initial commit / add all books

8 files changed, 288 insertions, 0 deletions

diff --git a/1943/CH3/EX3.1/Ex3_1.sce b/1943/CH3/EX3.1/Ex3_1.sce
new file mode 100755
index 000000000..540215eb9
--- /dev/null
+++ b/1943/CH3/EX3.1/Ex3_1.sce
@@ -0,0 +1,34 @@
+
+clc

+clear

+//Input data

+p=40//Pressure in bar

+T1=400+273//Temperature in K

+T2=40+273//Temperature in K

+x=[10,515.5,72.23,363.0,0.1478,0.5167,80.9*10^-6,0.0333]//Property values from table p(bar),t(degree C), hf,hg(kJ/kg),sf,sg(kJ/kg.K),vf,vg(m^3/kg)

+y=[0.2,277.3,38.35,336.55,0.0967,0.6385,77.4*10^-6,1.163]//Property values from table p(bar),t(degree C), hf,hg(kJ/kg),sf,sg(kJ/kg.K),vf,vg(m^3/kg)

+

+//Calculations

+h1=3216//Enthalpy in kJ/kg

+s1=6.7690//Entropy in kJ/kg.K

+s2=s1//Entropy in kJ/kg.K

+x2=(s2-0.5725)/(8.2570-0.5725)//Dryness fraction

+h2=167.57+x2*2406.7//Enthalpy in kJ/kg

+h3=167.57//Enthalpy in kJ/kg

+h4=(167.57+p*100*1.008*10^-3)//Enthalpy in kJ/kg

+h5=1087.31//Enthalpy in kJ/kg

+h6=2801.4//Enthalpy in kJ/kg

+ha=x(4)//Enthalpy in kJ/kg

+sa=x(6)//Entropy in kJ/kg.K

+sb=sa//Entropy in kJ/kg.K

+xb=(sb-y(5))/(y(6)-y(5))//Dryness fraction

+hb=(y(3)+xb*(y(4)-y(3)))//Enthalpy in kJ/kg

+hc=y(3)//Enthalpy in kJ/kg

+hd=hc//Enthalpy in kJ/kg

+m=(h6-h5)/(hb-hc)//Mass of mercury circulated per kg of steam

+Q1=m*(ha-hd)+(h1-h6)+(h5-h4)//Heat supplied in kJ/kg

+Q2=(h2-h3)//Heat rejected in kJ/kg

+nc=(1-(Q2/Q1))*100//Efficiency in percent

+

+//Output

+printf('(a) The amount of mercury circulated per kg of water is %3.4f kg \n (b) The efficiency of the combined cycle is %3.1f percent',m,nc)

diff --git a/1943/CH3/EX3.2/Ex3_2.sce b/1943/CH3/EX3.2/Ex3_2.sce
new file mode 100755
index 000000000..f2783e253
--- /dev/null
+++ b/1943/CH3/EX3.2/Ex3_2.sce
@@ -0,0 +1,40 @@
+
+clc

