initial commit / add all books

13 files changed, 235 insertions, 0 deletions

diff --git a/167/CH5/EX5.1/ex1.sce b/167/CH5/EX5.1/ex1.sce
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+//example 1

+// water flow through garden hose nozzle

+clear

+clc

+t=50 //time taken to fill the bucket in seconds

+v=10 //volume of bucket in gallon

+V=v*3.7854/t //volume flow rate in litres/second

+d=1 //density of water in kg/l

+M=V*d //mass flow rate in kg/s

+A=%pi*(0.4)^2*10^-4 //area of exit in m^2

+v1=V/(A*1000) //average velocity of water at exit in m/s

+printf("\n Hence, the volume flow rate of water through the hose is = %.3f L/s. \n",V);

+printf("\n The mass flow rate through the hose is = %.3f kg/s. \n",M);

+printf("\n The average velocity of water at the nozzle exit is = %.1f m/s. \n",v1);
\ No newline at end of file
diff --git a/167/CH5/EX5.10/ex10.sce b/167/CH5/EX5.10/ex10.sce
new file mode 100755
index 000000000..1d5c5fb50
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+++ b/167/CH5/EX5.10/ex10.sce
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+//example 10

+//cooling of refrigant 134-a by water

+clear

+clc

+disp('We take the entire heat exchanger as the system. This is a control volume since mass crosses the system boundary during the process.')

+disp('For each fluid stream since there is no mixing. Thus, m1=m2=mh and m3=m4=mr')

+mr=6 //mass flow rate of R-134a in kg/min

+h1=62.982 //specific enthalpy of water in kJ/kg

+h2=104.83 //specific enthalpy of water in kJ/kg

+P3=1 //pressure of R-134a at inlet in MPa

+T3=70 //temperature of R-134a at inlet in Celsius

+h3=303.85 //specific enthalpy corresponding to P3,T3 in kJ/kg

+P4=1 //pressure of R-134a at exit in MPa

+T4=35 // temp. of R-134a at exit in Celsius

+h4=100.87 // corresponding to P4,T4 in kJ/kg

+mw=mr*(h4-h3)/(h1-h2) //mass flow rate of the cooling water in kg/min

+qin=mw*(h2-h1) //the heat transfer rate from the refrigerant to water in kJ/min

+printf("\n Hence,mass flow rate of the cooling water required is = %.1f kg/min. \n",mw);

+printf("\n Heat transfer rate from refrigerant to water is = %.0f kJ/min. \n",qin);
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diff --git a/167/CH5/EX5.11/ex11.sce b/167/CH5/EX5.11/ex11.sce
new file mode 100755
index 000000000..ec348a7df
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+++ b/167/CH5/EX5.11/ex11.sce
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+//example 11

+// electric heating of air in house

+clear

+clc

+T1=290 //Initial temp. of air in K

+P1=100 //Initial pressure of air in kPa

+R=0.287 //Gas constant in KPa*m^3/kg-K

+V1=R*T1/P1 //Initial specific volume of air in m^3/kg

+v1=150 //volume flow rate in m^3/min

+m=v1/(V1*60) //mass flow rate in kg/s

+win=15 //Power of Electric heating system in kJ/s

+qout=0.2 //heat lost from air to surroundings in kJ/s

+cp=1.005 //heat capacity in kJ/kg-C

+T2=(win-qout)/(m*cp)+(T1-273) //Exit temp. of air in C

+printf("\n Hence,the exit temp. of air is = %.1f C. \n",T2);
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diff --git a/167/CH5/EX5.12/ex12.sce b/167/CH5/EX5.12/ex12.sce
new file mode 100755
index 000000000..01022a0b4
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+++ b/167/CH5/EX5.12/ex12.sce
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+//example 12

+//charging of rigid tank by system

+clear

+clc

+disp('This process can be analyzed as a uniform-flow process since the properties of the steam entering the control volume remain constant during the entire process.')

