initial commit / add all books

9 files changed, 343 insertions, 0 deletions

diff --git a/2705/CH7/EX7.1/Ex7_1.sce b/2705/CH7/EX7.1/Ex7_1.sce
new file mode 100755
index 000000000..3a5463218
--- /dev/null
+++ b/2705/CH7/EX7.1/Ex7_1.sce
@@ -0,0 +1,25 @@
+clear;

+clc;

+disp('Example 7.1');

+

+//  aim : To determine

+//  the specific enthalpy of water

+

+//  Given values

+Tf = 273+100;//  Temperature,[K]

+

+//  solution

+//  from steam table

+cpl = 4.187;// [kJ/kg K]

+//  using equation [8]

+sf = cpl*log(Tf/273.16);//  [kJ/kg*K]

+mprintf('\n The specific entropy of water is  =  %f kJ/kg K\n',sf);

+

+//  using steam table

+sf = 1.307;//  [kJ/kg K]

+mprintf('\n From table The accurate value of sf in this case is  =  %f  kJ/kg K\n',sf);

+

+// There is small error in book's final value of sf

+

+

+//  End

diff --git a/2705/CH7/EX7.2/Ex7_2.sce b/2705/CH7/EX7.2/Ex7_2.sce
new file mode 100755
index 000000000..38d974b14
--- /dev/null
+++ b/2705/CH7/EX7.2/Ex7_2.sce
@@ -0,0 +1,33 @@
+

+clear;

+clc;

+disp('Example 7.2');

+

+//  aim : To determine

+//  the specific entropy

+

+//  Given values

+P = 2;//  pressure,[MN/m^2]

+x = .8;//  dryness fraction

+

+//  solution

+//  from steam table at given pressure

+Tf = 485.4;//  [K]

+cpl = 4.187;//  [kJ/kg K]

+hfg = 1888.6;//  [kJ/kg]

+

+//  (a) finding entropy by calculation

+s = cpl*log(Tf/273.16)+x*hfg/Tf;//  formula for entropy calculation

+

+mprintf('\n (a) The specific entropy of wet steam is  =  %f kJ/kg K\n',s);

+

+//  (b) calculation of entropy using steam table

+//  from steam table at given pressure

+sf = 2.447;//  [kJ/kg K]

+sfg = 3.89;//  [kJ/kg K]

+//  hence

+s = sf+x*sfg;//  [kJ/kg K]

+

+mprintf('\n (b) The specific entropy using steam table is  =  %f kJ/kg K\n',s);

+

+//  End

diff --git a/2705/CH7/EX7.3/Ex7_3.sce b/2705/CH7/EX7.3/Ex7_3.sce
new file mode 100755
index 000000000..21e0a7087
--- /dev/null
+++ b/2705/CH7/EX7.3/Ex7_3.sce
@@ -0,0 +1,30 @@
+clear;

+clc;

+disp('Example 7.3');

+

+//  aim : To determine

+//  the specific entropy of steam

+

+//  Given values

+P = 1.5;//pressure,[MN/m^2]

+T = 273+300;//temperature,[K]

+

+//  solution

+

+//  (a)

+//  from steam table

+cpl = 4.187;//  [kJ/kg K]

+Tf = 471.3;//  [K]

+hfg = 1946;//  [kJ/kg]

+cpv = 2.093;//  [kJ/kg K]

+

+//  usung equation [2]

+s = cpl*log(Tf/273.15)+hfg/Tf+cpv*log(T/Tf);//  [kJ/kg K]

+mprintf('\n (a) The specific entropy of steam is  =  %f kJ/kg K\n',s);

+

+//  (b)

+//  from steam tables

+s = 6.919;//  [kJ/kg K]

+mprintf('\n (b) The accurate value of specific entropy from steam table is  =  %f kJ/kg K\n',s);

+

+//  End

diff --git a/2705/CH7/EX7.4/Ex7_4.sce b/2705/CH7/EX7.4/Ex7_4.sce
new file mode 100755
index 000000000..c908c859e
--- /dev/null
+++ b/2705/CH7/EX7.4/Ex7_4.sce
@@ -0,0 +1,30 @@
+clear;

+clc;

+disp('Example 7.4');

+

+//  aim : To determine

+//  the dryness fraction of steam

+

+//  Given values

+P1 = 2;//  initial pressure, [MN/m^2]

+t = 350;//  temperature, [C]

+P2 = .28;//  final pressure, [MN/m^2]

+

+//  solution

+//  at 2 MN/m^2 and 350 C,steam is superheated because the saturation temperature is 212.4 C

+//  From steam table

+s1 = 6.957;//  [kJ/kg K]

+

+//  for isentropic process

+s2 = s1;

+//  also

+sf2 = 1.647;//  [kJ/kg K]

+sfg2 = 5.368;//  [kJ/kg K]

+

+//  using

+//  s2 = sf2+x2*sfg2, where x2 is dryness fraction of steam

+//  hence

+x2 = (s2-sf2)/sfg2;

+mprintf('\n The final dryness fraction of steam is x2  =  %f\n',x2);

+

+//  End

diff --git a/2705/CH7/EX7.5/Ex7_5.sce b/2705/CH7/EX7.5/Ex7_5.sce
new file mode 100755
index 000000000..53cec3d7a
--- /dev/null
+++ b/2705/CH7/EX7.5/Ex7_5.sce
@@ -0,0 +1,53 @@
+clear;

+clc;

+disp('Example 7.5');

+

+//  aim : To determine

+//  the  final condition of steam...

