initial commit / add all books

8 files changed, 185 insertions, 0 deletions

diff --git a/2168/CH5/EX5.10/Chapter5_example10.sce b/2168/CH5/EX5.10/Chapter5_example10.sce
new file mode 100755
index 000000000..239d08320
--- /dev/null
+++ b/2168/CH5/EX5.10/Chapter5_example10.sce
@@ -0,0 +1,18 @@
+clc

+clear

+//Input data

+r=20//Compression ratio

+c=5//Cut off at 5%

+dc=1//Specific heat at constant volume increases by 1%

+Cv=0.171//pecific heat at constant volume in kJ/kg.K

+R=29.3//Characteristic gas constant in mkg/kg/degree C

+k=1.95//k can be obtained from relation de/e=-dcv/cv*(1-e/e)*(g-1)*((1/g)+ln(r)-(k^g*lnk)/(k^g-1))

+J=427//Mechanical equivalent of heat in kg.m/kcal

+

+//Calculations

+g=(R/(J*Cv))+1//Ratio of specific heats

+e=(1-((1/g)*(1/r^(g-1))*((k^g-1)/(k-1))))//Air standard efficiency of the cycle

+dee=((-(dc/100)*((1-e)/e)*(g-1)*((1/g)+log(r)-((k^g*log(k))/(k^g-1))))*100)//Change in efficiency due to 1% change in cv

+

+//Output

+printf('Percentage change in air standard efficiency is %3.3f percent \n This indicates that there is a decrease in efficiency',dee)

diff --git a/2168/CH5/EX5.2/Chapter5_example2.sce b/2168/CH5/EX5.2/Chapter5_example2.sce
new file mode 100755
index 000000000..f6e839545
--- /dev/null
+++ b/2168/CH5/EX5.2/Chapter5_example2.sce
@@ -0,0 +1,16 @@
+clc

+clear

+//Input data

+r=8//Compression ratio

+n=1.41//Adiabatic index of the medium

+cv=0.17//Mean Specific heat at constant volume in kcal/kg/degree C

+x=2//Percentage with which spcific heat at constant volume increases

+R=29.3//Characteristic gas constant in mkg/kg/degree C

+J=427//Mechanical equivalent of heat in kg.m/kcal

+

+//Calculations

+e=(1-(1/r^(n-1)))//Air standard efficiency neglecting the variation in specific heat

+debye=((x/100)*((1-e)/e)*(R/(J*cv))*log(r))*100//Ratio of de and e in percent

+

+//Output

+printf('The change in air standard efficiency of the cycle is %3.3f percent',debye)

diff --git a/2168/CH5/EX5.3/Chapter5_example3.sce b/2168/CH5/EX5.3/Chapter5_example3.sce
new file mode 100755
index 000000000..8275fb25c
--- /dev/null
+++ b/2168/CH5/EX5.3/Chapter5_example3.sce
@@ -0,0 +1,15 @@
+

+clc

+clear

+//Input data

+//Cv=0.125+0.000005T where Cv is Specific heat at constant volume and T is the temperature in K

+R=28.9//Characteristic gas constant in mkg/kg/degree C

+T=[100+273,50+273]//Temperature in K

+J=427//Mechanical equivalent of heat in kg.m/kcal

+

+//Calculations

+function y=f(x),y=0.125+(0.00005*x),endfunction

+I=J*intg(303,373,f)

+

+//Output

+printf('The work done is %i m.kg/kg of gas',I)

diff --git a/2168/CH5/EX5.4/Chapter5_example4.sce b/2168/CH5/EX5.4/Chapter5_example4.sce
new file mode 100755
index 000000000..ff76ebb48
--- /dev/null
+++ b/2168/CH5/EX5.4/Chapter5_example4.sce
@@ -0,0 +1,23 @@
+clc

+clear

+//Input data

+af=25//Air fuel ratio

+cv=[0.17,0.00004]//Cv=0.17+0.00004T where Cv is Specific heat at constant volume and T is the temperature in K

