diff options
Diffstat (limited to '1808/CH3')
38 files changed, 1065 insertions, 0 deletions
diff --git a/1808/CH3/EX3.1/Chapter3_Exampl1.sce b/1808/CH3/EX3.1/Chapter3_Exampl1.sce new file mode 100644 index 000000000..6f2a282b3 --- /dev/null +++ b/1808/CH3/EX3.1/Chapter3_Exampl1.sce @@ -0,0 +1,11 @@ +clc
+clear
+//INPUT DATA
+Tl=30+273;//engine temperature in K
+Th=500+273;//maximum temperature in K
+
+//CALCULATIONS
+nc=((Th-Tl)/Th)*100;//Efficiency of carnot cycle in percentage
+
+//OUTPUT
+printf('Efficiency of carnot cycle is %3.2f percentage ',nc)
diff --git a/1808/CH3/EX3.10/Chapter3_Exampl10.sce b/1808/CH3/EX3.10/Chapter3_Exampl10.sce new file mode 100644 index 000000000..49ff9828f --- /dev/null +++ b/1808/CH3/EX3.10/Chapter3_Exampl10.sce @@ -0,0 +1,25 @@ +clc
+clear
+//INPUT DATA
+t1=300;//temperature in K
+p1=1;//pressure in atm
+t3=1700;//temperature in K
+Rc=7;//compression ratio
+R=0.287;//gas constant
+cv=0.7234;//calorific value
+
+//CALCULATIONS
+t2=t1*(Rc^(1.4-1));//temperature in K
+p2=p1*(Rc^(1.4));//pressure in atm
+p3=p2*t3/(t2);//pressure in atm
+t4=t3/((Rc)^(1.4-1));//temperature in K
+p4=p3/(Rc^(1.4));//pressure in atm
+no=1-((1/Rc)^(1.4-1))*100;//Thermal efficiency in percentage
+Rwo=1-((t1/t3)*((Rc)^(1.4-1)));//work ratio
+v1=R*t1/p1;//specific volume in m^3/kg
+wn=(cv*(t3-t2))-((cv*(t4-t1)));//net work
+pm=(wn/(v1*(1-(1/Rc))));//mean effective pressure in Bar
+
+//OUTPUT
+printf('(a)pressure at state point 2 is %3.2f atm \n temperature at point 2 is %3.2f K \n pressure at state point 3 is %3.2f atm \n temperature at point 4 is %3.2f K \n pressure at state point 3 is %3.2f atm \n (b)Thermal efficiency is %3.2f percentage \n (c)work ratio is %3.5f \n (d)mean effective pressure is %3.2f Bar',p2,t2,p3,t4,p4,no,Rwo,pm)
+
diff --git a/1808/CH3/EX3.11/Chapter3_Exampl11.sce b/1808/CH3/EX3.11/Chapter3_Exampl11.sce new file mode 100644 index 000000000..09e1637ef --- /dev/null +++ b/1808/CH3/EX3.11/Chapter3_Exampl11.sce @@ -0,0 +1,22 @@ +clc
+clear
+//INPUT DATA
+d=0.26;//bore of the engine in m
+L=0.38;//stroke of the engine in m
+vc=0.0025;//clearence volume in m^3
+p1=1;//pressure in bar
+t1=313;//temperature in K
+p3=25;//pressure in bar
+v12=9.07;//volume in m^3
+
+//CALCULATIONS
+vs=(3.14*d^2*L)/4;//swept volume in m^3
+Rc=((vs+vc)/vc);//compression ratio
+no=(1-((1/Rc)^(1.4-1)))*100;//Air standard efficiency of the cycle
+p2=p1*(v12^1.4);//pressure in bar
+Rp=p3/p2;//compression pressure
+pm=(p1*Rc*((Rc^(1.4-1)-1)*(Rp-1)))/((1.4-1)*(Rc-1));//mean effective pressure in bar
+
+//OUTPUT
+printf('(a)The air standard efficiency of the cycle is %3.1f percentage \n (b)The mean effective pressure is %3.4f bar',no,pm)
+
diff --git a/1808/CH3/EX3.13/Chapter3_Exampl13.sce b/1808/CH3/EX3.13/Chapter3_Exampl13.sce new file mode 100644 index 000000000..d7b43d7b2 --- /dev/null +++ b/1808/CH3/EX3.13/Chapter3_Exampl13.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+v1=0.5;//volume in m^3
+p1=1;//pressure in bar
+t1=303;//temperature in K
+p2=12;//pressure in bar
+Qs=250;//heat is added in kJ
+wc=200;//working cycles in cycles/min
+v2=0.085;//volume in m^3
+m=1;//mass of air
+cv=0.7243;//calorific value
+
+//CALCULATIONS
+Rc=(p2/p1)^(1/1.4);//compression ratio
+t2=t1*((Rc)^(1.4-1));//temperature in K
+pc=((v2/(v1-v2))*100);//percentage clearance
+t3=(Qs/(m*cv))+t2;//temperature in K
+t4=((1/Rc)^(1.4-1))*t3;//temperature in K
+Qr=m*cv*(t4-t1);//heat rejected in kJ/kg
+no=((Qs-Qr)/Qs)*100;//thermal efficiency in percentage
+pm=((Qs-Qr)/(v1-v2));//mean effective pressure
+p=((Qs-Qr)*wc)/60;//power developed in kJ/s
+
+//OUTPUT
+printf('(a)percentage clearance is %3.2f percentage \n (b)the thermal efficiency is %3.2f percentage \n (c)mean effective pressure is %3.2f \n (d)power developed is %3.2f kJ/s',pc,no,pm,p)
+
diff --git a/1808/CH3/EX3.15/Chapter3_Exampl15.sce b/1808/CH3/EX3.15/Chapter3_Exampl15.sce new file mode 100644 index 000000000..b08bde254 --- /dev/null +++ b/1808/CH3/EX3.15/Chapter3_Exampl15.sce @@ -0,0 +1,18 @@ +clc
+clear
+//INPUT DATA
+t1=300;//temperature in K
+t3=1300;//temperature in K
+cp=5.22;//specific pressure
+cv=3.13;//specific volume
+g=1.688;//for helium as working medium
+
+//CALCULATIONS
+Rc=((t3/t1)^(1/(2*(1.4-1))));//compression ratio
+no1=(1-((1/Rc)^(1.4-1)))*100;//efficiency of air
+Rcn=((t3/t1)^(1/(2*(g-1))));//compression ratio
+no2=(1-((1/Rcn)^(g-1)))*100;//efficiency of helium
+
+//OUTPUT
+printf('(a)air as working medium efficiency is %3.2f percentage \n (b)Helium as working medium efficiency is %3.2f percentage \n Hence the change in efficiency is zero',no1,no2)
+
diff --git a/1808/CH3/EX3.16/Chapter3_Exampl16.sce b/1808/CH3/EX3.16/Chapter3_Exampl16.sce new file mode 100644 index 000000000..c0ed714e7 --- /dev/null +++ b/1808/CH3/EX3.16/Chapter3_Exampl16.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+t1=300;//temperature in K
+t3=1500;//temperature in K
+cv=0.7243;//calorific value
+m=0.4;//air flow rate in kg/min
+
+//CALCULATIONS
+t2=sqrt(t1*t3);//temperature in K
+Wnmax=cv*((t3-t2)-(t2-t1));//maximum workdone in kJ/kg
+Pnmax=m*Wnmax/60;//maximum power developed in kJ/s
+
+//OUTPUT
+printf('(a)Maximum power developed is %3.3f kJ/s ',Pnmax)
+
diff --git a/1808/CH3/EX3.17/Chapter3_Exampl17.sce b/1808/CH3/EX3.17/Chapter3_Exampl17.sce new file mode 100644 index 000000000..18a0d0eaf --- /dev/null +++ b/1808/CH3/EX3.17/Chapter3_Exampl17.sce @@ -0,0 +1,32 @@ +clc
+clear
+//INPUT DATA
+Rc=18;//compression ratio
+t1=300;//temperature in K
+t3=1700;//temperature in K
+p1=101.325;//Pressure in kN/m^2
+g=1.4;//constant
+cp=1.005;//specific pressure in kJ/kgK
