initial commit / add all books

15 files changed, 498 insertions, 0 deletions

diff --git a/3751/CH7/EX7.1/Ex7_1.sce b/3751/CH7/EX7.1/Ex7_1.sce
new file mode 100644
index 000000000..cc766d7c3
--- /dev/null
+++ b/3751/CH7/EX7.1/Ex7_1.sce
@@ -0,0 +1,27 @@
+//Fluid Systems by Shiv Kumar

+//Chapter 7 - Performance of water turbine

+//Example 7.1

+//To Calculate specific speed of turbine and state the type of turbine

+   clc

+   clear

+//Given

+   P=8000;       //Power developed, KW

+   H=30;       //Head, m

+   N=140;        //Speed, rpm

+//Computations

+       Ns=N*P^(1/2)/(H^(5/4));       // specific speed of turbine , in SI UNITS

+

+//Results

+      printf("The Specific speed of Turbine is %.2f (SI Units)\n",Ns)

+   //To Determine the type of turbine

+       if Ns>51 & Ns<=255 then

+              printf(" The type of turbine is Francis") 

+     elseif  Ns>=8.5 & Ns<=30 then

+            printf("The type of turbine is Pelton Wheel with single jet") 

+     elseif Ns>30& Ns<=51 then

+            printf("The type of turbine is Pelton Wheel with multi jet")

+     elseif Ns>255 & Ns<=860 then

+             printf("The type of turbine is Kaplan or Propeller Turbine")

