diff options
Diffstat (limited to '2417')
220 files changed, 5117 insertions, 0 deletions
diff --git a/2417/CH1/EX1.1/Ex1_1.sce b/2417/CH1/EX1.1/Ex1_1.sce new file mode 100755 index 000000000..234b2302a --- /dev/null +++ b/2417/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,16 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 1.1\n\n\n");
+// Chapter 1: Fundamental Concepts
+// Problem 1.1 (page no. 8)
+// Solution
+
+//C=(5/9)*(F-32);
+//F=32+(9*C/5);
+//Letting C=F in equation;
+//F=(5/9)*(F-32);
+//Therefore
+F=-160/4; //fahrenheit
+disp(F,"F=");
+printf("Both fahrenheit and celsius temperature scales indicate same temperature at %f",F);
diff --git a/2417/CH1/EX1.10/Ex1_10.sce b/2417/CH1/EX1.10/Ex1_10.sce new file mode 100755 index 000000000..d72fc597e --- /dev/null +++ b/2417/CH1/EX1.10/Ex1_10.sce @@ -0,0 +1,15 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 1.10\n\n\n");
+// Chapter 1: Fundamental Concepts
+// Problem 1.10 (page no. 35)
+// Solution
+
+//Given
+Rho=2000; //Unit: kg/m^3 //The density of fluid
+h=-10; //Unit: mm //Height of column of fluid //the height is negative because it is measured up from the base
+g=9.6 //Unit:m/s^2 //the local acceleration of gravity
+//Solution
+p=-Rho*g*h; //P=Pressure at the base of a column of fluid //Unit:Pa
+printf("Pressure at the base of a column of fluid is %f Pa",p);
diff --git a/2417/CH1/EX1.11/Ex1_11.sce b/2417/CH1/EX1.11/Ex1_11.sce new file mode 100755 index 000000000..f555d7f6b --- /dev/null +++ b/2417/CH1/EX1.11/Ex1_11.sce @@ -0,0 +1,15 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 1.11\n\n\n");
+// Chapter 1: Fundamental Concepts
+// Problem 1.11 (page no. 35)
+// Solution
+
+//Given
+Patm=30.0 //in. //pressure of mercury at standard temperature
+Vacuum=26.5 //in. //vaccum pressure
+Pabs=Patm-Vacuum; //Absolute pressure of mercury //in.
+// (3.5 inch* (ft/12 inch) * (13.6*62.4)LBf/ft^3 * kg/2.2 LBf * 9.806 N/kg)/((12 inch^2/ft^2) * (0.0254 m/inch)^2)
+p=(3.5*(1/12)*13.6*62.4*(1/2.2)*9.806)/(12^2*0.0254^2*1000); //kPa //Absolute pressure in psia
+printf("Absolute pressure of mercury is %f kPa",p)
diff --git a/2417/CH1/EX1.2/Ex1_2.sce b/2417/CH1/EX1.2/Ex1_2.sce new file mode 100755 index 000000000..e81e34141 --- /dev/null +++ b/2417/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,23 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 1.2\n\n\n");
+// Chapter 1: Fundamental Concepts
+// Problem 1.2 (page no. 18)
+// Solution
+
+//Given
+Mm=0.0123//Unit:lb //Mass of the moon;
+Me=1 //Unit:lb //Mass of the earth;
+Dm=0.273 //Unit:feet //Diameter of the moon;
+De=1 //Unit:feet //Diameter of the earth;
+Rm=Dm/2; //Radius of the moon; //Unit:feet
+Re=De/2; //Radius of the earth; //Unit:feet
+
+//F=(K*M1*M2)/d^2 //Law of universal gravitation;
+//Fe=(K*Me*m)/Re^2; //Fe=Force exerted on the mass;
+//Fm=(K*Mm*m)/Rm^2; //Fm=Force exerted on the moon;
+F=(Me/Mm)*(Rm/Re)^2; //F=Fe/Fm;
+printf("Relation of force exerted on earth to mass is")
+disp(F,"Fe/Fm =");
+
diff --git a/2417/CH1/EX1.3/Ex1_3.sce b/2417/CH1/EX1.3/Ex1_3.sce new file mode 100755 index 000000000..fd00437aa --- /dev/null +++ b/2417/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 1.3\n\n\n");
+// Chapter 1: Fundamental Concepts
+// Problem 1.3 (page no. 20)
+// Solution
+
+//Given
+M=5; //Unit:kg //mass of body;
+g=9.81; //Unit:m/s^2 //the local acceleration of gravity
+W=M*g; //W=the weight of the body //Unit:Newton // 1 N= 1 kg*m/s^2
+printf("The weight of the body is %f N",W);
diff --git a/2417/CH1/EX1.4/Ex1_4.sce b/2417/CH1/EX1.4/Ex1_4.sce new file mode 100755 index 000000000..807ae99e9 --- /dev/null +++ b/2417/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,20 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 1.4\n\n\n");
+// Chapter 1: Fundamental Concepts
+// Problem 1.4 (page no. 21)
+// Solution
+
+printf("Solution for (a)\n");
+//given
+M=10 //Unit:kg //mass of body;
+g=9.5 //Unit:m/s^2 //the local acceleration of gravity
+W=M*g; //W=the weight of the body; //Unit:Newton // 1 N= 1 kg*m/s^2
+printf("The weight of the body is %f N\n\n",W);
+
+printf("Solution for (b)\n");
+//Given
+F=10; //Unit:Newton //Horizontal Force
+a=F/M; //newton's second law of motion
+printf("The horizontal acceleration of the body is %f m/s^2\n",a);
diff --git a/2417/CH1/EX1.5/Ex1_5.sce b/2417/CH1/EX1.5/Ex1_5.sce new file mode 100755 index 000000000..60aa042b0 --- /dev/null +++ b/2417/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,15 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 1.5\n\n\n");
+// Chapter 1: Fundamental Concepts
+// Problem 1.5 (page no. 25)
+// Solution
+
+//Conversion Problem
+// 1 inch=0.0254 meter so, 1=0.0254 meter/inch //Eq.1
+// 1 ft=12 inch so, 1=12 inch/ft.........//Eq.2
+//Multiplying Eq.1 & Eq.2 // We get 1=0.0254*12 meter/ft
+//Taking Square both side
+//1^2=(0.0254*12)^2 meter^2/ft^2
+printf("1 ft^2=%f meter^2\n",(0.0254*12)^2);
diff --git a/2417/CH1/EX1.7/Ex1_7.sce b/2417/CH1/EX1.7/Ex1_7.sce new file mode 100755 index 000000000..207215a02 --- /dev/null +++ b/2417/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,23 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 1.7\n\n\n");
+// Chapter 1: Fundamental Concepts
+// Problem 1.7 (page no. 33)
+// Solution
+
+//The Specific gravity of mercury is 13.6 //Given
+//Converting the unit of weight of grams per cubic centimeter to pounds per cubic foot
+// 1 lbf=454 gram //1 inch= 2.54 cm
+//So 1 gram=1/454 lbf and 1 ft=12*2.54 cm
+//Gamma=(gram/cm^3)*(lb/gram)*(cm^3/ft^3)=lb/ft^3
+//Gamma=(1 gram/cm^3)*(1 lbf/454 gram)*(2.54*12)^3 *cm^3/ft^3
+Gamma=(1/454)*(2.54*12)^3; //lbf/ft^3 //conversion factor
+disp(Gamma,"Conversion Factor=");
+p=(1/12)*(Gamma*13.6); //lbf/ft^2 //gage pressure
+p=(1/12)*Gamma*13.6*(1/144) //ft^2/inch^2 //gage pressure
+printf("Guage Pressure is %f psi\n",p);
+printf("Local atmospheric pressure is 14.7 psia\n");
+P=p+14.7; //Pressure on the base of the column //Unit:psia
+printf(" So Pressure on the base of the column is %f psia",P);
+
diff --git a/2417/CH1/EX1.8/Ex1_8.sce b/2417/CH1/EX1.8/Ex1_8.sce new file mode 100755 index 000000000..4f12ee5bf --- /dev/null +++ b/2417/CH1/EX1.8/Ex1_8.sce @@ -0,0 +1,15 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 1.8\n\n\n");
+// Chapter 1: Fundamental Concepts
+// Problem 1.8 (page no. 34)
+// Solution
+
+//Given
+Rho=13.595; //Unit: kg/m^3 //The density of mercury
+h=25.4; //Unit: mm //Height of column of mercury
+g=9.806; //Unit:m/s^2 //the local acceleration of gravity
+//Solution
+p=Rho*g*h; //P=Pressure at the base of a column of mercury //Unit:Pa
+printf("Pressure at the base of a column of mercury is %f Pa",p);
diff --git a/2417/CH1/EX1.9/Ex1_9.sce b/2417/CH1/EX1.9/Ex1_9.sce new file mode 100755 index 000000000..ff62d1ccf --- /dev/null +++ b/2417/CH1/EX1.9/Ex1_9.sce @@ -0,0 +1,16 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 1.9\n\n\n");
+// Chapter 1: Fundamental Concepts
+// Problem 1.9 (page no. 34)
+// Solution
+
+//Given
+Patm=30.0; //in. //pressure of mercury at standard temperature
+Vacuum=26.5; //in. //vaccum pressure
+Pabs=Patm-Vacuum; //Absolute pressure of mercury //in.