+clear

+//Input data

+m=5//Mass flow rate in kg/s

+p1=40//Pressure in bar

+T1=440+273//Temperature in K

+p2=1.5//Pressure in bar

+p3=1//Pressure in bar

+T3=60+273//Temperature in K

+p4=16//Pressure in bar

+T4=100+273//Temperature in K

+p5=9//Pressure in bar

+

+//Calculations

+h1=3307.1//Enthalpy in kJ/kg

+s1=6.9041//Entropy in kJ/kg.K

+s2=s1//Entropy in kJ/kg.K

+h2=2570.8//Enthalpy in kJ/kg

+h3=417.46//Enthalpy in kJ/kg

+h6=(251.13+(1.0172*10^-3)*(p3-0.1994)*100)//Enthalpy in kJ/kg

+m3=(m/2)//Mass flow rate in kg/s

+m6=m3//Mass flow rate in kg/s

+h4=(m3*h3+m6*h6)/m//Enthalpy in kJ/kg

+h5=(h4+(1.0291*10^-3)*(p1-p3)*100)//Enthalpy in kJ/kg

+ha=241.58//Enthalpy in kJ/kg

+sa=0.7656//Entropy in kJ/kg.K

+sb=sa//Entropy in kJ/kg.K

+hb=229.43//Enthalpy in kJ/kg

+hc=71.93//Enthalpy in kJ/kg

+hd=hc+(0.7914*10^-3*(p4-p5)*100)//Enthalpy in kJ/kg

+Q1=(m*(h1-h5))/1000//Heat supplied in kW

+Wnets=(m*((h1-h2)-(h5-h4)))//Net workdone by steam in kW

+mR12=(m3*(h2-h3))/(ha-hd)//Mass of R12 in kg/s

+WnetR=(mR12*((ha-hb)-(hd-hc)))//Net workdone by R12 in kW

+T=Wnets+WnetR//Total output in kW

+Qh=(m6*(h2-h6))//Heat rejected in kW

+

+//Output

+printf('(a) Rate of heat transfer in the steam generator is %3.3f kW \n (b) The net power output of the binary cycle is %i kW \n (c) The rate of heat transfer to the industrial process is %3.0f kW',Q1,T,Qh)

diff --git a/1943/CH3/EX3.3/Ex3_3.sce b/1943/CH3/EX3.3/Ex3_3.sce
new file mode 100755
index 000000000..7594f5463
--- /dev/null
+++ b/1943/CH3/EX3.3/Ex3_3.sce
@@ -0,0 +1,41 @@
+
+clc

+clear

+//Input data

+rp=7.5//Pressure ratio 

+T1=15+273//Inlet air temperature in K

+T3=750+273//Maximum temperature in K

+T6=100+273//Temperature in K

+p1=50//Pressure in bar

+T7=600+273//Temperature in K

+p2=0.1//Pressure in bar

+P=200//Total power in MW

+CV=43.3//calorific value in MJ/kg

+cpg=1.11//Specific heat for gas in kJ/kg.K

+g=1.33//Ratio of specific heats for gas

+cpa=1.005//Specific heat for air in kJ/kg.K

+g1=1.4//Ratio of specific heats for air

+

+//Calculations

+T2=(T1*rp^((g1-1)/g1))//Temperature in K

+T4=(T3/rp^((g-1)/g))//Temperature in K

+ha=3670//Enthalpy in kJ/kg

+hb=2305//Enthalpy in kJ/kg

+hc=192//Enthalpy in kJ/kg

+hd=hc//Enthalpy in kJ/kg

+//ma*cpg*(T3-T6)=ms*(ha-hd)

+//ma*cpg*(T3-T4)-ma*cpa*(T2-T1)+ms*(ha-hb)=P*1000

+//Solving these two equations

+A=[cpg*(T3-T6) (hd-ha)

+   cpg*(T3-T4)-cpa*(T2-T1) (ha-hb)]//Coefficient matrix

+B=[0 

+   (P*10^3)]//Constant matrix

+X=inv(A)*B//Variable matrix

+Wgt=(cpg*(T3-T4)-cpa*(T2-T1))*X(1)*10^-3//Net workdone by Gas turbine in MW

+Wst=(P-Wgt)//Net workdone by steam turbine in MW

+Q1=(X(1)*cpg*(T3-T2+T3-T4))//Heat supplied in MW

+nth=(P/(Q1*10^-3))*100//Thermal efficiency in percent

+af=(CV*10^3)/(cpg*(T3-T2+T3-T4))//Air fuel ratio

+

+//Output

+printf('(a) The flow rate of air is %3.2f kg/s and steam is %3.2f kg/s \n (b) The power outputs of the gas turbine is %3.2f MW and steam turbine is %3.2f MW \n (c) The thermal efficiency of the combined plant is %3.0f percent \n (d) The air fuel ratio is %3.1f',X(1),X(2),Wgt,Wst,nth,af)

diff --git a/1943/CH3/EX3.4/Ex3_4.sce b/1943/CH3/EX3.4/Ex3_4.sce
new file mode 100755
index 000000000..77a3b0487
--- /dev/null
+++ b/1943/CH3/EX3.4/Ex3_4.sce
@@ -0,0 +1,90 @@
+
+clc