+cv=0.718;//specific volume in kJ/kgK
+R=0.287;//gas constant in kJ/kgK
+
+//CALCULATIONS
+t2=t1*(Rc^(g-1));//temperature at point 2 in K
+p2=p1*Rc;//Pressure at point 2 in atm
+r=t3/t2;//cut off ratio
+t4=t3*((r/Rc)^(g-1));//temperature in at point 4 in K
+p4=1*(t4/t1);//Pressure at point 4 in atm
+w12=cv*(t2-t1);//workdone 1-2 statein kJ/kg
+w23=R*(t3-t2);//workdoneat 2-3 statein kJ/kg
+w34=cv*(t3-t4);//workdone at 3-4 state in kJ/kg
+wn=w23+w34-w12;//workdone by the system
+Qs=cp*(t3-t2);//heat added in kJ/kg
+Qr=cv*(t4-t1);//heat rejected in kJ/kg
+nd=wn*100/Qs;//Thermal efficiency in percentage
+Rw=wn/(w23+w34);//work ratio
+v1=R*t1/p1;//specific volume in m^3/kg
+pm=wn/(v1*(1-(1/Rc)))/100;//Mean effective pressure in bar
+
+//OUTPUT
+printf('(a)pressure in each cycle is %3.3f atm \n (b)Specific work output is %3.2f kJ/kg \n (c)the Thermal efficiency is %3.3f percentage \n (d)The work ratio is %3.4f \n (e)Mean effective pressure is %3.3f bar',p4,wn,nd,Rw,pm)
+
diff --git a/1808/CH3/EX3.18/Chapter3_Exampl18.sce b/1808/CH3/EX3.18/Chapter3_Exampl18.sce new file mode 100644 index 000000000..e91c3483e --- /dev/null +++ b/1808/CH3/EX3.18/Chapter3_Exampl18.sce @@ -0,0 +1,30 @@ +clc
+clear
+//INPUT DATA
+p1=101.325;//Pressure in kN/m^2
+t1=303;//Temperature in K
+g=1.4;//constant
+cp=1.005;//specific pressure in kJ/kgK
+cv=0.718;//specific volume in kJ/kgK
+R=0.287;//gas constant in kJ/kgK
+r=2.5;//cut off ratio
+v2=0.1;//clearance volume in m^3/kg
+
+
+//CALCULATIONS
+v1=R*t1/p1;//volume at state 1 in m^3/kg
+p2=p1*((v1/v2)^(g));//pressure at state 2 in kN/m^2
+t2=t1*((v1/v2));//temperature at state 2 in K
+v3=r*v2;//volume at state 3 in m^3/kg
+t3=t2*(v3/v2);//Temperature at state 3 in K
+t4=t3*((v3/v1)^(g-1));//Temperature at state 4 in K
+p4=p2*((v3/v1)^g);//Pressure at state 4 in kN/m^2
+c=(v2/(v1-v2))*100;//percentage clearance
+Qs=cp*(t3-t2);//heat added in kJ/kg
+Qr=cv*(t4-t1);//heat rejected in kJ/kg
+nd=((Qs-Qr)/Qs)*100;//Thermal efficiency in percentage
+pm=((Qs-Qr)/(v1-v2));//Mean effective pressure in kN/m^2
+
+//OUTPUT
+printf('(a)pressure at state 2 is %3.2f kN/m^2 \n temperature at state 2 is %3.2f K \n volume at state 3 is %3.2f m^3/kg\n Temperature at state 3 is %3.2f K \n Temperature at state 4 is %3.2f K \n Pressure at state 4 is %3.2f kN/m^2 \n (b)percentage clearance is %3.2f percentage \n(c)Thermal efficiency is %3.2f percentage \n (d)Mean effective pressure is %3.2f kN/m^2',p2,t2,v3,t3,t4,p4,c,nd,pm)
+
diff --git a/1808/CH3/EX3.19/Chapter3_Exampl19.sce b/1808/CH3/EX3.19/Chapter3_Exampl19.sce new file mode 100644 index 000000000..ea0fdfa85 --- /dev/null +++ b/1808/CH3/EX3.19/Chapter3_Exampl19.sce @@ -0,0 +1,35 @@ +clc
+clear
+//INPUT DATA
+d=0.12;////bore of the engine in m
+l=0.13;//stroke of the engine in m
+p1=101.325;//pressure in atm
+t1=298;//temperature in K
+t3=1773;//temperature in K
+n=2000;//speed in rpm
+g=1.4;//constant
+cp=1.005;//specific pressure in kJ/kgK
+cv=0.718;//specific volume inkJ/kgK
+R=0.287;//gas constant inkJ/kgK
+
+//CALCULATIONS
+Rc=1.1/0.1;//compression ratio
+v1=R*t1/p1;//specific volume in m^3/kg
+v2=v1/Rc;//specific volume in m^3/kg
+t2=t1*((v1/v2)^(g-1));//temperature in K
+p2=p1*(v1/v2)^(g);//pressure in kN/m^2
+v3=v2*(t3/t2);//specific volume in m^3/kg
+t4=t3*((v3/v1)^(g-1));//temperature in K
+p4=p2*(v3/v1)^g;//pressure in kN/m^2
+Q3=cp*(t3-t2);//heat added in kJ/kg
+Qr=cv*(t4-t1);//heat rejected in kJ/kg
+nd=((Q3-Qr)/Q3)*100;//Thermal efficiency in percentage
+V1=1.1*(3.14*d^2*l)/4;//volume in m^3
+m=4*(n/2)*(V1/(v1*240));//flow rate in kg/s
+P=(Q3-Qr)*m;//Power of the engine in kg/s
+
+//OUTPUT
+printf('(a)compression ratio is %3.i \n (b)pressure and temperature at the end of compression is %3.2f kN/m^2 \n (c)Thermal efficiency is %3.2f percentage \n (d)Power of the engine is %2.3f kg/s ',Rc,p2,nd,P)
+
+
+
diff --git a/1808/CH3/EX3.2/Chapter3_Exampl2.sce b/1808/CH3/EX3.2/Chapter3_Exampl2.sce new file mode 100644 index 000000000..8b0234ade --- /dev/null +++ b/1808/CH3/EX3.2/Chapter3_Exampl2.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+Tl=150;//engine temprature in Degree C
+Th=1100;//engine temprature in Degree C
+Qs=4000;//Heat is added in kJ/min
+
+//CALCULATIONS
+nc=((Th-Tl)/(Th+273))*100;//Efficiency of carnot cycle in percentage
+wd=nc*Qs/100;//workdone in kJ/min
+P=wd/(60);//power developed in kJ/s
+Qr=Qs-wd;//Quality of heat rejected in kJ/min
+ds=(Qs-wd)/(Tl+273);//Change in entropy during heat rejection in kJ/min
+
+//OUTPUT
+printf('(a)power developed in the engine is %3.2f kJ/s \n (b)Quality of heat rejected is %3.2f kJ/min \n (c)Change in entropy during heat rejection is %3.2f kJ/min',P,Qr,ds)
+
diff --git a/1808/CH3/EX3.20/Chapter3_Exampl20.sce b/1808/CH3/EX3.20/Chapter3_Exampl20.sce new file mode 100644 index 000000000..d480c85f7 --- /dev/null +++ b/1808/CH3/EX3.20/Chapter3_Exampl20.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+Rc=15;//compression ratio
+r1=1.84;//cutoff ratio
+r2=1.98;//cutoff ratio
+g=1.4;//constant
+p1=101.325;//Pressure in kN/m^2
+Rc3=17;//compression ratio
+r3=1.84;//cutoff ratio
+Rc4=18;//compression ratio
+r4=1.88;//cutoff ratio
+
+//CALCULATIONS
+nd1=(1-(((1/(Rc^(g-1))))*(((r1^g)-1)/((r1-1)*g))))*100;//Air standard efficiency in precentage
+pm1=(p1/((Rc-1)*(g-1)))*(((Rc^g)*g*(r1-1))-(Rc*((r1^g)-1)));//Mean effective pressure in kN/m^2
+nd2=1-(((1/(Rc^(g-1))))*(((r2^g)-1)/((r2-1)*g)));//change in efficiency in precentage
+ndd1=nd1-nd2;//change in efficiency in precentage
+pm2=(p1/((Rc-1)*(g-1)))*(((Rc^g)*g*(r2-1))-(Rc*((r2^g)-1)));//Mean effective pressure in kN/m^2
