initial commit / add all books

10 files changed, 484 insertions, 0 deletions

diff --git a/2705/CH10/EX10.1/Ex10_1.sce b/2705/CH10/EX10.1/Ex10_1.sce
new file mode 100755
index 000000000..563d4d450
--- /dev/null
+++ b/2705/CH10/EX10.1/Ex10_1.sce
@@ -0,0 +1,31 @@
+clear;

+clc;

+disp('Example 10.1');

+

+//  aim : To determine

+//   the equivalent evaporation

+

+//  Given

+P = 1.4;//  [MN/m^2]

+m = 8;//  mass of water,[kg]

+T1 = 39;//  entering temperature,[C]

+T2 = 100;//  [C]

+x = .95;//dryness fraction 

+

+//  solution

+hf = 830.1;//  [kJ/kg]

+hfg = 1957.7;//  [kJ/kg]

+//  steam is wet so specific enthalpy of steam is

+h = hf+x*hfg;//  [kJ/kg]

+

+//  at 39 C

+h1 = 163.4;//  [kJ/kg]

+//  hence

+q = h-h1;//  [kJ/kg]

+Q = m*q;//  [kJ]

+

+evap = Q/2256.9;//  equivalent evaporation[kg steam/(kg coal)]

+

+mprintf('\n The equivalent evaporation, from and at 100 C is  =  %f  kg steam/kg coal\n ',evap);

+

+// End

diff --git a/2705/CH10/EX10.10/Ex10_10.sce b/2705/CH10/EX10.10/Ex10_10.sce
new file mode 100755
index 000000000..c0d346f7f
--- /dev/null
+++ b/2705/CH10/EX10.10/Ex10_10.sce
@@ -0,0 +1,64 @@
+clear;

+clc;

+disp('Example 10.10');

+

+// aim : To determine

+// (a) the mass of steam bled to each feed heater in kg/kg of supply steam

+// (b) the thermal efficiency of the arrangement

+

+// given values

+P1 = 7;// steam initial pressure, [MN/m^2]

+T1 = 273+500;// steam initil temperature, [K]

+P2 = 2;// pressure at stage 1, [MN/m^2]

+P3 = .5;// pressure at stage 2, [MN/m^2]

+P4 = .05;// condenser pressure,[MN/m^2]

+SE = .82;// stage efficiency of turbine

+

+// solution

+// from the enthalpy-entropy chart(Fig10.23) values of specific enthalpies are

+h1 = 3410;// [kJ/kg]

+h2_prim = 3045;// [kJ/kg]

+// h1-h2=SE*(h1-h2_prim), so

+h2 = h1-SE*(h1-h2_prim);// [kJ/kg]

+

+h3_prim = 2790;// [kJ/kg]

+// h2-h3=SE*(h2-h3_prim), so

+h3 = h2-SE*(h2-h3_prim);// [kJ/kg]

+

+h4_prim = 2450;// [kJ/kg]

+// h3-h4 = SE*(h3-h4_prim), so

+h4 = h3-SE*(h3-h4_prim);// [kJ/kg]

+

+// from steam table

+// @ 2 MN/m^2

+hf2 = 908.6;// [kJ/kg]

+// @ .5 MN/m^2

+hf3 = 640.1;// [kJ/kg] 

+// @ .05 MN/m^2

+hf4 = 340.6;// [kJ/kg]

+

+// (a) 

+// for feed heater1

+m1 = (hf2-hf3)/(h2-hf3);// mass of bled steam, [kg/kg supplied steam]

+// for feed heater2

+m2 = (1-m1)*(hf3-hf4)/(h3-hf4);// 

+mprintf('\n (a) The mass of steam bled in feed heater 1 is  =  %f  kg/kg supply steam\n',m1);

+mprintf('\n      The mass of steam bled in feed heater 2 is  =  %f kg/kg supply steam\n',m2);

+

+// (b)

+W = (h1-h2)+(1-m1)*(h2-h3)+(1-m1-m2)*(h3-h4);// theoretical work done, [kJ/kg]

