initial commit / add all books

13 files changed, 187 insertions, 0 deletions

diff --git a/172/CH6/EX6.1/ex1.sce b/172/CH6/EX6.1/ex1.sce
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+//example 1

+//calculating mass flow rate in kg/s

+clear

+clc

+R=0.287 //in kJ/kg-K

+T=25 //temperature in celsius

+P=150 //pressure in kPa

+v=R*(T+273.2)/P //specific volume in m^3/kg

+D=0.2 //diameter of pipe in metre

+A=%pi*D^2/4 //cross sectional area in m^2

+V=0.1 //velocity of air in m/s

+m=V*A/v //mass flow rate in kg/s

+printf("\n hence,the mass flow rate is m=%.4f kg/s.\n",m)
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diff --git a/172/CH6/EX6.10/ex10.sce b/172/CH6/EX6.10/ex10.sce
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index 000000000..64067f988
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+//example 10

+//analysis of refrigerator

+clear

+clc

+hf4=167.4 //in kJ/kg

+hfg4=215.6 //in kJ/kg

+h3=241.8 //specific heat of enthalpy of R-134a entering expansion valve

+h4=h3 //specific heat of enthalpy of R-134a leaving expansion valve

+h1=387.2 //in kJ/kg

+h2=435.1 //in kJ/kg

+x4=(h3-hf4)/hfg4 //quality of R-134a at evaporator inlet

+m=0.1 //mass flow rate in kg/s

+Qevap=m*(h1-h4) //rate of heat transfer to the evaporator

+Wcomp=-5 //power input to compressor in kW

+Qcomp=m*(h2-h1)+Wcomp //rate of heat transfer from compressor

+printf("\n hence, the quality at the evaporator inlet is x4=%.3f. \n",x4')

+printf("\n hence, the rate of heat transfer to the evaporator is Qevap=%.2f kW. \n",Qevap')

+printf("\n hence, rate of heat transfer from the compressor is Qcomp=%.2f kW. \n",Qcomp')
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diff --git a/172/CH6/EX6.11/ex11.sce b/172/CH6/EX6.11/ex11.sce
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index 000000000..748c5dd58
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+++ b/172/CH6/EX6.11/ex11.sce
@@ -0,0 +1,9 @@
+//example 11

+//Determining the final temperature of steam

+clear

+clc

+u2=3040.4 //final internal energy in kJ/kg

+hi=u2 //in kJ/kg

+P2=1.4 //final Pressure in MPa

+disp('Since, the final pressure is given as 1.4 MPa,we know two properties at the final state and hence,final state can be determined.The temperature corresponding to a pressure of 1.4 MPa and an internal energy of 3040.4 kJ/kg is found to be ')

+T2=452 //final temperature in Celsius
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diff --git a/172/CH6/EX6.12/ex12.sce b/172/CH6/EX6.12/ex12.sce
new file mode 100755
index 000000000..9e1d5283e
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+++ b/172/CH6/EX6.12/ex12.sce
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+//example 12

+//Calculating mass flow of steam in tank

+clear

+clc

+V1=0.4 //initial volume fo tank in m^3

+v1=0.5243 //initial specific volume in m^3/kg

+h1=3040.4 //initial specific enthalpy in kJ/kg

+u1=2548.9 //initial specific internal energy in kJ/kg

+m1=V1/v1 //initial mass of steam in tank in kg

+V2=0.4 //final volume in m^3

+disp('let x=V*(h1-u2)/v2-m1*(h1-u1).If we assume T2=300C, then v2=0.1823m^3/kg ,u2=2785.2kJ/kg and x=+ve.If we assume T2=350C,then v2=0.2003 m^3/kg, u2=2869.1kJ/kg and x=-ve.Hence,actualt T2 must be between these two assumed values in order that x=0.By interplotation,') 

+T2=342 //final temperature in Celsius

+v2=0.1974 //final specific volume in m^3/kg

+m2=V2/v2 //final mass of the steam in the tank in kg

+m=m2-m1 //mass of steam that flowsinto the tank

+printf(" \n Hence,mass of the steam that flows into the tank is m=%.3f kg. \n",m)
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diff --git a/172/CH6/EX6.13/ex13.sce b/172/CH6/EX6.13/ex13.sce
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index 000000000..b28e7e13e
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+++ b/172/CH6/EX6.13/ex13.sce
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+//example 13

