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+clear;

+clc;

+disp('Example 18.1');

+

+// aim : To determine

+// (a) the coefficient of performance

+// (b) the mass flow of the refrigerant

+// (c) the cooling water required by the condenser

+

+// given values

+P1 = 462.47;// pressure limit, [kN/m^2]

+P3 = 1785.90;// pressure limit, [kN/m^2]

+T2 = 273+59;// entering saturation temperature, [K]

+T5 = 273+32;// exit temperature of condenser, [K]

+d = 75*10^-3;// bore, [m]

+L = d;// stroke, [m]

+N = 8;// engine speed, [rev/s]

+VE = .8;// olumetric efficiency

+cpL = 1.32;// heat capacity of liquid, [kJ/kg K]

+c = 4.187;// heat capacity of water, [kj/kg K]

+

+// solution

+// from given table

+// at P1

+h1 = 231.4;// specific enthalpy, [kJ/kg]

+s1 = .8614;// specific entropy,[ kJ/kg K

+v1 = .04573;// specific volume, [m^3/kg]

+

+// at P3

+h3 = 246.4;// specific enthalpy, [kJ/kg]

+s3 = .8093;// specific entropy,[ kJ/kg K

+v3 = .04573;// specific volume, [m^3/kg]

+T3= 273+40;// saturation temperature, [K]

+h4 = 99.27;// specific enthalpy, [kJ/kg]

+// (a)

+s2 = s1;// specific entropy, [kJ/kg k]

+// using s2=s3+cpv*log(T2/T3)

+cpv = (s2-s3)/log(T2/T3);// heat capacity, [kj/kg k]

+

+// from Fig.18.8

+T4 = T3;

+h2 = h3+cpv*(T2-T3);// specific enthalpy, [kJ/kg]

+h5 = h4-cpL*(T4-T5);// specific enthalpy, [kJ/kg]

+h6 = h5;

+COP = (h1-h6)/(h2-h1);// coefficient of performance

+mprintf('\n (a) The coefficient of performance of the refrigerator is  =  %f\n',COP);

+

+// (b)

+SV = %pi/4*d^2*L;// swept volume of compressor/rev, [m^3]

+ESV = SV*VE*N*3600;// effective swept volume/h, [m^3]

+m = ESV/v1;// mass flow of refrigerant/h,[kg]

+mprintf('\n (b) The mass flow of refrigerant/h is  =  %f  kg\n',m);

+

+// (c)

+dT = 12;// temperature limit, [C]

+Q = m*(h2-h5);// heat transfer in condenser/h, [kJ]

+// using Q=m_dot*c*dT, so

+m_dot = Q/(c*dT);// mass flow of water required, [kg/h]

+mprintf('\n (c) The mass flow of water required is  =  %f kg/h\n',m_dot);

+

+//  End