+// 1 inch mercury exerts a pressure of 0.491 psi
+p=Pabs*0.491; //Absolute pressure in psia
+printf("Absolute pressure of mercury in is %f psia",p);
+
diff --git a/2417/CH10/EX10.1/Ex10_1.sce b/2417/CH10/EX10.1/Ex10_1.sce new file mode 100755 index 000000000..0d57bcd9f --- /dev/null +++ b/2417/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,20 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.1\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.1 (page no. 503)
+// Solution
+
+T1=70+460; //70F=70+460 R //Energy flows into the system at reservoir at constant temperature T1(unit:R)
+T2=32+460; //32F=32+460 R //Heat is rejected to the constant temperature T2(Unit:R)
+printf("Solution for (a),\n");
+COP=T2/(T1-T2); //Coefficient of performance
+printf("Coefficient of performance(COP) of the cycle is %f\n\n",COP);
+printf("Solution for (b),\n");
+Qremoved=1000; //Unit:Btu/min //heat removal
+WbyJ=Qremoved/COP; //The power required //Unit:Btu/min
+printf("The power required is %f Btu/min\n\n",WbyJ);
+printf("Solution for (c),\n");
+Qrej=Qremoved+WbyJ; //The rate of heat rejected to the room //Unit:Btu/min
+printf("The rate of heat rejected to the room is %f Btu/min",Qrej);
diff --git a/2417/CH10/EX10.10/Ex10_10.sce b/2417/CH10/EX10.10/Ex10_10.sce new file mode 100755 index 000000000..f78f22473 --- /dev/null +++ b/2417/CH10/EX10.10/Ex10_10.sce @@ -0,0 +1,52 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.10\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.10 (page no. 518)
+// Solution
+
+printf("Solution for (a)\n");
+//By defination,the efficiency of the compressor is the ratio of the ideal compression work to actual compression work.
+//Based on the points on fig.10.12, //n=(h4-h3)/(h4'-h3);
+//There is close correspondence between 5.3 psia and -60F for saturated conditions.Therefore,state 3 is a superheated vapour at 5.3 psia and approximately -20F,because the problem states
+//that state 3 has a 40F superheat.Interpolation in the Freon tables in Appendix 3 yields
+T=-20; //Unit:F //temperature
+// p h s
+//7.5 75.719 0.18371
+//5.3 76.885 0.18985 h3=75.886 Btu/lbm
+//5.0 75.990 0.19069
+
+//At 100 psia and s=0.18985,
+// t s h
+// 170F 0.18996 100.571
+// 169.6F 0.18985 100.5 h4=100.5 Btu/lbm
+// 160F 0.18726 98.884
+
+//The weight of refrigerant is given by
+// 200(tons)/(h3-h1) = (200*5)/(75.886-h1)
+//In the saturated tables,h1 is
+// p h
+// 101.86 26.832
+// 100psia 26.542
+// 98.87 26.365
+
+//m=mass flow/min
+h1=26.542; //enthalpy //Unit:Btu/lbm
+n=0.8; //Efficiency
+h4=100.5; //enthalpy //Unit:Btu/lbm
+h3=75.886; //enthalpy //Unit:Btu/lbm
+m=(200*5)/(75.886-h1); //mass
+h4dashminush3=(h4-h3)/n;
+//Total work of compression=m*(h4minush3)
+J=778; //J=Conversion factor
+work=(h4dashminush3*m*J)/33000; //1 horsepower = 33,000 ft*LBf/min //Unit:hp //work
+printf("%f horsepower is required to drive the compressor if it has a mechanical efficiency 100percentage\n\n",work);
+
+printf("Solution for (b)\n");
+//Assuming a specific heat of the water as unity,we obtain
+//From part (a),
+//h4'-h3=h4minush3
+h4dash=h4dashminush3+h3; //Unit:Btu/lbm
+mdot=(m*(h4dash-h1))/(70-60); //water enters at 60F and leaves at 70F //the required capacity in lbm/min
+printf("%f lbm/min of cooling water i.e. %f gal/min is the required capacity of cooling water to pump",mdot,mdot/8.3);
diff --git a/2417/CH10/EX10.11/Ex10_11.sce b/2417/CH10/EX10.11/Ex10_11.sce new file mode 100755 index 000000000..f411ec582 --- /dev/null +++ b/2417/CH10/EX10.11/Ex10_11.sce @@ -0,0 +1,24 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.11\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.11 (page no. 521)
+// Solution
+
+printf("Solution for (a)\n");
+//From appendix3,reading the p-h diagram directly,we have
+h3=76.2; //Unit:Btu/lbm //Enthalpy
+h4=100.5; //Unit:Btu/lbm //Enthalpy
+n=0.8; //Efficiency //From 10.10
+work=(h4-h3)/n; //Work of compression //Unit:Btu/lbm
+//The enthalpy of saturated liquid at 100 psia is given at 26.1 Btu/lbm.Proceeding as before yields
+m=(200*5)/(h3-26.1); //Unit:lbm/min //m=massflow/min
+J=778; //J=Conversion factor
+totalwork=(m*work*J)/33000; //1 horsepower = 33,000 ft*LBf/min //total ideal work //unit:hp
+printf("Total ideal work of compression is %f hp\n\n",totalwork);
+
+printf("Solution for (b)\n");
+h4dash=h3+work; //Btu/lbm
+mdot=(m*(h4dash-26.5))/(70-60); //water enters at 60F and leaves at 70F //the required capacity in lbm/min
+printf("%f lbm/min of cooling water i.e. %f gal/min is the required capacity of cooling water to pump",mdot,mdot/8.3);
diff --git a/2417/CH10/EX10.12/Ex10_12.sce b/2417/CH10/EX10.12/Ex10_12.sce new file mode 100755 index 000000000..7f47451e0 --- /dev/null +++ b/2417/CH10/EX10.12/Ex10_12.sce @@ -0,0 +1,20 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.12\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.12 (page no. 526)
+// Solution
+
+COP=2.5; //Coefficient of performance
+cp=0.24; //Unit:Btu/(lbm*R) //Specific heat constant for constant pressure process
+T1=-100+460; //temperatures converted to absolute temperatures; //Unit:R //lowest temperature of the cycle
+T3=150+460; //temperatures converted to absolute temperatures; //Unit:R //Upper temperature of the cycle
+//T1/T4-T1 = COP
+T4=(3.5*T1)/COP; //Unit:R //temperature at point 4
+//T2/T3-T2 =COP
+T2=(COP*T3)/3.5; //Unit:R //temperature at point 2
+printf("The work of the expander is %f Btu/lbm of air\n",cp*(T4-T1));
+printf("The work of the compressor is %f Btu/lbm of air\n",cp*(T3-T2));
+printf("The net work required by the cycle is %f Btu/lbm\n",(cp*(T3-T2))-(cp*(T4-T1)));
+printf("Per ton of refrigeration,the required airflow is %f lbm/min per ton\n",200/(cp*(T2-T1)));
diff --git a/2417/CH10/EX10.13/Ex10_13.sce b/2417/CH10/EX10.13/Ex10_13.sce new file mode 100755 index 000000000..f02de3f09 --- /dev/null +++ b/2417/CH10/EX10.13/Ex10_13.sce @@ -0,0 +1,21 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.13\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.13 (page no. 536)
+// Solution
+
+//A VACUUM REFRIGERATION SYSTEM
+//A vacuum refrigeration system is used to cool water from 90F to 45F
+h1=58.07; //Unit:Btu/lbm //enthalpy
+h2=13.04; //Unit:Btu/lbm //enthalpy
+h3=1081.1; //Unit:Btu/lbm //enthalpy
+m1=1; //mass //unit:lbm
+//m2=1-m3 //unit:lbm
+//Now, m1*h1 = m2*h2 + m3*h3
+//Putting the values and arranging the equation,
+m3=(m1*h1-h2)/(h3+h2); //The mass of vapour that must be removed per pound //unit:lbm
+printf("The mass of vapour that must be removed per pound of entering water is %f lbm",m3);
+
+
diff --git a/2417/CH10/EX10.14/Ex10_14.sce b/2417/CH10/EX10.14/Ex10_14.sce new file mode 100755 index 000000000..ae45de2e4 --- /dev/null +++ b/2417/CH10/EX10.14/Ex10_14.sce @@ -0,0 +1,25 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.14\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.14 (page no. 536)
