diff options
Diffstat (limited to '2417/CH3')
| -rwxr-xr-x | 2417/CH3/EX3.1/Ex3_1.sce | 14 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.10/Ex3_10.sce | 24 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.11/Ex3_11.sce | 44 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.12/Ex3_12.sce | 21 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.13/Ex3_13.sce | 20 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.14/Ex3_14.sce | 30 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.15/Ex3_15.sce | 35 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.16/Ex3_16.sce | 41 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.17/Ex3_17.sce | 43 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.18/Ex3_18.sce | 39 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.19/Ex3_19.sce | 13 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.21/Ex3_21.sce | 23 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.4/Ex3_4.sce | 17 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.5/Ex3_5.sce | 19 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.6/Ex3_6.sce | 19 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.7/Ex3_7.sce | 14 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.8/Ex3_8.sce | 21 | ||||
| -rwxr-xr-x | 2417/CH3/EX3.9/Ex3_9.sce | 21 |
18 files changed, 458 insertions, 0 deletions
diff --git a/2417/CH3/EX3.1/Ex3_1.sce b/2417/CH3/EX3.1/Ex3_1.sce new file mode 100755 index 000000000..a05877bbc --- /dev/null +++ b/2417/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.1\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.1 (page no. 91)
+// Solution
+
+//For a constant volume process, 10 Btu/lbm heat is added to the system
+//We can consider thet a tank having a fixed volume has heat added to it
+//Under these conditions,the mechanical work done on or by the system must be 0
+//u2-u1=q
+printf("Heat has been converted to internal energy of the working fluid\n");
+//So,
+printf(" So,Change in internal energy u2-u1=10 Btu/Lbm");
diff --git a/2417/CH3/EX3.10/Ex3_10.sce b/2417/CH3/EX3.10/Ex3_10.sce new file mode 100755 index 000000000..3403392d7 --- /dev/null +++ b/2417/CH3/EX3.10/Ex3_10.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.10\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.10 (page no. 105)
+// Solution
+
+Cp=0.22; //Unit:Btu/(LBm*R) //Specific heat for constant pressure process
+Cv=0.17; //Unit:Btu/(LBm*R) //Specific heat for constant volume process
+q=800/10; //data given:800 Btu as heat is added to 10 LBm //Unit:Btu/LBm
+T1=100; //Unit:Fahrenheit //Initial temperature //T2=Final temperature
+//For a non-flow,constant pressure process
+//q=deltah=h2-h1=Cp(T2-T1) //deltah=change in enthalpy
+//deltaT=T2-T1;
+deltaT=q/Cp; //Fahrenheit //change in temperature
+T2=deltaT+T1; //Fahrenheit //final temperature
+//For a constant volume pressure
+//u2-u1=Change in internal energy //w=workdone
+//q-w=u2-u1
+//-w=(u2-u1)-q = Cv*(T2-T1)-q
+w=-(Cv*(T2-T1)-q); //Unit:Btu/lbm //workdone
+printf("%f Btu/lbm work is taken out of the system due to workdone by gas\n",w);
+printf("As there is 10 lbm in the system\n")
+printf("%f Btu work is taken out of the system due to workdone by gas\n",w*10);
