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diff --git a/1775/CH6/EX6.1/Chapter6_Example1.sce b/1775/CH6/EX6.1/Chapter6_Example1.sce
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index 000000000..0e5a30cee
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@@ -0,0 +1,20 @@
+//Chapter-6, Illustration 1, Page 308
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+COP=8.5;//Co-efficient of performance
+T1=300;//Room temperature in K
+T2=267;//Refrigeration temperature in K
+
+//CALCULATIONS
+COPmax=T2/(T1-T2);//Maximum COP possible
+
+//OUTPUT
+mprintf('Maximum COP possible is %3.2f \n Since the COP claimed by the inventor is more than the maximum possible COP his claim is not correct',COPmax)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.10/Chapter6_Example10.sce b/1775/CH6/EX6.10/Chapter6_Example10.sce
new file mode 100755
index 000000000..ab81e67ea
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+++ b/1775/CH6/EX6.10/Chapter6_Example10.sce
@@ -0,0 +1,20 @@
+//Chapter-6, Illustration 10, Page 309
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+Tg=470;//Heating temperature in K
+T0=290;//Cooling temperature in K
+TL=270;//Refrigeration temperature in K
+
+//CALCULATIONS
+COP=((Tg-T0)/Tg)*(TL/(T0-TL));//Ideal COP of absorption refrigeration system
+
+//OUTPUT
+mprintf('Ideal COP of absorption refrigeration system is %3.2f',COP)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.11/Chapter6_Example11.sce b/1775/CH6/EX6.11/Chapter6_Example11.sce
new file mode 100755
index 000000000..96dbc4ae9
--- /dev/null
+++ b/1775/CH6/EX6.11/Chapter6_Example11.sce
@@ -0,0 +1,35 @@
+//Chapter-6, Illustration 11, Page 317
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T1=-18;//Temperature at point 1 in oC
+T3=27;//Temperature at point 3 in oC
+rp=4;//Pressure ratio
+m=0.045;//mass flow rate in kg/s
+y=1.4;//Ratio of specific heats
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+x=(y-1)/y;//Ratio
+T2=(rp^x)*(273+T1);//Temperature at point 2 in K
+Tmax=T2-273;//Maximum temperature in oC
+T4=((1/rp)^x)*(273+T3);//Temperature at point 4 in K
+Tmin=T4-273;//Minimum temperature in oC
+qL=Cp*(T1-Tmin);//Heat rejected
+Wcin=Cp*(Tmax-T1);//Compressor work
+Wtout=Cp*(T3-Tmin);//Turbine work
+Wnet=Wcin-Wtout;//Net work done
+COP=qL/Wnet;//Co-efficient of performance
+Qref=m*qL;//Rate of refrigeration in kW
+
+//OUTPUT
+mprintf('Maximum temperature in the cycle is %3.0foC \n Minimum temperature in the cycle is %3.0foC \n COP is %3.2f \n Rate of refrigeration is %3.2f kW',Tmax,Tmin,COP,Qref)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.12/Chapter6_Example12.sce b/1775/CH6/EX6.12/Chapter6_Example12.sce
new file mode 100755
index 000000000..077290251
--- /dev/null
+++ b/1775/CH6/EX6.12/Chapter6_Example12.sce
@@ -0,0 +1,31 @@
+//Chapter-6, Illustration 12, Page 318
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=1;//Pressure at point 1 in bar
+T1=268;//Temperature at point 1 in K
+P2=5;//Pressure at point 2 in bar
+T3=288;//Temperature at point 3 in K
+n=1.3;//Adiabatic gas constant
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+T2=((P2/P1)^x)*T1;//Temperature at point 2 in K
+T4=((P1/P2)^x)*T3;//Temperature at point 4 in K
+W=Cp*(T3-T4);//Work developed per kg of air in kJ/kg
+Re=Cp*(T1-T4);//Refrigerating effect per kg of air in kJ/kg