+pmii=((pm2-pm1)/pm1)*100;//Increase in mep in percentage
+nd3=(1-(((1/(Rc3^(g-1))))*(((r3^g)-1)/((r3-1)*g))))*100;//increased efficiency in precentage
+ni=nd3-nd1;//increased efficiency in percentage
+pm3=(p1/((Rc3-1)*(g-1)))*(((Rc3^g)*g*(r3-1))-(Rc*((r3^g)-1)));//Increase in Mean effective pressure in kN/m^2
+pmi=((pm3-pm1)/(2*pm1))*100;//Increase in Mean effective pressure in percentage
+K=((r4-1)/(Rc4-1))*100;//change in cutoff of stroke
+
+//OUTPUT
+printf('(a)Air standard efficiency is %3.2f precentage \n (b)Mean effective pressure is %3.2f kN/m^2 \n (II)\n (a1)Percentage change in efficiency is %3.2f percentage \n (b1)Increase in Mean effective pressure is %3.2f percentage \n (III) \n (a3)increased efficiency is %3.3f percentage \n (b3)Increase in Mean effective pressure is %3.2f kN/m^2 \n (IV)change in cutoff of stroke %3.1f percentage',nd1,pm1,ndd1,pmii,ni,pmi,K)
+
diff --git a/1808/CH3/EX3.21/Chapter3_Exampl21.sce b/1808/CH3/EX3.21/Chapter3_Exampl21.sce new file mode 100644 index 000000000..0768747a3 --- /dev/null +++ b/1808/CH3/EX3.21/Chapter3_Exampl21.sce @@ -0,0 +1,19 @@ +clc
+clear
+//INPUT DATA
+l=30;//Stroke in cm
+d=17;//Bore in cm
+vc=440*10^6;//Clearance volume in m^3
+r=0.05;//cutoff ratio
+g=1.4;//constant
+
+//CALCULATIONS
+vs=((3.14*(d^2)*l)*10^6)/4;//swept volume in m^3
+v1=vs+vc;//Total volume in m^3
+v3=vc+(r*(v1-vc));//volume at point of cutoff
+ro=v3/vc;//cutoff ratio
+Rc=(vs+vc)/vc;//compression ratio
+nd=(1-(((1/(Rc^(g-1))))*(((ro^g)-1)/((ro-1)*g))))*100;//Air standard efficiency in precentage
+
+//OUTPUT
+printf('Air standard efficiency is %3.2f precentage',nd)
diff --git a/1808/CH3/EX3.22/Chapter3_Exampl22.sce b/1808/CH3/EX3.22/Chapter3_Exampl22.sce new file mode 100644 index 000000000..a1967f94c --- /dev/null +++ b/1808/CH3/EX3.22/Chapter3_Exampl22.sce @@ -0,0 +1,37 @@ +clc
+clear
+//INPUT DATA
+Rc=17;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=303;//temperature in K
+ro=2.28;//cutoff ratio
+g=1.4;//constant
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+v1=0.06375;//specific volume in m^3/s
+v2=0.00375;//specific volume in m^3/s
+t4=960.62;//temperature in K
+
+//CALCULATIONS
+p3=p1*(Rc^(g));//maximum pressure in kN/m^2
+t2=t1*(Rc^(g-1));//temperature in K
+t3=t2*(ro);//maximum temperature in K
+nd=(1-(((1/(Rc^(g-1))))*(((ro^g)-1)/((ro-1)*g))))*100;//Air standard efficiency in precentage
+Qs=cp*(t3-t2);//heat supplied in kJ/kg
+wn=nd*Qs/100;//workdone in KJ/kg
+m=p1*v1/(R*t1);//mass flow rate in kg/s
+P=wn*m;//power developed in kJ/s
+wt=(((g*(ro-1)*(Rc^(g-1)))-((ro^g)-1))/((g*(ro-1)*(Rc^(g-1)))-((ro^g)-(Rc^(g-1)))));//workdone in kJ/kg
+Rw=wn/wt;//work ratio
+w12=cv*(t1-t2);//workdone in 1-2 process
+w23=R*t2*(ro-1);//workdone in 2-3 process
+w34=cv*(t3-t4);//orkdone in 3-4 process
+pw=w23+w34;//positive work
+Rw=(wn/pw);//Work ratio
+
+//OUTPUT
+printf('(a)The maximum pressure %3.2f kN/m^2 \n temperature is %3.2f K \n (b)The thermal efficiency is %3.2f percentage \n(c)The power developed is %3.4f kg/s \n (d)work ratio is %3.4f ',p3,t3,nd,P,Rw)
+
+
+
diff --git a/1808/CH3/EX3.23/Chapter3_Exampl23.sce b/1808/CH3/EX3.23/Chapter3_Exampl23.sce new file mode 100644 index 000000000..13274156a --- /dev/null +++ b/1808/CH3/EX3.23/Chapter3_Exampl23.sce @@ -0,0 +1,35 @@ +clc
+clear
+//INPUT DATA
+l=20;//Stroke in cm
+d=15;//Bore in cm
+N=400;//speed in rpm
+Rc=22;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=303;//temperature in K
+n1=1.3;//no of cycles
+n2=1.35;//no of cycles
+g=1.4;//constant
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constnat
+v1=0.003698;//specific volume in m^3/s
+vs=0.00353;//specific volume in m^3/s
+ro=2.68;//cutoff ratio
+
+//CALCULATIONS
+p2=p1*(Rc^(n1));//pressure in kN/m^2
+t2=t1*(Rc^(n1-1));//temperature in K
+t3=t2*(ro);//temperature in K
+p4=p2*((ro/Rc)^n2);//maximum pressure in kN/m^2
+t4=t3*(1/((Rc/ro)^(n2-1)));//maximum temperature in K
+m=p1*v1/(R*t1);//mass flow rate in kg/s
+wn=R*((t3-t2)+((t3-t4)/(n2-1))-((t2-t1)/(n1-1)));//work done in kJ/kg
+pm=wn*m/(vs);//mean effective pressure in kN/m^2
+Qs=cp*(t3-t2);//heat supplied in kJ/kg
+nd=(wn/Qs)*100;//thermal efficiency in percentage
+P=wn*m*N/60;//POWER DEVELOPED
+
+//OUTPUT
+printf('(a)The temperature and pressure at all corner points are \n pressure at point 1 is %3.2f kN/m^2 \n temperature at point 2 is %3.2f K \n temperature at point 3 is %3.2f K \n maximum pressure is %3.2f kN/m^2 \n temperature at point 4 is %3.2f K \n(b)The mean effective pressure is %3.3f kN/m^2 \n (c)Thermal efficiency is %3.2f percentage \n (d)Power developed is %3.2f kJ/s ',p2,t2,t3,p4,t4,pm,nd,P)
+
diff --git a/1808/CH3/EX3.24/Chapter3_Exampl24.sce b/1808/CH3/EX3.24/Chapter3_Exampl24.sce new file mode 100644 index 000000000..7d9fb3669 --- /dev/null +++ b/1808/CH3/EX3.24/Chapter3_Exampl24.sce @@ -0,0 +1,39 @@ +clc
+clear
+//INPUT DATA
+Rc=10;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=303;//temperature in K
+Qr=350;//heat rejected in kJ/kg
+Qs=450;//heat supplied in kJ/kg
+QR=452.92;//heat supplied in kJ/kg
+g=1.4;//constant
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+
+//CALCULATIONS
+Q=Qr+Qs;//total heat in kJ/kg
+p2=p1*(Rc^g);//pressure in kN/m^2
+t2=t1*(Rc^(g-1));//temperature in K
+t3=(Qs/cv)+t2;//temperature in K
+p4=p2*(t3/t2);//Maximum pressure in kN/m^2
+t4=(Qr/cp)+t3;//Maximum temperature in K
+wn=Q-QR;//workdone in kJ/kg
+v43=t4/t3;//volume ratio
+v12=t1/t3;//volume ratio
+v45=v43/Rc;//volume ratio
+t5=t4*(v45^(g-1));//temperature in K
+nd=(wn/Q)*100;//thermal efficiency in percentage
+v1=R*t1/p1;//specific volume in m^3/kg