+clear

+//Input data

+p1=1//Pressure in bar

+T1=25+273//Temperature in K

+rp=8//Pressure ratio of compressor

+Tm=900+273//Maximum temperature in K

+pd=3//pressure drop in combustion chamber in percent

+nc=0.88//Efficiency of compressor

+nt=0.88//Efficiency of turbine

+CV=44.43//Calorific value of fuel in MJ/kg

+cpa=1.006//Specific heat of air in kJ/kg.K

+cpg=1.148//Specific heat of gas in kJ/kg.K

+g1=1.333//Specific heat ratio of gas

+g=1.4//Specific heat ratio of air

+T3=425+273//Temperature in K

+p2=40//Pressure in bar

+p3=0.04//Condensor pressure in bar

+Th=170.4+273//Temperature of feed water to the HRSG in K

+nst=0.82//Efficiency of steam turbine

+pdh=5//Pressure drop in HRSG in kPa

+m=29.235//Steam flow rate in kg/s

+A=1.0401//si=1.0401+0.1728*(h/c)

+B=0.1728//si=1.0401+0.1728*(h/c)

+

+//Calculations

+//Gas turbine plant

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

+//Combustor

+pc=((pd/100)*rp)//Pressure loss in bar

+pcx=(rp-pc)//Pressure in bar

+f=((cpg*(Tm-T1))-(cpa*(T2-T1)))/((CV*10^3)-(cpa*(T2-T1)))//Fuel flow rate in kg/s

+af=(1-f)/f//Air fuel ratio

+//C8H18+12.5O2->8CO2+9H2O

+afc=(12.5*32)/(0.232*114)//Air fuel ratio for stoichiometric combustion

+ea=((af-afc)/afc)*100//Excess air in percent

+//Gas turbine

+p4=p1+0.05//Pressure in bar

+T4=(Tm/(pcx/p4)^(((g1-1)*nt)/g1))//Temperature in K

+//HRSG

+T5=250+30//Temeprature in K

+ha=3272//Enthalpy in kJ/kg

+hf=1087.31//Enthalpy in kJ/kg

+ws=(cpg*((T4-273)-T5))/(ha-hf)//Flow rate in kg/s

+he=721.1//Enthalpy in kJ/kg

+T6=(T4-273)-((ws*(ha-he))/cpg)//Temperature in degree C

+//Power output

+sa=6.853//Entropy in kJ/kg.K

+sbs=sa//Entropy in kJ/kg.K

+xbs=(sbs-0.4266)/8.052//Dryness fraction

+hbs=(121.46+xbs*2432.9)//ENthalpy in kJ/kg

+Wst=(m*(ha-hbs)*nst)//Workdone in kW

+wg=(m/ws)//gas flow rate in kg/s

+wa=(1-f)*wg//Air flow rate entering the compressor in kg/s

+Wgt=(wg*cpg*(Tm-T4))-(wa*cpa*(T2-T1))//Power output of gas turbine in kW

+TO=Wst+Wgt//Total power output in kW

+wf1=(f*wa)//Fuel mass flow rate in kg/s

+wf=4.466//Rounding off of wf1 for exact answers

+no=(TO/(wf*(CV*10^3)))*100//Overall efficiency of the combined plant in percent

+ns=((ha-hbs)/(ha-he))*nst//Efficiency of steam plant

+ngtp=(Wgt/(wf*(CV*10^3)))//Efficiency of the GT plant

+xL=((wg*cpg*(T6-(T1-273)))/(wf*(CV*10^3)))//Lost heat coefficient

+nov=(ns+ngtp-ns*ngtp-ngtp*xL)//The overall efficiency

+//Energy fluxes and irreversibilities

+si=(A+B*((18*1)/(8*12)))//si for octane C8H18

+dHo=(wf*CV*10^3)//Power in kW

+dGo=(si*dHo)//Power in kW

+TS=(dGo-dHo)//Power in kW

+//Compressor

+dS=(cpa*log(T2/T1))-(((cpa*(g-1))/g)*log(rp))//change in entropy in kJ/kg.K

+Ic=(wa*T1*dS)//power in kW

+Icx=((wg*T1*((cpg*log(Tm/T1))-(((cpg*(g1-1))/g1)*log(pcx))))-(wa*T1*((cpa*log(T2/T1))-(((cpa*(g-1))/g)*log(rp))))+TS)//Compressor in kW

