initial commit / add all books

6 files changed, 248 insertions, 0 deletions

diff --git a/2657/CH3/EX3.1/Ex3_1.sce b/2657/CH3/EX3.1/Ex3_1.sce
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+//Effect of variable specific heat on efficiency

+clc,clear

+//Given:

+r=7 //Compression ratio

+g=1.4 //Specific heat ratio(gamma)

+cv=0.718 //(Assume)Specific heat at constant volume in kJ/kgK

+dcv=1*cv/100 //Change in specific heat in kJ/kgK

+//Solution:

+R=cv*(g-1) //Specific gas constant in kJ/kgK

+eta=round(100*(1-1/r^(g-1)))/100 //Efficiency when there is no change in specific heat

+function [eta]=Otto(cv) //Defining efficiency as a function of specific heat

+    eta=1-1/r^(R/cv)

+endfunction

+funcprot(0)

+detaBydcv=derivative(Otto,cv) //Derivative of efficiency wrt to specific heat at initial value of specific heat

+detaByeta=detaBydcv*dcv/eta //Change in efficiency wrt to initial value of efficiency

+//Results:

+printf("\n The percentage change in the efficiency of Otto cycle = %.3f percent",detaByeta*100)

+if (detaByeta < 0) then

+    disp("decrease")

+end

diff --git a/2657/CH3/EX3.2/Ex3_2.sce b/2657/CH3/EX3.2/Ex3_2.sce
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index 000000000..351de2c9e
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+//Effect of variable specific heat on maximum pressure

+clc,clear

+//Given:

+r=6 //Compression ratio

+CV=44000 //Calorific value in kJ/kg of fuel

+A_F=15/1 //Air-fuel ratio

+P1=1 //Pressure at 1 in bar

+T1=60+273 //Temperature at 1 in K

+n=1.32 //Index of compression

+T=poly(0,"T") //Defining temperature(T) as unknown in K

+cv=0.71+20D-5*T //Specific heat at constant volume as a function of temperature(T) in kJ/kgK

+cv_c=0.71 //Constant specific heat in kJ/kgK

+//Solution:

+//Refer fig 3.19

+P2=P1*r^n //Pressure at 2 in bar

+T2=floor(T1*r^(n-1)) //Temperature at 2 in K

+T3=poly(0,"T3") //Defining temperature(T3) as unknown in K

+T_av=(T3+T2)/2 //Average temperature during combustion of charge in K

+cv_mean=horner(cv,T_av) //Mean specific heat in kJ/kgK

+//Assume cycle consumes 1 kg of air

+m_a=1,m_f=m_a/A_F,m_c=m_f+m_a //Mass of air, fuel, and charge in kg

+Q1=CV*m_f //Heat added per kg of air in kJ/kg

+T3_v=roots(Q1-cv_mean*m_c*(T3-T2)),T3_v=T3_v(2) //Temperature at 3 in K

+P3_v=P2*T3_v/T2 //Pressure at 3 in bar

+//For constant specific heat

+T3_c=roots(Q1-cv_c*m_c*(T3-T2)) //Temperature at 3 for constant specific heat in K

+P3_c=P2*T3_c/T2 //Pressure at 3 for constant specific heat in bar

+//Results:

+printf("\n The maximum pressure in the cycle for variable specific heat, P3 = %.1f bar",P3_v)

+printf("\n The maximum pressure in the cycle for constant specific heat, P3 = %.1f bar\n\n",P3_c)

diff --git a/2657/CH3/EX3.3/Ex3_3.sce b/2657/CH3/EX3.3/Ex3_3.sce
new file mode 100755
index 000000000..9d7fa463c
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+++ b/2657/CH3/EX3.3/Ex3_3.sce
@@ -0,0 +1,30 @@
+//Calculations on diesel engine

+clc,clear

+//Given:

+A_F=28/1 //Air-fuel ratio

+CV=42000 //Calorific value in kJ/kg

+cv='0.71+20D-5*T' //Specific heat at constant volume as a function of temperature(T) in kJ/kgK

+R='0.287' //Specific gas constant in kJ/kgK

+r=14/1 //Compression ratio

+T2=800 //Temperature at the end of the compression process (2) in K

+//Solution:

+//Refer fig 3.20

+//Assume cycle consumes 1 kg of fuel

+m_c=A_F*1+1 //Mass of charge in kg

+cp=addf(cv,R) //Specific heat at constant pressure as a function of temperature(T) in kJ/kgK

+//Since, heat transfer at constant pressure, Q1 = integration(cp*dt) from T2 to T3