+// Solution
+
+//In problem 10.13,
+//A VACUUM REFRIGERATION SYSTEM
+//A vacuum refrigeration system is used to cool water from 90F to 45F
+h1=58.07; //Unit:Btu/lbm //enthalpy
+h2=13.04; //Unit:Btu/lbm //enthalpy
+h3=1081.1; //Unit:Btu/lbm //enthalpy
+m1=1; //mass //lbm
+//m2=1-m3 //unit:lbm
+//Now, m1*h1 = m2*h2 + m3*h3
+//Putting the values and arranging the equation,
+m3=(m1*h1-h2)/(h3+h2); //The mass of vapour that must be removed per pound //unit:lbm
+m2=1-m3; //mass //unit:lbm
+printf("The mass of vapour that must be removed per pound of entering water is %f lbm\n",m3);
+//Now,in problem 10.14,
+//The refrigeration effect can be determined as m3*(h3-h1) or m2*(h1-h2)
+printf("The refrigeration effect using eqn m3*(h3-h1) is %f Btu/lbm\n",m3*(h3-h1));
+printf("The refrigeration effect using eqn m2*(h1-h2) is %f Btu/lbm\n",m2*(h1-h2));
diff --git a/2417/CH10/EX10.15/Ex10_15.sce b/2417/CH10/EX10.15/Ex10_15.sce new file mode 100755 index 000000000..4c9364a8d --- /dev/null +++ b/2417/CH10/EX10.15/Ex10_15.sce @@ -0,0 +1,17 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.15\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.15 (page no. 539)
+// Solution
+
+//THE HEAT PUMP
+T1=70+460; //70F=70+460 R //Energy flows into the system at reservoir at constant temperature T1(unit:R) //from problem 10.1
+T2=32+460; //32F=32+460 R //Heat is rejected to the constant temperature T2(Unit:R) //from problem 10.1
+COP=T1/(T1-T2); //Coefficient of performance for carnot heat pump
+printf("Coefficient of performance(COP) of the carnot cycle is %f\n",COP);
+printf("The COP can also be obtained from the energy items solved for in problem 10.1\n")
+//In problem 10.1, The power was found to be 77.2 Btu/min and the total tare of heat rejection was 1077.2 Btu/min
+//Therefore,
+printf("Coefficient of performance(COP) of the cycle is %f\n",1077.2/77.2);
diff --git a/2417/CH10/EX10.16/Ex10_16.sce b/2417/CH10/EX10.16/Ex10_16.sce new file mode 100755 index 000000000..d21b9d46c --- /dev/null +++ b/2417/CH10/EX10.16/Ex10_16.sce @@ -0,0 +1,22 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.16\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.16 (page no. 539)
+// Solution
+
+//Let us first consider the cycle as a refrigeration cycle
+//In problem 10.1
+T1=70+460; //70F=70+460 R //Energy flows into the system at reservoir at constant temperature T1(unit:R)
+T2=0+460; //0F=32+460 R //Heat is rejected to the constant temperature T2(Unit:R)
+COP=T2/(T1-T2); //Coefficient of performance
+printf("Coefficient of performance(COP) of the cycle is %f\n\n",COP);
+Qremoved=1000; //Unit:Btu/min //heat removal
+WbyJ=Qremoved/COP; //the power input //unit:Btu/min
+printf("The power input is %f Btu/min\n\n",WbyJ);
+Qrej=Qremoved+WbyJ; //The rate of heat rejected to the room //Unit:Btu/min
+printf("The rate of heat rejected to the room is %f Btu/min\n",Qrej);
+printf("The COP as a heat pump is %f\n",Qrej/WbyJ);
+printf("As a check,COP of heat pump is %f = 1 + COP of carnot cycle %f",Qrej/WbyJ,COP);
+
diff --git a/2417/CH10/EX10.2/Ex10_2.sce b/2417/CH10/EX10.2/Ex10_2.sce new file mode 100755 index 000000000..77d5e6ac7 --- /dev/null +++ b/2417/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,20 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.2\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.2 (page no. 504)
+// Solution
+
+T1=20+273; //20C=20+273 R //Energy flows into the system at reservoir at constant temperature T1(unit:R)
+T2=-5+273; //-5C=-5+273 R //Heat is rejected to the constant temperature T2(Unit:R)
+printf("Solution for (a),\n");
+COP=T2/(T1-T2); //Coefficient of performance
+printf("Coefficient of performance(COP) of the cycle is %f\n\n",COP);
+printf("Solution for (b),\n");
+Qremoved=30; //Unit:kW //heat removal
+W=Qremoved/COP; //power required //unit:kW
+printf("The power required is %f kW \n\n",W);
+printf("Solution for (c),\n");
+Qrej=Qremoved+W; //The rate of heat rejected to the room //Unit:kW
+printf("The rate of heat rejected to the room is %f kW",Qrej);
diff --git a/2417/CH10/EX10.3/Ex10_3.sce b/2417/CH10/EX10.3/Ex10_3.sce new file mode 100755 index 000000000..ca20b861d --- /dev/null +++ b/2417/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,18 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.3\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.3 (page no. 505)
+// Solution
+
+T1=70+460; //70F=70+460 R //Energy flows into the system at reservoir at constant temperature T1(unit:R)
+T2=20+460; //20F=20+460 R //Heat is rejected to the constant temperature T2(Unit:R)
+printf("Solution for (a),\n");
+COP=T2/(T1-T2); //Coefficient of performance
+printf("Coefficient of performance(COP) of the cycle is %f\n\n",COP);
+printf("Solution for (b),\n");
+HPperTOR=4.717/COP; //Horsepower per ton of refrigeration //Unit:hp/ton
+COPactual=2; //Actual Coefficient of performance(COP) is stated to be 2
+HPperTORactual=4.717/COPactual; //Horsepower per ton of refrigeration(actual) //Unit:hp/ton
+printf("The horsepower required by the actual cycle over the minimum is %f hp/ton",HPperTORactual-HPperTOR);
diff --git a/2417/CH10/EX10.4/Ex10_4.sce b/2417/CH10/EX10.4/Ex10_4.sce new file mode 100755 index 000000000..8849350a7 --- /dev/null +++ b/2417/CH10/EX10.4/Ex10_4.sce @@ -0,0 +1,20 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.4\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.4 (page no. 506)
+// Solution
+
+COP=4.5; //Coefficient of performance //From problem 10.1
+HPperTOR=4.717/COP; //Horsepower per ton of refrigeration //Unit:hp/ton
+Qremoved=1000; //Unit:Btu/min //From problem 10.1
+//1000 Btu/min /200 Btu/min ton = 5 tons of refrigeration
+HPrequired=HPperTOR*5; //The horsepower required //unit:hp
+printf("The horsepower required is %f hp\n",HPrequired);
+//In problem 10.1, 77.2 Btu/min was required
+printf("The power required is %f hp\n",77.2*778*inv(33000)); //1 Btu=778 ft*lbf //1 min*hp = 33000 ft*lbf
+//The ratio of the power required in each problem is the same as the inverse ratio of the COP value
+//Therefore,
+printf("The power required is %f hp\n",(COP/12.95)*HPrequired); //COP(in problem 10.1)=12.95
+printf("This checks our results")
diff --git a/2417/CH10/EX10.5/Ex10_5.sce b/2417/CH10/EX10.5/Ex10_5.sce new file mode 100755 index 000000000..b41b0a16e --- /dev/null +++ b/2417/CH10/EX10.5/Ex10_5.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.5\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.5 (page no. 506)
+// Solution
+
+COP=10.72; //In the problem 10.2 //Coefficient of performance
+P=2.8; //In the problem 10.2 //The power was 2.8 kW
+COPactual=3.8; //Actual Coefficient of performance(COP)