+v2=v1/10;//specific volume in m^3/kg
+pm=wn/(v1-v2);//mean effective pressure in kN/m^2
+w34=R*(t4-t3);//workdone in 3-4 process
+w45=R*(t4-t5);//workdone in 4-5 process
+Rw=(wn/(2*(w34+w45)));//work ratio
+
+//OUTPUT
+printf('(a)The maximum pressure is %3.2f kN/m^2 \n maxium temperature is %3.2f K \n (b)thermal efficiency is %3.2f percentage \n (c)mean effective pressure is %3.2f kN/m^2 \n (d)work ratio is %3.3f ',p4,t4,nd,pm,Rw)
+
+
+
diff --git a/1808/CH3/EX3.25/Chapter3_Exampl25.sce b/1808/CH3/EX3.25/Chapter3_Exampl25.sce new file mode 100644 index 000000000..ba73a018d --- /dev/null +++ b/1808/CH3/EX3.25/Chapter3_Exampl25.sce @@ -0,0 +1,32 @@ +clc
+clear
+//INPUT DATA
+Rc=20;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=300;//temperature in K
+p32=2;//pressure ratio of heating process
+v43=1.5;//volume ratio of heating process
+g=1.4;//constant
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+
+//CALCULATIONS
+t2=t1*(Rc^(g-1));//Temperature in K
+p2=p1*(Rc^g);//pressure in kN/m^2
+p3=p32*p2;//pressure in kN/m^2
+t3=p32*t2;//Temperature in K
+t4=v43*t3;//Temperature in K
+p5=p3/((Rc/v43)^g);//pressure in kN/m^2
+t5=t4/((Rc/v43)^(g-1));//Temperature in K
+nd=(((cv*(t3-t2))+(cp*(t4-t3))-(cv*(t5-t1)))/((cv*(t3-t2)+(cp*(t4-t3)))))*100;//The thermal efficiency in percentage
+x=((cv*(t3-t2)+(cp*(t4-t3))));//workdone
+y=(cv*(t5-t1));//workdone
+v1=R*t1/p1;//specific volume in m^3/kg
+pm=(x-y)/(v1*(1-(1/Rc)));//mean effective pressure in kN/m^2
+
+//OUTPUT
+printf('(a)The thermal efficiency is %3.2f percentage \n (b)The mean effective pressure is %3.1f kN/m^2',nd,pm)
+
+
+
diff --git a/1808/CH3/EX3.26/Chapter3_Exampl26.sce b/1808/CH3/EX3.26/Chapter3_Exampl26.sce new file mode 100644 index 000000000..c2fda1602 --- /dev/null +++ b/1808/CH3/EX3.26/Chapter3_Exampl26.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+p1=120;//Pressure in kN/m^2
+t1=303;//temperature in K
+v1=0.0708;//specific volume in m^3/s
+v2=0.004165;//specific volume in m^3/s
+t3=1423;//temperature in K
+t4=1873;//temperature in K
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+g=1.4;//constant
+
+//CALCULATIONS
+ro=t4/t3;//cutoff ratio
+Rc=v1/v2;//Compression ratio
+t2=t1*(Rc^(g-1));//temperature in K
+v45=(ro/Rc);//specific volume in m^3/s
+t5=t4*((v45)^(g-1));//temperature in K
+Qs=cv*(t3-t2)+cp*(t4-t3);//heat added in kJ/kg
+Qr=cv*(t5-t1);//heat rejected in kJ/kg
+nd=((Qs-Qr)/Qs)*100;//thermal efficiency in percentage
+
+//OUTPUT
+printf('(a)cutoff ratio %3.3f \n (b)Compression ratio is %3.1f \n (c)Heat added is %3.2f kJ/kg \n heat rejected is %3.2f kJ/kg \n (d)The thermal efficiency in %3.2f percentage',ro,Rc,Qs,Qr,nd)
+
+
+
diff --git a/1808/CH3/EX3.27/Chapter3_Exampl27.sce b/1808/CH3/EX3.27/Chapter3_Exampl27.sce new file mode 100644 index 000000000..cc18c4b24 --- /dev/null +++ b/1808/CH3/EX3.27/Chapter3_Exampl27.sce @@ -0,0 +1,38 @@ +clc
+clear
+//INPUT DATA
+l=22;//Stroke in cm
+d=15;//Bore in cm
+Rc=10;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=303;//temperature in K
+g=1.4;//constnat
+cp=1.005;//specific pressure
+cv=0.718;//specific volumespecific volume
+R=0.287;//gas constant
+n=1.3;//no of flows
+v1=0.00433;//volume in m^3
+
+//CALCULATIONS
+t2=t1*(Rc^(g-1));//Temperature in K
+p2=p1*(Rc^g);//pressure in kN/m^2
+ro=Rc/6;//cutoff ratio
+t3=1133.5;//temperature in K
+t4=(Rc/6)*t3;//temperature in K
+wd=3.43;//workdone per cycle in kN/m
+p3=p2*t3/t2;//pressure in kN/m^2
+p5=p3*(1/6)^n;//pressure in kN/m^2
+pm=((p3*(ro-1))+((p3*ro-p5*Rc)-(p2-p1*Rc))*(1/(n-1)))/(Rc-1);//mean effective pressure in kN/m^2
+pm1=pm/100;//mean effective pressure in bar
+vs=3.14*d^2*l/4;//stroke volume in m^3
+m=p1*v1/(R*t1);//mass flow rate in kg/s
+Qs=m*((cv*(t3-t2))+cp*(t4-t3));//heat supplied in kJ/cycle
+nd=(wd/Qs)*100;//thermal efficiency in percentage
+p=wd*400/60;//power of the engine in kJ/s
+Rw=((p3*(ro-1))+(1/(n-1))*((p3*ro-p5*Rc)-(p2-p1*Rc)))/((p3*(ro-1))+(1/(n-1))*(p3*ro-p5*Rc));//work ratio
+
+//OUTPUT
+printf('(a)The temperature and pressure are \n p2 %3.2f kN/m^2 \n p3 %3.2f kN/m^2 \n p5 %3.2f kN/m^2 \n t2 %3.2f K \n t3 %3.2f K \n t4 %3.2f K \n (b)mean effective pressure is %3.2f bar \n (c)thermal efficiency is %3.2f percentage \n (d)power of the engine is %3.2f kJ/s \n (e)The work ratio is %3.1f ',p2,p3,p5,t2,t3,t4,pm1,nd,p,Rw)
+
+
+
diff --git a/1808/CH3/EX3.28/Chapter3_Exampl28.sce b/1808/CH3/EX3.28/Chapter3_Exampl28.sce new file mode 100644 index 000000000..353e6574c --- /dev/null +++ b/1808/CH3/EX3.28/Chapter3_Exampl28.sce @@ -0,0 +1,40 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+vs=0.01;//Swept volume in m^3
+Rc=18;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=303;//temperature in K
+p3=80*10^2;//pressure in kN/m^2
+g=1.4;//constant
+v2=0.000588;//volume in m^3
+v43=0.0006;//difference in pressure
+
+//CALCULATIONS
+p2=p1*(Rc^g);//pressure in kN/m^2
+a=p3/p2;//Pressure ratio
+v4=v2+v43;//volume in m^3
+ro=v4/v2;//cutoff ratio
+v1=v2+0.01;//volume of cylinder in m^3
+m=p1*v1/(R*t1);//mass of air contained in cylinder in kg
+t2=t1*(Rc^(g-1));//temperature in K
+t3=t2*(p3/p2);//temperature in K
+t4=t3*(v4/v2);//temperature in K
+Qs=(cv*(t3-t2)+cp*(t4-t3))*0.01234;//heat added in kJ
+t5=t4/((v1/v4)^(g-1));//temperature in K
+Qr=cv*(t5-t1)*0.01234;//Heat rejected in kJ
+wn=(Qs-Qr);//workdone per cycle
+nd=(wn/Qs)*100;//Thermal efficiency in percentage
+pm=(wn/vs);//mean effective pressure in kN/m^2
+p5=p1*(t5/t1);//pressure in kN/m^2
+wp=p3*(v4-v2)+((p3*v4-p5*v1)/(g-1));//positive work done
+Rw=wn/wp;//work ratio
+
+//OUTPUT