+  end

+

+

diff --git a/3751/CH7/EX7.10/Ex7_10.sce b/3751/CH7/EX7.10/Ex7_10.sce
new file mode 100644
index 000000000..283f7c48a
--- /dev/null
+++ b/3751/CH7/EX7.10/Ex7_10.sce
@@ -0,0 +1,29 @@
+//Fluid Systems- By Shiv Kumar
+//Chapter 7- Performance of Water Turbine
+//Example 7.10
+// To Calculate Speed and Power Developed by the Prototype when working Under a Head of 8 m.
+   clc
+   clear
+
+//Given:-
+      Lr=1/5;        //Scale Ratio
+      DmbyDp=Lr;
+
+     //For Prototype
+            Hp=8;     //Head, m
+    
+      //For Model
+          Pm=5;        //Power, kW
+          Hm=2;       //Head, m
+          Nm=600;   //rpm
+
+//Computations
+    Np=Nm*DmbyDp*(Hp/Hm)^(1/2);        //rpm
+    Pp=Pm*(Np/Nm)^3/(DmbyDp^5);        //KW
+
+
+//Results
+    printf("For the Prototype (Working Under a Head of 8 m:\n")
+    printf("      Speed, Np=%.f rpm\n      Power Developed, Pp=%.f kW",Np,Pp)
+
+
diff --git a/3751/CH7/EX7.11/Ex7_11.sce b/3751/CH7/EX7.11/Ex7_11.sce
new file mode 100644
index 000000000..4c0f85df2
--- /dev/null
+++ b/3751/CH7/EX7.11/Ex7_11.sce
@@ -0,0 +1,33 @@
+//Fluid Systems- By Shiv Kumar
+//Chapter 7- Performance of Water Turbine
+//Example 7.11
+// To Find  (a)Power Developed by Model   (b)Ratio of Heads and Ratio of Mass Flow Rates between Prototype and Model.
+   clc
+   clear
+
+//Given:-
+      Pp=12;       //Power Developed by Prototype,MW
+      Lr=1/10;        //Scale Ratio
+      DmbyDp=Lr;
+       LmbyLp=Lr;
+
+
+//Computations:-
+
+     //(a)Power Developed by the Model
+                   //As Np=Nm and effeciencies of prototype and model are equal
+          Pm=Pp*10^6*(DmbyDp)^5;      //W
+          
+     //(b)Ratio of Heads and Ratio of Mass flow Rates
+         HpbyHm=DmbyDp^(-2);            //Dimensionless
+         QpbyQm=DmbyDp^(-3)
+        //As m=rho*Q   and  rho is Constant. So,
+         m_pbym_m=QpbyQm;
+
+//Results
+
+    printf("(a)Power Developed by the Model,Pm=%.f W\n",Pm)
+    printf(" (b)Ratio of Heads, Hp/Hm=%.f\n    Ratio of Mass flow rates, m_p/m_m=%.f\n",HpbyHm,m_pbym_m)
+
+
+
diff --git a/3751/CH7/EX7.12/Ex7_12.sce b/3751/CH7/EX7.12/Ex7_12.sce
new file mode 100644
index 000000000..1583c0430
--- /dev/null
+++ b/3751/CH7/EX7.12/Ex7_12.sce
@@ -0,0 +1,34 @@
+//Fluid Systems- By Shiv Kumar
+//Chapter 7- Performance of Water Turbine
+//Example 7.12
+// To Find    (a)The Speed ,Discharge and Power required for the Actual Pump    (b) The Discharge of the model.
+   clc
+   clear
+
+//Given:-
+      Lr=5;        //Scale Ratio
+      DpbyDm=Lr;
+       DmbyDp=1/DpbyDm;
+        //For Model
+             Pm=22;         //Power Required, kW
+             Hm=7;        //Head, m
+             Nm=410;     //Speed, rpm
+             eta_m=1;     //Assumption that  efficiency of the model is 100%
+      //For Prototype
+        Hp=35;         //Head, m
+  //Data Required:-
+       rho=1000;     //Density of Water, Kg/m^3
+       g=9.81;       //Acceleration due to gravity, m/s^2
+
+//Computations:-
+       Np=Nm*DmbyDp*(Hp/Hm)^(1/2);     //rpm
+       Pp=Pm*(Np/Nm)^3*DpbyDm^5;    //KW
+       Qm=Pm*1000*eta_m/(rho*g*Hm);    //m^3/s
+       Qp=Qm*(Np/Nm)^2*DpbyDm^2;       //m^3/s
+
+//Results:-
+       printf("(a)For the Actual Pump(Prototype):\n         Speed, Np=%.2f rpm  , \n          Discharge, Qp=%.3f m^3/s  and \n         Power,Pp=%.2fKW\n",Np,Qp,Pp)         //The Answer vary due to Round off Error(For Qp),  The Answer Provided in the Textbook is Wrong (For Np and Pp).
+
+       printf("(b)The Discharge of the Model, Qm=%.4f m^3/s",Qm)        //The Answer vary due to Round off Error
+
+
diff --git a/3751/CH7/EX7.13/Ex7_13.sce b/3751/CH7/EX7.13/Ex7_13.sce
new file mode 100644
index 000000000..648937bdb
--- /dev/null
+++ b/3751/CH7/EX7.13/Ex7_13.sce
@@ -0,0 +1,38 @@
+//Fluid Systems- By Shiv Kumar
+//Chapter 7- Performance of Water Turbine
+//Example 7.13
+// To Determine the maximum flow rate and specific speed for the Turbine and To Find Speed, Power Output and Water Consumption of the Model.
+   clc
+   clear
+
+//Given:-
+      Lr=10;        //Scale Ratio
+      DpbyDm=Lr;
+       DmbyDp=1/DpbyDm;
+        //For Prototype