+//Calculating mass flow of steam in tank

+clear

+clc

+vf1=0.001725 //in m^3/kg

+vf2=0.0016 //in m^3/kg

+uf1=368.7 //in kJ/kg

+uf2=226 //in kJ/kg

+vg1=0.08313 //in m^3/kg

+vfg2=0.20381

+ug1=1341 //in kJ/kg

+ufg2=1099.7 //in kJ/kg

+Vf=1 //initial volume of liquid in m^3

+Vg=1 //initial volume of vapor in m^3

+mf1=Vf/vf1 //initial mass of liquid in kg

+mg1=Vg/vg1 //initial mass of vapor in kg

+m1=mf1+mg1 //initial mass of liquid in kg

+he=1461.1 //in kJ/kg

+V=2 //volume of tank in m^3

+disp('m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.00160+0.20381*x2) and u2=uf2+x2*ufg2=226.0+1099.7*x2.')

+disp('Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.')

+x2=((2*1461.1)-(2*226)-(0.00160*634706))/((634706*0.20381)+(2*1099.7)) //quality of ammonia

+v2=0.00160+(0.20381*x2) //final specific volume in m^3/kg

+m2=V/v2 //final mass of ammonia in kg

+m=m1-m2 //mass of ammonia withdrawn

+printf(" \n Hence,mass of ammonia withdrawn is m=%.1f kg. \n",m)
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diff --git a/172/CH6/EX6.2/ex2.sce b/172/CH6/EX6.2/ex2.sce
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index 000000000..29798c1aa
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+//example 2

+//work done for adding the fluid

+clear

+clc

+P=600 //pressure in kPa

+m=1 //in kg

+v=0.001 //specific volume in m^3/kg

+W=P*m*v //necessary work in kJ for adding the fluid 

+printf(" \n hence,the work involved in this process is W=%.3f kJ. \n",W)
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diff --git a/172/CH6/EX6.3/ex3.sce b/172/CH6/EX6.3/ex3.sce
new file mode 100755
index 000000000..8c193fc60
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+//example 3

+//rate of flow of water

+clear

+clc

+hir=441.89 //in kJ/kg for refrigerant using steam table 

+her=249.10 //in kJ/kg for refrigerant using steam table

+hiw=42 //in kJ/kg for water using steam table

+hew=83.95 //in kJ/kg for water using steam table

+mr=0.2 //the rate at which refrigerant enters the condenser in kg/s

+mw=mr*(hir-her)/(hew-hiw) //rate of flow of water in kg/s

+printf("\n hence,the rate at which cooling water flows thorugh the condenser is mw=%.3f kg/s. \n", mw)
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diff --git a/172/CH6/EX6.4/ex4.sce b/172/CH6/EX6.4/ex4.sce
new file mode 100755
index 000000000..f149c0a87
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+++ b/172/CH6/EX6.4/ex4.sce
@@ -0,0 +1,12 @@
+//example 4

+//determining quality of steam

+clear

+clc

+hi=2850.1 //initial specific heat enthalpy for steam in kJ/kg

+Vi=50 //initial velocity of steam in m/s

+Ve=600 //final velocity of steam in m/s

+he=hi+Vi^2/(2*1000)-Ve^2/(2*1000) //final specific heat enthalpy for steam in kJ/kg

+hf=467.1 //at final state in kJ/kg

+hfg=2226.5 //at final state in kJ/kg

+xe=(he-hf)/hfg //quality of steam in final state

+printf(" \n hence, the quality is xe=%.3f. \n",xe)
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diff --git a/172/CH6/EX6.5/ex5.sce b/172/CH6/EX6.5/ex5.sce
new file mode 100755
index 000000000..b93f21942
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+++ b/172/CH6/EX6.5/ex5.sce
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+//example 5

+//quality of ammonia leaving expansion valve

+clear

+clc

+hi=346.8 //specific heat enthalpy for ammonia at initial state in kJ/kg

+he=hi //specific heat enthalpy for ammonia at final state will be equal that at initial state because it is a throttling process

+hf=134.4 //at final state in kJ/kg

+hfg=1296.4//at final state in kJ/kg

+xe=(he-hf)/hfg //quality at final state

+printf("\n hence,quality of the ammonia leaving the expansion valve is xe=%.4f. \n",xe')
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diff --git a/172/CH6/EX6.6/ex6.sce b/172/CH6/EX6.6/ex6.sce
new file mode 100755
index 000000000..cd8cadd97
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+++ b/172/CH6/EX6.6/ex6.sce
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+//example 6

