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-rw-r--r--3751/CH7/EX7.1/Ex7_1.sce27
-rw-r--r--3751/CH7/EX7.10/Ex7_10.sce29
-rw-r--r--3751/CH7/EX7.11/Ex7_11.sce33
-rw-r--r--3751/CH7/EX7.12/Ex7_12.sce34
-rw-r--r--3751/CH7/EX7.13/Ex7_13.sce38
-rw-r--r--3751/CH7/EX7.14/Ex7_14.sce36
-rw-r--r--3751/CH7/EX7.15/Ex7_15.sce34
-rw-r--r--3751/CH7/EX7.2/Ex7_2.sce34
-rw-r--r--3751/CH7/EX7.3/Ex7_3.sce36
-rw-r--r--3751/CH7/EX7.4/Ex7_4.sce26
-rw-r--r--3751/CH7/EX7.5/Ex7_5.sce47
-rw-r--r--3751/CH7/EX7.6/Ex7_6.sce23
-rw-r--r--3751/CH7/EX7.7/Ex7_7.sce39
-rw-r--r--3751/CH7/EX7.8/Ex7_8.sce28
-rw-r--r--3751/CH7/EX7.9/Ex7_9.sce34
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
+
+
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