diff --git a/2417/CH3/EX3.11/Ex3_11.sce b/2417/CH3/EX3.11/Ex3_11.sce new file mode 100755 index 000000000..4f1e00850 --- /dev/null +++ b/2417/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,44 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.11\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.11 (page no. 111)
+// Solution
+
+//Given data
+// Inlet Outlet
+//Pressure(psia) 1000 1
+//Temperature(F) 1000 101.74
+//Velocity(ft/s) 125 430
+//Inlet position(ft) +10 0
+//Enthalpy(Btu/LBm) 1505.4 940.0
+//Steam flow rate of 150000 LBm/hr
+
+//From the table,
+Z1=10; V1=125; h1=1505.4; Z2=0; V2=430; h2=940.0;
+
+//Energy equation is given by
+//((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + h1 + q = ((Z2/J)*(g/gc)) + (V2^2/(2*gc*J)) + h2 + w/J
+printf("Solution for (a) \n");
+q=0; //net heat
+J=778; //Conversion factor
+gc=32.174; //Unit: (LBm*ft)/(LBf*s^2) //gc is constant of proportionality
+g=gc; //Unit:ft/s^2 //g=The local gravity
+//W1=w/J;
+//Energy equation is given by
+W1=((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + h1 + q - ((Z2/J)*(g/gc)) - (V2^2/(2*gc*J)) - h2; //Unit:Btu/LBm
+printf("If heat losses are negligible,\n");
+printf("Total work of the turbine is %f Btu/LBm\n",W1);
+printf("Total work of the turbine is %f Btu/hr\n",W1*150000);
+//(W*150000*778)/(60*33000) //in terms of horsepower //1 hr=60 min //1 hp=33000 (ft*LBf)
+printf("Total work of the turbine is %f hp \n",(W1*150000*778)/(60*33000));
+//1 hp =0.746 kW
+printf("Total work of the turbine is %f kW \n\n",((W1*150000*778)/(60*33000))*0.746);
+
+
+printf("\nSolution for (b) \n");
+//Heat losses equal 50,000 Btu/hr
+q=50000/150000; //Unit:Btu/LBm //Heat loss
+W2=((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + h1 - q - ((Z2/J)*(g/gc)) - (V2^2/(2*gc*J)) - h2; //Unit:Btu/LBm
+printf("If heat losses equal 50,000 Btu/hr , Total work of the turbine is %f Btu/LBm\n",W2);
+
diff --git a/2417/CH3/EX3.12/Ex3_12.sce b/2417/CH3/EX3.12/Ex3_12.sce new file mode 100755 index 000000000..f30c7dc4c --- /dev/null +++ b/2417/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.12\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.12 (page no. 112)
+// Solution
+
+Z1=2; //Unit:m //Inlet position
+g=9.81 //Unit:m/s^2 //g=The local gravity
+V1=40; //Unit:m/s //Inlet velocity
+h1=3433.8; //Unit:kJ/kg //Inlet enthalpy
+q=1 //Unit:kJ/kg //Heat losses
+Z2=0; //Outlet position //unit:m
+V2=162; //Unit:m/s //Outlet velocity
+h2=2675.5; //Unit:kJ/kg //Outlet enthalpy
+
+//Energy equation is given by
+//((Z1*g)) + (V1^2/2) + h1 + q = ((Z2*g) + (V2^2/2) + h2 + w
+
+w= ((Z1*g)/1000) + ((V1^2/2)/1000) + h1 - q - ((Z2*g)/1000) - ((V2^2/2)/1000) - h2 ; //Unit:kJ/kg //Conersation: 1 kJ=1000 J
+printf("The work output per kilogram is %f kJ/kg\n",w);
diff --git a/2417/CH3/EX3.13/Ex3_13.sce b/2417/CH3/EX3.13/Ex3_13.sce new file mode 100755 index 000000000..925d4fcf8 --- /dev/null +++ b/2417/CH3/EX3.13/Ex3_13.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.13\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.13 (page no. 113)
+// Solution
+
+p1=150; //Unit:psia //Initial pressure
+T1=1000; //Unit:R //Temperature at pressure p1
+p2=15; //Unit:psia //Final pressure
+T2=600; //Unit:R //Temperature at pressure p2
+Cp=0.24; //Unit:Btu/(LBm*R) //Specific heat for constant pressure process
+v1=2.47; //Unit:ft^3/LBm //Specific volume at inlet conditions
+v2=14.8; //Unit:ft^3/LBm //Specific volume at outlet conditions
