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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+//Example 4_10
+clc;funcprot(0);
+//Given data
+h=36;// m
+p=135;// kW
+q=0.44;// m^3/sec
+H=100;// m
+N=428;// r.p.m
+d_r=4;// d_r=D/d
+
+// Calculation
+w=1000*9.81;// N
+n=N*d_r*sqrt(h/H);// r.p.m
+n_m=(p*1000)/(1000*9.81*q*h);// The efficiency of the model
+n_p=n_m+0.03;//The efficiency of the prototype
+P=p*(n_p/n_m)*(d_r)^2*(H/h)^(3/2);// kW
+printf('\n The power developed by the prototype,P=%0.0f kW',P);
+n_s=(n*sqrt(p))/h^(5/4);
+N_s=(N*sqrt(P))/(H)^(5/4);
+if(N_s~=n_s)
+ printf('\n The runner is of Francis type.');
+else
+ printf('\n Wrong');
+end
+// The answer vary due to round off error