- /dev/null
+++ b/3733/CH2/EX2.4/Ex2_4.sce
@@ -0,0 +1,22 @@
+// Example 2_4

+clc;funcprot(0);

+//Given data

+m_1=20;// The steam discharge during the monsoon season of four months in m^3/sec

+m_2=2.5;// The steam discharge during the remaining year in m^3/sec

+h_l=3;//The head loss in the pipe in %

+n_o=90;//Over all efficiency of the generation in %

+Tn=365;// Total number of days in a year

+H=80;// metres

+g=9.81;// m/s^2

+//Calculation

+N_m=30+31+31+30;//The number of days during which the discharge of 20 m^3/sec is available

+N_r=Tn-N_m;//The number of days during which the discharge of 2.5 m^3/sec is available

+Tf=(m_1*3600*24*N_m)+(m_2*3600*24*N_r);// Total flow during the year in m^3

+m_avg=(Tf)/(3600*24*Tn);//Average discharge in m^3/sec

+gradm=m_1-m_avg;// The difference between the maximum and average discharge in  m^3/sec

+Rc=(gradm*3600*24*N_m)/86400;// Reservoir capacity to store the excess water in day -sec-metre

+H_net=H*(1-(h_l/100));// metres

+P_avg=(m_avg*1000*g*H_net*(n_o/100))/(1000);//Average kW generated in kW

+P_avg=P_avg/1000;// MW

+printf('\nReservoir capacity to store the excess water=%0.0f day-sec-metre \nAverage kW generated=%0.2f MW',Rc,P_avg);

+// The answer provided in the textbook is wrong

diff --git a/3733/CH2/EX2.5/Ex2_5.sce b/3733/CH2/EX2.5/Ex2_5.sce
new file mode 100644
index 000000000..a84b14a19
--- /dev/null
+++ b/3733/CH2/EX2.5/Ex2_5.sce
@@ -0,0 +1,22 @@
+// Example 2_5

+clc;funcprot(0);

+//Given data

+A=2260;// The catchment area in km^2

+AAR=154;// The average annual rainfall in cm

+H=120;// The head drop in m

+n_t=85;// Turbine efficiency in %

+n_g=90;// Generation efficiency in %

+F_l=1;// Load factor

+N=240;// The speed of the runner in rpm

+PEL=20;// Percoalation and evaporation losses in %

+g=9.81;// m/s^2

+

+//Calculation

+V=A*10^6*(AAR/100)*(1-((PEL/100)));// The quantity of water available for power generation per year in cu.m

+Q=V/(365*24*3600);// Quantity of water available per second in m^3/sec

+m=Q*1000;// Discharge in kg/sec

+P=((m*g*H)/1000)*(n_t/100)*(n_g/100);// Power developed in kW

+P=P/1000;// MW

+N_a=(N*sqrt(P))/(H)^(5/4);

+printf('\nPower developed,P=%0.2f MW \nSingle pelton wheel with 4 jets can be used.',P)

+//The answer seems different due to calculation error occur in the book

diff --git a/3733/CH2/EX2.6/Ex2_6.sce b/3733/CH2/EX2.6/Ex2_6.sce
new file mode 100644
index 000000000..ec9d719a5
--- /dev/null
+++ b/3733/CH2/EX2.6/Ex2_6.sce
@@ -0,0 +1,20 @@
+// Example 2_6

+clc;funcprot(0);

+//Given data

+A=200;// The catchment area in km^2

+AAR=100;// The average annual rainfall in cm

+Tro=80;//Total run off in%

+H=80;// the mean head available in m

+n_g=75;// Over all efficiency of generation in %

+Apw=16;// The average period of working in hours

+g=9.81;// m/s^2

+F_l=1;// Load factor

+

+//Calculation

+V=A*10^6*(Tro/100);//Total water available in m^3/year

+Q=V/(365*24*3600);// m^3/sec

+m=Q*1000;// Discharge in kg/sec

+P=((m*g*H)/1000)*(n_g/100);// Capacity of the plant in kW

+E=(P/1000)*Apw*365*10^3;//Energy generated per year in kWh

+printf('\nThe energy generated per year =%0.3e kWh',E );