+
+//For a non-flow,constant pressure process
+//w/J=deltah=h2-h1=Cp(T2-T1) //deltah=change in enthalpy
+//W=w/J
+W=Cp*(T1-T2); //W=Work output //Unit:Btu/LBm
+printf("The work output of the turbine per pound of working fluid is %f Btu/LBm",W);
diff --git a/2417/CH3/EX3.14/Ex3_14.sce b/2417/CH3/EX3.14/Ex3_14.sce new file mode 100755 index 000000000..3dc849758 --- /dev/null +++ b/2417/CH3/EX3.14/Ex3_14.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.14\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.14 (page no. 114)
+// Solution
+
+//In problem 3.13 ,
+p1=150; //Unit:psia //Initial pressure
+T1=1000; //Unit:R //Temperature at pressure p1
+p2=15; //Unit:psia //Final pressure
+T2=600; //Unit:R //Temperature at pressure p2
+Cp=0.24; //Unit:Btu/(LBm*R) //Specific heat for constant pressure process
+v1=2.47; //Unit:ft^3/LBm //Specific volume at inlet conditions
+v2=14.8; //Unit:ft^3/LBm //Specific volume at outlet conditions
+
+//For a non-flow,constant pressure process
+//w/J=deltah=h2-h1=Cp(T2-T1) //deltah=change in enthalpy
+//W=w/J
+W=Cp*(T1-T2); //W=Work output //Unit:Btu/LBm //h2-h1
+printf("In problem 3.13,The work output of the turbine per pound of working fluid is %f Btu/LBm \n \n",W);
+
+//Now,In problem 3.14 ,
+q=1.1; //Unit:Btu/LBm //Heat losses
+//For a non-flow,constant pressure process
+//q-w/J=deltah=h2-h1=Cp(T2-T1) //deltah=change in enthalpy
+//W1=w/J
+W1=-q+W; //W=Work output //Unit:Btu/LBm //W=h2-h1 //Because q is out of the system,it is a negative quantity
+printf("In problem 3.14,heat loss equal to 1.1 Btu/LBm,\n");
+printf("The work output of the turbine per pound of working fluid is %f Btu/LBm \n",W1);
diff --git a/2417/CH3/EX3.15/Ex3_15.sce b/2417/CH3/EX3.15/Ex3_15.sce new file mode 100755 index 000000000..9c1e4ef1f --- /dev/null +++ b/2417/CH3/EX3.15/Ex3_15.sce @@ -0,0 +1,35 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.15\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.15 (page no. 115)
+// Solution
+
+p1=100; //Unit:psia //Initial pressure
+t1=950; //Unit:Fahrenheit //Temperature at pressure p1
+p2=76; //Unit:psia //Final pressure
+t2=580; //Unit:Fahrenheit //Temperature at pressure p2
+v1=4; //Unit:ft^3/LBm //Specific volume at inlet conditions
+v2=3.86; //Unit:ft^3/LBm //Specific volume at outlet conditions
+Cv=0.32; //Unit:Btu/(LBm*R) //Specific heat for constant volume process
+
+T1=t1+460; //Unit:R //Temperature at pressure p1
+T2=t2+460; //Unit:R //Temperature at pressure p2
+J=778; //J=Conversion factor
+
+//Z1=Inlet position //Unit:m
+//V1=Inlet velocity //Unit:m/s
+//Z2=Outlet position //Unit:m
+//V2=Outlet velocity Unit:m/s
+//u1=internal energy //energy in
+//u2=internal energy //energy out
+
+//Energy equation is given by
+//((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + u1 + ((p1*v1)/J) + q = ((Z2/J)*(g/gc)) + (V2^2/(2*gc*J)) + u2 + ((p2*v2)/J) + w/J; //Unit:Btu/LBm
+//Because pipe is horizontal and velocity terms are to be neglected,
+// Also no work crosses the boundaries of the system, the energy equation is reduced to
+//u1 + ((p1*v1)/J) + q = u2 + ((p2*v2)/J)
+//u2-u1=Cv*(T2-T1) //For a constant volume process //u2-u1=Chnage in internal energy
+//So,
+q=Cv*(T2-T1) + (p2*v2*144)/J - (p1*v1*144)/J; //q=heat transfer //1 ft^2=144 in^2 //Unit:Btu/LBm
