7 files changed, 109 insertions, 0 deletions
diff --git a/409/CH21/EX21.1/Example21_1.sce b/409/CH21/EX21.1/Example21_1.sce
new file mode 100755
index 000000000..6d8bdefe1
--- /dev/null
+++ b/409/CH21/EX21.1/Example21_1.sce
@@ -0,0 +1,25 @@
+clear ;
+clc;
+// Example 21.1
+printf('Example 21.1\n\n');
+//page no. 616
+// Solution Fig. E21.1a and E21.1b
+
+//Given
+V1 = 0.1 ;// Volume of gas initially -[cubic metres]
+V2 = 0.2 ;// Volume of gas finally -[cubic metres]
+T1 = 300 ;// Temperature of gas initially -[K]
+P1 = 200 ;// Pressure of gas finally -[kPa]
+R = 8.314 ;// Universal gas constant 
+n = (P1*V1)/(T1*R) ;// Moles of gas taken-[kg mol]
+//You are asked to calculate work by eqn. 21.1 , but you do not know the F(force) exerted by gas , so write F = P.A, multiply divide A and eqn 21.1 reduces to W= integate(P.dv)
+
+//(a)
+// Isobaric process see fig E21.1b to see the path followed
+W= integrate('-(P1)','V',V1,V2) ;// Work done by gas on piston -[kJ]
+printf('\n (a)Work done by gas on piston for isobaric process is %.0f kJ .\n ',W);
+
+//(b)
+// Isobaric process see fig E21.1b to see the path followed
+W= integrate('-(T1*R*n/V)','V',V1,V2) ;// Work done by gas on piston -[kJ]
+printf('(b)Work done by gas on piston for isothermal process is %.2f kJ .\n ',W);
+\ No newline at end of file
diff --git a/409/CH21/EX21.2/Example21_2.sce b/409/CH21/EX21.2/Example21_2.sce
new file mode 100755
index 000000000..b9c4d4af4
--- /dev/null
+++ b/409/CH21/EX21.2/Example21_2.sce
@@ -0,0 +1,18 @@
+clear ;
+clc;
+// Example 21.2
+printf('Example 21.2\n\n');
+//page no. 624
+// Solution 
+
+//Given
+id = 3 ;// Internal diameter of tube-[cm]
+Vf = 0.001 ;// Volume flow rate of water in tube-[cubic meter/s]
+rho = 1000 ;// Assumed density of water-[kg/cubic meter] 
+
+rad = id/2 ;// Radius of tube -[ cm]
+a = 3.14*rad^2 ;// Area of flow of tube -[squqre centimeter]
+v = Vf*(100)^2/a ;// Velocity of water in tube - [m/s]
+KE = v^2/2 ;// Specific(mass=1kg) kinetic energy of water in tube -[J/kg]
+
+printf('Specific kinetic energy of water in tube is %.2f J/kg .\n ',KE);
+\ No newline at end of file
diff --git a/409/CH21/EX21.3/Example21_3.sce b/409/CH21/EX21.3/Example21_3.sce
new file mode 100755
index 000000000..5a3e2fb7b
--- /dev/null
+++ b/409/CH21/EX21.3/Example21_3.sce
@@ -0,0 +1,14 @@
+clear ;
+clc;
+// Example 21.3
+printf('Example 21.3\n\n');
+//page no. 626
+// Solution 
+
+//Given
+// Let water level in first reservoir be the reference plane
+h = 40 ;// Difference of water-[ft]
+g = 32.2 ;// acceleration due to gravity-[ft/square second]
+PE=g*h/(32.2*778.2) ;//// Specific(mass=1kg) potential energy of water -[Btu/lbm]
+
+printf('Specific potential energy of water is %.4f Btu/lbm .\n ',PE);
+\ No newline at end of file
diff --git a/409/CH21/EX21.4/Example21_4.sce b/409/CH21/EX21.4/Example21_4.sce
new file mode 100755
index 000000000..1e4c13940
--- /dev/null
+++ b/409/CH21/EX21.4/Example21_4.sce
@@ -0,0 +1,18 @@
+clear ;
+clc;
+// Example 21.4
+printf('Example 21.4\n\n');
+//page no. 629
+// Solution
+
+//Given
+//Constant volume process 
+mol_air = 10 ;// Moles of air-[kg mol]
+T1 = 60+273 ;// Initial temperature of air-[K]
+T2 = 30+273 ;// final temperature of air-[K]
+// Additional data needed
+Cv = 2.1*10^4 ; // Specific heat capacity of air at constant volume-[J/(kg mol*C)]
+
+// Use eqn. 21.6 for del_U
+del_U = integrate('mol_air*Cv','T',T1,T2) ;//Change in internal energy-[J]
+printf('\nChange in internal energy is %.1e J .\n ',del_U);
+\ No newline at end of file
diff --git a/409/CH21/EX21.5/Example21_5.sce b/409/CH21/EX21.5/Example21_5.sce
new file mode 100755
index 000000000..42495a84d
--- /dev/null
+++ b/409/CH21/EX21.5/Example21_5.sce
@@ -0,0 +1,8 @@
+clear ;
+clc;
+// Example 21.5
+printf('Example 21.5\n\n');
+//page no. 629
+// Solution
+
+printf('\n As we know that internal energy(U) is state variable , therefore change in internal energy(del_U) depends only on initial and final state , independent of the path taken for process.\n Hence, change in internal energy for both paths A and B are same.  ');
+\ No newline at end of file
diff --git a/409/CH21/EX21.6/Example21_6.sce b/409/CH21/EX21.6/Example21_6.sce
new file mode 100755
index 000000000..8e3b9ae18
--- /dev/null
+++ b/409/CH21/EX21.6/Example21_6.sce
@@ -0,0 +1,8 @@
+clear ;
+clc;
+// Example 21.6
+printf('Example 21.6\n\n');
+//page no. 632
+// Solution
+
+printf('\n As we know that enthalpy(H) is state variable , therefore change in enthalpy(del_H) depends only on initial and final state , independent of the path taken for process.\n Hence, change in enthalpy for both paths A-B-D and A-C-D are same.  ');
+\ No newline at end of file
diff --git a/409/CH21/EX21.7/Example21_7.sce b/409/CH21/EX21.7/Example21_7.sce
new file mode 100755
index 000000000..44d32a00d
--- /dev/null
+++ b/409/CH21/EX21.7/Example21_7.sce
@@ -0,0 +1,18 @@
+clear;
+clc;
+// Example 21.7
+printf('Example 21.7\n\n');
+//page no. 633
+// Solution
+
+//Given
+//Constant pressure process 
+mol_air = 10 ;// Moles of air-[kg mol]
+T1 = 60+273 ;// Initial temperature of air-[K]
+T2 = 30+273 ;// final temperature of air-[K]
+// Additional data needed
+Cp = 2.9*10^4  ;// Specific heat capacity of air at constant pressure-[J/(kg mol*C)]
+
+// Use eqn. 21.11 for del_H
+del_H = integrate('mol_air*Cp','T',T1,T2) ;//Change in enthalpy-[J]
+printf('\nChange in enthalpy is %.1e J .\n ',del_H);
+\ No newline at end of file