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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+//Fluid system - By - Shiv Kumar
+//Chapter 4 - Pelton Turbine (Impulse Turbine)
+//Example 4.1
+ clc
+ clear
+
+//Given Data:-
+ P=735.75; //Power Developed, kW
+ H=200; //Head, m
+ N=800; //Speed, rpm
+ eta_O=86/100; //Overall Efficiency
+ d_by_D=1/10; //Ratio of Jet diameter to turbine diameter (d/D)
+ Cv=0.98; //Co-efficienct of velocity
+ Ku=0.45; //Speed ratio
+
+//Data Used:-
+ rho=1000; //Density of water, kg/m^3
+ g=9.81; //Acceleration due to gravity, m/s^2
+
+//Computations:-
+ Q=P*1000/(rho*g*H*eta_O); //Net discharge, m^3/s
+ //(a)Diameter of Turbine, D
+ D=60*Ku*sqrt(2*g*H)/(%pi*N); //m
+ d=D*d_by_D; //m
+ //(b)The no. of Jets required
+ q=(%pi/4)*d^2*Cv*sqrt(2*g*H); //Discharge of a single Jet, m^3/s
+ n=round(Q/q); //No. of Jets
+ //(c)Diameter of Jet, d
+ d=d_by_D*D; //m
+
+//Results:-
+ printf("(a)Diameter of Turbine, D=%.4f m \n", D) //The answer vary due to round off error
+ printf("(b)The number of Jets required, n=%.f \n", n)
+ printf("(c)Diameter of Jet, d=%.4f m \n", d)
+