+printf("%f Btu/LBm heat is transferred from the gas \n",q);
diff --git a/2417/CH3/EX3.16/Ex3_16.sce b/2417/CH3/EX3.16/Ex3_16.sce new file mode 100755 index 000000000..bc665751f --- /dev/null +++ b/2417/CH3/EX3.16/Ex3_16.sce @@ -0,0 +1,41 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.16\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.16 (page no. 116)
+// Solution
+
+//In problem 3.15,
+p1=100; //Unit:psia //Initial pressure
+t1=950; //Unit:Fahrenheit //Temperature at pressure p1
+p2=76; //Unit:psia //Final pressure
+t2=580; //Unit:Fahrenheit //Temperature at pressure p2
+v1=4; //Unit:ft^3/LBm //Specific volume at inlet conditions
+v2=3.86; //Unit:ft^3/LBm //Specific volume at outlet conditions
+Cv=0.32; //Unit:Btu/(LBm*R) //Specific heat for constant volume process
+
+T1=t1+460; //Unit:R //Temperature at pressure p1
+T2=t2+460; //Unit:R //Temperature at pressure p2
+J=778; //J=Conversion factor
+gc=32.174; //Unit: (LBm*ft)/(LBf*s^2) //gc is constant of proportionality
+g=gc; //Unit:ft/s^2 //g=The local gravity
+
+//Z1=Inlet position //Unit:m
+//V1=Inlet velocity //Unit:m/s
+//Z2=Outlet position //Unit:m
+//V2=Outlet velocity Unit:m/s
+//u1=internal energy //energy in
+//u2=internal energy //energy out
+
+//Energy equation is given by
+//((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + u1 + ((p1*v1)/J) + q = ((Z2/J)*(g/gc)) + (V2^2/(2*gc*J)) + u2 + ((p2*v2)/J) + w/J; //Unit:Btu/LBm
+//In 3.15, the elevation of the pipe at section 1 makes Z1 = 0
+// Also no work crosses the boundaries of the system, the energy equation is reduced to
+//u1 + ((p1*v1)/J) + q = u2 + ((p2*v2)/J) + ((Z2/J)*(g/gc))
+//In problrm 3.16,
+Z2=100; //Given //Unit:ft //Outlet position
+//u2-u1=Cv*(T2-T1) //For a constant volume process //u2-u1=Chnage in internal energy
+//So,
+q=Cv*(T2-T1) + (p2*v2*144)/J - (p1*v1*144)/J + ((Z2/J)*(g/gc)) ; //q=heat transfer //1 ft^2=144 in^2 //Unit:Btu/LBm
+printf("%f Btu/LBm heat is transferred from the gas \n",q);
+//For this problem , neglecting the elevation term leads to an insignificant error
diff --git a/2417/CH3/EX3.17/Ex3_17.sce b/2417/CH3/EX3.17/Ex3_17.sce new file mode 100755 index 000000000..858b2d948 --- /dev/null +++ b/2417/CH3/EX3.17/Ex3_17.sce @@ -0,0 +1,43 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.17\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.17 (page no. 117)
+// Solution
+
+p1=1000; //Unit:psia //Initial pressure
+t1=100; //Unit:Fahrenheit //Temperature at pressure p1
+p2=1000; //Unit:psia //Final pressure
+t2=1000; //Unit:Fahrenheit //Temperature at pressure p2
+// feed in 10,000 LBm/hr
+h1=70.68 //Unit:Btu/LBm //Inlet enthalpy
+h2=1505.9 //Unit:Btu/LBm //Outlet enthalpy
+
+T1=t1+460; //Unit:R //Temperature at pressure p1
+T2=t2+460; //Unit:R //Temperature at pressure p2
+//Energy equation is given by
+J=778; //J=Conversion factor
+
+//Z1=Inlet position //Unit:m
+//V1=Inlet velocity //Unit:m/s
+//Z2=Outlet position //Unit:m
+//V2=Outlet velocity Unit:m/s
+//u1=internal energy //energy in
+//u2=internal energy //energy out
+//h=enthalpy
+
+//Energy equation is given by
+//((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + u1 + ((p1*v1)/J) + q = ((Z2/J)*(g/gc)) + (V2^2/(2*gc*J)) + u2 + ((p2*v2)/J) + w/J; //Unit:Btu/LBm
+
+//we can consider this system as a single unit with feed water entering ans steam leaving.