+Eb = h1-hf2;// energy input in the boiler, [kJ/kg]

+TE1 = W/Eb;// thermal efficiency

+mprintf('\n (b) The thermal efficiency of the arrangement is  =  %f percent\n',TE1*100);

+

+// If there is no feed heating

+hf5 = hf4;

+h5_prim = 2370;// [kJ/kg]

+// h1-h5 = SE*(h1-h5_prim), so

+h5 = h1-SE*(h1-h5_prim);// [kJ/kg]

+Ei = h1-hf5;//energy input, [kJ/kg]

+W = h1-h5;//  theoretical work, [kJ/kg]

+TE2 = W/Ei;// thermal efficiency

+mprintf('\n      The thermal efficiency if there is no feed heating is  =  %f percent\n',TE2*100);

+

+//  End 

diff --git a/2705/CH10/EX10.2/Ex10_2.sce b/2705/CH10/EX10.2/Ex10_2.sce
new file mode 100755
index 000000000..7993e77ac
--- /dev/null
+++ b/2705/CH10/EX10.2/Ex10_2.sce
@@ -0,0 +1,47 @@
+clear;

+clc;

+disp('Example 10.2');

+

+// aim : To determine 

+//  the mass of oil used per hour and the fraction of enthalpy drop through the turbine

+//  heat transfer available per kilogram of exhaust steam

+

+//  Given values

+ms_dot = 5000;// generation of steam, [kg/h]

+P1 = 1.8;// generated steam pressure, [MN/m^2]

+T1 = 273+325;// generated steam temperature, [K]

+Tf = 273+49.4;// feed temperature, [K]

+neta = .8;// efficiency of boiler plant 

+c = 45500;// calorific value, [kJ/kg]

+P = 500;// turbine generated power, [kW]

+Pt = .18;// turbine exhaust pressure, [MN/m^2]

+x = .98;// dryness farction of steam

+

+//  solution

+//  using steam table at 1.8 MN/m^2

+hf1 = 3106;// [kJ/kg]

+hg1 = 3080;// [kJ/kg]

+//  so

+h1 = hf1-neta*(hf1-hg1);// [kJ/kg]

+//  again using steam table specific enthalpy of feed water is

+hwf = 206.9;// [kJ/kg]

+h_rais = ms_dot*(h1-hwf);// energy to raise steam, [kJ]

+

+h_fue = h_rais/neta;// energy from fuel per hour, [kJ]

+m_oil = h_fue/c;// mass of fuel per hour, [kg]

+

+//  from steam table at exhaust

+hf = 490.7;// [kJ/kg]

+hfg = 2210.8;//  [kJ/kg]

+//  hence

+h = hf+x*hfg;// [kJ/kg]

+//  now

+h_drop = (h1-h)*ms_dot/3600;// specific enthalpy drop in turbine [kJ]

+f = P/h_drop;// fraction ofenthalpy drop converted into work

+//  heat transfer available in exhaust is

+Q = h-hwf;// [kJ/kg]

+mprintf('\n The mass of oil used per hour is  =  %f  kg\n',m_oil);

+mprintf('\n The fraction of the enthalpy drop through the turbine that is converted into useful work is  =  %f\n',f);

+mprintf('\n The heat transfer available in exhaust steam above 49.4 C is  =  %f  kJ/kg\n',Q);

+

+//  End

diff --git a/2705/CH10/EX10.3/Ex10_3.sce b/2705/CH10/EX10.3/Ex10_3.sce
new file mode 100755
index 000000000..5ea7c08db
--- /dev/null
+++ b/2705/CH10/EX10.3/Ex10_3.sce
@@ -0,0 +1,44 @@
+clear;

+clc;

+disp('Example 10.3');

+

+//  aim : To determine

+//  (a) the thermal efficiency of the boiler

+//  (b) the equivalent evaporation of the boiler

+//  (c) the new coal consumption 

+

+//  given values

+ms_dot = 5400;// steam feed rate, [kg/h]