+power=P*COP/COPactual; //The power required //unit:kW
+printf("The power required is %f kW",power)
diff --git a/2417/CH10/EX10.6/Ex10_6.sce b/2417/CH10/EX10.6/Ex10_6.sce new file mode 100755 index 000000000..967100470 --- /dev/null +++ b/2417/CH10/EX10.6/Ex10_6.sce @@ -0,0 +1,31 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.6\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.6 (page no. 509)
+// Solution
+
+//From Appendix 3,at 120psia,the corresponding saturation temperature is 66 F, enthalpies are
+h1=116.0; //Unit:Btu/lbm //enthalpy
+h2=116.0; //Unit:Btu/lbm //Throttling gives h1=h2 //enthalpy
+h3=602.4; //Unit:Btu/lbm //enthalpy
+//From the consideration that s3=s4,h4 is found at 15 psia,
+s3=1.3938; //s=entropy //Unit:Btu/(lbm*F)
+//Therefore by interpolation in the superheat tables at 120 psia,
+t4=237.4; //Unit:fahrenheit //temperature
+h4=733.4; //Unit:Btu/lbm //enthalpy
+printf("Solution for (a),\n");
+COP=(h3-h1)/(h4-h3); //Coefficient of performance
+printf("Coefficient of performance is %f\n\n",COP);
+printf("Solution for (b),\n");
+printf("The work of compression is %f Btu/lbm\n\n",h4-h3);
+printf("Solution for (c),\n");
+printf("The refrigatering effect is %f Btu/lbm\n\n",h3-h1);
+printf("Solution for (d),\n");
+tons=30; //capacity of 30 tons is desired
+printf("The pounds per minute of ammonia required for ciculation is %f lbm/min\n\n",(200*tons)/(h3-h1));
+printf("Solution for (e),\n");
+printf("The ideal horsepower per ton of refrigeration is %f hp/ton\n\n",4.717*((h4-h3)/(h3-h1)));
+
+
diff --git a/2417/CH10/EX10.7/Ex10_7.sce b/2417/CH10/EX10.7/Ex10_7.sce new file mode 100755 index 000000000..8299885a1 --- /dev/null +++ b/2417/CH10/EX10.7/Ex10_7.sce @@ -0,0 +1,30 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.7\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.7 (page no. 510)
+// Solution
+
+//From Appendix 3,110 psig corresponds to 96 F, enthalpies are
+h1=30.14; //Unit:Btu/lbm //enthalpy
+h2=30.14; //Unit:Btu/lbm //Throttling gives h1=h2 //enthalpy
+h3=75.110; //Unit:Btu/lbm //enthalpy
+//From the consideration that s3=s4,at -20F,
+s3=0.17102; //Unit:Btu/(lbm*F) //s=entropy
+//Therefore by interpolation in the Freon-12 superheat table at these values,
+h4=89.293; //Unit:Btu/lbm //enthalpy
+printf("Solution for (a),\n");
+COP=(h3-h1)/(h4-h3); //Coefficient of performance
+printf("Coefficient of performance is %f\n\n",COP);
+printf("Solution for (b),\n");
+printf("The work of compression is %f Btu/lbm\n\n",h4-h3);
+printf("Solution for (c),\n");
+printf("The refrigatering effect is %f Btu/lbm\n\n",h3-h1);
+printf("Solution for (d),\n");
+tons=30; //capacity of 30 tons is desired
+printf("The pounds per minute of ammonia required for ciculation is %f lbm/min\n\n",(200*tons)/(h3-h1));
+printf("Solution for (e),\n");
+printf("The ideal horsepower per ton of refrigeration is %f hp/ton\n\n",4.717*((h4-h3)/(h3-h1)));
+
+
diff --git a/2417/CH10/EX10.8/Ex10_8.sce b/2417/CH10/EX10.8/Ex10_8.sce new file mode 100755 index 000000000..6b54adcdc --- /dev/null +++ b/2417/CH10/EX10.8/Ex10_8.sce @@ -0,0 +1,21 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.8\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.8 (page no. 517)
+// Solution
+
+//From Appendix 3,using the Freon-12 tables, enthalpies are
+h1=28.713; //Unit:Btu/lbm //enthalpy
+h2=28.713; //Unit:Btu/lbm //Throttling gives h1=h2 //enthalpy
+h3=78.335; //Unit:Btu/lbm //enthalpy
+//From the consideration that s3=s4,
+s3=0.16798; //Unit:Btu/(lbm*F) //s=entropy
+//Therefore by interpolation in the superheat tables at 90 F,
+s=0.16798; //entropy at 90F //Btu/lbm*F
+h4=87.192; //Unit:Btu/lbm //enthalpy
+printf("The heat extracted is %f Btu/lbm\n\n",h3-h1);
+printf("The work required is %f Btu/lbm\n\n",h4-h3);
+COP=(h3-h1)/(h4-h3); //Coefficient of performance
+printf("The Coefficient of performance(COP) of this ideal cycle is %f",COP);
diff --git a/2417/CH10/EX10.9/Ex10_9.sce b/2417/CH10/EX10.9/Ex10_9.sce new file mode 100755 index 000000000..c282b6864 --- /dev/null +++ b/2417/CH10/EX10.9/Ex10_9.sce @@ -0,0 +1,19 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 10.9\n\n\n");
+// Chapter 10 : Refrigeration
+// Problem 10.9 (page no. 518)
+// Solution
+
+//From Appendix 3,using the HFC-134a tables, enthalpies are
+h1=41.6; //Unit:Btu/lbm //enthalpy
+h2=41.6; //Unit:Btu/lbm //Throttling gives h1=h2 //enthalpy
+h3=104.6; //Unit:Btu/lbm //enthalpy
+//From the consideration that s3=s4,
+s3=0.2244; //Unit:Btu/(lbm*F) //s=entropy
+h4=116.0; //Unit:Btu/lbm //enthalpy
+printf("The heat extracted is %f Btu/lbm\n\n",h3-h1);
+printf("The work required is %f Btu/lbm\n\n",h4-h3);
+COP=(h3-h1)/(h4-h3); //Coefficient of performance
+printf("The Coefficient of performance(COP) of this ideal cycle is %f",COP);
diff --git a/2417/CH11/EX11.1/Ex11_1.sce b/2417/CH11/EX11.1/Ex11_1.sce new file mode 100755 index 000000000..8b7f92223 --- /dev/null +++ b/2417/CH11/EX11.1/Ex11_1.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 11.1\n\n\n");
+// Chapter 11: Heat Transfer
+// Problem 11.1 (page no. 553)
+// Solution
+
+
+deltaX=6/12; //6 inch = 6/12 feet //deltaX=length //unit:feet
+k=0.40; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table
+T1=150; //temperature maintained at one face //fahrenheit
+T2=80; //tempetature maintained at other face //fahrenheit
+deltaT=T2-T1; //fahrenheit //Change in temperature
+Q=(-k*deltaT)/deltaX; //Heat transfer per square foot of wall //Unit:Btu/hr*ft^2
+printf("Heat transfer per square foot of wall is %f Btu/hr*ft^2",Q);
diff --git a/2417/CH11/EX11.10/Ex11_10.sce b/2417/CH11/EX11.10/Ex11_10.sce new file mode 100755 index 000000000..d43f2bc55 --- /dev/null +++ b/2417/CH11/EX11.10/Ex11_10.sce @@ -0,0 +1,19 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.10\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.10 (page no. 566)
+// Solution
+
+//A bare steel pipe
+ro=90; //Outside diameter //Unit:mm
+ri=75; //inside diameter //Unit:mm
+Ti=110; //Inside temperature //Unit:Celcius
+To=40; //Outside temperature //Unit:Celcius
+L=2; //Length //Unit:m
+deltaT=Ti-To; //Change in temperature //Unit:Celcius
+k=45 //Unit:W/(m*C) //k=proportionality constant //k=thermal conductivity
+Q=(2*%pi*k*L*deltaT)/log(ro/ri); //The heat loss from the pipe //unit:W
+printf("The heat loss from the pipe is %f W",Q);
+
diff --git a/2417/CH11/EX11.11/Ex11_11.sce b/2417/CH11/EX11.11/Ex11_11.sce new file mode 100755 index 000000000..dd04cd66b --- /dev/null +++ b/2417/CH11/EX11.11/Ex11_11.sce @@ -0,0 +1,30 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.11\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.11 (page no. 567)
+// Solution
+
+//From problem 11.9,
+//A bare steel pipe
+r2=3.50; //Outside diameter //Unit:in.
