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+// Example 14_11
+clc;funcprot(0);
+// Given data
+m=4.00;// lbm/s
+T_1=530;// R
+p_1=1;// psia
+p_2s=3.00// psia
+p_3=3.00;// psia
+p_4s=1;// psia
+T_3=600;// R
+c_p=0.240;// Btu/(lbm.R)
+k=1.40;// The specific heat ratio
+
+// Calculation
+// (a)
+T_4s=T_3*((p_4s/p_3)^((k-1)/k));// R
+W_expander=m*c_p*(T_3-T_4s);// Btu/s
+// (b)
+T_2s=T_1*((p_2s/p_1)^((k-1)/k));// R
+W_compressor=m*c_p*(T_1-T_2s);// Btu/s
+// (c)
+PR=p_3/p_4s;// Pressure ratio
+COP=(((PR)^((k-1)/k))-1)^-1;// The coefficient of performance of the unit
+// (d)
+W_net=(abs(W_compressor)-W_expander);// Btu/s
+Q_L=(COP*(W_net)*60*(1/200));// The refrigeration capacity of the unit in tons
+printf("\n(a)The expander power,W_expander=%3.0f Btu/s \n(b)The compressor power,W_compressor=%3.0f Btu/s \n(c)The coefficient of performance of the unit,COP=%1.2f \n(d)The refrigeration capacity of the unit in tons,Q_L=%2.1f tons of refrigeration",W_expander,W_compressor,COP,Q_L);