+Icg=(-cpg*log(Tm/T4))-(((cpg*(g1-1))/g1)*log(p4/pcx))//Difference in entropy in kJ/kg.K

+IGT=(Icg*T1*wg)//Gas turbine in kW

+se=2.046//Enntropy in kJ/kg.K

+sae=(sa-se)//Difference in entropy in kJ/kg.K

+s64=(cpg*log((T6+273)/T4))-(((cpg*(g1-1))/g1)*log(p4/p1))//Difference in entropy in kJ/kg.K

+Ih=(T1*m*sae)+(wg*T1*s64)//For HRSG in kW

+hb=(ha-(nst*(ha-hbs)))//Enthalpy in kJ/kg

+xb=(hb-121.46)/2432.9//Dryness Fraction

+sb=(0.4226+xb*8.052)//Entropy in kJ/kg.K

+Ist=(m*(sb-sa)*T1)//For steam turbine in kW

+Iexh=(wg*cpg*((T6-(T1-273))-(T1*log((T6+273)/T1))))//For exhaust in kW

+Tl=Icx+Icg+IGT+Ih+Ist+Iexh//Exergy losses in kW

+T=Tl+Wgt+Wst//Total exergy output and exergy destruction in kW

+ee=((Wst+Wgt)/T)*100//Exergy efficiency in percent

+

+//Output

+printf('(a) Total power output is %3.2f kW and overall efficiency is %3.2f percent lost heat coefficient is %3.3f\n Exergy efficiency is %3.0f percent \n\n Input is %3.0f kW \n Total Output is %3.0f kW \n Total losses is %3.0f kW \n Exergy outut + exergy destruction = %3.0f kW which is 1.3 percent gretter than the exergy input',TO,no,xL,ee,dGo,(Wgt+Wst),Tl,T)

diff --git a/1943/CH3/EX3.5/Ex3_5.sce b/1943/CH3/EX3.5/Ex3_5.sce
new file mode 100755
index 000000000..b8a2dfab8
--- /dev/null
+++ b/1943/CH3/EX3.5/Ex3_5.sce
@@ -0,0 +1,13 @@
+
+clc

+clear

+//Input data

+n1=0.5//Efficiency of mercury

+n2=0.4//Efficiency of steam

+n3=0.25//Efficiency of composite cycle

+

+//Calculations

+n=(1-(1-n1)*(1-n2)*(1-n3))*100//Overall efficiency of the combined cycle in percent

+

+//Output

+printf('The overall efficiency of the combined cycle is %3.1f percent',n)

diff --git a/1943/CH3/EX3.6/Ex3_6.sce b/1943/CH3/EX3.6/Ex3_6.sce
new file mode 100755
index 000000000..f45a7ccc9
--- /dev/null
+++ b/1943/CH3/EX3.6/Ex3_6.sce
@@ -0,0 +1,12 @@
+
+clc

+clear

+//Input data

+z=30//Percentage of total energy of fuel

+n=40//Cycle efficiency in percent

+

+//Calculations

+on=((z/100)+(1-(z/100))*(n/100))*100//Overall efficiency in percent

+

+//Output

+printf('The overall efficiency of the combined plant is %3.0f percent',on)

diff --git a/1943/CH3/EX3.7/Ex3_7.sce b/1943/CH3/EX3.7/Ex3_7.sce
new file mode 100755
index 000000000..60ca65b42
--- /dev/null
+++ b/1943/CH3/EX3.7/Ex3_7.sce
@@ -0,0 +1,21 @@
+
+clc