+// The answer vary due to round off error

diff --git a/3733/CH2/EX2.7/Ex2_7.sce b/3733/CH2/EX2.7/Ex2_7.sce
new file mode 100644
index 000000000..5998d6e9f
--- /dev/null
+++ b/3733/CH2/EX2.7/Ex2_7.sce
@@ -0,0 +1,20 @@
+// Example 2_7

+clc;funcprot(0);

+// Given data

+A=1200;// The catchment area in km^2

+AR=160;// The annual rainfall in cm

+H=360;// The head available in m

+n_o=75;// Over all efficiency of the plant in %

+F_l=0.5;// Load factor

+PEL=25;// Percoalation and evaporation losses in %

+g=9.81;// m/s^2

+

+// Calculation

+V=A*10^6*(AR/100)*(1-((PEL/100)));// The quantity of water available for power generation per year in cu.m

+Q=V/(365*24*3600);// Average flow per second in m^3/sec

+m=Q*1000;// Discharge in kg/sec

+P_avg=((m*g*H)/1000)*(n_o/100);// Average power developed in kW

+P_avg=P_avg/1000;// MW

+MD=(P_avg/F_l);// Maximum demand in MW

+printf('\nThe average power developed=%0.2f MW \nMaximum demand=%0.1f MW',P_avg,MD);

+// The answer vary due to round off error

diff --git a/3733/CH2/EX2.8/Ex2_8.sce b/3733/CH2/EX2.8/Ex2_8.sce
new file mode 100644
index 000000000..f6bf15b38
--- /dev/null
+++ b/3733/CH2/EX2.8/Ex2_8.sce
@@ -0,0 +1,21 @@
+// Example 2_8

+clc;funcprot(0);

+//Given data

+A=50;// Area in sq.km

+H_1=100;// Head in m

+E=13.5*10^6;// The energy utilised by the customer in kWh

+n_g=0.75;// The over all generation efficiency

+rho_w=1000;// kg/m^3

+g=9.81;// m/s^2

+

+//Calculation

+// V=A*H;// Water used during 5 hours in m^3

+// Q=(A*H)/(5*3600);(discharge/sec)

+function[X]=head(y)

+    X(1)=E-(5*(rho_w*((A*10^6*y(1))/(5*3600))*g*(H_1/1000)*n_g));

+endfunction

+y=[10];

+z=fsolve(y,head);

+H=z(1);// metres

+printf('\nThe fall in the height of water in the reservoir=%0.2f metres',H);

+// The answer provided in the textbook is wrong

diff --git a/3733/CH2/EX2.9/Ex2_9.sce b/3733/CH2/EX2.9/Ex2_9.sce
new file mode 100644
index 000000000..a2a012ff7
--- /dev/null
+++ b/3733/CH2/EX2.9/Ex2_9.sce
@@ -0,0 +1,19 @@
+//Example 2_9

+clc;funcprot(0);

+// Given values

+A=250*10^6;// Catchment area in m^2

+Ar=1.25;// Annual rainfall in m

+H=60;// Average head in m

+P_w=70;// Percentage of water in the dam

+n_t=0.9// Turbine efficiency

+n_g=0.95// Generator efficiency

+g=9.81;// The acceleration due to gravity in m/s^2

+

+//Calculation

+V=(A*Ar*(P_w/100));// Total water used for power generation in m^3

+printf('Total water used for power generation=%0.3e m^3\n',V);

+q=(V/(365*24*3600));

+printf('Water flow rate =%0.2f m^3/sec\n',q);

+P=((q*1000*9.81*60)/1000)*n_t*n_g*(1/1000);

+printf('The capacity of the power plant,P=%0.1f MW\n',P);

+// The answer vary due to round off error