+printf('(a)Pressure ratio is %3.2f \n (b)cutoff ratio is %3.4f \n (c)mass of air contained in cylinder is %3.5f kg \n (d)Heat added is %3.2f kJ \n (e)Heat rejected is %3.3f kJ \n (f)workdone per cycle is %3.2f kJ\n (g)Thermal efficiency is %3.2f percentage \n (h)Mean effective pressure is %3.1f kN/m^2 \n (i)The work ratio is %3.4f ',a,ro,m,Qs,Qr,wn,nd,pm,Rw)
+
+
+
diff --git a/1808/CH3/EX3.29/Chapter3_Exampl29.sce b/1808/CH3/EX3.29/Chapter3_Exampl29.sce new file mode 100644 index 000000000..a89bcb8c3 --- /dev/null +++ b/1808/CH3/EX3.29/Chapter3_Exampl29.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+p1=100;//Pressure in kPa
+t1=303;//temperature in Degree C
+g=1.4;//constant
+t2=700;//temperature in Degree C
+v1=0.05;//volume in m^3
+Rc=10;//compression ratio
+nr=0.9;//regenerator efficiency in percentage
+t=30
+//CALCULATIONS
+m=p1*v1/(R*t1);//mass flow rate
+
+wn=m*R*log(Rc*(t2-t));//Net workdone in kJ
+ns=(1-((t+273)/(t2+273)))*100;//Thermal efficiency with 100% refrigerator in percentage
+Qs=m*cv*(t2-t)+(m*R*(273+t2)*log(Rc));//heat added in kJ
+Qr=m*cv*(t2-t)+(m*R*(273+t)*log(Rc));//heat added in kJ
+nso=(1-(Qr/Qs))*100;//Thermal efficiency without refrigerator in percentage
+nsa=(((R*(t2-t)*log(Rc)))/((R*(273+t2)*log(Rc))+((1-nr)*cv*(t2-t))))*100;//Thermal efficiency with 90% refrigerator in percentage
+
+//OUTPUT
+printf('(i)net workdone is %3.2f kJ \n (ii)Thermal efficiency with 100 percentage efficiency is %3.2f percentage \n (iii)Thermal efficiency without regenerator is %3.2f percentage \n (iv)Thermal efficiency with 90percentage efficiency is %3.2f percentage \n',wn,ns,nso,nsa )
diff --git a/1808/CH3/EX3.3/Chapter3_Exampl3.sce b/1808/CH3/EX3.3/Chapter3_Exampl3.sce new file mode 100644 index 000000000..7e7c863b4 --- /dev/null +++ b/1808/CH3/EX3.3/Chapter3_Exampl3.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+Tl=300;//engine temprature in Degree C
+Th=1500;//engine temprature in Degree C
+Fc=0.45;//Fuel consumption in kg/hr
+cv=40000//kJ/kg
+wd=4;//workdone in kW
+
+//CALCULATIONS
+nc=((Th-Tl)/(Th+273))*100;//Efficiency of carnot cycle in percentage
+Qs=Fc*cv;//Heat is added in kJ/min
+na=(wd/(Qs))*(3600*100);//efficiency developed by scientist in percentage
+
+//OUTPUT
+printf('(a)Efficiency of carnot cycle is %3.2f percentage \n (b)efficiency developed by scientist is %3.i percentage',nc,na)
diff --git a/1808/CH3/EX3.30/Chapter3_Exampl30.sce b/1808/CH3/EX3.30/Chapter3_Exampl30.sce new file mode 100644 index 000000000..eb1e72a8f --- /dev/null +++ b/1808/CH3/EX3.30/Chapter3_Exampl30.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+p1=100;//Pressure in kPa
+t1=30;//temperature in Degree C
+t2=800;//temperature in Degree C
+g=1.4;//constant
+Rc=5;//compression ratio
+Qs=900;//heat supplied in kJ/kg
+nr=0.75;//regenerator efficiency in percentage
+
+
+//CALCULATIONS
+Qs1=R*(t2+273)*log(Rc)+(1-nr)*cv*(t2-t1);//heat supplied in kJ/kg
+m=Qs/Qs1;//mass flow rate in kg/min
+wn=(m/60)*R*log(Rc)*(t2-t1);//net work done in kW
+ns=(wn/(Qs/60))*100;//Thermal efficiency in percentage
+vs=((m)*R*(t1+273)*(1-(1/Rc)))/(p1*60);//swept volume in m^3/s
+pm=wn/vs;//mean effective pressure in kN/m^2
+P=wn*1;//Power developed by the engine in kW
+
+//OUTPUT
+printf('(i)The net work done is %3.2f kW \n (ii)Thermal efficiency is %3.2f percentage \n (iii)mean effective pressure is %3.2f kN/m^2 \n (iv)Power developed by the engine is %3.2f kW ',wn,ns,pm,P)
+
diff --git a/1808/CH3/EX3.31/Chapter3_Exampl31.sce b/1808/CH3/EX3.31/Chapter3_Exampl31.sce new file mode 100644 index 000000000..e00d2b8f2 --- /dev/null +++ b/1808/CH3/EX3.31/Chapter3_Exampl31.sce @@ -0,0 +1,20 @@ +clc
+clear
+//INPUT DATA
+t1=300;//temperature in Degree C
+t2=700;//temperature in Degree C
+p1=1;//pressure in bar
+p3=12;//pressure in bar
+R=0.287;//gas constant
+
+//CALCULATIONS
+ns=(1-(t1/t2))*100;//Thermal efficiency in percentage
+Rc=((p3/p1)*(t1/t2));//compression ratio
+wn=R*log(Rc)*(t2-t1);//net work done in kJ/kg
+vs=(R*t1*(1-(1/Rc)))/(p1);//swept volume in m^3/kg
+pm=wn/vs;//mean effective pressure in bar
+
+//OUTPUT
+printf('(i)Thermal efficiency is %3.2f percentage \n (ii)The mean effective pressure is %3.2f bar',ns,pm)
+
+
diff --git a/1808/CH3/EX3.32/Chapter3_Exampl32.sce b/1808/CH3/EX3.32/Chapter3_Exampl32.sce new file mode 100644 index 000000000..b18fd13b8 --- /dev/null +++ b/1808/CH3/EX3.32/Chapter3_Exampl32.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+p1=100;//pressure in kPa
+t1=300;//temperature in K
+t3=1500;//temperature in K
+g=1.4;//constant
+v1=6;//volume in m^3/s
+Rp=10;//compression ratio
+
+//CALCULATIONS
+wn=cp*(1-((1/(Rp^((g-1)/g)))))*(t3-t1*(Rp^((g-1)/g)));//Net work done in kJ/kg
+t2=t1*(Rp^((g-1)/g));//temperature in K
+ng=(wn/(cp*(t3-t2)))*100;//Thermal efficiency in percentage
+t4=t3/(Rp^((g-1)/g));//temperature in K
+Rw=((cp*(t2-t1))/(cp*(t3-t4)))*100;//back work ratio
+m=p1*v1/(R*t1);//mass flow rate in kg/s
+P=m*wn;//Power developed in kW
+
+//OUTPUT
+printf('(i))Thermal efficiency is %3.2f percentage \n(ii)The back work ratio is %3.1f percentage \n(iii)Power developed is %3.1f kW',ng,Rw,P)
+
+
diff --git a/1808/CH3/EX3.33/Chapter3_Exampl33.sce b/1808/CH3/EX3.33/Chapter3_Exampl33.sce new file mode 100644 index 000000000..32039fff5 --- /dev/null +++ b/1808/CH3/EX3.33/Chapter3_Exampl33.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gs constant
+p1=100;//pressure in kPa
+t1=300;//temperature in K
+t3=1300;//temperature in K
+g=1.4;//index of expansion
+
+
+//CALCULATIONS
+Rp=(t3/t1)^(g/(g-1));//pressure ratio
+Rpo=sqrt(Rp);//Pressure ratio which will give maximum net work output
+t2=t1*(Rpo^((g-1)/g));//temperature in K