+//Thus, Q1 is the function of T3. Defining the function Q1 of T3

+function [Q1toCV]=difference(T3)

+    Q1=integrate(cp,'T',T2,T3)

+    Q1toCV=Q1-CV/m_c

+endfunction

+//Since, heat transfer at constant pressure must be equal to calorific value per kg of charge

+//Thus, their difference must be zero, function Q1toCV is solve for zero

+T3=fsolve(1,difference)

+T3=round(T3) //Temperature at the end of constant pressure proces (3) in K

+rho=T3/T2 //Cut off ratio

+V2=1 //Assume clearance volume in unit

+V3=rho //Volume at 3 in units

+p=(V3-V2)*100/(r-V2) //percentage of stroke at which constant pressure process ends

+//Results:

+printf("\n At %.2f percentage of stroke combustion is completed.\n\n",p)

diff --git a/2657/CH3/EX3.4/Ex3_4.sce b/2657/CH3/EX3.4/Ex3_4.sce
new file mode 100755
index 000000000..6ad3afa39
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+++ b/2657/CH3/EX3.4/Ex3_4.sce
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+//Calculations on dual combustion cycle

+clc,clear

+//Given:

+P1=1 //Pressure at 1 in bar

+T1=90+273 //Temperature at 1 in K

+r=13 //Compression ratio

+Q1=1675 //Heat supplied per kg of air in kJ/kg

+Q1_v=Q1/2,Q1_p=Q1/2 //Heat supplied at constant volume and pressure per kg of air in kJ/kg

+g=1.4 //Specific heat ratio(gamma)

+R='0.287' //Specific gas constant in kJ/kgK

+cv='0.71+20D-5*T' //Specific heat at constant volume as a function of temperature(T) in kJ/kgK

+//Solution:

+//Refer fig 3.21

+P2=P1*r^g //Pressure at 2 in bar

+T2=T1*r^(g-1) //Temperature at 2 in K

+//Since, heat transfer at constant volume, Q1_v = integration(cv*dt) from T2 to T3

+//Thus, Q1_v is the function of T3. Defining the function Q1_v of T3

+function [Q1_vtoQ1]=Volume(T3)

+    Q1_v=integrate(cv,'T',T2,T3)

+    Q1_vtoQ1=Q1_v-Q1/2

+endfunction

+//Since, heat transfer at constant volume must be equal to half of total heat added

+//Thus, their difference must be zero, function Q1_vtoQ1 is solve for zero

+T3=fsolve(1,Volume) //Temperature at 3 in K

+P3=P2*T3/T2 //Pressure at 3 in bar

+cp=addf(cv,R) //Specific heat at constant pressure as a function of temperature(T) in kJ/kgK

+//Since, heat transfer at constant pressure, Q1_p = integration(cp*dt) from T3 to T4

+//Thus, Q1_p is the function of T4. Defining the function Q1_p of T4

+function [Q1_ptoQ1]=Pressure(T4)

+    Q1_p=integrate(cp,'T',T3,T4)

+    Q1_ptoQ1=Q1_p-Q1/2

+endfunction

+//Since, heat transfer at constant pressure must be equal to half of total heat added

+//Thus, their difference must be zero, function Q1_ptoQ1 is solve for zero

+T4=fsolve(1,Pressure) //Temperature at 4 in K

+rho=T4/T3 //Cut off ratio

+p=(rho-1)*100/(r-1) //Percentage of stroke at which cut off occurs

+//Results:

+printf("\n The maximum pressure in the cycle, P3 = %.1f bar",P3)

+printf("\n The percentage of stroke at which cut off occurs = %.2f percent\n\n",p)