+r=14//Compression ratio

+p1=1//Pressure at the beginning of compression in kg/cm^2

+T1=153+273//Temperature at the beginning of compression in K

+CV=10000//Heating value of fuel in kcal/kg

+n=1.35//Adiabatic constant

+R=29//Characteristic gas constant in mkg/kg.K

+J=427//Mechanical equivalent of heat in kg.m/kcal

+

+//Calculations

+T2=(T1*r^(n-1))//Temperature at the end of compression in K

+a=(cv(2)/2)//For solving T3

+b=cv(1)+(R/J)//For solving T3

+c=(-T2*cv(1))-((cv(2)/2)*T2^2)-((R/J)*T2)-(CV/(af+1))//Foe solving T3

+T3=(-b+sqrt(b^2-(4*a*c)))/(2*a)//Soving for T3 in K

+pc=(((T3/T2)-1)/(r-1))*100//Percentage cut off

+

+//Output

+printf('The percentage of stroke at which the constant pressure combustion stops is %i percent',pc)

diff --git a/2168/CH5/EX5.5/Chapter5_example5.sce b/2168/CH5/EX5.5/Chapter5_example5.sce
new file mode 100755
index 000000000..ef75811e5
--- /dev/null
+++ b/2168/CH5/EX5.5/Chapter5_example5.sce
@@ -0,0 +1,22 @@
+clc

+clear

+//Input data

+af=25//Air fuel ratio

+CV=10000//Calorific value in kcal/kg

+cv=[0.17,0.00004]//Cv=0.17+0.00004T where Cv is Specific heat at constant volume and T is the temperature in K

+r=14//Compression ratio

+T2=800+273//Temperature at the end of compression in K

+R=29//Characteristic gas constant in mkg/kg/degree C

+J=427//Mechanical equivalent of heat in kg.m/kcal

+

+//Calculations

+CVm=(CV/(af+1))//Calorific value of mixture in kcal/kg

+cpv=(R/J)//Difference in mean specific heats in kcal/kg mol.K

+a=(cv(2)/2)//For solving T3

+b=cpv+cv(1)//For solving T3

+c=(-T2*(cpv+cv(1)))-((cv(2)/2)*T2^2)-CVm//Foe solving T3

+T3=(-b+sqrt(b^2-(4*a*c)))/(2*a)//Soving for T3 in K

+s=((T3/T2)/(r-1))*100//Percentage of the stroke

+

+//Output

+printf('The percentage of the stroke at which the combustion will be complete is %3.2f percent',s)

diff --git a/2168/CH5/EX5.6/Chapter5_example6.sce b/2168/CH5/EX5.6/Chapter5_example6.sce
new file mode 100755
index 000000000..bfbd98bb6
--- /dev/null
+++ b/2168/CH5/EX5.6/Chapter5_example6.sce
@@ -0,0 +1,14 @@
+clc

+clear

+//Input data

+T=[500,2000]//Change in temperature in K

+x=[11.515,-172,1530]//Cp=11.515-172/sqrt(T)+1530/T in kcal/kg mole.K

+mO2=32//Molecular weight of oxygen

+

+//Calculations

+function y=f(T),y=(x(1)+(x(2)/sqrt(T))+(x(3)/T)),endfunction

+I=-intg(T(2),T(1),f)//Integration

+dh=(I/mO2)//Change in enthalpy in kcal/kg

+

+//Output

+printf('The change in enthalpy is %3.1f kcal/kg',dh)

diff --git a/2168/CH5/EX5.7/Chapter5_example7.sce b/2168/CH5/EX5.7/Chapter5_example7.sce
new file mode 100755
index 000000000..e92f9a45f
--- /dev/null
+++ b/2168/CH5/EX5.7/Chapter5_example7.sce
@@ -0,0 +1,34 @@
+clc

+clear

+//Input data

+r=14//Compression ratio

+s=5//Fuel injection stops at 5% stroke after inner head centre

+pm=50//Maximum pressure in kg/cm^2

+p4=1//Pressure at the end of suction stroke in kg/cm^2

+T4=90+273//Temperature at the end of suction stroke in K

+R=29.3//Characteristic gas constant in mkg/kg/degree C

+cv=[0.171,0.00003]//Cv=0.171+0.00003T where Cv is Specific heat at constant volume and T is the temperature in K

+J=427//Mechanical equivalent of heat in kg.m/kcal

+

+//Calculations

+a=(R/J)+cv(1)//a value in kcal/kg.mole.K

+g=(a+cv(2)*T4)/(cv(1)+cv(2)*T4)//Adiabatic index of compression

+z=1.3//Rounding off 'z' value to one decimal.