+//It well designed,this unit will be thoroughly insulated,and heat losse will be reduced to a negligible amount
+//Alos,no work will be added to the fluid during the time it is passing through the unit, and kinetic energy differences will be assumed to be negligibly small
+//Differennces in elevation also be considered negligible
+//So,the energy equation is reduced to
+//u1 + ((p1*v1)/J) + q = u2 + ((p2*v2)/J)
+//Because h=u+(p*v/J)
+q=h2-h1; //q=net heat losses //Unit:Btu/LBm
+printf("Net heat losses is %f Btu/LBm \n",q);
+printf("For 10000 LBm/hr,\n");
+printf("%f Btu/hr energy has been added to the water to convert it to steam",q*10000)
+
diff --git a/2417/CH3/EX3.18/Ex3_18.sce b/2417/CH3/EX3.18/Ex3_18.sce new file mode 100755 index 000000000..d7d829e8a --- /dev/null +++ b/2417/CH3/EX3.18/Ex3_18.sce @@ -0,0 +1,39 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.18\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.18 (page no. 119)
+// Solution
+
+h1=1220 //Unit:Btu/LBm //Inlet enthalpy
+h2=1100 //Unit:Btu/LBm //Outlet enthalpy
+
+//Z1=Inlet position //Unit:m
+//V1=Inlet velocity //Unit:m/s
+//Z2=Outlet position //Unit:m
+//V2=Outlet velocity Unit:m/s
+//u1=internal energy //energy in
+//u2=internal energy //energy out
+J=778; //J=Conversion factor
+gc=32.174; //Unit: (LBm*ft)/(LBf*s^2) //gc is constant of proportionality
+
+//Energy equation is given by
+//((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + u1 + ((p1*v1)/J) + q = ((Z2/J)*(g/gc)) + (V2^2/(2*gc*J)) + u2 + ((p2*v2)/J) + w/J; //Unit:Btu/LBm
+
+//For this device,differences in elevation are negligible.No work is done on or by the fluid,friction is negligible
+//And due to the speed of the fluid flowing and the short length of the nozzle,heat transfer to or from the surroundings is also negligible.
+//So,the energy equation is reduced to
+//u1 + ((p1*v1)/J) +(V1^2/(2*gc*J) = u2 + ((p2*v2)/J) + (V2^2/(2*gc*J)
+// h1-h2 = ((V2^2-V1^2)/(2*gc*J))
+
+printf("Solution for (a)\n");
+//For neglegible entering velocity, V1=0
+//So,
+V2=sqrt((2*gc*J)*(h1-h2)); //the final velocity //ft/s
+printf("It the initial velocity of the system is negligible,the final velocity is %f ft/s \n \n",V2);
+
+printf("Solution for (b)\n");
+//If the initial velocity is appreciable,
+V1=1000; //Unit:ft/s //the initial velocity
+V2=sqrt(((h1-h2)*(2*gc*J)) + V1^2 ) ;
+printf("It the initial velocity of the system is appreciable,the final velocity is %f ft/s \n \n",V2);
diff --git a/2417/CH3/EX3.19/Ex3_19.sce b/2417/CH3/EX3.19/Ex3_19.sce new file mode 100755 index 000000000..b9271887a --- /dev/null +++ b/2417/CH3/EX3.19/Ex3_19.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.19\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.19 (page no. 120)
+// Solution
+
+h1=3450*1000 //Unit:J/kg //Enthalpy of steam when it enters a nozzle
+h2=2800*1000 //Unit:J/kg //Enthalpy of steam when it leaves a nozzle
+
+//V2^2/2=h1-h2;
+V2=sqrt(2*(h1-h2)); //V2=Final velocity //Unit:m/s
+printf("Final velocity = %f m/s\n",V2);