+clear

+//Input data

+Tc=1250+273//Cathode temperature in K

+Ta=500+273//Anode temperature in K

+e=1.602*10^-19//Charge in coloumb

+K=1.38*10^-23//Boltzmann constant in J/molecule.K

+b=18//Constant

+

+//Calculations

+Va=((b*K*Ta)/e)//Voltage of anode in V

+Vc=((b*K*Tc)/e)//Voltage of cathode in V

+Vo=Vc-Va//Output voltage in V

+Ja=(120*Ta^2*exp(-b))//Current density in Cathode in A/cm^2

+Jc=(120*Tc^2*exp(-b))//Current density in Anode in A/cm^2

+P=Vo*(Jc-Ja)//Power output per unit area in /cm^2

+nth=(((Tc-Ta)/Tc)*(b/(b+2)))*100//Thermal efficiency in percent

+

+//Output

+printf('(a) The output voltage is %3.4f V \n (b) The current density in the cathode is %3.3f A/cm^2 and anode is %3.3f A/cm^2 \n (c) Power output per unit area is %3.2f W/cm^2 \n (d) Thermal efficiency is %3.1f percent',Vo,Jc,Ja,P,nth)

diff --git a/1943/CH3/EX3.8/Ex3_8.sce b/1943/CH3/EX3.8/Ex3_8.sce
new file mode 100755
index 000000000..70a6aa650
--- /dev/null
+++ b/1943/CH3/EX3.8/Ex3_8.sce
@@ -0,0 +1,37 @@
+
+clc

+clear

+//Input data

+P=100//Power in kW

+V=115//Voltage in V

+To=1500//Outer temperature in K

+Te=1000//Exit temperature in K

+Ta=350//Ambient temperature in K

+nth=30//Thermal efficiency in percent

+nge=92//Generator efficiency in percent

+//Properties of thermoelectrons 

+a=0.0012//At 1250K in V/K

+kp=0.02//In W/cm.K

+kn=0.03//In W/cm.K

+dp=0.01//In ohm.cm

+dn=0.012//In ohm.cm

+J=20//Current density in A/cm^2

+

+//Calculations

+zmax=(a^2/(sqrt(dp*kp)+sqrt(dn*kn))^2)//Maximum value of figure of merit in K^-1

+mo=sqrt(1+(zmax*((To+Te)/2)))//Optimum value of the resistance ratio

+nmax=(((To-Te)/To)*((mo-1)/(mo+(Te/To))))*100//Maximum thermal efficiency in percent

+Vl=(a*(To-Te)*(mo/(mo+1)))//Voltage per couple in V

+nc=(V/Vl)//Number of couples in series

+L=((a*(To-Te))/((1+mo)*(dp+dn)))/J//Length in cm

+A=((P*Te)/V)/J//Area in cm^2

+I=(J*A)//Current in A

+Vo=(a*(To-Te))//Voltage in V

+Q1=((a*I*To)-((1/2)*(L/A)*I^2*(dp+dn))+((A/L)*(kp+kn)*(To-Te)))/1000//Heat input to the thermoelectric generator in kW

+Q2=((a*I*Te)+((A/L)*(kp+kn)*(To-Te))+P)/1000//Heat rejected at full load in kW

+Q1n=(((A/L)*(kp+kn)*(To-Te)))/1000//At no load heat input in kW

+Q2n=Q1n//At no load heat rejected in kW

+no=((nmax/100)+(1-(nmax/100))*(nth/100)*(nge/100))*100//Overall efficiency in percent

+

+//Output

+printf('(a) The thermal efficiency of thermocouple generator is %3.1f percent \n (b) The number of thermo couples in series is %i \n (c) The lenght of the thermal elements is %3.3f cm and area is %3.2f cm^2 \n (d) The output open-circuit voltage is %3.1f V \n (e) At full load: \n The heat input is %3.3f kW \n The heat rejected is %3.3f kW \n At no load: \n The heat input is %3.3f kW \n The heat rejected is %3.3f kW \n (f) The overall efficiency of the combined thermo-electric steam power plant is %3.2f percent',nmax,nc,L,A,Vo,Q1,Q2,Q1n,Q2n,no)