7 files changed, 139 insertions, 0 deletions
diff --git a/3843/CH10/EX10.1/Ex10_1.sce b/3843/CH10/EX10.1/Ex10_1.sce
new file mode 100644
index 000000000..d96c553c6
--- /dev/null
+++ b/3843/CH10/EX10.1/Ex10_1.sce
@@ -0,0 +1,12 @@
+// Example 10_1
+clc;funcprot(0);
+// Given data
+T_1=25+273;// The initial temperature in K
+P_1=122;// The initial pressure in kPa
+T_2=29+273;// The final temperature in K
+P_2=120;// The final pressure in kPa
+R=0.287;// kJ/kg.K
+
+// Calculation
+dv=((R*T_2)/P_2)-((R*T_1)/P_1);// The change in the specific volume of air in m^3/kg
+printf("\nThe change in the specific volume of air,dv=%0.5f m^3/kg",dv);
diff --git a/3843/CH10/EX10.10/Ex10_10.sce b/3843/CH10/EX10.10/Ex10_10.sce
new file mode 100644
index 000000000..da876ef19
--- /dev/null
+++ b/3843/CH10/EX10.10/Ex10_10.sce
@@ -0,0 +1,56 @@
+// Example 10_10
+clc;funcprot(0);
+// Given data
+T_1=-50;// °C
+P_1=2;// MPa
+T_2=40;// °C
+P_2=6;// MPa
+c_p=1.042;// kJ/kg.K
+c_v=0.745;// kJ/kg.K
+R=0.297;// kJ/kg.K
+M=28;// The molecular weight of nitrogen in kg/kmol
+
+// Calculation
+// (a)
+dh=c_p*(T_2-T_1);// The enthalpy change in kJ/kg
+du=c_v*(T_2-T_1);// The change in internal energy in kJ/kg
+ds=(c_p*log((T_2+273)/(T_1+273)))-(R*log(P_2/P_1));// The entropy change in kJ/kg.K
+printf("\n(a)The enthalpy change,dh=%2.1f kJ/kg \n   The change in internal energy,du=%2.0f kJ/kg \n   The entropy change,ds=%0.2f kJ/kg.K",dh,du,ds);
+// (b)
+// Interpolating in the ideal gas table (Table F-2) gives
+h_1=6479;// kJ/kmol
+h_2=9102;// kJ/kmol
+dh=(h_2-h_1)/M;// The enthalpy change in kJ/kg
+u_1=4625;// kJ/kmol
+u_2=6499;// kJ/kmol
+du=(u_2-u_1)/M;// The change in internal energy in kJ/kg
+phi_1=183.0;// kJ/kmol.K
+phi_2=192.9;// kJ/kmol.K
+ds=((phi_2-phi_1)/M)-(R*log(P_2/P_1));// The entropy change in kJ/kg.K
+printf("\n(b)The enthalpy change,dh=%2.1f kJ/kg \n   The change in internal energy,du=%2.0f kJ/kg \n   The entropy change,ds=%0.2f kJ/kg.K",dh,du,ds);
+// (c)
+// Using (10.69) and the enthalpy departure chart in Appendix I we find
+T_c=126.2;// K
+T_R1=(T_1+273)/T_c;// The reduced temperature at state 1
+T_R2=(T_2+273)/T_c;// The reduced temperature at state 2
+P_c=3.39;// MPa
+P_R1=P_1/P_c;// The reduced pressure at state 1
+P_R2=P_2/P_c;// The reduced pressure at state 2
+// The enthalpy departure chart(Appendix I) provides us with
+// Assume dh_s1=(hbar*_1-hbar_1)/T_c,dh_s2=(hbar*_2-hbar_2)/T_c,dh_1=h*_1-h_1,dh_2=h*_2-h_2,
+dh_s1=1.6;// kJ/kmol.K
+dh_s2=2.5;// kJ/kmol.K
+dh_1=(dh_s1*T_c)/M;// kJ/kg
+dh_2=(dh_s2*T_c)/M;// kJ/kg
+dh=-dh_1+dh_2+[c_p*(T_2-T_1)];// The enthalpy change in kJ/kg
+// Using Compressibility chart,
+Z_1=0.99;// The Compressibility factor at state 1
+Z_2=0.985;// The Compressibility factor at state 2
+du=dh-[R*((Z_2*(T_2+273))-(Z_1*(T_1+273)))];// The change in internal energy in kJ/kg
+// Assume ds_s1=(sbar*_1-sbar_1),ds_s2=(sbar*_2-sbar_2),ds_1=s*_1-s_1,ds_2=s*_2-s_2,
+ds_s1=1.0;// kJ/kmol.K