+Wnet=Cp*((T2-T1)-(T3-T4));//Net work output in kJ/kg
+COP=Re/Wnet;//Co-efficient of performance
+
+//OUTPUT
+mprintf('Work developed per kg of air is %3.3f kJ/kg \n Refrigerating effect per kg of air is %3.3f kJ/kg \n COP of the cycle is %3.2f',W,Re,COP)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.13/Chapter6_Example13.sce b/1775/CH6/EX6.13/Chapter6_Example13.sce
new file mode 100755
index 000000000..15cf6908e
--- /dev/null
+++ b/1775/CH6/EX6.13/Chapter6_Example13.sce
@@ -0,0 +1,36 @@
+//Chapter-6, Illustration 13, Page 319
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T1=277;//Temperature at point 1 in K
+T3=328;//Temperature at point 3 in K
+P1=0.1;//Pressure at point 1 in MPa
+P2=0.3;//Pressure at point 2 in MPa
+nc=0.72;//Isentropic efficiency of compressor
+nt=0.78;//Isentropic efficiency of turbine
+y=1.4;//Adiabatic gas constant
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+m=3;//Cooling load in tonnes
+
+//CALCULATIONS
+x=(y-1)/y;//Ratio
+T2s=T1*((P2/P1)^x);//Temperature at point 2s in K
+T2=((T2s-T1)/nc)+T1;//Temerature at point 2 in K
+T4s=T3*((P1/P2)^x);//Temperature at point 4s in K
+T4=T3-((T3-T4s)*nt);//Temperature at point 4 in K
+Re=Cp*(T1-T4);//Refrigerating effect in kJ/kg
+Wnet=Cp*((T2-T1)-(T3-T4));//Net work output in kJ/kg
+COP=Re/Wnet;//Co-efficient of performance
+P=(m*3.52)/COP;//Driving power required in kW
+ma=(m*3.52)/Re;//Mass flow rate of air in kg/s
+
+//OUTPUT
+mprintf('COP of refrigerator is %3.2f \n Driving power required is %3.0f kW \n Mass flow rate of air is %3.3f kg/s',COP,P,ma)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.14/Chapter6_Example14.sce b/1775/CH6/EX6.14/Chapter6_Example14.sce
new file mode 100755
index 000000000..f2fa9f579
--- /dev/null
+++ b/1775/CH6/EX6.14/Chapter6_Example14.sce
@@ -0,0 +1,35 @@
+//Chapter-6, Illustration 14, Page 321
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=2.5;//Pressure at point 1 in bar
+P3=9;//Pressure at point 3 in bar
+COPr=0.65;//Ratio of actual COP to the theoretical COP
+m=5;//Refrigerant flow in kg/min
+T1=309;//Temperature at point 1 in K
+T2s=300;//Temperature at point 2s in K
+h1=570.3;//Enthalpy at P1 from the given tables in kJ/kg
+h4=456.4;//Enthalpy at P3 from the given tables in kJ/kg
+h2g=585.3;//Enthalpy at P3 from the given tables in kJ/kg
+s2=4.76;//Entropy at P1 from the given tables in kJ/kg-K
+s2g=4.74;//Entropy at P3 from the given tables in kJ/kg-K
+Cp=0.67;//Specific heat at P3 in kJ/kg-K
+
+//CALCULATIONS
+T2=(2.718^((s2-s2g)/Cp))*T2s;//Temperature at point 2 in K
+h2=h2g+(Cp*(T2-T2s));//Enthalpy at point 2 in kJ/kg
+COPR=(h1-h4)/(h2-h1);//Refrigerant COP
+COPact=COPr*COPR;//Actual COP
+qL=COPact*(h2-h1);//Heat rejected in kJ/kg
+QL=((m*qL*60)/3600)/3.516;//Cooling produced per kg of refrigerant in tonnes of refrigeration
+
+//OUTPUT
+mprintf('Theoretical COP is %3.2f \n Net cooling produced per hour is %3.2f TR',COPR,QL)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.15/Chapter6_Example15.sce b/1775/CH6/EX6.15/Chapter6_Example15.sce
new file mode 100755
index 000000000..761db224d
--- /dev/null
+++ b/1775/CH6/EX6.15/Chapter6_Example15.sce
@@ -0,0 +1,31 @@
+//Chapter-6, Illustration 15, Page 322
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T2=298;//Temperature at point 2 in K
+T1=268;//Temperature at point 1 in K