+             Pp=1000;         //Power , kW
+             Hp=14;            //Head, m
+             Np=130;     //Speed, rpm
+             eta_o=91/100;     //Overall efficiency 
+      //For  Model
+        Hm=6;         //Head, m
+  //Data Required:-
+       rho=1000;     //Density of Water, Kg/m^3
+       g=9.81;       //Acceleration due to gravity, m/s^2
+
+//Computations:-
+        Qp=Pp*1000/(rho*g*Hp *eta_o );    //m^3/s
+        Ns=Np*Pp^(1/2)/(Hp^(5/4));        //Specific Speed, In SI UNITS      
+        Nm=Np*DpbyDm*(Hm/Hp)^(1/2);     //rpm
+        Qm=Qp*(Nm/Np)*(DmbyDp)^3;       //m^3/s
+        Pm=Pp*(Nm/Np)^3*DmbyDp^5;    //KW
+       
+//Results:-
+       printf("For the Turbine : \n\t")
+        printf("Maximum Flow Rate, Qp=%.f m^3/s\n\t",Qp)
+        printf("Specific Speed, Ns=%.2f (SI Units)\n\n",Ns)
+        printf("For the Model : \n\t")
+        printf("Speed, Nm=%.2f rpm\n      Power Output, Pm=%.2f kW\n         Water Consumption, Qm=%.4f m^3/s \n", Nm,Pm,Qm)      //The Answer vary due to Round off Error
+        
+        
+        
diff --git a/3751/CH7/EX7.14/Ex7_14.sce b/3751/CH7/EX7.14/Ex7_14.sce
new file mode 100644
index 000000000..9a4c66d2e
--- /dev/null
+++ b/3751/CH7/EX7.14/Ex7_14.sce
@@ -0,0 +1,36 @@
+//Fluid Systems- By Shiv Kumar
+//Chapter 7- Performance of Water Turbine
+//Example 7.14
+//To Determine the Size (Scale Ratio) of the Model and To Find the Model Speed and  Power.
+
+   clc
+    clear
+
+//Given:-
+     TP=240000;       //Total Power Produced, kW
+      n=4;       //No. of Turbines
+      eta_o=91/100;        //Effeciency of each turbine
+       Np=120;        //Speed of each Turbine, rpm
+       Hp=70;        //Head for each Turbine, m 
+        
+       Qm=0.45;        //Discharge for Model, m^3/s
+       Hm=5;          //Head for testing the Model, m
+
+//Data Required:-
+         rho=1000;     //Density of Water, Kg/m^3
+         g=9.81;        //Acceleratrion due to gravity, m/s^2
+
+//Calculations:-
+       Pp=TP/n;       //Power produced from each Turbine, kW
+       Qp=Pp*1000/(rho*g*Hp*eta_o);      //Discharge passing through one Turbine, m^3/s
+       DmbyDp=(Qm/Qp)^(1/2)*(Hp/Hm)^(1/4);          //From Relation of Flow Coefficient
+        Lr=DmbyDp;       //Scale Ratio
+        Nm=(Np/DmbyDp)*(Hm/Hp)^(1/2);     //rpm
+        Pm=Pp*(Nm/Np)^3*DmbyDp^5;        //KW
+
+//Results
+    printf("The Scale Ratio is  1:%.2f\n ",1/Lr)
+    printf("Model Speed, Nm=%.2f rpm\n",Nm)
+    printf("Model Power, Pm=%.2f kW\n",Pm)       //The Answer vary due to Round off Error
+
+
diff --git a/3751/CH7/EX7.15/Ex7_15.sce b/3751/CH7/EX7.15/Ex7_15.sce
new file mode 100644
index 000000000..b508479a7
--- /dev/null
+++ b/3751/CH7/EX7.15/Ex7_15.sce
@@ -0,0 +1,34 @@
+//Fluid Systems- By Shiv Kumar
+//Chapter 7- Performance of Water Turbine
+//Example 7.15
+//To Determine the rpm of Prototype, Ratio of Power Developed by Model and Prototype and Efficiency of Prototytpe .
+
+   clc
+    clear
+
+//Given:-
+        Lr=1/8;         //Scale Ratio
+     //For Model,
+         Hm=5;      //Head, m
+         Nm=350;      //Speed,rpm
+          eta_m=78/100;       //Effiency of model
+      //For Prototype,
+            Hp=100;     //Head, m
+
+//Calculations:-
+      DpbyDm=1/Lr;    //  Dp/Dm
+      //(a) Speed of Prototype, Np
+               Np=Nm*Lr*(Hp/Hm)^(1/2);     //rpm
+      //(b)Ratio of Power Developed,  Pp/Pm
+               PpbyPm=DpbyDm^5*(Np/Nm)^3;     
+      //(c)Efficiency of Prototype when Scale Effect is Considered
+            //From Moody's Equation,
+                eta_p=(1-Lr^0.2*(1-eta_m))*100;      //In Percentage
+
+//Results
+
+printf(" (a)The Speed of Prototype, Np=%.2f rpm\n",Np)       //The Answer vary due to Round off Error
+printf(" (b)Ratio of Power Developed,  Pp/Pm =%.2f \n",PpbyPm)       //The Answer vary due to Round off Error
+printf(" (c)Efficiency of Prototype, eta_p =%.2f Percent\n",eta_p)       //The Answer vary due to Round off Error
+
+
diff --git a/3751/CH7/EX7.2/Ex7_2.sce b/3751/CH7/EX7.2/Ex7_2.sce
new file mode 100644
index 000000000..8f75a0621
--- /dev/null
+++ b/3751/CH7/EX7.2/Ex7_2.sce
@@ -0,0 +1,34 @@
+//Fluid Systems by Shiv Kumar