+//power output of turbine in kW

+clear

+clc

+hi=3137 //initial specific heat of enthalpy in kJ/kg

+he=2675.5 //final specific heat of enthalpy in kJ/kg

+Vi=50 //initial velocity of steam in m/s

+Ve=100 //final velocity of steam in m/s

+Zi=6 //height of inlet conditions in metres

+Ze=3 //height of exit conditions in metres

+m=1.5 //mass flow rate of steam in kg/s

+g=9.8066 //acc. due to gravity in m/s^2

+Qcv=-8.5 //heat transfer rate from turbine in kW

+Wcv=Qcv+m*(hi+Vi^2/(2*1000)+g*Zi/1000)-m*(he+Ve^2/(2*1000)+g*Ze/1000) //power output of turbine in kW

+printf("\n hence,the power output of the turbine is Wcv=%.3f kW. \n",Wcv)
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diff --git a/172/CH6/EX6.7/ex7.sce b/172/CH6/EX6.7/ex7.sce
new file mode 100755
index 000000000..3eb6e1a06
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+++ b/172/CH6/EX6.7/ex7.sce
@@ -0,0 +1,14 @@
+//example 7

+//heat transfer rate in aftercooler

+clear

+clc

+V1=0 //we assume initial velocity to be zero because its given that it enters with a low velocity

+V2=25 //final velocity with which carbon dioxide exits in m/s

+h2=401.52 //final specific enthalpy of heat when carbon dioxide exits in kJ/kg

+h1=198 //initial specific enthalpy of heat in kJ/kg

+w=h1-h2-V2^2/(2*1000) //in kJ/kg

+Wc=-50 //power input to the compressor in kW

+m=Wc/w //mass flow rate of carbon dioxide in kg/s

+h3=257.9 //final specific enthalpy of heat when carbon dioxide flows into a constant pressure aftercooler

+Qcool=-m*(h3-h2) //heat transfer rate in the aftercooler in kW

+printf(" \n hence,heat transfer rate in the aftercooler is Qcool=%.3f kW. \n",Qcool)

diff --git a/172/CH6/EX6.8/ex8.sce b/172/CH6/EX6.8/ex8.sce
new file mode 100755
index 000000000..49e52f5b6
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+++ b/172/CH6/EX6.8/ex8.sce
@@ -0,0 +1,13 @@
+//example 8

+//Required pump work

+clear

+clc

+m=1.5 //mass flow rate of water in kg/s

+g=9.807 //acceleration due to gravity in m/s^2

+Zin=-15 //depth of water pump in well in metres

+Zex=0 //in metres

+v=0.001001 //specific volume in m^3/kg

+Pex=400+101.3 //exit pressure in kPa

+Pin=90 //in kPa

+W=m*(g*(Zin-Zex)*0.001-(Pex-Pin)*v) //power input in kW

+printf(" \n Hence, the pump requires power input of W=%.0f W. \n",W*1000)
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diff --git a/172/CH6/EX6.9/ex9.sce b/172/CH6/EX6.9/ex9.sce
new file mode 100755
index 000000000..c74622274
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+++ b/172/CH6/EX6.9/ex9.sce
@@ -0,0 +1,21 @@
+//example 9

+//heat tranfer in simple steam power plant

+clear

+clc

+h1=3023.5 //specific heat of enthalpy of steam leaving boiler in kJ/kg

+h2=3002.5 //specific heat of enthalpy of steam entering turbine in kJ/kg

+x=0.9 //quality of steam entering condenser

+hf=226 //in kJ/kg

+hfg=2373.1 //in kJ/kg

+h3=hf+x*hfg //specific heat of enthalpy of steam entering condenser in kJ/kg

+h4=188.5 //specific heat of enthalpy of steam entering pump in kJ/kg

+q12=h2-h1 //heat transfer in line between boiler and turbine in kJ/kg

+w23=h2-h3 //turbine work in kJ/kg

+q34=h4-h3 //heat transfer in condenser

+w45=-4 //pump work in kJ/kg

+h5=h4-w45 //in kJ/kg

+q51=h1-h5 //heat transfer in boiler in kJ/kg

+printf("\n hence, heat transfer in line between boiler and turbine is q12=%.1f kJ/kg. \n",q12')

+printf("\n hence, turbine work is w23=%.1f kJ/kg. \n",w23')

+printf("\n hence, heat transfer in condenser is q34=%.1f kJ/kg. \n",q34')

+printf("\n hence, heat transfer in boiler is q51=%.0f kJ/kg. \n",q51')
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