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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+// Example 4_14
+clc;funcprot(0);
+// Given data
+P_1=4000;// kPa
+T_1=500;// °C
+d_1=50;// mm
+V_1=200;// m/s
+d_2=250;// mm
+x_2=1.0;// The quality of steam at state 2
+
+// Calculation
+// (a)
+v_1=0.08643;// m^3/kg
+A_1=(%pi/4)*(d_1/1000)^2;// m^2
+mdot=(1/v_1)*A_1*V_1;// kg/s
+// The enthalpies are found from Tables C-3 and C-2 to be
+h_1=3445.2;// kJ/kg
+h_2=2665.7;// kJ/kg
+W_T=-(h_2-h_1)*mdot;// The turbine power output in kJ/s
+// (b)
+v_2=2.087;// m^3/kg
+A_2=(%pi/4)*(d_2/1000)^2;/// m^2
+V_2=(A_1*V_1*(1/v_1))/(A_2*(1/v_2));// m/s
+d_KE=mdot*((V_2^2-V_1^2)/2);// The kinetic energy change in J/s
+printf("\n(a)The turbine power output,W_T=%4.0f kJ/s or %1.3f MW \n(b)The kinetic energy change,delKE=%4.0f J/s or %1.2f kJ/s",W_T,W_T/10^3,d_KE,d_KE/10^3);