+//  the change in specific entropy during hyperbolic process

+

+//  Given values

+P1 = 2;//  pressure, [MN/m^2]

+t = 250;//  temperature, [C]

+P2 = .36;//  pressure, [MN/m^2]

+P3 = .06;//  pressure, [MN/m^2]

+

+//  solution

+

+//  (a)

+//  from steam table

+s1 = 6.545;//  [kJ/kg K]

+//  at .36 MN/m^2

+sg = 6.930;//  [kJ/kg*K]

+

+sf2 = 1.738;//  [kJ/kg K]

+sfg2 = 5.192;//  [kJ/kg K]

+vg2 = .510;//  [m^3]

+

+//  so after isentropic expansion, steam is wet

+//  hence, s2=sf2+x2*sfg2, where x2 is dryness fraction

+//  also

+s2 = s1;

+//  so

+x2 = (s2-sf2)/sfg2;

+//  and

+v2 = x2*vg2;//  [m^3]

+

+//  for  hyperbolic process

+//  P2*v2=P3*v3

+//  hence

+v3 = P2*v2/P3;//  [m^3]

+	

+mprintf('\n (a) From steam table at .06 MN/m^2 steam is superheated and has temperature of 100 C with specific volume is  =  %f  m^3/kg\n',v3);

+

+//  (b)

+//  at this condition

+s3 = 7.609;//  [kJ/kg*K]

+//  hence

+change_s23 = s3-sg;//  change in specific entropy during the hyperblic process[kJ/kg*K]

+mprintf('\n (b) The change in specific entropy during the hyperbolic process is  =  %f kJ/kg K\n',change_s23);

+

+// In the book they have taken sg instead of s2 for part (b), so answer is not matching

+

+//  End

+

diff --git a/2705/CH7/EX7.6/Ex7_6.sce b/2705/CH7/EX7.6/Ex7_6.sce
new file mode 100755
index 000000000..b9e9a3beb
--- /dev/null
+++ b/2705/CH7/EX7.6/Ex7_6.sce
@@ -0,0 +1,67 @@
+

+clear;

+clc;

+disp('Example 7.6');

+

+//  aim : To determine the

+//  (a) heat transfer during the expansion and

+//  (b) work done durind the expansion

+

+//  given values

+m = 4.5;  //  mass of steam,[kg]

+P1 = 3;  //  initial pressure,[MN/m^2]

+T1 = 300+273;  //  initial temperature,[K]

+

+P2 = .1;  //  final pressure,[MN/m^2]

+x2 = .96;  //  dryness fraction at final stage

+

+//  solution

+//  for state point 1,using steam table

+s1 = 6.541;//  [kJ/kg/K]

+u1 = 2751;// [kJ/kg]

+ 

+ //  for state point 2

+ sf2 = 1.303;//  [kJ/kg/K]

+ sfg2 = 6.056;//  [kJ/kg/k]

+ T2 = 273+99.6;//  [K]

+ hf2 = 417;//  [kJ/kg]

+ hfg2 = 2258;//  [kJ/kg]

+ vg2 = 1.694;//   [m^3/kg]

+ 

+ //  hence

+ s2 = sf2+x2*sfg2;// [kJ/kg/k]

+ h2 = hf2+x2*hfg2;// [kJ/kg]

+ u2 = h2-P2*x2*vg2*10^3;// [kJ/kg]

+ 

+ //  Diagram of example 7.6

+ x = [s1 s2];

+ y = [T1 T2];

+plot2d(x,y);

+ xtitle('Diagram for example 7.6(T vs s)');

+ xlabel('Entropy (kJ/kg K)');

+ ylabel('Temperature (K)');

+

+x = [s1,s1];

+y = [0,T1];

+plot2d(x,y);

+

+x = [s2,s2];

+y = [0,T2];

+plot2d(x,y);

+

+ //  (a)

+ //  Q_rev is area of T-s diagram

+ Q_rev = (T1+T2)/2*(s2-s1);// [kJ/kg]

+ //  so total heat transfer is

+ Q_rev = m*Q_rev;// [kJ]

+ 

+ //  (b)

+ del_u = u2-u1;//  change in internal energy, [kJ/kg]

+ //  using 1st law of thermodynamics

+ W = Q_rev-m*del_u;//  [kJ]