+disp('We take the tank as the system. This is a control volume since mass crosses the system boundary during the process.We observe that this is an unsteady-flow process since changes occur within the control volume')

+m1=0 //since system is initially evacuated

+disp('The properties of the steam at the inlet state are')

+P1=1 //pressure in MPa

+T1=300 //temp. in Celsius

+h1=3051.6 //especific enthalpy in kJ/kg

+P2=1 // pressure at final state in MPa

+u2=h1 //final internal energy of the steam in kJ/kg

+disp('From steam table,the temp. corresponding to final properties are')

+T2=456.1//final temp. in Celsius

+printf("\n The final temp. of the steam in the tank is = %.1f C. \n",T2);
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diff --git a/167/CH5/EX5.13/ex13.sce b/167/CH5/EX5.13/ex13.sce
new file mode 100755
index 000000000..39e527c27
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+++ b/167/CH5/EX5.13/ex13.sce
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+//example 13

+//cooking with a pressure cooker

+clear

+clc

+disp('This process can be analyzed as a uniform-flow process since the properties of the steam leaving the control volume remain constant during the entire cooking process')

+disp('We take the pressure cooker as the system. This is a control volume since mass crosses the system boundary during the process.We observe that this is an unsteady-flow process since changes occur within the control volume. Also, there is one exit and no inlets for mass flow.')

+Pgage=75 //gage pressure inside cooker in kPa

+Patm=100 //atmospheric pressure in kPa

+Pabs=Pgage+Patm //absolute pressure inside pressure cooker in kPa

+disp('Since saturation conditions exist in the cooker at all times , the cooking temperature must be the saturation temperature corresponding to this pressure.From steam table, it is')

+Tsat=116.04 // Saturation Temp. at 175 kPa in Celsius

+T=Tsat //Temp. at which cooking takes place

+Qin=0.5 //Heat supplied to the pressure cooker in kJ/s

+t=30*60 //time for which het is supplied to pressure cooker in seconds

+qin=Qin*t //total heat supplied to pressure cooker in kJ

+m1=1 //initial mass of water in kg

+V=0.006 //volume of pressure cooker in m^3

+V1=V/m1 // initial specific volume in kg/m^3

+Vf=0.001 //in kg/m^3

+Vfg=1.004-0.001 //in kg/m^3

+x1=(V1-Vf)/Vfg //quality

+uf=486.82 //in kJ/kg

+ufg=2037.7 //in kJ/kg

+u1=uf+x1*ufg //specific internal energy in kJ/kg

+U1=m1*u1 //total internal energy

+disp('V2=Vf+x2*Vfg  and u2=uf+x2*ufg.upon substituting various values,we get ')

+x2=0.009 //quality of steam in final state

+V2=Vf+x2*Vfg //final specific volume in m^3/kg

+m2=V/V2 //amount of water left in the pressure cooker in kg

+printf("\n Hence,the temperature at whih cooking takes place is = %.2f C. \n",T);

+printf("\n The amount of water left in the pressure cooker at the end of the process = %.1f kg. \n",m2);
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diff --git a/167/CH5/EX5.2/ex2.sce b/167/CH5/EX5.2/ex2.sce
new file mode 100755
index 000000000..9dc69c0a5
--- /dev/null
+++ b/167/CH5/EX5.2/ex2.sce
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+//example 2

+//discharge of water from a tank

+clear

+clc

+h0=4 //height of cylindrical water tank in ft

+h2=2 //final water level in tank in ft

+g=32.2 //acc. due to gravity in ft/s^2

+Dt=3*12 //diameter of tank in inches

+Djet=0.5 //diameter of water jet in inches

+t=(h0^0.5-h2^0.5)*(Dt)^2/((Djet)^2*(g/2)^0.5) //time taken for water level to fall to half of its initial value in seconds

+printf("\n Hence, the time taken for water level to fall to half of its initial value is = %.1f min. \n",t/60);
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diff --git a/167/CH5/EX5.3/ex3.sce b/167/CH5/EX5.3/ex3.sce
new file mode 100755
index 000000000..10e2794ae
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+++ b/167/CH5/EX5.3/ex3.sce
@@ -0,0 +1,22 @@
+//example 3

+//energy transport by mass

+clear

+clc

+vf=0.001053 //specific volume of saturated liquid water  in m3/kg

+vg=1.1594  //specific volume of water vapour in m3/kg

+ug=2519.2  //specific internal energy of water vapour kJ/kg

+hg=2693.1  //specific enthalpy of water vapour kJ/kg

+disp('Saturation conditions exist in a pressure cooker at all times after the steady operating conditions are established')

+disp(' Therefore, the liquid has the properties of saturated liquid and the exiting steam has the properties of saturated vapor at the operating pressure.')