+hf1=-7.54;//Liquid Enthalpy at T1 in kJ/kg
+x1=0.6;//Quality factor 1
+hfg1=245.3;//Latent heat at T1 in kJ/kg
+sf1=0.251;//Liquid Entropy at T1 in kJ/kg-K
+s1=0.507;//Entropy at point 1 in kJ/kg-K
+hfg2=121.4;//Latent heat at T2 in kJ/kg
+hf2=81.3;//Liquid Enthalpy at T2 in kJ/kg
+h4=hf2;//Enthalpy at point 4 in kJ/kg
+
+//CALCULATIONS
+h1=hf1+(x1*hfg1);//Enthalpy at point 1 in kJ/kg
+x2=((s1-sf1)*T2)/hfg2;//Quality factor 2
+h2=hf2+(x2*hfg2);//Enthalpy at point 2 in kJ/kg
+COP=(h1-h4)/(h2-h1);//COP of the machine
+
+//OUTPUT
+mprintf('COP of the machine is %3.2f',COP)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.16/Chapter6_Example16.sce b/1775/CH6/EX6.16/Chapter6_Example16.sce
new file mode 100755
index 000000000..57c2150e4
--- /dev/null
+++ b/1775/CH6/EX6.16/Chapter6_Example16.sce
@@ -0,0 +1,31 @@
+//Chapter-6, Illustration 16, Page 323
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=25;//Pressure at point 1 in bar
+P2=60;//Pressure at point 2 in bar
+h2=208.1;//Vapour enthalpy at P2 in kJ/kg
+h3=61.9;//Liquid enthalpy at P2 in kJ/kg
+h4=h3;//Liquid enthalpy at P2 in kJ/kg
+s2=0.703;//Vapour entropy at P2 in kJ/kg-K
+sf1=-0.075;//Liquid entropy at P1 in kJ/kg-K
+sfg1=0.971;//Entropy in kJ/kg-K
+hf1=-18.4;//Liquid Enthalpy at P1 in kJ/kg
+hfg1=252.9;//Latent heat at P1 in kJ/kg
+m=5;//Refrigerant flow in kg/min
+
+//CALCULATIONS
+x1=(s2-sf1)/sfg1;//Quality factor 1
+h1=hf1+(x1*hfg1);//Enthalpy at point 1 in kJ/kg
+COP=(h1-h4)/(h2-h1);//Co-efficient of performance
+QL=(m*(h1-h4))/60;//Capacity of the refrigerator in kW
+
+//OUTPUT
+mprintf('COP of refrigerator is %3.2f \n Capacity of refrigerator is %3.2f kW',COP,QL)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.17/Chapter6_Example17.sce b/1775/CH6/EX6.17/Chapter6_Example17.sce
new file mode 100755
index 000000000..1699e40ee
--- /dev/null
+++ b/1775/CH6/EX6.17/Chapter6_Example17.sce
@@ -0,0 +1,45 @@
+//Chapter-6, Illustration 17, Page 324
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T1=271;//Temperature at point 1 in K
+T=265;//Temperature at point 1' in K
+Ta=303;//Temperature at point 2' in K
+Cpv=0.733;//Specific heat of vapour in kJ/kg
+Cpl=1.235;//Specific heat of liquid in kJ/kg
+h=184.07;//Liquid enthalpy at T in kJ/kg
+s=0.7;//Entropy at point 1' in kJ/kg-K
+sa=0.685;//Vapour entropy at Ta in kJ/kg-K
+ha=199.62;//Enthalpy at point 2' in kJ/kg
+hfb=64.59;//Liquid enthalpy at Ta in kJ/kg
+DT3=5;//Temperature difference in oC
+Q=2532;//Refrigeration capacity in kJ/min
+
+//CALCULATIONS
+s2=s+(Cpv*((log(T1/T))/(log(2.718))));//Entropy at point 1 in kJ/kg-K
+h1=h+(Cpv*(T1-T));//Enthalpy at point 1 in kJ/kg-K
+T2=(2.718^((s2-sa)/Cpv))*Ta;//Temperature at point 2 in K
+h2=ha+(Cpv*(T2-Ta));//Enthalpy at point 2 in kJ/kg
+h4=hfb-(Cpl*DT3);//Enthalpy at point 4 in kJ/kg
+COP=(h1-h4)/(h2-h1);//Co-efficient of performance
+m=Q/(h1-h4);//Mass flow rate of refrigerant in kJ/min
+P=(m*(h2-h1))/(60*12);//Power required in kW/TR
+
+//OUTPUT
+mprintf('COP is %3.2f \n Theoretical power required per tonne of refrigeration is %3.3f kW/TR',COP,P)
+
+
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.2/Chapter6_Example2.sce b/1775/CH6/EX6.2/Chapter6_Example2.sce
new file mode 100755
index 000000000..014616dfc
--- /dev/null
+++ b/1775/CH6/EX6.2/Chapter6_Example2.sce
@@ -0,0 +1,30 @@
