From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 10 Oct 2017 12:27:19 +0530 Subject: initial commit / add all books --- 3751/CH7/EX7.1/Ex7_1.sce | 27 ++++++++++++++++++++++++++ 3751/CH7/EX7.10/Ex7_10.sce | 29 ++++++++++++++++++++++++++++ 3751/CH7/EX7.11/Ex7_11.sce | 33 ++++++++++++++++++++++++++++++++ 3751/CH7/EX7.12/Ex7_12.sce | 34 +++++++++++++++++++++++++++++++++ 3751/CH7/EX7.13/Ex7_13.sce | 38 +++++++++++++++++++++++++++++++++++++ 3751/CH7/EX7.14/Ex7_14.sce | 36 +++++++++++++++++++++++++++++++++++ 3751/CH7/EX7.15/Ex7_15.sce | 34 +++++++++++++++++++++++++++++++++ 3751/CH7/EX7.2/Ex7_2.sce | 34 +++++++++++++++++++++++++++++++++ 3751/CH7/EX7.3/Ex7_3.sce | 36 +++++++++++++++++++++++++++++++++++ 3751/CH7/EX7.4/Ex7_4.sce | 26 +++++++++++++++++++++++++ 3751/CH7/EX7.5/Ex7_5.sce | 47 ++++++++++++++++++++++++++++++++++++++++++++++ 3751/CH7/EX7.6/Ex7_6.sce | 23 +++++++++++++++++++++++ 3751/CH7/EX7.7/Ex7_7.sce | 39 ++++++++++++++++++++++++++++++++++++++ 3751/CH7/EX7.8/Ex7_8.sce | 28 +++++++++++++++++++++++++++ 3751/CH7/EX7.9/Ex7_9.sce | 34 +++++++++++++++++++++++++++++++++ 15 files changed, 498 insertions(+) create mode 100644 3751/CH7/EX7.1/Ex7_1.sce create mode 100644 3751/CH7/EX7.10/Ex7_10.sce create mode 100644 3751/CH7/EX7.11/Ex7_11.sce create mode 100644 3751/CH7/EX7.12/Ex7_12.sce create mode 100644 3751/CH7/EX7.13/Ex7_13.sce create mode 100644 3751/CH7/EX7.14/Ex7_14.sce create mode 100644 3751/CH7/EX7.15/Ex7_15.sce create mode 100644 3751/CH7/EX7.2/Ex7_2.sce create mode 100644 3751/CH7/EX7.3/Ex7_3.sce create mode 100644 3751/CH7/EX7.4/Ex7_4.sce create mode 100644 3751/CH7/EX7.5/Ex7_5.sce create mode 100644 3751/CH7/EX7.6/Ex7_6.sce create mode 100644 3751/CH7/EX7.7/Ex7_7.sce create mode 100644 3751/CH7/EX7.8/Ex7_8.sce create mode 100644 3751/CH7/EX7.9/Ex7_9.sce (limited to '3751/CH7') 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 + + -- cgit