+ 

+mprintf('\n (a) The heat transfer during the expansion is  =  %f kJ  (received)\n',Q_rev);

+

+ mprintf('\n (b) The work done during the expansion is  =  %f kJ\n',W);

+ 

+ //  End

diff --git a/2705/CH7/EX7.7/Ex7_7.sce b/2705/CH7/EX7.7/Ex7_7.sce
new file mode 100755
index 000000000..9a68eddbd
--- /dev/null
+++ b/2705/CH7/EX7.7/Ex7_7.sce
@@ -0,0 +1,45 @@
+

+clear;

+clc;

+disp('Example 7.7');

+

+//  aim : To determine the  

+//  (a) change of entropy

+//  (b)  The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression

+

+//  Given values

+P1 = 140;// initial pressure,[kN/m^2]

+V1 = .14;// initial volume, [m^3]

+T1 = 273+25;// initial temperature,[K]

+ P2 = 1400;// final pressure [kN/m^2]

+ n = 1.25; // polytropic index

+ cp = 1.041;// [kJ/kg K]

+ cv = .743;// [kJ/kg K]

+ 

+ //  solution

+ // (a)

+ R = cp-cv;// [kJ/kg/K]

+ //  using ideal gas equation 

+ m = P1*V1/(R*T1);// mass of gas,[kg]

+ //  since gas is following law P*V^n=constant ,so 

+ V2 = V1*(P1/P2)^(1/n);//  [m^3]

+ 

+ //  using eqn [9]

+ del_s = m*(cp*log(V2/V1)+cv*log(P2/P1));//  [kJ/K]

+ mprintf('\n (a) The change of entropy is  =  %f kJ/K\n',del_s);

+ 

+ //  (b)

+ W = (P1*V1-P2*V2)/(n-1);// polytropic work,[kJ]

+ Gamma = cp/cv;// heat capacity ratio

+ Q = (Gamma-n)/(Gamma-1)*W;// heat transferred,[kJ]

+ 

+ // Again using polytropic law

+ T2 = T1*(V1/V2)^(n-1);// final temperature, [K]

+ T_avg = (T1+T2)/2;// mean absolute temperature, [K]

+ 

+ // so approximate change in entropy is

+ del_s = Q/T_avg;// [kJ/K]

+ 

+ mprintf('\n (b) The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression  =  %f  kJ/K\n',del_s);

+ 

+ //  End

diff --git a/2705/CH7/EX7.8/Ex7_8.sce b/2705/CH7/EX7.8/Ex7_8.sce
new file mode 100755
index 000000000..711505b0c
--- /dev/null
+++ b/2705/CH7/EX7.8/Ex7_8.sce
@@ -0,0 +1,39 @@
+clear;

+clc;

+disp('Example 7.8');

+

+//  aim : To determine

+//  the change of entropy

+

+//  Given values

+m = .3;//  [kg]

+P1 = 350;//  [kN/m^2]

+T1 = 273+35;//  [K]

+P2 = 700;//  [kN/m^2]

+V3 = .2289;//  [m^3]

+cp = 1.006;//  [kJ/kg K]

+cv = .717;//  [kJ/kg K]

+

+//  solution

+//  for constant volume process

+R = cp-cv;//  [kJ/kg K]

+//  using PV=mRT

+V1 = m*R*T1/P1;//  [m^3]

+

+//  for constant volume process P/T=constant,so

+T2 = T1*P2/P1;//  [K]

+s21 = m*cv*log(P2/P1);//  formula for entropy change for constant volume process

+mprintf('\n The change of entropy in constant volume process is  =  %f  kJ/kg K\n',s21);

+

+// 'For the above part result given in the book is wrong

+

+V2 = V1;

+// for constant pressure process

+T3 = T2*V3/V2;//  [K]

+s32 = m*cp*log(V3/V2);//  [kJ/kg K]

+

+mprintf('\n The change of entropy in constant pressure process is  =  %f  kJ/kg K\n',s32);

+

+// there is misprint in the book's result

+

+// End

diff --git a/2705/CH7/EX7.9/Ex7_9.sce b/2705/CH7/EX7.9/Ex7_9.sce
new file mode 100755
index 000000000..05f58353b
--- /dev/null
+++ b/2705/CH7/EX7.9/Ex7_9.sce
@@ -0,0 +1,21 @@
+clear;

+clc;

+disp('Example 7.9');

+

+//  aim : To determine

+//  the  change of entropy

+

+// Given values

+P1 = 700;// initial pressure, [kN/m^2]

+T1 = 273+150;//  Temperature ,[K]

+V1 = .014;// initial volume, [m^3]

+V2 = .084;//  final volume, [m^3]

+

+//  solution

+//  since process is isothermal so

+T2 = T1;

+// and using fig.7.10

+del_s = P1*V1*log(V2/V1)/T1 ;//  [kJ/K]

+mprintf('\n The change of entropy is  =  %f  kJ/kg K\n',del_s);

+

+//  End