+r1=3.00; //inside diameter //Unit:in.
+Ti=240; //Inside temperature //unit:fahrenheit
+L=5; //Length //Unit:ft
+k1=26; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity
+ans1=(inv(k1)*log(r2/r1));
+
+//Now,in problem 11.11,
+//Mineral wool
+r3=5.50; //inside diameter //Unit:in.
+r2=3.50; //outside diameter //Unit:in.
+To=85; //Outside temperature //unit:fahrenheit
+deltaT=Ti-To; //Change in temperature //unit:fahrenheit
+k2=0.026 //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity
+ans2=(inv(k2)*log(r3/r2));
+
+Q=(2*%pi*L*deltaT)/(ans1+ans2); //The heat loss from the pipe //unit:Btu/hr
+printf("The heat loss from the pipe is %f Btu/hr",Q);
+
+
diff --git a/2417/CH11/EX11.12/Ex11_12.sce b/2417/CH11/EX11.12/Ex11_12.sce new file mode 100755 index 000000000..ebe3a6e28 --- /dev/null +++ b/2417/CH11/EX11.12/Ex11_12.sce @@ -0,0 +1,30 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.12\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.12 (page no. 569)
+// Solution
+
+//From problem 11.9,
+//The bare pipe
+r2=3.50; //Outside diameter //Unit:in.
+r1=3.00; //inside diameter //Unit:in.
+Ti=240; //Inside temperature //unit:fahrenheit
+L=5; //Length //Unit:ft
+k=26; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity
+Rpipe=log(r2/r1)/(2*%pi*k*L); //the resistance of pipe //Unit:(hr*F)/Btu
+printf("The resistance of pipe is %f (hr*F)/Btu\n",Rpipe);
+
+//Now,in problem 11.12,
+To=70; //Outside temperature //unit:fahrenheit
+deltaT=Ti-To; //Change in temperature //unit:fahrenheit
+h=0.9; //Coefficient of heat transfer //Unit:Btu/(hr*ft^2*F)
+A=(%pi*r2)/12*L; //Area //Unit:ft^2 //1 inch = 1/12 feet //unit:ft^2
+Rconvection=inv(h*A); //The resistance due to natural convection to the surrounding air //Unit:(hr*F)/Btu
+printf("The resistance due to natural convection to the surrounding air is %f (hr*F)/Btu\n",Rconvection);
+
+Rtotal=Rpipe+Rconvection; //The total resistance //unit:(hr*F)/Btu
+printf("The total resistance is %f (hr*F)/Btu\n\n",Rtotal);
+Q=deltaT/Rtotal; //ohm's law (fourier's equation) //The heat transfer from the pipe to the surrounding air //unit:Btu/hr
+printf("The heat transfer from the pipe to the surrounding air is %f Btu/hr\n",Q);
diff --git a/2417/CH11/EX11.15/Ex11_15.sce b/2417/CH11/EX11.15/Ex11_15.sce new file mode 100755 index 000000000..ced8a28f2 --- /dev/null +++ b/2417/CH11/EX11.15/Ex11_15.sce @@ -0,0 +1,17 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.15\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.15 (page no. 574)
+// Solution
+
+D=3.5/12; //3.5 inch = 3.5/12 feet//Unit:ft //Outside diameter
+Ti=120; //Inside temperature //unit:fahrenheit
+To=70; //Outside temperature //unit:fahrenheit
+deltaT=Ti-To; //unit:fahrenheit //Change in temperature
+h=0.9; //Coefficient of heat transfer //Unit:Btu/(hr*ft^2*F)
+L=5; //Length //Unit:ft //From problem 11.10
+A=(%pi*D)*L; //Area //Unit:ft^2
+Q=h*A*deltaT; //The heat loss due to convection //Unit:Btu/hr //Newton's law of cooling
+printf("The heat loss due to convection is %f Btu/hr",Q);
diff --git a/2417/CH11/EX11.16/Ex11_16.sce b/2417/CH11/EX11.16/Ex11_16.sce new file mode 100755 index 000000000..48ec9d9ca --- /dev/null +++ b/2417/CH11/EX11.16/Ex11_16.sce @@ -0,0 +1,47 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.16\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.16 (page no. 575)
+// Solution
+
+//This problem can not be solved directly,because the individual film resistances aree functions of unknown temperature differences.Therefore,
+//From the first approximation,
+h=1/2; //Coefficient of heat transfer //unit:Btu/(hr*ft^2*F)
+//For area 1 ft^2,
+R=(3/12)/0.07; //The wall resistance is deltax/(k*A) //k=0.07 //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity
+Roverall=inv(1/2)+inv(1/2)+R; //the overall series resistance //Unit:Btu/(hr*ft*F)
+printf("For h=0.5,the overall series resistance is %f Btu/(hr*ft*F)\n",Roverall);
+//Using the value of Roverall,we can now obtain Q and individual temperature differences,
+Ti=80; //warm air temperature //unit:fahrenheit
+To=50; //cold air temperature //unit:fahrenheit
+deltaT=Ti-To; //unit:fahrenheit //Change in temperature
+Q=deltaT/Roverall; //Unit:Btu/(hr*ft^2) //heat transfer //ohm's law (fourier's equation)
+printf("For h=0.5,heat transfer is %f Btu/(hr*ft^2)\n",Q);
+printf("For h=0.5,\n");
+//deltaT through the hot air film is Q/(1/2)
+printf("Temperaure difference through the hot air film is %f F\n",Q/(1/2));
+//Throught the wall deltaT is R*Q
+printf("Temperaure difference through the wall is %f F\n",Q*R);
+//deltaT through the cold air film is Q/(1/2)
+printf("Temperaure difference through the cold air film is %f F\n\n",Q/(1/2));
+
+//With these temperature differences,we can now enter figures 11.12 and 11.14 to verify our approximation.From figure 11.14,we find h=0.42 Btu/(hr*ft*2*F)
+//Using h=0.42,we have for the overall resistance (1/0.42)+(1/0.42)+R
+h=0.42; //Coefficient of heat transfer //unit:Btu/(hr*ft^2*F)
+Roverall=inv(h)+inv(h)+R; //the overall series resistance //Unit:Btu/(hr*ft*F)
+printf("For h=0.42,the overall series resistance is %f Btu/(hr*ft*F)\n",Roverall);
+Q=deltaT/Roverall; //Unit:Btu/(hr*ft^2) //heat transfer //ohm's law (fourier's equation)
+printf("For h=0.42,heat transfer is %f Btu/(hr*ft^2)\n",Q);
+printf("For h=0.42,\n");
+// deltat through both air films is Q/h
+printf("Temperaure difference through the hot and cold air film is %f F\n",Q/h);
+//and through the wall,deltat is Q*R
+printf("Temperaure difference through the wall is %f F\n\n",Q*R);
+
+//Entering figure 11.14,we find that h stays essentially 0.42,and our solution is that the heat flow is Q,the "hot" side of the wall is at Ti-(Q/h),the "cold" side is at To+(Q/h) ,and temperature drop in the wall is Ti-(Q/h)-(To+(Q/h)).
+printf("The temperature drop on the hot side of the wall is %f F\n",Ti-(Q/h));
+printf("The temperature drop on the cold side of the wall is %f F\n",To+(Q/h));
+printf("The temperature drop in the wall is %f F\n",Ti-(Q/h)-(To+(Q/h)));
+//Which checks our wall deltat calculation.
diff --git a/2417/CH11/EX11.17/Ex11_17.sce b/2417/CH11/EX11.17/Ex11_17.sce new file mode 100755 index 000000000..b453bf07b --- /dev/null +++ b/2417/CH11/EX11.17/Ex11_17.sce @@ -0,0 +1,24 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.17\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.17 (page no. 578)
+// Solution
+
+//The first step is to check Reynolds number.It will be recalled that the Reynolds number is given by (D*V*rho)/mu and is dimensionless.Therefore,we can use D, diameter in feet;V velocity in ft/hr;rho density in lbm/ft^3 and mu viscosity in lbm/(ft*hr).
+//Alternatively,the Reynolds number is given by (D*G)/mu,where G is the mass flow rate per unit area (lbm/(hr*ft^2)).