+//Chapter 7 - Performance of water turbine

+//Example 7.2

+//To Find (a) Specific speed of turbine (b) Power Developed   (c) Type of turbine

+    clc

+     clear

+//Given:

+   

+   H=28;       //Head, m

+   N=225;      //Speed, rpm

+   Q=10;       //Discharge, cumec=m^3/s

+   eta_o=90/100;       //Overall Efficiency

+//Data Required

+    rho=1000   // Density of Water, Kg/m^3

+    g=9.81;    // Acceleration due to gravity,  m/ s^2

+//Computations

+   P=eta_o*rho*Q*g*H/1000;     //Power developed, KW

+   Ns=N*P^(1/2)/(H^(5/4));       // specific speed of turbine , in SI UNITS

+

+//Result1

+   

+    printf("(a)The Specific speed of Turbine = %.2f (SI Units)\n",Ns)       //The Answer Vary due to Round off Error

+  printf("(b)Power developed =%.2f kW\n",P)

+//To Determine the type of turbine, Result2

+       if Ns>51 & Ns<=255 then

+              printf("(c)The type of turbine is Francis.") 

+     elseif  Ns>=8.5  &  Ns<=30 then

+            printf("(c)The type of turbine is Pelton Wheel with single jet.") 

+     elseif  Ns>30 & Ns<=51 then

+            printf("(c)The type of turbine is Pelton Wheel with multi jet.")

+     elseif  Ns>255 & Ns<=860 then

+             printf("(c)The type of turbine is Kaplan or Propeller turbine.")

+

+    end

diff --git a/3751/CH7/EX7.3/Ex7_3.sce b/3751/CH7/EX7.3/Ex7_3.sce
new file mode 100644
index 000000000..d044d619e
--- /dev/null
+++ b/3751/CH7/EX7.3/Ex7_3.sce
@@ -0,0 +1,36 @@
+//Fluid Systems by Shiv Kumar

+//Chapter 7 - Performance of water turbine

+//Example 7.3

+//To Find (a) The Discharge required   (b) The Diameter of Wheel  (c) The Diameter and number of jets required    (d)The Specific Speed

+       clc

+       clear

+//Given:

+    P=8200;    //Power Developed, kW

+    H=128;     // Head , m

+    N=210;    // Speed, rpm

+    Cv=0.98;   // Co-efficient of Velocity

+    eta_H=89/100;    //Hydraulic Efficiency

+    Ku=0.45;     //	Speed Ratio

+   dbyD=1/10;   //Ratio of jet diameter to wheel Diameter

+   eta_m=92/100;   //Mechanical Efficiency

+//Data required

+    rho=1000;   //Density of Water, Kg/m^3

+    g=9.81;    // Acceleration due to gravity, m/s^2

+//Assumptions:

+    eta_v=100/100;    // Volumetric efficiency is 100%

+

+//Computations

+    D=Ku*sqrt(2*g*H)*60/(%pi*N);   //Wheel Diameter, m

+    d=D*dbyD;      // Jet diameter, m

+    eta_o=eta_H*eta_m*eta_v;   //Overall Efficiency

+    Q=P*1000/(rho*g*H*eta_o);    //Net Discharge,  m^3/s

+    q=(%pi/4)*d^2*Cv*sqrt(2*g*H);    //Discharge through one jet, m^3/s

+     n=round(Q/q);    //Number of jets

+     Ns= N*P^(1/2)/(H^(5/4));    //Specific Speed, SI Units

+

+//Results

+printf("(a) The Discharge required, Q =%.3f m^3/s\n",Q)

+printf("(b) The Diameter of Wheel, D =%.2f m\n",D)

+printf("(c) The Diameter,  d=%.3f m  and\n     number of jets required =%.f \n",d,n)

+printf("(d)The Specific Speed, Ns=%.2f  (SI Units)\n",Ns)     //The Answer Vary due to Round off Error

+   

diff --git a/3751/CH7/EX7.4/Ex7_4.sce b/3751/CH7/EX7.4/Ex7_4.sce
new file mode 100644
index 000000000..bbfd9ddbc
--- /dev/null
+++ b/3751/CH7/EX7.4/Ex7_4.sce
@@ -0,0 +1,26 @@
+//Fluid Systems by Shiv Kumar

+//Chapter 7 - Performance of water turbine

+//Example 7.4

+//To Find (a) The Diameter of Runner  (b) The Diameter of jet 

+      clc

+       clear

+//Given:

+     P=3200;    //Power Developed, kW

+     H=310;     // Effective Head , m

+     eta_o=82/100;    //Overall Efficiency

+     Ku=0.46;     //	Speed Ratio

+     Cv=0.98   // Co-efficient of Velocity

+     Ns=18;    //Specific Speed (SI Units)

+//Data required

+    rho=1000;   //Density of Water, Kg/m^3

+    g=9.81    // Acceleration due to gravity, m/s^2

+

+//Computations

+     N=Ns*H^(5/4)/sqrt(P);    //Speed, rpm

+     D=Ku*sqrt(2*g*H)*60/(%pi*N);     //Diameter of runner, m

+     Q=P*1000/(rho*g*H*eta_o);     //Discharge, m^3/s

+     d=sqrt(Q/((%pi/4)*Cv*sqrt(2*g*H)));    // Diameter of Jet, m

+

+//Results

+  printf("(a) The Diameter of Runner, D =%.2f m\n",D)       //The Answer Vary due to Round Off Error

+  printf("(c) The Diameter of Jet,  d=%.3f m \n",d)

diff --git a/3751/CH7/EX7.5/Ex7_5.sce b/3751/CH7/EX7.5/Ex7_5.sce
new file mode 100644
index 000000000..f2d86c55b
--- /dev/null
+++ b/3751/CH7/EX7.5/Ex7_5.sce
@@ -0,0 +1,47 @@
+//Fluid Systems by Shiv Kumar

+//Chapter 7 - Performance of water turbine

+//Example 7.5

+//To Find (a) Number of Units to be installed (b) Diameter of Wheel  (c) Diameter of Jet

+    clc

+     clear

+

+//Given:

+

+     H_G= 310;  //Gross Head,m

+     l=2.5;     // Length, km

+     h_f=25;    // Friction Loses, J/N=m

+     TO=20;   //Total Output Power, MW

+     N=660;   // Speed, rpm

+     ubyVi=0.46    //Ratio of bucket to jet speed

+     eta_o=88/100;   //Overall Efficiency

+     Ns=28;     //Specific Speed, SI Units

+     Cv=0.97;   