+m=0.6/(vf*1000) //reduction in mass of liquid in pressure cooker in kg

+M=m/(40*60) //mass flow rate of steam in kg/s

+A=8*10^-6 //exit area in m^2

+V=M*vg/A //exit velocity in m/s

+e=hg-ug //flow energy of steam in kJ/kg

+TE=hg //total nergy of steam in kJ/kg

+E=M*hg //energy flow rate of steam leaving cooker in kW

+printf("\n Hence,The mass flow rate of the steam is = %.6f kg/s. \n",M); 

+printf("\n The exit velocity is = %.1f m/s. \n",V);

+printf("\n The total energy of the steam is = %.1f kJ/kf. \n",TE);

+printf("\n The flow energy of the steam is = %.1f kJ/kg. \n",e);

+printf("\n The rate at which energy leaves the cooker by steam is = %.3f kW. \n",E);
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diff --git a/167/CH5/EX5.4/ex4.sce b/167/CH5/EX5.4/ex4.sce
new file mode 100755
index 000000000..9a4cd0c7a
--- /dev/null
+++ b/167/CH5/EX5.4/ex4.sce
@@ -0,0 +1,19 @@
+//example 4

+//deceleration of air in diffuser

+clear

+clc

+disp('we assume that Air is an ideal gas since it is at a high temperature and low pressure relative to its critical-point values ')

+T1=283 //Initial temp. of air in kelvins  

+P1=80 //initial pressure of air in kPa

+R=0.287 //gas constant in kPa-m3/kg-K

+A1=0.4 //inlet area in m^2

+v1=200 //inintial velocity of air in m/s

+V1=R*T1/P1 //specific volume of air in m^3/kg

+m=v1*A1/V1 //mass flow rate in kg/s

+h1=283.14 // specific enthalpy of air in kJ/kg

+v2=0 //exit velocity is very small compared to initial velocity

+h2=h1-(v2^2-v1^2)/2000 //final specific enthalpy of air in kJ/kg

+disp('from steam table , the temperature corresponding to this value of enthalpy is')

+T2=303 //Temp. of air leaving the diffuser in K

+printf("\n Hence,The mass flow rate of the air is = %.1f kg/s. \n",m);

+printf("\n The temp. of air leaving the diffuser is = %.0f K. \n",T2);
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diff --git a/167/CH5/EX5.5/ex5.sce b/167/CH5/EX5.5/ex5.sce
new file mode 100755
index 000000000..f9cfa2056
--- /dev/null
+++ b/167/CH5/EX5.5/ex5.sce
@@ -0,0 +1,19 @@
+//example 5

+//acceleration of steam in nozzle

+clear

+clc

+P1=250  //initial pressure of steam in psia

+T1=700 //initial temperature of steam in °F 

+disp('The specific volume and enthalpy of steam at the nozzle inlet are ')

+v1=2.6883  //specific volume of steam at the nozzle inlet in ft3/lbm

+h1=1371.4 // specific enthalpy of steam at the nozzle inletin Btu/lbm

+A1=0.2 //inlet area in ft^2

+M=10 // mass flow rate of steam through nozzle in lbm/s

+V1=M*v1/A1 // inlet velocity in ft/s

+qout=1.2 // heat loss in Btu/lbm

+v2=900 //exit velocity in ft/s

+h2=h1-qout-(v2^2-v1^2)/(2*25037) // enthalpy after exit 

+disp('from steam table, the temp. corresponding to this pressure and specifi enthalpy is ')

+T2=662.0 ///in °F

+printf("\n Hence,the inlet velocity is = %.1f ft/s. \n",V1);

+printf("\n The exit temp. of the seam  is = %.0f F. \n",T2);
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diff --git a/167/CH5/EX5.6/ex6.sce b/167/CH5/EX5.6/ex6.sce
new file mode 100755
index 000000000..6fd6e6660
--- /dev/null
+++ b/167/CH5/EX5.6/ex6.sce
@@ -0,0 +1,10 @@
+//example 6

+//compressing air by compressor

+clear

+clc

+m=0.02 //mass flow rate of the air in kg/s

+qout=16 //heat loss during the process in kJ/kg

+h1=280.13 //specific enthalpy of air at 280K in kJ/kg

+h2=400.98 //specific enthalpy of air at 400K in kJ/kg

+win=m*qout + m*(h2-h1) //power input to compressor in kW

+printf("\n Hence,the power input to the compressor is = %.2f kW. \n",win);
\ No newline at end of file
diff --git a/167/CH5/EX5.7/ex7.sce b/167/CH5/EX5.7/ex7.sce
new file mode 100755
index 000000000..58de44e1f
--- /dev/null
+++ b/167/CH5/EX5.7/ex7.sce
@@ -0,0 +1,28 @@
+//example 7