+t4=t3/(Rpo^((g-1)/g));//temperature in K
+wn=cp*t3*((1-(sqrt(t1/t3)))^2);//maximum net work output in kJ/kg
+ng=(1-sqrt(t1/t3))*100;//Thermal efficiency in percentage
+Rw=(1-sqrt(t1/t3));//work ratio at maximum work output
+nc=(1-(t1/t3))*100;//Carnot efficiency for the same temperature limits in percentage
+
+//OUTPUT
+printf('(i)The pressure ratio which will give maximum net work output is %3.2f \n (ii)maximum net work output is %3.2f kJ/kg \n (iii)Thermal efficiency at maximum output is %3.2f percentage \n (iv)work ratio at maximum work output is %3.4f percentage \n (v)Carnot efficiency for the same temperature limits is %3.2f percentage',Rpo,wn,ng,Rw,nc)
+
+
+
diff --git a/1808/CH3/EX3.34/Chapter3_Exampl34.sce b/1808/CH3/EX3.34/Chapter3_Exampl34.sce new file mode 100644 index 000000000..8d219cfdd --- /dev/null +++ b/1808/CH3/EX3.34/Chapter3_Exampl34.sce @@ -0,0 +1,15 @@ +clc
+clear
+//INPUT DATA
+t1=300;//temperature in K
+t3=1300;//temperature in K
+g=1.4;//constant
+
+//CALCULATIONS
+Rpm=(t3/t1)^(g/(g-1));//Solution pressure ratio
+ng=(1-(t1/t3))*100;//thermal efficiency corresponds to maximum pressure ratio
+
+//OUTPUT
+printf('(i)Solution pressure ratio is %3.2f \n (ii)net workdone corresponds to maximum pressure ratio is zero \n (iii)thermal efficiency corresponds to maximum pressure ratio is %3.2f percntage \n (iv)work ratio is zero',Rpm,ng)
+
+
diff --git a/1808/CH3/EX3.35/Chapter3_Exampl35.sce b/1808/CH3/EX3.35/Chapter3_Exampl35.sce new file mode 100644 index 000000000..da803cfd5 --- /dev/null +++ b/1808/CH3/EX3.35/Chapter3_Exampl35.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+nt=0.8;//Thermal efficiency in percentage
+nc=0.8;//compressor efficiency in percentage
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+g=1.4;//constant
+t1=300;//temperature in K
+t3=1500;//temperature in K
+Rp=10;//pressure ratio
+
+//CALCULATIONS
+t2=t1*((Rp)^((g-1)/g));//temperature in K
+t21=t1+((t2-t1)/(nc));//temperature in K
+t4=t3/((Rp)^((g-1)/g));//temperature in K
+t41=t3-(nt*(t3-t4));//temperature in K
+wna=cp*((t3-t41)-(t21-t1));//net work done in kJ/kg
+ng=wna/(cp*(t3-t21))*100;//Thermal efficiency in percentage
+Rw=wna/(cp*(t3-t41));//work ratio
+
+//OUTPUT
+printf('(i)Net work done is %3.4f kJ/kg \n (ii)Thermal efficiency is %3.2f percentage \n (iii)Work ratio is %3.4f ',wna,ng,Rw)
+
+
+
diff --git a/1808/CH3/EX3.36/Chapter3_Exampl36.sce b/1808/CH3/EX3.36/Chapter3_Exampl36.sce new file mode 100644 index 000000000..d5b9f2ae0 --- /dev/null +++ b/1808/CH3/EX3.36/Chapter3_Exampl36.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+p1=100;//Pressure in kPa
+t1=300;//temperature in K
+g=1.35//constant
+t3=1000;//temperature in K
+nc=0.85;//compressor efficiency in percentage
+nt=0.9;//Thermal efficiency in percentage
+
+
+//CALCULATIONS
+Rp=(t3/t1)^(g/(g-1));//maximum pressure ratio
+Rpo=sqrt(Rp*nc*nt);//Pressure ratio for maximum work
+t2=t1*(Rpo)^((g-1)/g);//temperature in K
+t21=t1+((t2-t1)/nc);//temperature in K
+t4=t3/(Rpo^((g-1)/g));//temperature in K
+t41=t3-(nt*(t3-t4));//temperature in K
+nbt=(((t3-t41)-(t21-t1))/(t3-t21))*100;//Thermal efficiency in percentage
+
+//OUTPUT
+printf('(i)Pressure ratio for maximum work is %3.2f \n (ii)Thermal efficiency is %3.2f percentage ',Rpo,nbt)
+
+
+
+
diff --git a/1808/CH3/EX3.37/Chapter3_Exampl37.sce b/1808/CH3/EX3.37/Chapter3_Exampl37.sce new file mode 100644 index 000000000..fad7410fa --- /dev/null +++ b/1808/CH3/EX3.37/Chapter3_Exampl37.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=42000;//specific volume
+R=0.287;//gas constant
+g=1.4;//constant
+t1=300;//temperature in K
+t3=1000;//temperature in K
+Rp=5;//Pressure ratio
+ma=42.93;//mass of air
+mf=0.5;//mass of fuel
+nc=0.8;//compressor efficiency
+nt=0.85;//turbine efficiency
+
+//CALCULATIONS
+t2=t1*(Rp^((g-1)/g));//Temperature in K
+t4=t3/(Rp^((g-1)/g));//Temperature in K
+t21=t1+((t2-t1)/nc);//Temperature in K
+t41=t3-((t3-t4)*nt);//Temperature in K
+wna=-ma*cp*(t21-t1)+(ma+mf)*(t3-t41);//Actual power available in kJ
+ng=(wna/(mf*cv))*100;//Actual thermal efficiency in percentage
+
+//OUTPUT
+printf('(a)Actual power available is %3.2f kJ \n (b)Actual thermal efficiency is %3.2f percentage',wna,ng)
+
diff --git a/1808/CH3/EX3.38/Chapter3_Exampl38.sce b/1808/CH3/EX3.38/Chapter3_Exampl38.sce new file mode 100644 index 000000000..1ca7bcf58 --- /dev/null +++ b/1808/CH3/EX3.38/Chapter3_Exampl38.sce @@ -0,0 +1,31 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=42000;//calorific value
+R=0.287;//gas constant
+g=1.35;//constant
+t1=293;//temperature in K
+t3=973;//temperature in K
+nc=0.85;//compressor efficiency in percentage
+Rp=5;//pressure ratio
+nt=0.8;//turbine efficiency in percentage
+p=1000;//power in mW
+
+//CALCULATIONS
+t2=t1*(Rp^((g-1)/g));//Temperature in K
+t4=t3/(Rp^((g-1)/g));//Temperature in K
+t21=t1+((t2-t1)/nc);//Temperature in K
+t41=t3-((t3-t4)*nt);//Temperature in K
+wna=cp*((t3-t41)-(t2-t1));//net work done in kJ/kg
+m=p/wna;//Quantity of air circulation
+Qsa=cp*(t3-t21)/(nc);//Heat supplied in kJ/kg
+nba=(wna/Qsa)*100;//Thermal efficiency in percentage
+
+//OUTPUT
+printf('(i)Quantity of air circulation is %3.3f kg/s \n(ii)Heat supplied is %3.2f kJ/kg \n (iii)Thermal efficiency is %3.2f percentage',m,Qsa,nba)
+
+
+
+
+
diff --git a/1808/CH3/EX3.39/Chapter3_Exampl39.sce b/1808/CH3/EX3.39/Chapter3_Exampl39.sce new file mode 100644 index 000000000..45dbea28f --- /dev/null +++ b/1808/CH3/EX3.39/Chapter3_Exampl39.sce @@ -0,0 +1,39 @@ +clc
+clear
+//INPUT DATA
+t1=300;//Initial temperature in K
+t21=523;//intermmediate temperature in K