+T5=(T4*r^(z-1))//Temperature in K

+p5=(p4*r^g)//Pressure in kg/cm^2

+T1=T5*(pm/p5)//Tmperature in K

+T2=(T1*(1+(s/100)*(r-1)))//Temperature in K

+T3=(T2*((1+(s/100)*(r-1))/r)^(g-1))//Temperature in K

+p3=(p4*(T3/T4))//Pressure in kg/cm^2

+function y=f1(T),y=cv(1)+(cv(2)*T),endfunction

+I1=intg(T5,T1,f1)

+function y=f2(T),y=(a+(cv(2)*T)),endfunction

+I2=intg(T1,T2,f2)//I2 answer is given wrong in the textbook

+qs=(I1+I2)//Heat supplied per kg of air in kcal/kg

+function y=f3(T),y=a+(cv(2)*T),endfunction

+qre=intg(T4,T3,f3)//Heat required per kg of air in kcal/kg

+nth=((qs-qre)/qs)*100//Thermal efficiency in percent

+

+//Output

+printf('The tempertautes and pressures at salient points of the cycle are : \n T1 = %3.0f K  p1 = %3.1f kg/cm^2 \n T2 = %3.0f K  p2 = %3.1f kg/cm^2 \n T3 = %3.0f K  p3 = %3.1f kg/cm^2 \n T4 = %3.0f K  p4 = %3.1f kg/cm^2 \n T5 = %3.0f K  p5 = %3.1f kg/cm^2 \n\n Heat supplied per kg of air is %3.1f kcal/kg \n\n The thermal efficiency of the cycle is %3.1f percent',T1,pm,T2,pm,T3,p3,T4,p4,T5,p5,qs,nth)

+//Textbook answers are given wrong

diff --git a/2168/CH5/EX5.8/Chapter5_example8.sce b/2168/CH5/EX5.8/Chapter5_example8.sce
new file mode 100755
index 000000000..6a6656729
--- /dev/null
+++ b/2168/CH5/EX5.8/Chapter5_example8.sce
@@ -0,0 +1,43 @@
+clc

+clear

+//Input data

+r=14//Compression ratio

+c=5//Cut off takes place at 5% of the stroke

+p1=1//Pressure at the beginning of compression in kg/cm^2. In texbook, it is given wrong as 10

+T1=90+273//Temperature at the beginning of compression in K

+p3=50//Maximum pressure in kg/cm^2

+R=29.3//Characteristic gas constant in mkg/kg/degree C

+cv=[0.171,0.00003]//Cv=0.171+0.00003T where Cv is Specific heat at constant volume and T is the temperature in K

+g1=1.4//Ratio of specific heats

+J=427//Mechanical equivalent of heat in kg.m/kcal

+

+//Calculations

+T2x=(T1*r^(g1-1))//Temperature in K

+function y=f1(T),y=cv(1)+(cv(2)*T),endfunction

+I1=intg(T1,T2x,f1)

+Cv=(1/(T2x-T1))*I1//Mean value of Cv in kJ/kg.K

+Cp=(Cv+(R/J))//Mean value of Cp in kJ/kg.K

+g=1.35//(Cp/Cv) value and rounded off to 2 decimal places for calculation purpose. Ratio of specific heats

+T2=(T1*r^(g-1))//Temperature in K

+I2=intg(T1,T2,f1)

+CV=(1/(T2-T1))*I2//Maen value of Cv in kJ/kg.K

+CP=(Cv+(R/J))//Mean value of Cp in kJ/kg.K

+g2=1.36//(Cp/Cv) value and rounded off to 2 decimal places for calculation purpose.Ratio of specific heats

+T2a=(T1*r^(g2-1))//Temperature in K

+p2=(p1*r*(T2a/T1))//Pressure in kg/cm^2

+T3=(T2a*(p3/p2))//Temperature in K

+T4=(((r-1)*(c/100))+1)*T3//Temperature in K

+g3=1.3//Assuming gamma as 1.3 for process 4-5

+T5=(T4/(r/(((r-1)*(c/100))+1))^(g3-1))//Temperature in K

+cV=cv(1)+(cv(2)/2)*(T5+T4)//Mean value of Cv in kJ/kg.K

+cP=cV+(R/J)//Mean value of Cp in kJ/kg.K

+g4=(cP/cV)//Ratio of specific heats

+T5a=(T4/(r/(((r-1)*(c/100))+1))^(g4-1))

+I3=intg(T2a,T3,f1)

+function y=f2(T),y=cv(1)+(R/J)+(cv(2)*T),endfunction

+I4=intg(T3,T4,f2)//Textbook answer is wrong

+q=I3+I4//Heat supplied per kg of working substance in kcal/kg

+

+//Output

+printf('(a) Temperatures at all the points of the cycle are: \n T1 = %i K \n T2 = %3.0f K \n T3 = %3.0f K \n T4 = %3.0f K \n T5 = %i K \n\n (b) heat supplied per kg of the working substance is %3.1f kcal/kg',T1,T2a,T3,T4,T5a,q)

+//Textbook answer is wrong