+P = 750;// steam pressure, [kN/m^2]

+x = .98;// steam dryness fraction

+Tf1 = 41.5;// feed water temperature, [C]

+CV = 31000;// calorific value of coal used in the boiler, [kJ/kg]

+mc1 = 670;// rate of burning of coal/h, [kg]

+Tf2 = 100;// increased water temperature, [C]

+

+// solution

+//   (a)

+SRC = ms_dot/mc1;// steam raised/kg coal, [kg]

+hf = 709.3;// [kJ/kg]

+hfg = 2055.5;// [kJ/kg]

+h1 = hf+x*hfg;// specific enthalpy of steam raised, [kJ/kg]

+//  from steam table 

+hfw = 173.9;// specific enthalpy of feed water, [kJ/kg]

+EOB = SRC*(h1-hfw)/CV;// efficiency of boiler

+mprintf('\n (a) The thermal efficiency of the boiler is  =  %f percent\n',EOB*100);

+

+// (b)

+he = 2256.9;// specific enthalpy of evaporation, [kJ/kg]

+Ee = SRC*(h1-hfw)/he;// equivalent evaporation[kg/kg coal]

+mprintf('\n (b) The equivalent evaporation of boiler is  =  %f  kg/kg coal\n',Ee);

+

+// (c)

+hw = 419.1;// specific enthalpy of feed water at 100 C, [kJ/kg]

+Eos = ms_dot*(h1-hw);// energy of steam under new condition, [kJ/h]

+neb = EOB+.05;// given condition new efficiency of boiler if 5%more than previous

+Ec = Eos/neb;// energy from coal, [kJ/h]

+mc2 = Ec/CV;// mass of coal used per hour in new condition, [kg]

+mprintf('\n (c) Mass of coal used in new condition is  =  %f kg\n',mc2);

+mprintf('\n      The saving in coal per hour is  =  %f  kg\n',mc1-mc2);

+

+//  End

diff --git a/2705/CH10/EX10.4/Ex10_4.sce b/2705/CH10/EX10.4/Ex10_4.sce
new file mode 100755
index 000000000..c7d9ea745
--- /dev/null
+++ b/2705/CH10/EX10.4/Ex10_4.sce
@@ -0,0 +1,46 @@
+clear;

+clc;

+disp('Example 10.4');

+

+//  aim : To determine the

+//  (a) Heat transfer in the boiler

+//   (b) Heat transfer in the superheater

+//  (c) Gas used

+

+//  given values

+P = 100;// boiler operating pressure, [bar]

+Tf = 256;// feed water temperature, [C]

+x = .9;// steam dryness fraction.

+Th = 450;// superheater exit temperature, [C]

+m = 1200;// steam generation/h, [tonne]

+TE = .92;// thermal efficiency

+CV = 38;// calorific value of fuel, [MJ/m^3]

+

+// solution

+// (a)

+// from steam table

+hw = 1115.4;// specific enthalpy of feed water, [kJ/kg]

+// for wet steam

+hf = 1408;// specific enthalpy, [kJ/kg]

+hg = 2727.7;// specific enthalpy, [kJ/kg]

+//  so

+h = hf+x*(hg-hf);// total specific enthalpy of wet steam, [kJ/kg]

+//  hence

+Qb = m*(h-hw);// heat transfer/h for wet steam, [MJ]

+mprintf('\n (a) The heat transfer/h in producing wet steam in the boiler is  =  %f  MJ\n',Qb);

+

+// (b)

+// again from steam table

+// specific enthalpy of superheated stem at given condition is,

+hs = 3244;// [kJ/kg]

+

+Qs = m*(hs-h);// heat transfer/h in superheater, [MJ]

+mprintf('\n (b) The heat transfer/h in superheater is  =  %f  MJ\n',Qs);

+

+// (c)

+V = (Qb+Qs)/(TE*CV);// volume of gs used/h, [m^3]

+mprintf('\n (c) The volume of gas used/h is  =  %f m^3\n',V);