+//Chapter-6, Illustration 2, Page 309
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+TL=268;//Low temperature in K
+TH=293;//High temperature in K
+t=24;//time in hrs
+C=2100;//Capacity of refrigerator in kJ/s
+Tw=10;//Water temperature in oC
+L=335;//Latent heat of ice in kJ/kg
+
+//CALCULATIONS
+COP=TL/(TH-TL);//Co-efficient of performance
+Pmin=C/COP;//Minimum power required in kW
+Qr=(4.187*(Tw-0))+L;//Heat removed from water in kJ/kg
+m=C/Qr;//mass of ice formed in kg/s
+W=(m*t*3600)/1000;//Weight of ice formed in tons
+
+//OUTPUT
+mprintf('Minimum power required is %3.2f kW \n Weight of ice formed in 24 hours is %3.2f tons',Pmin,W)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.3/Chapter6_Example3.sce b/1775/CH6/EX6.3/Chapter6_Example3.sce
new file mode 100755
index 000000000..ee4bcdf7b
--- /dev/null
+++ b/1775/CH6/EX6.3/Chapter6_Example3.sce
@@ -0,0 +1,25 @@
+//Chapter-6, Illustration 3, Page 309
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+TL=-10;//Temperature of brine in oC
+TH=20;//Temperature of water in oC
+L=335;//Latent heat of ice in kJ/kg
+
+//CALCULATIONS
+Qr=(4.187*(TH-0))+L;//Heat removed from water in kJ/kg
+COP=(TL+273)/(TH-TL);//Co-efficient of performance
+mi=(COP*3600)/Qr;//mass of ice formed per kWh in kg
+
+//OUTPUT
+mprintf('Mass of ice formed per kWh is %3.1f kg',mi)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.4/Chapter6_Example4.sce b/1775/CH6/EX6.4/Chapter6_Example4.sce
new file mode 100755
index 000000000..645c9248a
--- /dev/null
+++ b/1775/CH6/EX6.4/Chapter6_Example4.sce
@@ -0,0 +1,30 @@
+//Chapter-6, Illustration 4, Page 310
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=1.2;//Pressure at point 1 in bar
+P2=7;//Pressure at point 2 in bar
+m=0.05;//mass flow rate of refrigerant in kg/s
+h1=340.1;//Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
+s1=1.57135;//Entropy at point 1 from refrigerant-12 tables in kJ/kg-K
+s2=1.57135;//Entropy at point 2 from refrigerant-12 tables in kJ/kg-K
+h2=372;//Enthalpy at point 2 from refrigerant-12 tables in kJ/kg
+h3=226.575;//Enthalpy at point 3 from refrigerant-12 tables in kJ/kg
+h4=226.575;//Enthalpy at point 4 from refrigerant-12 tables in kJ/kg
+
+//CALCULATIONS
+Q2=m*(h1-h4);//Rate of heat removed from the refrigerated space in kW
+W=m*(h2-h1);//Power input to the compressor in kW
+Q1=m*(h2-h3);//Rate of heat rejection to the environment in kW
+COP=Q2/W;//Co-efficient of performance
+
+//OUTPUT
+mprintf('Rate of heat removed from the refrigerated space is %3.2f kW \n Power input to the compressor is %3.3f kW \n Rate of heat rejection to the environment is %3.2f kW \n Co-efficient of performance is %3.2f',Q2,W,Q1,COP)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.5/Chapter6_Example5.sce b/1775/CH6/EX6.5/Chapter6_Example5.sce
new file mode 100755
index 000000000..abfbbad8e
--- /dev/null
+++ b/1775/CH6/EX6.5/Chapter6_Example5.sce
@@ -0,0 +1,30 @@
+//Chapter-6, Illustration 5, Page 311
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T2=40;//Temperature at point 2 in oC
+T1=-10;//Temperature at point 1 in oC
+h2=367.155;//Enthalpy at point 2 from refrigerant-12 tables in kJ/kg
+s2=1.54057;//Entropy at point 2 from refrigerant-12 tables in kJ/kg-K
+s1=1.54057;//Entropy at point 1 from refrigerant-12 tables in kJ/kg-K
+sg=1.56004;//Entropy from refrigerant-12 tables in kJ/kg-K