diff --git a/2657/CH3/EX3.5/Ex3_5.sce b/2657/CH3/EX3.5/Ex3_5.sce
new file mode 100755
index 000000000..920b01c13
--- /dev/null
+++ b/2657/CH3/EX3.5/Ex3_5.sce
@@ -0,0 +1,44 @@
+//Effect of molecular contraction
+clc,clear
+//Given:
+r=7 //Compression ratio
+CV=44000 //Calorific value of the fuel in kJ/kg
+A_F=13.67 //Air fuel ratio of the mixture
+cv=0.718 //Specific heat at constant volume in kJ/kgK
+n=1.3 //Polytropic index
+P1=1,T1=67+273 //Pressure and temperature at the beginning in bar and K
+//Solution:
+//Refer fig 3.22
+C=12 //Atomic mass of Carbon(C)
+H=1 //Atomic mass of Hydrogen(H)
+O=16 //Atomic mass of Oxygen(O)
+p=23 //Percentage of oxygen in air by mass
+//Stoichiometric equation of combustion of fuel (C6H14)
+//   [C6H14] + x[O2]   =   y[CO2] + z[H2O]
+//Equating coefficients
+x=9.5,y=6,z=7 //Coefficients of stoichiometric equation
+A_F_g=x*2*O/(6*C+14*H)*100/p //Gravimetric air fuel ratio
+MS=A_F_g/A_F*100 //Actual mixture strength in percent
+//Combustion is incomplete
+//Stoichiometric equation of incomplete combustion of fuel (C6H14)
+//    MS/100[C6H14] + x[O2] = a[CO2] + b[CO] + c[H2O]
+//Equating coefficients
+a=4.39,b=2.36,c=7.87 //Coefficients of stoichiometric equation
+//Stoichiometric equation of combustion of fuel (C6H14) by adding Nitrogen
+//   MS/100[C6H14] + x[O2] + x*79/21[N2] = a[CO2] + b[CO] + c[H2O] + x*79/21[N2]
+m1=MS/100+x+x*79/21 //Moles before combustion
+m2=a+b+c+x*79/21 //Moles after combustion
+Me=(m2-m1)/m1*100 //Molecular expansion in percent
+T2=T1*r^(n-1) //Temperature at 2 in K
+m_c=A_F+1 //Mass of charge in kg
+T3=CV/(m_c*cv)+T2 //Temperature at 3 in K
+T3=round(T3)
+P3=P1*r*(T3/T1) //Pressure at 3 in bar (printing error)
+//Temperature and pressure considering molecular expansion
+T3!=T3 //Temperature remains same at 3 in K
+P3!=P3*m2/m1 //Pressure at 3 in bar
+//Results:
+printf("\n\t The molecular expansion = %.2f percent\n",Me)
+printf("\n (a)Without considering the molecular contraction\n\t The maximum pressure, P3 = %.2f bar\n\t The maximum temperature, T3 = %.0f K",P3,T3)
+printf("\n (b)Considering the molecular contraction\n\t The maximum pressure, P3 = %.2f bar\n\t The maximum temperature, T3 = %.0f K",P3!,T3!)
+//Answer in the book is wrong
diff --git a/2657/CH3/EX3.6/Ex3_6.sce b/2657/CH3/EX3.6/Ex3_6.sce
new file mode 100755
index 000000000..fff92ea3e
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+++ b/2657/CH3/EX3.6/Ex3_6.sce
@@ -0,0 +1,83 @@
+//Calculations on Otto cycle
+clc,clear
+//Given:
+p=15 //Clearance volume in percentage of displacement volume
+V_s=2.8 //Swept volume in litres
+N=2500 //Engine speed in rpm
+Q1=1400 //Heat added in kJ/kg
+T1=27+273 //Temperature at inlet in K
+P1=100 //Pressure at inlet in kPa
+R=0.287 //Specific gas constant in kJ/kgK
+//Solution:
+//Refer fig 3.23
+//By using gas tables
+//Refer Ideal-gas properties of air
+V2=(p/100)*(V_s/1000) //Volume at 2 (Clearance volume) in m^3
+V3=V2 //Volume at 3 in m^3
+V1=V_s/1000+V2,V4=V1 //Volume at 1, 4 in m^3
+//  Process 1-2
+vr1=621.2,pr1=1.3860,u1=214.09,phi1=5.7016 //Relative specific volume, relative pressure, specific internal energy(kJ/kg), specific entropy(kJ/kgK) at 1 (from air tables)
+vr2=vr1*(V2/V1) //Relative specific volume at 2
+vr=[81.89 78.61],T=[660 670],pr=[23.13 24.46],u=[481.01 488.81] //Relative specific volume, temperature(K), relative pressure, specific internal energy(kJ/kg) (extracted from air tables)
+//Finding the corresponding temperature at vr2 by interpolation
+T2=interpln([vr;T],vr2) //Temperature at 2 in K
+//Finding the corresponding relative pressure at T2 by interpolation