+

+//  There is calculation mistake in the book so our answer is not matching

+

+//  End

diff --git a/2705/CH10/EX10.5/Ex10_5.sce b/2705/CH10/EX10.5/Ex10_5.sce
new file mode 100755
index 000000000..e7086049a
--- /dev/null
+++ b/2705/CH10/EX10.5/Ex10_5.sce
@@ -0,0 +1,31 @@
+clear;

+clc;

+disp('Example 10.5');

+

+//aim : To determine 

+// the flow rate of cooling water

+

+//Given values

+P=24;//pressure, [kN/m^2]

+ms_dot=1.8;//steam condense rate,[tonne/h]

+x=.98;//dryness fraction

+T1=21;//entrance temperature of cooling water,[C]

+T2=57;//outlet temperature of cooling water,[C]

+

+//solution

+//at 24 kN/m^2, for steam

+hfg=2616.8;//[kJ/kg]

+hf1=268.2;//[kJ/kg]

+//hence

+h1=hf1+x*(hfg-hf1);//[kJ/kg]

+

+//for cooling water

+hf3=238.6;//[kJ/kg]

+hf2=88.1;//[kJ/kg]

+

+//using equation [3]

+//ms_dot*(hf3-hf2)=mw_dot*(h1-hf1),so

+mw_dot=ms_dot*(h1-hf1)/(hf3-hf2);//[tonne/h]

+disp('tonne/h',mw_dot,'The flow rate of the cooling water is =')

+

+//End

diff --git a/2705/CH10/EX10.6/Ex10_6.sce b/2705/CH10/EX10.6/Ex10_6.sce
new file mode 100755
index 000000000..b4d0e04fe
--- /dev/null
+++ b/2705/CH10/EX10.6/Ex10_6.sce
@@ -0,0 +1,49 @@
+clear;

+clc;

+disp('Example 10.6');

+

+//  aim : To determine

+//  (a) the energy supplied in the boiler

+//  (b) the dryness fraction of the steam entering the condenser

+//  (c) the rankine efficiency

+

+//  given values

+P1 = 3.5;// steam entering pressure, [MN/m^2]

+T1 = 273+350;// entering temperature, [K]

+P2 = 10;//steam exhaust pressure, [kN/m^2]

+

+// solution

+//  (a)

+//  from steam table, at P1 is,

+hf1 = 3139;// [kJ/kg]

+hg1 = 3095;// [kJ/kg]

+h1 = hf1-1.5/2*(hf1-hg1);

+// at Point 3

+h3 = 191.8;// [kJ/kg]

+Es = h1-h3;// energy supplied, [kJ/kg]

+mprintf('\n (a) The energy supplied in boiler/kg steam is  =  %f kJ/kg\n',Es);

+

+// (b)

+// at P1

+sf1 = 6.960;// [kJ/kg K]

+sg1 = 6.587;// [kJ/kg K]

+s1 = sf1-1.5/2*(sf1-sg1);// [kJ/kg K]

+// at P2

+sf2 = .649;// [kJ/kg K] 

+ sg2 = 8.151;// [kJ/kg K]

+ // s2=sf2+x2(sg2-sf2)

+ // theoretically expansion through turbine is isentropic so s1=s2

+ // hence

+ s2 = s1;

+ x2 = (s2-sf2)/(sg2-sf2);// dryness fraction

+ mprintf('\n (b) The dryness fraction of steam entering the condenser is =  %f \n',x2);

+ 

+ // (c)

+ // at point 2

+ hf2 = 191.8;// [kJ/kg]

+ hfg2 = 2392.9;// [kJ/kg]

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

+ Re = (h1-h2)/(h1-h3);// rankine efficiency

+ mprintf('\n (c) The Rankine efficiency is  =  %f percent\n',Re*100);