+G=((20*60)*(4*144)/(%pi*0.87^2)); //Unit:lbm/(hr*ft^2) //Inside diameter=0.87 inch ////1 in.^2=144 ft^2 //20 lbm/min of water(min converted to second)
+//the viscosity of air at these conditions is obtained from figure 11.17 as 0.062 lbm/(ft*hr).So,
+mu=0.33; //the viscosity of air //unit:lbm/(ft*hr)
+D=0.87/12; //Inside diameter //1 in^2=144 ft^2
+//Therefore Reynolds number is
+Re=(D*G)/mu; //Reynolds number
+//which is well into the turbulent flow regime.
+printf("The Reynolds number is %f\n",Re);
+//The next step is to enter Figure 11.18 at W/1000 of 20*(60/1000)=1.2 and 400F to obtain h1=630.
+//From the figure 11.20,we obtain F=1.25 for an inside diameter of 0.87 inch.So,
+h1=630; //basic heat transfer coefficient //unit:Btu/(hr*ft^2*F)
+F=1.25; //correction factor
+h=h1*F; //heat transfer coefficient //the inside film coefficient //unit:Btu/(hr*ft^2*F)
+printf("The heat-transfer coefficient is %f Btu/(hr*ft^2*F)\n",h);
diff --git a/2417/CH11/EX11.18/Ex11_18.sce b/2417/CH11/EX11.18/Ex11_18.sce new file mode 100755 index 000000000..164f9344a --- /dev/null +++ b/2417/CH11/EX11.18/Ex11_18.sce @@ -0,0 +1,23 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.18\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.18 (page no. 579)
+// Solution
+
+//We first check the Reynolds number and note that G is same as for problem 11.17.So,
+//G is the mass flow rate per unit area (lbm/(hr*ft^2)).
+G=((20*60)*(4*144))/(%pi*(0.87^2)); //Unit:lbm/(hr*ft^2) //Inside diameter=0.87 inch ////1 in.^2=144 ft^2 //20 lbm/min of water(min converted to second)
+//the viscosity of air at these conditions is obtained from figure 11.17 as 0.062 lbm/(ft*hr).So,
+mu=0.062; //the viscosity of air //unit:lbm/(ft*hr)
+D=0.87/12; //Inside diameter //1 in^2=144 ft^2
+//Reynolds number is DG/mu,therefore
+Re=(D*G)/mu; //Reynolds number
+printf("The Reynolds number is %f\n",Re);
+//which places the flow in the turbulent regime.Because W/1000(W=weight flow) is same as for problem 11.17 and equals 1.2,we now enter figure 11.19 at 1.2 and 400F to obtain h1=135.Because the inside tube diameter is same as before,F=1.25.Therefore,
+h1=135; //basic heat transfer coefficient //unit:Btu/(hr*ft^2*F)
+F=1.25; //correction factor
+h=h1*F; //heat transfer coefficient //the inside film coefficient //unit:Btu/(hr*ft^2*F)
+printf("The inside film coefficient is %f Btu/(hr*ft^2*F)\n",h);
+//It is interesting that for equal mass flow rates,water yields a heat-transfer coefficient almost five times greater than air
diff --git a/2417/CH11/EX11.19/Ex11_19.sce b/2417/CH11/EX11.19/Ex11_19.sce new file mode 100755 index 000000000..6552fdee6 --- /dev/null +++ b/2417/CH11/EX11.19/Ex11_19.sce @@ -0,0 +1,20 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.19\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.19 (page no. 586)
+// Solution
+
+//A bare steel pipe
+//From the Table 11.5,case 2,
+Fe=0.79; //Emissivity factor to allow for the departure of the surfaces interchanging heat from complete blackness;Fe is a function of the surface emissivities and configurations
+FA=1; //geometric factor to allow for the average solid angle through which one surface "sees" the other
+sigma=0.173*10^-8; //Stefan-Boltzmann constant //Unit:Btu/(hr*ft^2*R^4)
+T1=120+460; //outside temperature //Unit:R //fahrenheit converted to absolute temperature
+T2=70+460; //inside temperature //Unit:R //fahrenheit converted to absolute temperature
+D=3.5/12; //3.5 inch = 3.5/12 feet//Unit:ft //Outside diameter
+L=5; //Length //Unit:ft //From problem 11.10
+A=(%pi*D)*L; //Area //Unit:ft^2
+Q=sigma*Fe*FA*A*(T1^4-T2^4); //The net interchange of heat by radiation between two bodies at different temperatures //Unit:Btu/hr ////Stefan-Boltzmann law
+printf("The heat loss by radiation is %f Btu/hr\n",Q);
diff --git a/2417/CH11/EX11.2/Ex11_2.sce b/2417/CH11/EX11.2/Ex11_2.sce new file mode 100755 index 000000000..f45333f9b --- /dev/null +++ b/2417/CH11/EX11.2/Ex11_2.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 11.2\n\n\n");
+// Chapter 11: Heat Transfer
+// Problem 11.2 (page no. 553)
+// Solution
+
+deltaX=0.150; //Given,150 mm =0.150 meter // //deltaX=length //Unit:meter
+k=0.692; //Unit:W/(m*celcius) //k=proportionality constant //k=thermal conductivity
+T1=70; //temperature maintained at one face //celcius
+T2=30; //tempetature maintained at other face //celcius
+deltaT=T2-T1; //celcius //change in temperature
+Q=(-k*deltaT)/deltaX; //Heat transfer per square foot of wall //unit:W/m^2
+printf("Heat transfer per square foot of wall is %f W/m^2",Q);
diff --git a/2417/CH11/EX11.20/Ex11_20.sce b/2417/CH11/EX11.20/Ex11_20.sce new file mode 100755 index 000000000..16828c969 --- /dev/null +++ b/2417/CH11/EX11.20/Ex11_20.sce @@ -0,0 +1,27 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.20\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.20 (page no. 588)
+// Solution
+
+//The upper temperature is given as 120 F and the temperature difference is
+Ti=120; //Inside temperature //unit:fahrenheit
+To=70; //Outside temperature //unit:fahrenheit
+deltaT=120-70; //unit:fahrenheit //Change in temperature
+//Using figure 11.28,
+hrdash=1.18; //factor for radiation coefficient //Unit:Btu/(hr*ft^2*F)
+Fe=1; //Emissivity factor to allow for the departure of the surfaces interchanging heat from complete blackness;Fe is a function of the surface emissivities and configurations
+FA=0.79; //geometric factor to allow for the average solid angle through which one surface "sees" the other
+hr=Fe*FA*hrdash; //The radiation heat-transfer coefficient for the pipe //Unit:Btu/(hr*ft^2*F)
+printf("The radiation heat-transfer coefficient for the pipe is %f Btu/(hr*ft^2*F)\n",hr);
+
+//As a check,Using the results of problem 11.17,
+printf("As a check,using the results of problem 11.17,\n");
+D=3.5/12; //3.5 inch = 3.5/12 feet//Unit:ft //Outside diameter
+L=5; //Length //Unit:ft //From problem 11.10
+A=(%pi*D)*L; //Area //Unit:ft^2
+Q=214.5; //heat loss //Unit:Btu/hr
+hr=Q/(A*deltaT); //The radiation heat-transfer coefficient for the pipe //Unit:Btu/(hr*ft^2*F) //Newton's law of cooling
+printf("The radiation heat-transfer coefficient for the pipe is %f Btu/(hr*ft^2*F)\n",hr);
diff --git a/2417/CH11/EX11.21/Ex11_21.sce b/2417/CH11/EX11.21/Ex11_21.sce new file mode 100755 index 000000000..6e1c17a05 --- /dev/null +++ b/2417/CH11/EX11.21/Ex11_21.sce @@ -0,0 +1,25 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.21\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.21 (page no. 589)
+// Solution
+
+//Because the conditions of illustrative problem 11.15 are the same as for problem 11.19 and 11.20,we can solve this problem in two ways to obtain a check.