+     Cd=0.94;   

+     

+     

+//Data Required:

+     rho=1000;      //Density of water, kg/m^3

+     g=9.81;       //Acceleration due to gravity, m/s^2

+      

+

+//Computations:

+   H=H_G-h_f;   //Effective Head, m

+   P=(Ns*H^(5/4)/N)^2;   //Power Output of each Unit, KW

+  //(a) The no. of units to be lnstalled,n

+   n=round(TO*1000/P);   

+  //(b)Diameter of Wheel,D

+    Vi=Cv*sqrt(2*g*H);    //m/s

+    D=ubyVi*Vi*60/(%pi*N);  //m

+

+//(c) Diameter of Jet,  d

+    Q=TO*10^6/(rho*g*H*eta_o);   //Net Discharge, m^3/s

+    q=Q/n;  // Discharge through one unit, m^/s

+    d=sqrt(q/((%pi/4)*Cd*sqrt(2*g*H)))*1000;  //mm

+

+

+//Results

+   printf("(a) The no. of units to be Installed=%.f Units\n",n)

+   printf("(b) Diameter of Wheel, D=%.3f m\n",D)

+   printf("(c) Diameter of Jet, d=%.1f mm\n",d)      //The Answer Vary due to round off Error

+

+

diff --git a/3751/CH7/EX7.6/Ex7_6.sce b/3751/CH7/EX7.6/Ex7_6.sce
new file mode 100644
index 000000000..1599b5c04
--- /dev/null
+++ b/3751/CH7/EX7.6/Ex7_6.sce
@@ -0,0 +1,23 @@
+//Fluid Systems by Shiv Kumar

+//Chapter 7 - Performance of water turbine

+//Example 7.6

+//To Find Speed and Power Developed by the Turbine

+      clc

+       clear

+

+//Given:

+   //Ist Condition

+      P1=8500;  //Power Developed, kW

+      N1=120;   //Speed, rpm

+      H1=32;    //Head, m

+

+   //2nd Condition

+       H2=25;    //Head, m

+

+//Computations:

+  P2=P1*(H2/H1)^(3/2);      //kW

+  N2=N1*(H2/H1)^(1/2);     //rpm

+

+//Results

+   printf("The Speed Developed by the Turbine,N2=%.f rpm\n",N2)

+   printf("The power developed= %.2f kW",P2)

diff --git a/3751/CH7/EX7.7/Ex7_7.sce b/3751/CH7/EX7.7/Ex7_7.sce
new file mode 100644
index 000000000..86ba6ebaa
--- /dev/null
+++ b/3751/CH7/EX7.7/Ex7_7.sce
@@ -0,0 +1,39 @@
+//Fluid Systems by Shiv Kumar

+//Chapter 7 - Performance of water turbine

+//Example 7.7

+//To Determine unit power, unit speed and unit discharge and also find speed, discharge and power at condition 2

+       clc

+       clear

+

+//Given:

+   //At Condition 1

+      P1=7200;  //Power Developed, KW

+      N1=300;   //Speed, rpm

+      H1=350;    //Head, m

+      eta_o=85/100;    // Overall efficiency

+

+   //At Condition 2

+       H2=300;  //Head, m

+   

+//Data Used:

+      rho=1000;      //Density of Water, kg/m^3

+      g=9.81;      //Acceleration due to gravity, m/s^@

+      

+      

+//Computations:

+   Q1=P1*1000/(rho*g*H1*eta_o);    //m^3/s

+   Pu=P1/(H1^(3/2));    //Unit Power, KW

+   Nu=N1/sqrt(H1);       //Unit Speed, rpm

+   Qu=Q1/sqrt(H1);    //Unit Discharge, m^3/s

+  P2=P1*(H2/H1)^(3/2);      //KW

+  N2=N1*(H2/H1)^(1/2);     //rpm

+  Q2=Q1*sqrt(H2/H1);       //m^3/s

+

+//Results

+   printf("Unit Power, Pu= %.3f kW\n Unit Speed, Nu=%.2f rpm\n Unit Discharge, Qu=%.4f m^/s\n",Pu, Nu, Qu)     //The Answer vary due to Round off Error

+   

+ printf("At head of 300 m:\n\t")

+   printf("The Speed,N2=%.2f rpm\n\t",N2)      //The Answer vary due to Round off Error  

+   printf("The power,P2= %.2f kW\n\t",P2)

+   printf("The Discharge, Q2=%.3f m^3/s\n",Q2)      //The Answer vary due to Round off Error

+

diff --git a/3751/CH7/EX7.8/Ex7_8.sce b/3751/CH7/EX7.8/Ex7_8.sce
new file mode 100644
index 000000000..b4b94bb30
--- /dev/null
+++ b/3751/CH7/EX7.8/Ex7_8.sce
@@ -0,0 +1,28 @@
+//Fluid Systems- By Shiv Kumar
+//Chapter 7- Performance of Water Turbine
+//Example 7.8
+// To Find Model Runner Speed and Prototype to Model Scale ratio
+   clc
+   clear
+
+//Given:-
+     //For Prototype
+            Pp=30;        //Power Developed, MW
+            Hp=55;        //Head,  m
+            Np=100;      //Speed, rpm
+            Pp=Pp*1000;      //KW
+      //For Model
+            Pm=25 ;        //Power Developed, KW
+            Hm=6;           //Head, m
+ 
+//Computations:-
+        Nm=Np*(Hm/Hp)^(5/4)*(Pp/Pm)^(1/2);              //rpm
+         DpbyDm=((Pp/Pm)*(Nm/Np)^3)^(1/5);              //A Ratio(Dimensionless)
+         Lr= DpbyDm;        //Scale Ratio
+
+
+//Results
+   printf("The Model Runner Speed, Nm=%.2f rpm And\n",Nm)
+   printf("Prototype to Model Scale Ratio,Lr=%.2f",Lr)      //The Answer vary due to Round off Error
+
+
diff --git a/3751/CH7/EX7.9/Ex7_9.sce b/3751/CH7/EX7.9/Ex7_9.sce
new file mode 100644
index 000000000..e86f4aeb8
--- /dev/null
+++ b/3751/CH7/EX7.9/Ex7_9.sce
@@ -0,0 +1,34 @@
+//Fluid Systems- By Shiv Kumar
+//Chapter 7- Performance of Water Turbine
+//Example 7.9
+// To Determine the Performance of the Turbine Under a Head of 20 m
+   clc
+   clear
+
+//Given:-
+       //Condition 1:
+            H1=25;     //Head, m
+            N1=200;    //Speed, rpm
+            Q1=9;        //Discharge, m^3/s
+            eta_o=90/100;     //Overall Efficiency
+
+        //Condition 2:
+           H2=20;     //Head, m
+
+//Data Required:-
+    rho=1000;        //Density of Water, Kg/m^3
+    g=9.81;          //Acceleration due to Gravity, m/s^2
+
+
+//Calculations:-
+         P1=rho*Q1*g*H1*eta_o/1000;           //KW
+         N2=N1*sqrt(H2/H1);          //rpm
+         Q2=Q1*sqrt(H2/H1);          //m^3/s
+         P2=P1*(H2/H1)^(3/2);       //KW
+
+
+//Results:-
+   printf("At Condition 2 (Under a Head of 20 m):\n")  
+   printf("\tSpeed, N2=%.2f rpm\n     Discharge, Q2=%.2f m^3/s\n      Power Developed, P2=%.2f kW",N2,Q2,P2)          //The Answer vary due to Round off Error
+
+