diff --git a/2417/CH3/EX3.21/Ex3_21.sce b/2417/CH3/EX3.21/Ex3_21.sce new file mode 100755 index 000000000..b729c0085 --- /dev/null +++ b/2417/CH3/EX3.21/Ex3_21.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.21\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.21 (page no. 125)
+// Solution
+
+m=400; //Unit:LBm/min //mass of lubricating oil
+Cp=0.85; //Unit:Btu/LBm*R //Specific heat of the oil
+T1=215; //Temperature when hot oil is entering //Unit:Fahrenheit
+T2=125; //Temperature when hot oil is leaving //Unit:Fahrenheit
+DeltaT=T2-T1; //Unit:Fahrenheit //change in temperature
+Qoil=m*Cp*DeltaT; //Heat out of oil //Btu/min
+printf("Heat out of oil is %f Btu/min (Out of oil)\n",Qoil);
+//Heat out of oil is the heat into the water
+//Mw=Water flow rate
+//M*Cpw*DeltaTw=Qoil
+Cpw=1.0; //Unit:Btu/LBm*R //Specific heat of the water
+T3=60; //Temperature when water is entering //Unit:Fahrenheit
+T4=90; //Temperature when water is leaving //Unit:Fahrenheit
+DeltaTw=T4-T3; //Unit:Fahrenheit //change in temperature
+Mw=Qoil/(Cpw*DeltaTw); //The Required water flow rate //Unit;lbm/Min
+printf("The Required water flow rate is %f lbm/Min\n",abs(Mw));
diff --git a/2417/CH3/EX3.4/Ex3_4.sce b/2417/CH3/EX3.4/Ex3_4.sce new file mode 100755 index 000000000..d764e2189 --- /dev/null +++ b/2417/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.1\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.1 (page no. 91)
+// Solution
+
+printf("Solution For (a)\n");
+m=10; //Unit:lbm //mass of water
+//delataU=U2-U1
+Heat=100; //Unit:Btu //heat added
+deltaU=Heat/m; //Change in internal energy //unit:Btu/lbm
+printf("Change in internal energy per pound of water is %f Btu/lbm\n",deltaU);
+
+printf("Solution For (b)\n");
+printf("In this process,energy crosses the boundary of the system by means of fractional work\n");
+printf("The contents of the tank will not distinguish between the energy if it is added as heat or the energy added as fraction work\n");
diff --git a/2417/CH3/EX3.5/Ex3_5.sce b/2417/CH3/EX3.5/Ex3_5.sce new file mode 100755 index 000000000..74430dfda --- /dev/null +++ b/2417/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.5\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.5 (page no. 96)
+// Solution
+
+P1=100 //Unit:psia //Pressure at the entrance to a steady-flow device
+Rho1=62.4 //Unit:lbm/ft^3 //the density of the fluid
+A1V1=10000 //Unit:ft^3/min //Entering fluid
+A2=2 //Unit:ft^2 //Exit area
+m=Rho1*A1V1; //Unit:lbm/min //mass rate of flow per unit time
+printf("Mass flow rate is %f LBm/min\n",m);
+
+Rho2=Rho1; //Unit:lbm/ft^3 //the density of the fluid
+//m=Rho2*A2*V2
+//So,
+V2=m/(Rho2*A2); //velocity at exit //Unit:ft/min
+printf("The exit velocity is %f ft/min",V2);
diff --git a/2417/CH3/EX3.6/Ex3_6.sce b/2417/CH3/EX3.6/Ex3_6.sce new file mode 100755 index 000000000..b45c41aa0 --- /dev/null +++ b/2417/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.6\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.6 (page no. 97)
+// Solution
+