+ds_s2=1.2;// kJ/kmol.K
+ds_1=ds_s1/M;// kJ/kg.K
+ds_2=ds_s2/M;// kJ/kg.K
+ds=-ds_1+ds_2+((c_p*log((T_2+273)/(T_1+273)))-(R*log(P_2/P_1)));// The entropy change in kJ/kg.K
+printf("\n(c)The enthalpy change,dh=%2.1f kJ/kg \n   The change in internal energy,du=%2.0f kJ/kg \n   The entropy change,ds=%0.2f kJ/kg.K",dh,du,ds);
diff --git a/3843/CH10/EX10.2/Ex10_2.sce b/3843/CH10/EX10.2/Ex10_2.sce
new file mode 100644
index 000000000..30f9f0d1d
--- /dev/null
+++ b/3843/CH10/EX10.2/Ex10_2.sce
@@ -0,0 +1,12 @@
+// Example 10_2
+clc;funcprot(0);
+// Given data
+T=400;// °C
+P=4;// MPa
+
+// Calculation
+// From steam tables
+dh=3330-3092;// kJ/kg
+ds=6.937-6.583;// kJ/kg.K
+dhbyds=dh/ds;// K
+printf("\n(dh/ds)_P=%3.0f K or %3.0f°C",dhbyds,dhbyds-273);
diff --git a/3843/CH10/EX10.3/Ex10_3.sce b/3843/CH10/EX10.3/Ex10_3.sce
new file mode 100644
index 000000000..666e006a3
--- /dev/null
+++ b/3843/CH10/EX10.3/Ex10_3.sce
@@ -0,0 +1,16 @@
+// Example 10_3
+clc;funcprot(0);
+// Given data
+T=200;// °C
+P=1554;// kPa
+R=0.462;// kJ/kg.K
+
+// Calculation
+v_g=(R*(T+273))/P;// m^3/kg
+rho=1000;// kg/m^3
+v_f=0.001;// m^3/kg
+dPbydT=(1906-1254)/(210-190);// kN/m^2.K
+h_fg=(T+273)*(v_g-v_f)*dPbydT;// kJ/kg
+h_fga=1941;// kJ/kg (From steam tables)
+error=((h_fg-h_fga)/h_fga)*100;// The percentage error
+printf("\nThe percent error=%2.1f percentage",error);
diff --git a/3843/CH10/EX10.4/Ex10_4.sce b/3843/CH10/EX10.4/Ex10_4.sce
new file mode 100644
index 000000000..b9c555be6
--- /dev/null
+++ b/3843/CH10/EX10.4/Ex10_4.sce
@@ -0,0 +1,12 @@
+// Example 10_4
+clc;funcprot(0);
+// Given data
+P_1=2;// kPa
+T_1=17.5+273;// K
+P_2=1;// kPa
+h_fg=2480;// kJ/kg
+R=0.462;// kJ/kg.K
+
+// Calculation
+T_2=1/((1/T_1)-((R/h_fg)*log(P_2/P_1)));// K
+printf("\nT_2=%3.0f K or %1.0f°C",T_2,T_2-273);
diff --git a/3843/CH10/EX10.8/Ex10_8.sce b/3843/CH10/EX10.8/Ex10_8.sce
new file mode 100644
index 000000000..a1ddb349b
--- /dev/null
+++ b/3843/CH10/EX10.8/Ex10_8.sce
@@ -0,0 +1,13 @@
+// Example 10_8
+clc;funcprot(0);
+// Given data
+m=10;// kg
+P_1=100;// The initial pressure in kPa
+P_2=50;// The final pressure in MPa
+beta=5*10^-5;// K^-1
+rho=8770;// kg/m^3
+
+// Calculation
+// ds=s_2-s_1;
+ds=-(1/rho)*beta*[(P_2-(P_1/10^3))*10^6];// J/kg.K
+printf("\nThe entropy change,s_2-s_1=%0.3f J/kg.K",ds);
diff --git a/3843/CH10/EX10.9/Ex10_9.sce b/3843/CH10/EX10.9/Ex10_9.sce
new file mode 100644
index 000000000..c6144f7f4
--- /dev/null
+++ b/3843/CH10/EX10.9/Ex10_9.sce
@@ -0,0 +1,18 @@
+// Example 10_9
+clc;funcprot(0);
+// Given data
+T=400;// °C
+P=1;// MPa
+v=0.3066;// m^3/kg
+
+// Calculation
+ds=7.619-7.302;// kJ/kg.K
+dT=450-350;// K
+c_p=(T+273)*(ds/dT);// kJ/kg.K
+dv=0.3304-0.2825;// m^3/kg
+mu_j=(1/(c_p*10^3))*[((T+273)*(dv/dT))-v];// K/Pa
+printf("\nThe Joule thomson coefficient,mu_j=%1.2e K/Pa",mu_j);
+dT=403.7-396.2;// K
+dP=(1.5-0.5)*10^6;// Pa
+mu_j=dT/dP;// K/Pa
+printf("\nThe Joule thomson coefficient,mu_j=%1.2e K/Pa",mu_j);