+ 

+ //  End

diff --git a/2705/CH10/EX10.7/Ex10_7.sce b/2705/CH10/EX10.7/Ex10_7.sce
new file mode 100755
index 000000000..b3e5be740
--- /dev/null
+++ b/2705/CH10/EX10.7/Ex10_7.sce
@@ -0,0 +1,61 @@
+clear;

+clc;

+disp('Example 10.7');

+

+// aim : To determine

+//  the specific work done and compare this with that obtained when determining the rankine effficiency

+

+// given values

+P1 = 1000;// steam entering pressure, [kN/m^2]

+x1 = .97;// steam entering dryness fraction

+P2 = 15;//steam exhaust pressure, [kN/m^2]

+n = 1.135;// polytropic index

+

+// solution

+// (a)

+// from steam table, at P1 is

+hf1 = 762.6;// [kJ/kg]

+hfg1 = 2013.6;// [kJ/kg]

+h1 = hf1+hfg1; // [kJ/kg]

+

+sf1 = 2.138;// [kJ/kg K]

+sg1 = 6.583;// [kJ/kg K]

+s1 = sf1+x1*(sg1-sf1);// [kJ/kg K]

+

+// at P2

+sf2 = .755;// [kJ/kg K] 

+ sg2 = 8.009;// [kJ/kg K]

+// s2 = sf2+x2(sg2-sf2)

+// since expansion through turbine is isentropic so s1=s2

+ // hence

+ s2 = s1;

+ x2 = (s2-sf2)/(sg2-sf2);// dryness fraction

+ 

+  // at point 2

+ hf2 = 226.0;// [kJ/kg]

+ hfg2 = 2373.2;// [kJ/kg]

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

+ 

+// at Point 3

+h3 = 226.0;// [kJ/kg]

+

+// (a)

+ Re = (h1-h2)/(h1-h3);// rankine efficiency

+ mprintf('\n (a) The Rankine efficiency is  =  %f percent\n',Re*100);

+ 

+// (b)

+vg1 = .1943;// specific volume at P1, [m^3/kg]

+vg2 = 10.02;// specific volume at P2, [m^3/kg]

+V1 = x1*vg1;// [m^3/kg]

+V2 = x2*vg2;// [m^3/kg]

+

+W1 = n/(n-1)*(P1*V1-P2*V2);// specific work done, [kJ/kg]

+

+//  from rankine cycle

+W2 = h1-h2;// [kJ/kg]

+mprintf('\n (b) The specific work done is  =  %f kJ/kg\n',W1);

+mprintf('\n     The specific work done (from rankine) is  =  %f kJ/kg\n',W2);

+

+// there is calculation mistake in the book so our answer is not matching

+

+//  End

diff --git a/2705/CH10/EX10.8/Ex10_8.sce b/2705/CH10/EX10.8/Ex10_8.sce
new file mode 100755
index 000000000..f65703173
--- /dev/null
+++ b/2705/CH10/EX10.8/Ex10_8.sce
@@ -0,0 +1,63 @@
+clear;

+clc;

+disp('Example 10.8');

+

+//  aim : To determine

+//  (a) the rankine fficiency

+//  (b) the specific steam consumption

+//  (c) the carnot efficiency of the cycle

+

+// given values

+P1 = 1100;// steam entering pressure, [kN/m^2]

+T1 = 273+250;// steam entering temperature, [K]

+P2 = 280;// pressure at point 2, [kN/m^2]

+P3 = 35;// pressure at point 3, [kN/m^2]

+

+// solution

+// (a)

+// from steam table, at P1  and T1 is

+hf1 = 2943;// [kJ/kg]

+hg1 = 2902;// [kJ/kg]

+h1 = hf1-.1*(hf1-hg1); // [kJ/kg]

+

+sf1 = 6.926;// [kJ/kg K]

+sg1 = 6.545;// [kJ/kg K]

+s1 = sf1-.1*(sf1-sg1);// [kJ/kg K]

+

+// at P2

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

+ sg2 = 7.014;// [kJ/kg K]

+// s2=sf2+x2(sg2-sf2)

+// since expansion through turbine is isentropic so s1=s2

+ // hence

+ s2  = s1;

+ x2 = (s2-sf2)/(sg2-sf2);// dryness fraction

+ 

+ // at point 2

+ hf2 = 551.4;// [kJ/kg]

+ hfg2 = 2170.1;// [kJ/kg]

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

+ vg2 = .646;// [m^3/kg]

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

+ 

+ // by Fig10.20.