+Rho1=1000 //Unit:kg/m^3 //the density of the fluid at entrance
+A1V1=2000 //Unit:m^3/min //Entering fluid
+A2=0.5 //Unit:ft^2 //Exit area
+m=Rho1*A1V1; //Unit:kg/min //mass rate of flow per unit time
+printf("Mass flow rate is %f kg/min\n",m);
+
+Rho2=Rho1; //Unit:kg/m^3 //the density of the fluid at exit
+//m=Rho2*A2*V2
+//So,
+V2=m/(Rho2*A2); //The exit velocity //Unit:m/min
+printf("The exit velocity is %f m/min",V2);
+
diff --git a/2417/CH3/EX3.7/Ex3_7.sce b/2417/CH3/EX3.7/Ex3_7.sce new file mode 100755 index 000000000..e007560b0 --- /dev/null +++ b/2417/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.7\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.7 (page no. 97)
+// Solution
+
+Rho=62.4 //Unit:lbm/ft^3 //the density of the fluid
+V=100 //Unit:ft/s //Velocity of fluid
+d=1 //Unit:in //Diameter
+//1 ft^2=144 in^2 //A=(%pi/4)*d^2
+A=(%pi*d^2)/(4*144) //Unit:ft^2 //area
+m=Rho*A*V; //Unit:lbm/s //mass rate of flow per unit time
+printf("Mass flow rate is %f lbm/s\n",m);
diff --git a/2417/CH3/EX3.8/Ex3_8.sce b/2417/CH3/EX3.8/Ex3_8.sce new file mode 100755 index 000000000..8dcc2d9af --- /dev/null +++ b/2417/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.8\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.8 (page no. 98)
+// Solution
+
+m1=50000; //Unit:LBm/hr //An inlet steam flow
+v1=0.831 //Unit:ft^3/LBm //Specific volume of inlet steam
+d1=6 //Unit:in //Inlet diameter
+A1=(%pi*d1^2)/(4*144) //1 ft^2=144 in^2 //Entering area
+V1=(m1*v1)/(A1*60*60) //(60 min/hr * 60 s/min) //To convert hours into seconds //velocity at inlet
+printf("The velocity at inlet is %f ft/s\n",V1);
+
+
+m2=m1; //Unit:LBm/hr //m2=An outlet steam flow
+v2=1.825 //Unit:ft^3/LBm //Specific volume of outlet steam
+d2=8 //Unit:in //Outlet diameter
+A2=(%pi*d2^2)/(4*144) //1 ft^2=144 in^2 //Exit area
+V2=(m1*v2)/(A2*60*60) //(60 min/hr * 60 s/min) //To convert hours into seconds //velocity at outlet
+printf("The velocity at outlet is %f ft/s",V2);
diff --git a/2417/CH3/EX3.9/Ex3_9.sce b/2417/CH3/EX3.9/Ex3_9.sce new file mode 100755 index 000000000..8d9af4f1d --- /dev/null +++ b/2417/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 3.9\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.9 (page no. 99)
+// Solution
+
+m1=10000; //Unit:kg/hr //An inlet steam flow
+v1=0.05 //Unit:m^3/kg //Specific volume of inlet steam
+d1=0.1 //Unit:m //Inlet diameter //100 mm =0.1 m
+A1=(%pi/4)*d1^2 //Unit:m^2 //Entering area
+V1=(m1*v1)/(A1*60*60) //(60 min/hr * 60 s/min) //To convert hours into seconds //velocity at inlet //Unit:m/s
+printf("The velocity at inlet is %f m/s\n",V1);
+
+
+m2=m1; //Unit:kg/hr //m2=An outlet steam flow
+v2=0.10 //Unit:m^3/kg //Specific volume of outlet steam
+d2=0.2 //Unit:m //Outlet diameter //200 mm = 0.2 m
+A2=(%pi/4)*(d2^2) //Unit:m^2 //Exit area
+V2=(m1*v2)/(A2*60*60) //(60 min/hr * 60 s/min) //To convert hours into seconds //velocity at outlet //Unit:m/s
+printf("The velocity at outlet is %f m/s",V2);
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