+sf=0.96601;//Entropy from refrigerant-12 tables in kJ/kg-K
+hf=190.822;//Enthalpy from refrigerant-12 tables in kJ/kg-K
+hfg=156.319;//Enthalpy from refrigerant-12 tables in kJ/kg-K
+h3=238.533;//Enthalpy at point 3 from refrigerant-12 tables in kJ/kg-K
+h4=h3;//Enthalpy at point 4 from refrigerant-12 tables in kJ/kg-K
+
+//CALCULATIONS
+x1=(s1-sf)/(sg-sf);//Quality factor
+h1=hf+(x1*hfg);//Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
+COP=(h1-h4)/(h2-h1);//Co-efficient of performance
+
+//OUTPUT
+mprintf('COP of the system is %3.2f',COP)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.6/Chapter6_Example6.sce b/1775/CH6/EX6.6/Chapter6_Example6.sce
new file mode 100755
index 000000000..03638b1bb
--- /dev/null
+++ b/1775/CH6/EX6.6/Chapter6_Example6.sce
@@ -0,0 +1,41 @@
+//Chapter-6, Illustration 6, Page 311
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+Tc=35;//Temperature of condenser in oC
+Te=-15;//Temperature of evaporator in oC
+m=10;//Mass of ice per day in tons
+Tw=30;//Temperature of water in oC
+Ti=-5;//Temperature of ice in oC
+nv=0.65;//Volumetric efficiency
+N=1200;//Speed in rpm
+x=1.2;//Stroke to bore ratio
+na=0.85;//Adiabatic efficiency
+nm=0.95;//Mechanical efficiency
+S=4.187;//Specific heat of water in kJ/kg
+L=335;//Latent heat of ice in kJ/kg
+h1=1667.24;//Enthalpy at Te from Ammonia chart in kJ/kg
+h2=1925;//Enthalpy at Te from Ammonia chart in kJ/kg
+h4=586.41;//Enthalpy at Tc from Ammonia chart in kJ/kg
+v1=0.508;//Specific humidity at Te from Ammonia chart in (m^3)/kg
+
+//CALCULATIONS
+Qr=(((m*1000)/24)*((S*(Tw-0))+L+(1.94*(0-Ti))))/3600;//Refrigerating capacity in kW
+mr=Qr/(h1-h4);//Refrigerant mass flow rate in kg/s
+T2=112;//Discharge temperature in oC
+D=((mr*v1*4*60)/(nv*3.14*x*N))^(1/3);//Cylinder diameter in m
+L=x*D;//Stroke length in m
+W=(mr*(h2-h1))/(na*nm);//Compressor motor power in kW
+COPth=(h1-h4)/(h2-h1);//Theoretical COP
+COPact=Qr/W;//Actual COP
+
+//OUTPUT
+mprintf('Refrigerating capacity of plant is %3.2f kW \n Refrigerant mass flow rate is %3.4f kg/s \n Discharge temperature is %3.0f oC \n Cylinder diameter is %3.3f m \n Stroke length is %3.3f m \n Compressor motor power is %3.2f kW \n Theoretical COP is %3.2f \n Actual COP is %3.2f',Qr,mr,T2,D,L,W,COPth,COPact)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.7/Chapter6_Example7.sce b/1775/CH6/EX6.7/Chapter6_Example7.sce
new file mode 100755
index 000000000..561ecb6bc
--- /dev/null
+++ b/1775/CH6/EX6.7/Chapter6_Example7.sce
@@ -0,0 +1,40 @@
+//Chapter-6, Illustration 7, Page 313
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T1=-5;//Temperature at point 1 in oC
+T2=30;//Temperature at point 2 in oC
+m=13500;//mass of ice per day in kg
+Tw=20;//Temperature of water in oC
+COP=0.6;//Co-efficient of performance
+h2=1709.33;//Enthalpy at point 2 in kJ/kg
+s2=6.16259;//Entropy at point 2 in kJ/kg-K
+s1=6.16259;//Entropy at point 1 in kJ/kg-K
+sf=1.8182;//Entropy in kJ/kg-K
+sg=6.58542;//Entropy in kJ/kg-K
+hf=400.98;//Enthalpy in kJ/kg
+hfg=1278.35;//Enthalpy in kJ/kg
+h4=562.75;//Enthalpy at point 4 in kJ/kg
+S=4.187;//Specific heat of water in kJ/kg
+L=336;//Latent heat of ice in kJ/kg
+
+//CALCULATIONS
+x1=(s1-sf)/(sg-sf);//Quality factor
+h1=hf+(x1*hfg);//Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
