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authorpriyanka2015-06-24 15:03:17 +0530
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-rwxr-xr-x2705/CH18/EX18.1/Ex18_1.sce61
-rwxr-xr-x2705/CH18/EX18.2/Ex18_2.sce54
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diff --git a/2705/CH18/EX18.1/Ex18_1.sce b/2705/CH18/EX18.1/Ex18_1.sce
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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
diff --git a/2705/CH18/EX18.2/Ex18_2.sce b/2705/CH18/EX18.2/Ex18_2.sce
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+clear;
+clc;
+disp('Example 18.2');
+
+// aim : To determine
+// (a) the mass flow of R401
+// (b) the dryness fraction of R401 at the entry to the evaporator
+// (c) the power of driving motor
+// (d) the ratio of heat transferred from condenser to the power required to the motor
+
+// given values
+P1 = 411.2;// pressure limit, [kN/m^2]
+P3 = 1118.9;// pressure limit, [kN/m^2]
+Q = 100*10^3;// heat transfer from the condenser,[kJ/h]
+T2 = 273+60;// entering saturation temperature, [K]
+
+// given
+// from given table
+// at P1
+h1 = 409.3;// specific enthalpy, [kJ/kg]
+s1 = 1.7431;// specific entropy,[ kJ/kg K
+
+// at P3
+h3 = 426.4;// specific enthalpy, [kJ/kg]
+s3 = 1.7192;// specific entropy,[ kJ/kg K
+T3 = 273+50;// saturation temperature, [K]
+h4 = 265.5;// 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
+h2 = h3+cpv*(T2-T3);// specific enthalpy, [kJ/kg]
+Qc = h2-h4;// heat transfer from condenser, [kJ/kg]
+mR401 = Q/Qc;// mass flow of R401, [kg]
+ mprintf('\n (a) The mass flow of R401 is = %f kg/h\n',mR401);
+
+// (b)
+hf1 = 219;// specific enthalpy, [kJ/kg]
+h5 = h4;
+// using h5=hf1+s5*(h1-hf1),so
+x5 = (h5-hf1)/(h1-hf1);// dryness fraction
+mprintf('\n (b) The dryness fraction of R401 at the entry to the evaporator is = %f\n',x5);
+
+// (c)
+P = mR401*(h2-h1)/3600/.7;// power to driving motor, [kW]
+ mprintf('\n (c) The power to driving motor is = %f kW\n',P);
+
+// (d)
+r = Q/3600/P;// ratio
+mprintf('\n (d) The ratio of heat transferred from condenser to the power required to the motor is = %f : 1\n',r);
+
+// End