+pr2=interpln([T;pr],T2) //Relative pressure at 2
+//Finding the corresponding specific internal energy at T2 by interpolation
+u2=interpln([T;u],T2) //specific internal energy at 2 in kJ/kg
+P2=P1*(pr2/pr1) //Pressure at 2 in kPa
+//  Process 2-3
+u3=Q1+u2 //Specific internal energy at 3 in kJ/kg
+vr=[2.356 2.175 2.012],T=[2100 2150 2200],pr=[2559 2837 3138],u=[1775.3 1823.8 1872.8] //Relative specific volume, temperature(K), relative pressure, specific internal energy(kJ/kg) (extracted from air tables)
+//Finding the corresponding relative specific volume at u3 by interpolation
+vr3=interpln([u;vr],u3) //Relative specific volume at 3
+//Finding the corresponding relative pressure at u3 by interpolation
+pr3=interpln([u;pr],u3) //Relative pressure at 3
+//Finding the corresponding temperature at u3 by interpolation
+T3=interpln([u;T],u3) //Temperature at 3(maximum) in K (Round off error)
+P3=P2*(T3/T2) //Pressure at 3(maximum) in kPa
+//  Process 3-4
+vr4=vr3*(V4/V3) //Relative specific volume at 4
+vr=[15.241 14.470],T=[1180 1200],pr=[222.2 238.0],u=[915.57 933.33],phi=[7.1586 7.1684] //Relative specific volume, temperature(K), relative pressure, specific internal energy(kJ/kg), specific entropy(kJ/kgK) (extracted from air tables)
+//Finding the corresponding temperature at vr4 by interpolation
+T4=interpln([vr;T],vr4) //Temperature at 4 in K
+//Finding the corresponding specific internal energy at T4 by interpolation
+u4=interpln([T;u],T4) //Specific internal energy at 4 in kJ/kg
+//Finding the corresponding relative pressure at T4 by interpolation
+pr4=interpln([T;pr],T4) //Relative pressure at 4
+P4=P3*(pr4/pr3) //Pressure at 4 in kPa
+//Finding the corresponding specific entropy at T4 by interpolation
+phi4=interpln([T;phi],T4) //Specific entropy at 4 in kJ/kgK
+//  Process 4-1
+Q2=u1-u4 //Heat rejected in kJ/kg
+W=Q1+Q2 //Work done in kJ/kg
+eta=W/Q1 //Efficiency
+m=P1*V1/(R*T1) //Mass of air in cycle in kg
+W=m*W*N/60 //Rate of work in kW
+Delta_s=phi1-phi4-R*log(P1/P4) //Change in specific entropy between 1 and 4 in kJ/kgK
+AE=Q2-T1*(Delta_s) //Available portion of energy of Q2 in kJ/kg (Round off error)
+p_AE=AE/Q2 //Available energy in percentage of Q2
+//  Without using gas tables
+g=1.4 //Specific heat ratio(gamma)
+cv=0.718 //Specific heat at constant volume in kJ/kgK
+r=V1/V2 //Compression ratio
+eta!=1-1/r^(g-1) //Efficiency
+//  Process 1-2
+T2=T1*(r)^(g-1) //Temperature at 2 in K
+P2=P1/100*(r)^g //Pressure at 2 in bar
+//  Process 2-3
+T3!=Q1/cv+T2 //Temperature at 3(maximum) in K
+P3!=P2*T3!/T2 //Pressure at 3(maximum) in bar
+//  Process 3-4
+T4=T3!*(1/r)^(g-1) //Temperature at 4 in K
+Q2=cv*(T1-T4) //Heat rejected in kJ/kg
+W!=Q1+Q2 //Work done in kJ/kg
+eta!=W!/Q1 //Efficiency
+power=m*W!*N/60 //Power in kW
+Delta_s=cv*log(T1/T4) //Change in specific entropy between 1 and 4 in kJ/kgK
+AE!=Q2-T1*Delta_s //Available portion of energy of Q2 in kJ/kg (Round off error)
+p_AE!=AE!/Q2 //Available energy in percentage of Q2 (Round off error)
+//Results:
+printf("\n Constant specific heat:\n\t Maximum temperature, Tmax = %.1f K\n\t Maximum pressure, Pmax = %.1f bar\n\t Thermal efficiency, eta = %.2f percent\n\t Power = %.1f kW\n\t Available portion of heat rejected = %.1f kJ/kg (%.1f percent)",T3!,P3!,eta!*100,power,abs(AE!),p_AE!*100)
+printf("\n Variable specific heat:\n\t Maximum temperature, Tmax = %.0f K\n\t Maximum pressure, Pmax = %.1f bar\n\t Thermal efficiency, eta = %.1f percent\n\t Power = %.1f kW\n\t Available portion of heat rejected = %.1f kJ/kg (%.1f percent)",T3,P3/100,eta*100,W,abs(AE),p_AE*100)
+//Round off error in 'T3', 'AE', 'AE!', 'p_AE!'