+COPi=(h1-h4)/(h2-h1);//Ideal COP
+COPact=COP*COPi;//Actual COP
+Qr=((m*S*(Tw-0))+(m*L))/(24*3600);//Total amount of heat removed in kJ/s
+mr=Qr/(h1-h4);//Circulation rate of ammonia in kg/s
+W=mr*(h2-h1);//Power required in kW
+
+//OUTPUT
+mprintf('Circulation rate of ammonia is %3.3f kg/s \n Power required is %3.3f kW \n COP is %3.3f',mr,W,COPact)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.8/Chapter6_Example8.sce b/1775/CH6/EX6.8/Chapter6_Example8.sce
new file mode 100755
index 000000000..d65de8ae1
--- /dev/null
+++ b/1775/CH6/EX6.8/Chapter6_Example8.sce
@@ -0,0 +1,40 @@
+//Chapter-6, Illustration 8, Page 314
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+Tc=20;//Temperature of condenser in oC
+Te=-25;//Temperature of evaporator in oC
+m=15;//Mass of ice per day in tons
+Ts=5;//Subcooled temperature in oC
+Tsh=10;//Superheated temperature in oC
+n=6;//No. of cylinders
+N=950;//Speed of compressor in rpm
+x=1;//Stroke to bore ratio
+h1=402;//Enthalpy at point 1 from R-22 tables in kJ/kg
+h2=442;//Enthalpy at point 2 from R-22 tables in kJ/kg
+h3=216;//Enthalpy at point 3 from R-22 tables in kJ/kg
+h4=216;//Enthalpy at point 4 from R-22 tables in kJ/kg
+v1=2.258;//Specific volume at point 1 in (m^3)/min
+
+//CALCULATIONS
+Re=h1-h4;//Refrigerating effect in kJ/kg
+mr=(m*14000)/(Re*60);//Mass flow of refrigerant in kg/min
+Pth=(mr*(h2-h1))/60;//Theoretical power in kW
+COP=(h1-h4)/(h2-h1);//Co-efficient of performance
+Dth=v1/n;//Theoretical displacement per cylinder
+D=(((Dth*4)/(3.147*N))^(1/3))*1000;//Theoretical bore of compressor in mm
+L=D;//Theoretical stroke of compressor in mm
+
+//OUTPUT
+mprintf('Refrigerating effect is %3.0f kJ/kg \n Mass flow of refrigerant per minute is %3.2f kg/min \n Theoretical input power is %3.2f kW \n COP is %3.2f \n Theoretical bore of compressor is %3.2f mm \n Theoretical stroke of compressor is %3.2f mm',Re,mr,Pth,COP,D,L)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.9/Chapter6_Example9.sce b/1775/CH6/EX6.9/Chapter6_Example9.sce
new file mode 100755
index 000000000..2e12a43a1
--- /dev/null
+++ b/1775/CH6/EX6.9/Chapter6_Example9.sce
@@ -0,0 +1,35 @@
+//Chapter-6, Illustration 9, Page 316
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T2=40;//Temperature at point 2 in oC
+T1=-5;//Temperature at point 1 in oC
+h2=367.155;//Enthalpy at point 2 from F-12 tables in kJ/kg
+sg=1.55717;//Entropy from F-12 tables in kJ/kg-K
+s1=1.54057;//Entropy at point 1 from F-12 tables in kJ/kg-K
+sf=0.98311;//Entropy from F-12 tables in kJ/kg-K
+hf=195.394;//Enthalpy from F-12 tables in kJ/kg
+hfg=153.934;//Enthalpy from F-12 tables in kJ/kg
+h4=238.533;//Enthalpy at point 4 from F-12 tables in kJ/kg
+h4s=218;//Enthalpy at point 4 with subcooling from F-12 tables in kJ/kg
+
+//CALCULATIONS
+x1=(s1-sf)/(sg-sf);//Quality factor
+h1=hf+(x1*hfg);//Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
+COPns=(h1-h4)/(h2-h1);//Co-efficient of performance with no subcooling
+COPs=(h1-h4s)/(h2-h1);//Co-efficient of performance with subcooling
+
+//OUTPUT
+mprintf('COP with no subcooling is %3.3f \n COP with subcooling is %3.3f',COPns,COPs)
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
author	priyanka	2015-06-24 15:03:17 +0530
committer	priyanka	2015-06-24 15:03:17 +0530
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