+ A6125 = h1-h2;// area of 6125, [kJ/kg]

+ A5234 = v2*(P2-P3);// area 5234, [kJ/kg]

+ W = A6125+A5234;// work done 

+ hf = 304.3;// specific enthalpy of water at condenser pressuer, [kJ/kg]

+ ER = h1-hf;// energy received, [kJ/kg]

+ Re = W/ER;// rankine efficiency

+ mprintf('\n (a) The rankine efficiency is  =  %f percent\n',Re*100);

+ 

+ // (b)

+ kWh = 3600;// [kJ]

+ SSC = kWh/W;// specific steam consumption, [kJ/kWh]

+ mprintf('\n (b) The specific steam consumption is = %f kJ/kWh\n',SSC);

+ 

+ // (c)

+ // from steam table 

+T3 = 273+72.7;// temperature at point 3

+CE = (T1-T3)/T1;// carnot efficiency

+mprintf('\n (c) The carnot efficiency of the cycle is  =  %f percent\n',CE*100);

+

+//  End

diff --git a/2705/CH10/EX10.9/Ex10_9.sce b/2705/CH10/EX10.9/Ex10_9.sce
new file mode 100755
index 000000000..cd4a60ca0
--- /dev/null
+++ b/2705/CH10/EX10.9/Ex10_9.sce
@@ -0,0 +1,48 @@
+clear;

+clc;

+disp('Example 10.9');

+

+// aim : To determine

+// (a) the theoretical power of steam passing through the turbine

+// (b) the thermal efficiency of the cycle

+// (c) the thermal efficiency of the cycle assuming there is no reheat

+

+// given values

+P1 = 6;// initial pressure, [MN/m^2]

+T1 = 450;// initial temperature, [C]

+P2 = 1;// pressure at stage 1, [MN/m^2]

+P3 = 1;// pressure at stage 2, [MN/m^2]

+T3 = 370;// temperature, [C]

+P4 = .02;// pressure at stage 3, [MN/m^2]

+P5 = .02;// pressure at stage 4, [MN/m^2]

+T5 = 320;// temperature, [C]

+P6 = .02;// pressure at stage 5, [MN/m^2]

+P7 = .02;// final pressure , [MN/m^2]

+

+// solution

+// (a) 

+// using Fig 10.21

+h1 = 3305;// specific enthalpy, [kJ/kg]

+h2 = 2850;// specific enthalpy, [kJ/kg]

+h3 = 3202;// specific enthalpy, [kJ/kg]

+h4 = 2810;// specific enthalpy, [kJ/kg]

+h5 = 3115;// specific enthalpy, [kJ/kg]

+h6 = 2630;// specific enthalpy, [kJ/kg]

+h7 = 2215;// specific enthalpy, [kJ/kg]

+W = (h1-h2)+(h3-h4)+(h5-h6);// specific work through the turbine, [kJ/kg]

+mprintf('\n (a) The theoretical power/kg steam/s is  =  %f kW\n',W);

+

+// (b)

+// from steam table

+hf6 = 251.5;// [kJ/kg]

+

+TE1 = ((h1-h2)+(h3-h4)+(h5-h6))/((h1-hf6)+(h3-h2)+(h5-h4));// thermal efficiency

+mprintf('\n (b) The thermal efficiency of the cycle is  =  %f percent\n',TE1*100);

+

+// (c)

+// if there is no heat

+hf7 = hf6;

+TE2 = (h1-h7)/(h1-hf7);// thermal efficiency

+mprintf('\n (c) The thermal efficiency of the cycle if there is no heat is  =  %f percent\n',TE2*100);

+

+//  End