+//Thus,adding the results of these problems yields,
+printf("Adding the results of the problems yields,\n")
+Qtotal=206.2+214.5; //Unit:Btu/hr //total heat loss
+printf("The heat loss due to convection is %f Btu/hr\n",Qtotal);
+
+//We can also approach this solution by obtaining radiation and convection heat-transfer co-efficcient.Thus,
+hcombined=0.9+0.94; //Coefficient of heat transfer //Unit:Btu/(hr*ft^2*F)
+D=3.5/12; //3.5 inch = 3.5/12 feet //Unit:ft //Outside diameter
+Ti=120; //Inside temperature //unit:fahrenheit
+To=70; //Outside temperature //unit:fahrenheit
+deltaT=Ti-To; //unit:fahrenheit //Change in temperature
+L=5; //Length //Unit:ft //From problem 11.10
+A=(%pi*D)*L; //Area //Unit:ft^2
+Qtotal=hcombined*A*deltaT; //Unit:Btu/hr //total heat loss due to convection //Newton's law of cooling
+printf("By obtaining radiation and convection heat-transfer co-efficcient,\n")
+printf("The heat loss due to convection is %f Btu/hr",Qtotal);
diff --git a/2417/CH11/EX11.22/Ex11_22.sce b/2417/CH11/EX11.22/Ex11_22.sce new file mode 100755 index 000000000..5bd77c11f --- /dev/null +++ b/2417/CH11/EX11.22/Ex11_22.sce @@ -0,0 +1,50 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.22\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.22 (page no. 595)
+// Solution
+
+//For brick,concrete,plaster,hot film and cold film,
+A=1; //area //Unit:ft^2
+//For a plane wall,the areas are all the same,and if we use 1 ft^2 of wall surface as the reference area,
+//For Brick,
+deltax=6/12; //6 inch = 6/12 feet //deltax=length //unit:ft
+k=0.40; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table
+brickResistance=deltax/(k*A); //Thermal resistance //Unit:(hr*f)/Btu
+printf("For brick,");
+printf("The resistance is %f (hr*F)/Btu\n",brickResistance);
+
+//For Concrete,
+deltax=(1/2)/12; //(1/2) inch = (1/2)/12 feet //deltax=length //unit:ft
+k=0.80; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table
+concreteResistance=deltax/(k*A); //Thermal resistance //Unit:(hr*f)/Btu
+printf("For Concrete,");
+printf("The resistance is %f (hr*F)/Btu\n",concreteResistance);
+
+//For plaster,
+deltax=(1/2)/12; // (1/2) inch = 6/12 feet //deltax=length //unit:ft
+k=0.30; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table
+plasterResistance=deltax/(k*A); //Thermal resistance //Unit:(hr*f)/Btu
+printf("For plaster,");
+printf("The resistance is %f (hr*F)/Btu\n",plasterResistance);
+
+//For "hot film",
+h=0.9; //Coefficient of heat transfer //Unit:Btu/(hr*ft^2*F)
+hotfilmResistance=inv(h*A); //Thermal resistance //Unit:(hr*f)/Btu
+printf("For hot film,");
+printf("The resistance is %f (hr*F)/Btu\n",hotfilmResistance);
+
+//For "cold film",
+h=1.5; //Coefficient of heat transfer //Unit:Btu/(hr*ft^2*F)
+coldfilmResistance=inv(h*A); //Thermal resistance //Unit:(hr*f)/Btu
+printf("For cold film,");
+printf("The resistance is %f (hr*F)/Btu\n\n",coldfilmResistance);
+
+totalResistance=brickResistance+concreteResistance+plasterResistance+hotfilmResistance+coldfilmResistance; //the overall resistance //Unit:(hr*f)/Btu
+printf("The overall resistance is %f (hr*F)/Btu\n",totalResistance);
+
+U=inv(totalResistance); //Unit:Btu/(hr*ft^2) //The overall conductance(or overall heat-transfer coefficient)
+printf("The overall conductance(or overall heat-transfer coefficient) is %f Btu/(hr/ft^2)\n",U);
+//In problem 11.21,the solution is straightforward,because the heat-transfer area is constant for all series resistances.
diff --git a/2417/CH11/EX11.23/Ex11_23.sce b/2417/CH11/EX11.23/Ex11_23.sce new file mode 100755 index 000000000..3d34c8c33 --- /dev/null +++ b/2417/CH11/EX11.23/Ex11_23.sce @@ -0,0 +1,21 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.23\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.23 (page no. 596)
+// Solution
+
+hi=45; //Film coefficient on the inside of the pipe //Unit:Btu/(hr*ft^2*F)
+r1=3.0/2; //Inside radius //Unit:inch
+k1=26; //Unit:Btu/(hr*ft^2*F) //k=proportionality constant for steel pipe //k=thermal conductivity for fir //From the table
+r2=3.5/2; //outide radius //Unit:inch
+k2=0.026; //Unit:Btu/(hr*ft^2*F) //k=proportionality constant for mineral wool //k=thermal conductivity for fir //From the table
+r3=5.50/2; //radius //Unit:inch
+ho=0.9; //Film coefficient on the outside of the pipe //Unit:Btu/(hr*ft^2*F)
+//Results of problem 11.23,
+Ui=1/((1/hi)+((r1/(k1*12))*log(r2/r1))+((r1/(k2*12))*log(r3/r2))+(1/(ho*(r3/r1)))); //Unit:Btu/(hr*ft^2*F) //1 in.=12 ft //Heat transfer coefficient based on inside surface
+printf("Heat transfer coefficient based on inside surface is %f Btu/(hr*ft^2*F)\n",Ui);
+//Because Uo*Ao=Ui*Ai
+Uo=Ui*(r1/r3); //Heat transfer coefficient based on outside surface //Unit:Btu/(hr*ft^2*F)
+printf("Heat transfer coefficient based on outside surface is %f Btu/(hr*ft^2*F)\n",Uo);
diff --git a/2417/CH11/EX11.24/Ex11_24.sce b/2417/CH11/EX11.24/Ex11_24.sce new file mode 100755 index 000000000..4553db0b0 --- /dev/null +++ b/2417/CH11/EX11.24/Ex11_24.sce @@ -0,0 +1,25 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.24\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.24 (page no. 601)
+// Solution
+
+//A COUNTERFLOW HEAT EXCHANGER
+//Hot oil enters at 215 F and leaves at 125 F
+//Water enters the unit at 60 F and leaves at 90 F
+//Therefore,From figure 11.34,
+thetaA=215-90; //the greatest temperature difference between the fluids(at either inlet or outlet) //Unit:fahrenheit
+thetaB=125-60; //the least temperature difference between the fluids(at either inlet or outlet) //Unit:fahrenheit
+deltaTm=(thetaA-thetaB)/log(thetaA/thetaB); //logarithmic mean temperature difference //Unit:fahrenheit
+//From the oil data,
+m=400*60; //mass //Unit:lb/sec //1 min=60 sec
+Cp=0.85; //Specific heat of the oil //Unit:Btu/(lb*F)
+deltaT=215-125; //Change in temperature //Unit:fahrenheit
+Q=m*Cp*deltaT //The heat transfer //Unit:Btu/hr
+//Q=U*A*deltaTm
+U=40;//The overall coefficient of heat transfer of the unit //Unit:Btu/(hr*ft^2*F)
+A=Q/(U*deltaTm); //Umit:ft^2 //The outside surface area
+printf("The outside surface area required is %f ft^2",A);
+
diff --git a/2417/CH11/EX11.25/Ex11_25.sce b/2417/CH11/EX11.25/Ex11_25.sce new file mode 100755 index 000000000..dea04a5e5 --- /dev/null +++ b/2417/CH11/EX11.25/Ex11_25.sce @@ -0,0 +1,29 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.25\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.25 (page no. 602)
+// Solution
+
+//In problem 11.24, A COUNTERFLOW HEAT EXCHANGER is operated in the parallel flow
+//Hot oil enters at 215 F and leaves at 125 F
+//Water enters the unit at 60 F and leaves at 90 F
+//Therefore,From figure 11.35,
+thetaA=215-60; //the greatest temperature difference between the fluids(at either inlet or outlet) //Unit:fahrenheit
+thetaB=125-90; //the least temperature difference between the fluids(at either inlet or outlet) //Unit:fahrenheit
+deltaTm=(thetaA-thetaB)/log(thetaA/thetaB); //logarithmic mean temperature difference //Unit:fahrenheit
+//From the oil data,
+m=400*60; //mass //Unit:lb/sec //1 min=60 sec
+Cp=0.85; //Specific heat of the oil //Unit:Btu/(lb*F)