+t3=1073;//final temperature in K
+t41=723;//turbine outlet temperature in K
+p1=1;//pressure in bar
+p2=6;//final pressure in bar
+Rp=6;//pressure ratio
+g=1.4;//constant
+cp=1.005;//specific pressure
+
+//CALCULATIONS
+t2=t1*(Rp^((g-1)/g));//Temperature in K
+t4=t3/(Rp^((g-1)/g));//Temperature in K
+
+nc=((t2-t1)/(t21-t1))*100;//compressor efficiency in percentage
+nt=((t3-t41)/(t3-t4))*100;//Turbine efficiency in percentage
+
+ngt=(1-(1/Rp)^((g-1)/g))*100;//Thermal efficiency in percentage
+ngt1=((((nt/100)*t3*(ngt/100))-((t1/(nc/100))*((Rp^((g-1)/g))-1)))/(t1*((t3/t1)-((Rp^((g-1)/g))))))*100;//Thermal efficiency in percentage
+
+Rw=((cp*((t3-t4)-(t2-t1)))/(cp*(t3-t4)));//Work ratio
+Rw1=((cp*((t3-t41)-(t21-t1)))/(cp*(t3-t41)));//Work ratio
+
+Rpo=sqrt((t3/t1)^(g/(g-1)));//pressure ratio for maximum output
+Rpo1=sqrt(((t3/t1)^(g/(g-1)))*(nc/100)*(nt/100));//pressure ratio for maximum output
+
+Rpm=(t3/t1)^(g/(g-1));//pressure ratio for maximum efficiency
+Rpm1=(t3/t1)^(g/(g-1))*(1/((nc/100)*(nt/100)));//pressure ratio for maximum efficiency
+
+//OUTPUT
+printf('(A)The compressor efficiency is %3.3f percentage \n turbine efficiency is %3.2f percentage \n',nc,nt)
+
+printf('(B)IDEAL CYCLE \n (i)thermal efficiency is %3.2f percentage \n(ii)Work ratio is %3.4f \n (iii)Pressure ratio for maximum output is %3.2f \n (iv)pressure ratio for maximum efficiency is %3.2f \n ',ngt,Rw,Rpo,Rpm)
+
+printf('(B)ACTUAL CYCLE \n (i)thermal efficiency is %3.2f percentage \n(ii)Work ratio is %3.4f \n (iii)Pressure ratio for maximum output is %3.2f \n (iv)pressure ratio for maximum efficiency is %3.1f ',ngt1,Rw1,Rpo1,Rpm1)
+
diff --git a/1808/CH3/EX3.4/Chapter3_Exampl4.sce b/1808/CH3/EX3.4/Chapter3_Exampl4.sce new file mode 100644 index 000000000..9308e465a --- /dev/null +++ b/1808/CH3/EX3.4/Chapter3_Exampl4.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+n=1/5;//Fraction of heat input converted into power
+ts=100;//Reduced sink temperature in Degree C
+
+//CALCULATIONS
+//4Th-5Tl=0
+//3Th-5Tl=-500
+A=[4 -5
+ 3 -5]//Coefficient matrix
+B=[0
+ -500]//Constant matrix
+X=inv(A)*B//Variable matrix
+
+//Output
+printf('(a)The temperature of the source is %3.f K \n (b)The temperature of sink is %3.f K',X(1),X(2))
diff --git a/1808/CH3/EX3.40/Chapter3_Exampl40.sce b/1808/CH3/EX3.40/Chapter3_Exampl40.sce new file mode 100644 index 000000000..87c3661b3 --- /dev/null +++ b/1808/CH3/EX3.40/Chapter3_Exampl40.sce @@ -0,0 +1,30 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+R=0.287;//gas constant
+g=1.4;//constant
+t1=303;//temperature in K
+t3=1773;//temperature in K
+t5=1123;//temperature in K
+nc=0.85;//compressor efficiency in percentage
+Rp=6;//pressure ratio
+nt=0.8;//turbine efficiency in percentage
+
+//CALCULATIONS
+t2=t1*(Rp^((g-1)/g));//Temperature in K
+t4=t3/(Rp^((g-1)/g));//Temperature in K
+t21=t1+((t2-t1)/nc);//Temperature in K
+t41=t3-((t3-t4)*nt);//Temperature in K
+wc=cp*(t21-t1);//compressor work in kJ/kg
+wt=cp*(t3-t41);//turbine work in kJ/kg
+nb=((wt-wc)/(cp*(t3-t2)))*100;//Thermal efficiency in percentage
+wn=wt-wc;//net work in kJ/kg
+Qs=cp*(t3-t5);//Heat supplied in kJ/kg
+ns=((wt-wc)/Qs)*100;//Thermal efficiency in percentage
+e=((t5-t21)/(t41-t21))*100;//Effectiveness of the regenerator
+
+//OUTPUT
+printf('(a)compressor work is %3.2f kJ/kg \n turbine work is %3.2f kJ/kg \n (b)Thermal efficiency is %3.3f percentage \n (c)Thermal efficiency with regenerator is %3.2f percentage \n (d)Effectiveness of the regenerator is %3.1f percentage ',wc,wt,nb,ns,e)
+
+
diff --git a/1808/CH3/EX3.41/Chapter3_Exampl41.sce b/1808/CH3/EX3.41/Chapter3_Exampl41.sce new file mode 100644 index 000000000..60fd69fb0 --- /dev/null +++ b/1808/CH3/EX3.41/Chapter3_Exampl41.sce @@ -0,0 +1,78 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+R=0.287;//gas constant
+g=1.4;//constant
+t1=303;//temperature in K
+t3=1073;//temperature in K in case I
+t5=1123;//temperature in K
+Rp=4;//pressure ratio
+p1=1;//atmospheric pressure in bar
+p2=4;//exit pressure in bar
+
+
+//CALCULATIONS
+//case 1
+t2=t1*(Rp^((g-1)/g));//Temperature in K
+t4=t3/(Rp^((g-1)/g));//Temperature in K
+Qs=cp*(t3-t2);//Heat supplied in kJ/kg
+wc=cp*(t2-t1);//compressor work in kJ/kg
+wt=cp*(t3-t4);//turbine work in kJ/kg
+ng=((wt-wc)/(cp*(t3-t2)))*100;//Thermal efficiency in percentage
+
+//case 2
+//a regenerator of effectiveness 0.6 is added
+t51=0.6*(t4-t2)+t2;//temperature in K
+nbr=(((wt-wc)/(cp*(t3-t51))))*100;//Thermal efficiency eith regenerator in percentage
+//case 3
+pi=(p1*p2)^(1/2);//intermediate pressure
+t21=t1*(pi)^((g-1)/g);//temperature in K
+t41=t1*(pi)^((g-1)/g);//temperature in K
+t61=t3/((Rp)^((g-1)/g));//temperature in K
+t7=0.6*(t61-t2)+t21;//temperature in K
+Qs1=cp*(t3-t7);//heat added in kJ/kg
+wt1=cp*(t3-t61);//turbine work in kJ/kg
+wc1=cp*((t41-t1)+(t41-t1));//compressor work in kJ/kg
+nt=((wt1-wc1)/Qs1)*100;//Thermal efficiency in percentage
+//case 4
+t22=t1*(Rp)^((g-1)/g);//temperature in K
+t42=t3/(pi)^((g-1)/g);//temperature in K
+t62=t3/((pi)^((g-1)/g));//temperature in K
+t72=t22+(0.6*(t62-t22));//temperature in K
+wc2=cp*(t22-t1);//compressor work in kJ/kg
+wt2=cp*((t3-t42)+(t3-t62));//turbine work in kJ/kg
+Qs2=cp*((t3-t72)+(t3-t42));//heat added in kJ/kg
+ns=((wt2-wc2)/Qs2)*100;//Thermal efficiency in percentage
+//case 5
+t23=t1*(pi)^((g-1)/g);//temperature in K
+t43=t1*(pi)^((g-1)/g);//temperature in K
+t73=t3/(pi)^((g-1)/g);//temperature in K
+t93=t3/(pi)^((g-1)/g);//temperature in K
+t53=0.6*(t93-t43)+t43;//temperature in K
+Qs3=cp*((t3-t53)+(t3-t73));//heat added in kJ/kg
