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+clear;
+clc;
+disp('Example 17.2');
+
+// aim : To determine
+// (a) bp
+// (b) ip
+// (c) mechanical efficiency
+// (d) indicated thermal efficiency
+// (e) brake specific steam consumption
+// (f) draw up complete energy account for the test one-minute basis taking 0 C as datum
+
+// given values
+d = 200*10^-3;// cylinder diameter, [mm]
+L = 250*10^-3;// stroke, [mm]
+N = 5;// speed, [rev/s]
+r = .75/2;// effective radious of brake wheel, [m]
+Ps = 800;// stop valve pressure, [kN/m^2]
+x = .97;// dryness fraction of steam
+BL = 136;// brake load, [kg]
+SL = 90;// spring balance load, [N]
+PM = 232;// mean effective pressure, [kN/m^2]
+Pc = 10;// condenser pressure, [kN/m^2]
+m_dot = 3.36;// steam consumption, [kg/min]
+CC = 113;// condenser cooling water, [kg/min]
+Tr = 11;// temperature rise of condenser cooling water, [K]
+Tc = 38;// condensate temperature, [C]
+C = 4.18;// heat capacity of water, [kJ/kg K]
+g = 9.81;// gravitational acceleration, [m/s^2]
+
+// solution
+// from steam table
+// at 800 kN/m^2
+tf1 = 170.4;// saturation temperature, [C]
+hf1 = 720.9;// [kJ/kg]
+hfg1 = 2046.5;// [kJ/kg]
+hg1 = 2767.5;// [kJ/kg]
+vg1 = .2403;// [m^3/kg]
+
+// at 10 kN/m^2
+tf2 = 45.8;// saturation temperature, [C]
+hf2 = 191.8;// [kJ/kg]
+hfg2 = 2392.9;// [kJ/kg]
+hg2 = 2584.8;// [kJ/kg]
+vg2 = 14.67;// [m^3/kg]
+
+// (a)
+T = (BL*g-SL)*r;// torque, [Nm]
+bp = 2*%pi*N*T*10^-3;// brake power,[W]
+mprintf('\n (a) The brake power is  =   %f  kW\n',bp);
+
+// (b)
+A = %pi*d^2/4;// area, [m^2]
+ip = PM*L*A*N*2;// double-acting so*2, [kW]
+mprintf('\n (b) The indicated power is  =  %f kW\n',ip);
+
+// (c)
+n_mec = bp/ip;// mechanical efficiency
+mprintf('\n (c) The mechanical efficiency is  =  %f  percent\n',n_mec*100);
+
+// (d)
+h = hf1+x*hfg1;// [kJ/kg]
+hf = hf2;
+ITE = ip/((m_dot/60)*(h-hf));// indicated thermal efficiency
+mprintf('\n (d) The indicated thermal efficiency is  =  %f percent\n',ITE*100);
+// (e)
+Bsc=m_dot*60/bp;// brake specific steam consumption, [kg/kWh]
+mprintf('\n (e) The brake steam consumption is  =  %f  kg/kWh\n',Bsc);
+
+// (f)
+// energy balanvce reckoned from 0 C
+Es = m_dot*h;// energy supplied, [kJ]
+Eb = bp*60;// energy to bp, [kJ]
+Ecc = CC*C*Tr;// energy to condensate cooling water, [kJ]
+Ec = m_dot*C*Tc;// energy to condensate, [kJ]
+Ese = Es-Eb-Ecc-Ec;// energy to surrounding,etc, [kJ]
+
+mprintf('\n (f) Energy supplied/min is  =  %f  kJ\n',Es);
+
+mprintf('\n    Energy to bp/min is  =  %f  kJ\n',Eb);
+mprintf('\n    Energy to condenser cooling water/min is  =  %f kJ\n',Ecc);
+mprintf('\n    Energy to condensate/min is  =  %f kJ\n',Ec);
+mprintf('\n    Energy to surrounding, etc/min is  =  %f  kJ\n',Ese);
+
+// answer in the book is misprinted
+
+//  End