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//Fluid Systems - By - Shiv Kumar
//Chapter 4 - Pelton Turbine (Impulse Turbine)
//Example 4.16
clc
clear
//Given Data:-
P=4900; //Shaft Power, kW
P_mr=100; //Power absorbed in mechanical resistance, kW
eta_H=92/100; //Hydraulic Efficiency
P_n=415; //Power lost in Nozzle, kW
//Computations:-
P_rd=P+P_mr; //Power Devrloped by Runner, kW
P_rs=P_rd/eta_H; //Power Supplied to Runner, kW
P_an=P_n+P_rs; //Power Available at base of Nozzle, kW
eta_o=P/P_an*100; //Overall Efficiency in Percentage
//Results:-
printf("(a)Power Available at the Base of Nozzle=%.3f kW\n",P_an) //The answer vary due to round off error
printf("(b)Overall Efficiency, eta_o=%.2f Percent\n",eta_o)
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