+wt3=cp*((t3-t93)+(t3-t73));//turbine work in kJ/kg
+wc3=cp*((t23-t1)+(t43-t1));//compressor work in kJ/kg
+ns1=((wt3-wc3)/Qs3)*100;//Thermal efficiency in percentage
+
+
+//OUTPUT
+printf('CASE I \n (i)Compressor work %3.2f kJ/kg \n (ii)Turbine work %3.2f kJ/kg \n (iii)Thermal efficiency %3.1f percentage \n ',wc,wt,ng)
+printf('CASE II \n (i)Compressor work %3.2f kJ/kg \n (ii)Turbine work %3.2f kJ/kg \n (iii)Thermal efficiency %3.1f percentage \n ',wc,wt,nbr)
+printf('CASE III \n (i)Compressor work %3.2f kJ/kg \n (ii)Turbine work %3.2f kJ/kg \n (iii)Thermal efficiency %3.1f percentage \n ',wc,wt1,nt)
+printf('CASE IV \n (i)Compressor work %3.2f kJ/kg \n (ii)Turbine work %3.2f kJ/kg \n (iii)Thermal efficiency %3.1f percentage \n ',wc2,wt2,ns)
+printf('CASE V \n (i)Compressor work %3.2f kJ/kg \n (ii)Turbine work %3.2f kJ/kg \n (iii)Thermal efficiency %3.1f percentage \n ',wc3,wt3,ns1)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/1808/CH3/EX3.5/Chapter3_Exampl5.sce b/1808/CH3/EX3.5/Chapter3_Exampl5.sce new file mode 100644 index 000000000..6d44da775 --- /dev/null +++ b/1808/CH3/EX3.5/Chapter3_Exampl5.sce @@ -0,0 +1,18 @@ +clc
+clear
+//INPUT DATA
+Tl=20;//engine temprature in Degree C
+Th=500;//engine temprature in Degree C
+g=1.4;//gas constant
+v13=25;//expansion ratio
+
+//CALCULATIONS
+v14=((Th+273)/(Tl+273))^(1/(g-1));//Isentropic volume expansion
+v43=v13/v14;//Overall expansion ratio
+
+//OUTPUT
+printf('(a)Isentropic volume expansion ratio (4-1)is %3.2f \n (b)Isentropic volume expansion ratio(4-3) is %3.2f',v14,v43)
+
+
+
+
diff --git a/1808/CH3/EX3.6/Chapter3_Exampl6.sce b/1808/CH3/EX3.6/Chapter3_Exampl6.sce new file mode 100644 index 000000000..9937b5475 --- /dev/null +++ b/1808/CH3/EX3.6/Chapter3_Exampl6.sce @@ -0,0 +1,24 @@ +clc
+clear
+//INPUT DATA
+Th=600;//engine temprature in Degree C
+p3=5;//Pressure of air in bar
+v3=4;//volume of air in m^3
+v43=3;//Isentropic volume expansion
+v23=6;//Isentropic compression ratio
+
+//CALCULATIONS
+p4=p3*(1/(v43));//pressure of carnot cycle
+p1=p4*(1/(v23))^(1.4);//pressure of carnot cycle
+t1=(Th+273)*(1/(v23))^(0.4);//Temperature of carnot cycle
+p2=p1*(v43);//pressure of carnot cycle
+Qs=p3*10^2*v3*log(v43);//heat supplied to the cycle
+Qr=p3*v3*(t1/Th)*log(v43);//Heat rejected by the system
+nc=(((Th+273)-t1)/(273+Th))*100;//Thermal efficiency in percentage
+w=Qs-Qr;//work done in kJ
+pm=w/(17*2*v3*100);//Mean effective pressure in bar
+
+//OUTPUT
+printf('(a)The pressure of carnot cycle is %3.3f bar \n temperature of carnot cycle is %3.4f K \n (b)Heat supplied to the cycle %3.2f kJ \n (c)Thermal efficiency is %3.2f percentage \n (d)Mean effective pressure is %3.4f bar ',p1,t1,Qs,nc,pm)
+
+
diff --git a/1808/CH3/EX3.8/Chapter3_Exampl8.sce b/1808/CH3/EX3.8/Chapter3_Exampl8.sce new file mode 100644 index 000000000..f3cfb30c4 --- /dev/null +++ b/1808/CH3/EX3.8/Chapter3_Exampl8.sce @@ -0,0 +1,28 @@ +clc
+clear
+//INPUT DATA
+p3=20;//Pressure of air in bar
+v3=0.2;//volume of air in m^3
+Th=500;//engine temprature in Degree C
+v23=7;//Isentropic compression ratio
+v43=2;//Isentropic volume expansion
+v3=0.2;//volume in m^3
+
+//CALCULATIONS
+Tl=(Th+273)/((v23)^(1.4-1));//minimum temperature in K
+p2=p3/((v23)^(1.4));//pressure in bar
+p4=p3*10^2*(1/(v43));//isentropic expansion pressure
+p1=((1/(v23))^1.4)*p4;//isentropic compression
+s43=(p3*10^2*v3/(Th+273))*log(v43);//Change in entropy in kJ/K
+nc=(((Th+273)-Tl)/(Th+273))*100;//Efficiency of carnot cycle in percentage
+v1=v43*7*v3;//volume in m^3
+vs=v1-v3;//swept volume in m^3
+wd=(p3*10^2*v3-p1*v1)*log(2);//workdone in kJ/cycle
+P=wd/2.6;//Mean effective pressure in kN/m^2
+p=wd*(200/60);//power of engine in kW
+
+//OUTPUT
+printf('(a)The minimum temperature in the cycle is %3.2f K \n (b)Change in entropy during isothermal expansion is %3.4f kJ/K \n (c)Thermal efficiency of the cycle is %3.2f percentage \n (d)The mean effective pressure is %3.2f kN/m^2 \n (e)Power of the engine is %3.2f kW',Tl,s43,nc,P,p)
+
+
+
diff --git a/1808/CH3/EX3.9/Chapter3_Exampl9.sce b/1808/CH3/EX3.9/Chapter3_Exampl9.sce new file mode 100644 index 000000000..eeea2a66a --- /dev/null +++ b/1808/CH3/EX3.9/Chapter3_Exampl9.sce @@ -0,0 +1,30 @@ +clc
+clear
+//INPUT DATA
+p1=101;//Pressure in kPa
+t1=293;//temperature in K
+v12=7;//compression ratio
+Qs=1000;//heat added in kJ
+Ra=0.287;//gas constant
+m=1;//mass of air in kg/min
+cv=0.7243;//calorific value
+
+//CALCULATIONS
+v1=(Ra*t1)/p1;//specific volume in m^3/kg
+v2=v1/(v12);//specific volume in m^3/kg
+p2=p1*((v12)^(1.4));//pressure of cycle
+t2=t1*((v12)^(1.4-1));//temperature in K
+t3=(Qs/(m*cv))+t2;//temperature in K
+p3=Ra*t3/v2;//pressure in Kn/m^2
+t4=t3*((1/v12)^(1.4-1));//temperature in K
+Qr=m*cv*(t4-t1);//heat rejected in kJ
+no=((Qs-Qr)/Qs)*100;//otto cycle efficiency in percentage
+pm=((Qs-Qr)/(v1-v2));//mean effective pressure in kN/m^2
+P=m*(Qs-Qr);//power developed in kJ/min
+//CASE B
+nc=((t3-t1)/t3)*100;//Carnot cycle efficiency in percentage
+
+//OUTPUT
+printf('(a)Specific volume of cycle is %3.2f m^3/kg \n pressure in the cycle is %3.2f kW/m^2 \n temperature in the cycle is %3.2f K \n Specific volume of cycle at point 3 is %3.3f m^3/kg \n pressure in the cycle at point 3 is %3.2f kW/m^2 \n temperature in the cycle at point 4 is %3.2f K \n(b)the efficiency of the otto cycle is %3.2f percentage \n (c)mean effective pressure is %3.2f kN/m^2 \n (d)power developed is %3.2f kJ/min \n CASE B \n Carnot cycle efficiency is %3.2f percentage \n Carnot cylce efficiency is high compared to Otto cycle efficiency',v1,p2,t2,v2,p3,t4,no,pm,P,nc)
+
+
|