+// power generation by steam turbine

+clear

+clc

+P1= 2 //initial pressure of steam in  MPa

+T1= 400 //initial temp. of steam in °C

+V1= 50 //initial velocity of steam in m/s

+z1= 10 //height of inlet in metres

+h1=3248.4 //initial specific enthalpy of air in kJ/kg

+P2= 15 //final pressure of air in kPa

+V2= 180 //final velocity of air in m/s

+z2= 6 //exit height in m

+x2=0.9 //quality of steam after exit 

+disp('At turbine exit,we obviously have liquid vapour mixture at 15 kPa.')

+hf=225.94 //in kJ/kg 

+hfg=2372.3 //in  kJ/kg

+h2=hf+x2*hfg //final specific enthalpy of mixture in kJ/kg

+dh=h2-h1 // change in enthalpy of steam in kJ/kg

+dke=(V2^2-V1^2)/2000 //change in kinetic energy of steam in kJ/kg

+g=9.8 //acc. due to gravity in m/s^2

+dpe=g*(z2-z1)/1000//change in potential energy of steam in kJ/kg

+wout=-((h2-h1)+(V2^2-V1^2)/2000+g*(z2-z1)/1000) //work done per unit mass of the steam flowing through the turbine in kJ/kg

+m=5000/wout //mass flow rate of steam in kg/s

+printf("\n The value of dh is = %.2f kJ/kg. \n",dh);

+printf("\n The value of dke is = %.2f kJ/kg. \n",dke);

+printf("\n The value of dpe is = %.2f kJ/kg. \n",dpe);

+printf("\n The work done per unit mass of the steam flowing through the turbine is = %.2f kJ/kg. \n",wout);

+printf("\n The mass flow rate of the steam is = %.2f kg/s. \n",m);
\ No newline at end of file
diff --git a/167/CH5/EX5.8/ex8.sce b/167/CH5/EX5.8/ex8.sce
new file mode 100755
index 000000000..2b1dbeee6
--- /dev/null
+++ b/167/CH5/EX5.8/ex8.sce
@@ -0,0 +1,21 @@
+//example 8

+// expansion of refrigant 134-a in refrigerator

+clear

+clc

+disp('Refrigerant-134a that enters a capillary tube as saturated liquid.Therefore,from table of refrigerant-134a ')

+P1=0.8  // initial pressure in MPa

+T1=31.31 //initial temp. in Celsius

+h1=95.47 //initial specific enthalpy in kJ/kg

+disp('Flow through a capillary tube is a throttling process. Thus, the enthalpy of the refrigerant remains constant')

+h2=h1 //final specific enthalpy 

+P2=0.12 //final pressure in MPa

+T2=-22.32 //final temp. in Celsius

+hf=22.49 // in kJ/kg

+hg=236.97 // in kJ/kg

+hfg=hg-hf //in kJ/kg

+disp('Obviously hf<h2<hg  thus, the refrigerant exists as a saturated mixture at the exit state')

+disp('Thus quality at this state is')

+x=(h2-hf)/hfg

+dT=T2-T1 //in Celsius

+printf("\n The quality of the refrigerant at the final state is = %.3f. \n",x);

+printf("\n The temp. drop during this process is = %.2f C. \n",dT);
\ No newline at end of file
diff --git a/167/CH5/EX5.9/ex9.sce b/167/CH5/EX5.9/ex9.sce
new file mode 100755
index 000000000..092859f35
--- /dev/null
+++ b/167/CH5/EX5.9/ex9.sce
@@ -0,0 +1,10 @@
+//example 9

+//Mixing of Hot and Cold Waters in a Shower

+clear

+clc

+disp('We take the chamber as system.Then,there are two inlets and one exit ')

+h3=78.02 //enthalpy at 110 F in Btu/lbm

+h2=18.07 //enthalpy at 50 F in Btu/lbm

+h1=107.99 //enthalpy at 140F in Btu/lbm

+y=(h3-h2)/(h1-h3) //mass ratio of hot to cold water 

+printf("\n The mass ratio of hot to cold water is = %.1f . \n",y);
\ No newline at end of file