+deltaT=215-125; //Change in temperature //Unit:fahrenheit
+Q=m*Cp*deltaT //The heat transfer //Unit:Btu/hr
+//Q=U*A*deltaTm
+U=40;//The overall coefficient of heat transfer of the unit //Unit:Btu/(hr*ft^2*F)
+A=Q/(U*deltaTm); //Umit:ft^2 //The outside surface area
+printf("The outside surface area required is %f ft^2",A);
+
+
+
+
+
diff --git a/2417/CH11/EX11.26/Ex11_26.sce b/2417/CH11/EX11.26/Ex11_26.sce new file mode 100755 index 000000000..56f359ab6 --- /dev/null +++ b/2417/CH11/EX11.26/Ex11_26.sce @@ -0,0 +1,22 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.26\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.26 (page no. 603)
+// Solution
+
+//From the table 11.7,
+//For the oil side,a resistance(fouling factor) of 0.005 (hr*F*ft^2)/Btu can be used
+//and for the water side,a fouling factor of 0.001 (hr*F*ft^2)/Btu can be used
+//From problem 11.25,
+U=40;//The coefficient of heat transfer of the unit //Unit:Btu/(hr*ft^2*F)
+//therefore,
+Roil=0.005; //unit:(hr*ft^2*F)/Btu //resistance at oil side
+Rwater=0.001; //unit:(hr*ft^2*F)/Btu //resistance for water side
+Rcleanunit=inv(U); //unit:(hr*ft^2*F)/Btu //resistance at clean unit
+Roverall=Roil+Rwater+Rcleanunit; //unit:(hr*ft^2*F)/Btu //overall resistance
+Uoverall=inv(Roverall); //Unit:Btu/(hr*ft^2*F) //The overall coefficient of heat transfer of the unit
+//Because all the parameters are the same,the surface area required will vary inversely as U
+A=569*(U/Uoverall); //A=569 ft^2 in the problem 11.25 //unit:ft^2 //The outside surface area
+printf("The outside surface area required is %f ft^2",A);
diff --git a/2417/CH11/EX11.27/Ex11_27.sce b/2417/CH11/EX11.27/Ex11_27.sce new file mode 100755 index 000000000..7c19e7c2e --- /dev/null +++ b/2417/CH11/EX11.27/Ex11_27.sce @@ -0,0 +1,24 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.27\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.27 (page no. 605)
+// Solution
+
+//HEAT EXCHANGER
+//Oil flows in the tube side and is cooled from 280 F to 140 F
+//Therefore,
+t2=140; //Unit:fahrenheit
+t1=280; //Unit:fahrenheit
+//On the shell side,water is heated from 85 F to 115 F
+T1=85; //Unit:fahrenheit
+T2=115; //Unit:fahrenheit
+P=(t2-t1)/(T1-t1);
+R=(T1-T2)/(t2-t1);
+//From the figure,
+F=0.91;//Correction factor
+LMTD=((t1-T2)-(t2-T1))/log((t1-T2)/(t2-T1)); //LMTD=Log mean temperature difference //Unit:fahrenheit
+TMTD=F*LMTD; //TMTD=True mean temperature difference //Unit:fahrenheit
+printf("The true mean temperature is %f fahrenheit",TMTD);
+
diff --git a/2417/CH11/EX11.3/Ex11_3.sce b/2417/CH11/EX11.3/Ex11_3.sce new file mode 100755 index 000000000..d7b6132d0 --- /dev/null +++ b/2417/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,30 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.3\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.3 (page no. 556)
+// Solution
+
+//From example 11.1,
+deltaX=6/12; //6 inch = 6/12 feet //deltaX=length //unit:feet
+A=1; //area //ft^2
+k=0.40; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table
+
+Rt=deltaX/(k*A); //Thermal resistance //Unit:(hr*f)/Btu
+
+//Q=deltaT/Rt //Q=heat transfer //ohm's law (fourier's equation)
+//i=deltaE/Re //i=current in amperes //deltaE=The potential difference //Re=the electrical resistance //ohm's law
+// Q/i = (deltaT/Rt)*(deltaE/Re)
+//Q/i=100; //Given // 1 A correspond to 100 Btu/(hr*ft^2)
+deltaE=9; //Unit:Volt //potential difference
+T1=150; //temperature maintained at one face //fahrenheit
+T2=80; //tempetature maintained at other face //fahrenheit
+deltaT=T2-T1; //fahrenheit //Change in temperature
+Re=(100*deltaE*Rt)/deltaT; //Unit:Ohms //The electrical resistance needed
+printf("The electrical resistance needed is %f ohms\n",abs(Re));
+i=deltaE/Re; //current //Unit:amperes
+Q=100*i; //Heat transfer per square foot of wall //Unit:Btu/hr*ft^2
+printf("Heat transfer per square foot of wall is %f Btu/hr*ft^2",abs(Q));
+
+
diff --git a/2417/CH11/EX11.4/Ex11_4.sce b/2417/CH11/EX11.4/Ex11_4.sce new file mode 100755 index 000000000..a1ac89fa9 --- /dev/null +++ b/2417/CH11/EX11.4/Ex11_4.sce @@ -0,0 +1,42 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.4\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.4 (page no. 558)
+// Solution
+
+//For Brick,
+deltaX=6/12; //6 inch = 6/12 feet //deltaX=length //unit:ft
+A=1; //area //unit:ft^2
+k=0.40; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:(hr*f)/Btu
+printf("For brick,\n");
+printf("The resistance is %f (hr*F)/Btu\n\n",R);
+R1=R;
+
+//For Concrete,
+deltaX=(1/2)/12; //(1/2) inch = (1/2)/12 feet //deltaX=length //unit:ft
+A=1; //area //ft^2
+k=0.80; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:(hr*f)/Btu
+printf("For Concrete,\n");
+printf("The resistance is %f (hr*F)/Btu\n\n",R);
+R2=R;
+
+//For plaster,
+deltaX=(1/2)/12; // (1/2) inch = 6/12 feet //deltaX=length //unit:ft
+A=1; //area //ft^2
+k=0.30; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:(hr*f)/Btu
+printf("For plaster,\n");
+printf("The resistance is %f (hr*F)/Btu\n\n",R);
+R3=R;
+
+Rot=R1+R2+R3; //Rot=The overall resistance //unit:(hr*F)/Btu
+printf("The overall resistance is %f (hr*F)/Btu\n\n",Rot);
+T1=70; //temperature maintained at one face //fahrenheit
+T2=30; //tempetature maintained at other face //fahrenheit
+deltaT=T2-T1; //fahrenheit //Change in temperature
+Q=deltaT/Rot; //Q=Heat transfer //Unit:Btu/(hr*ft^2); //ohm's law (fourier's equation)
+printf("Heat transfer per square foot of wall is %f Btu/hr*ft^2",abs(Q));
diff --git a/2417/CH11/EX11.5/Ex11_5.sce b/2417/CH11/EX11.5/Ex11_5.sce new file mode 100755 index 000000000..442c91a81 --- /dev/null +++ b/2417/CH11/EX11.5/Ex11_5.sce @@ -0,0 +1,63 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.5\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.5 (page no. 558)
+// Solution
+
+printf("In problem 11.4,\n");
+//From example 11.4,,,
+//For Brick,
+deltaX=6/12; //6 inch = 6/12 feet //deltaX=length //unit:ft
+A=1; //area //unit:ft^2
+k=0.40; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:(hr*f)/Btu
+printf("For brick,\n");
+printf("The resistance is %f (hr*F)/Btu\n\n",R);
+R1=R;
+
+//For Concrete,
+deltaX=(1/2)/12; //(1/2) inch = (1/2)/12 feet //deltaX=length //unit:ft
+A=1; //area //ft^2
+k=0.80; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:(hr*f)/Btu
+printf("For Concrete,\n");
+printf("The resistance is %f (hr*F)/Btu\n\n",R);
+R2=R;
+
+//For plaster,
+deltaX=(1/2)/12; // (1/2) inch = 6/12 feet //deltaX=length //unit:ft
+A=1; //area //ft^2
+k=0.30; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:(hr*f)/Btu
+printf("For plaster,\n");
+printf("The resistance is %f (hr*F)/Btu\n\n",R);
+R3=R;
+
+Rot=R1+R2+R3; //Rot=The overall resistance //unit:(hr*F)/Btu
+printf("The overall resistance is %f (hr*F)/Btu\n\n",Rot);
+T1=70; //temperature maintained at one face //fahrenheit
+T2=30; //tempetature maintained at other face //fahrenheit
+deltaT=T2-T1; //fahrenheit //Change in temperature
+Q=deltaT/Rot; //Q=Heat transfer //Unit:Btu/(hr*ft^2);
+printf("Heat transfer per square foot of wall is %f Btu/hr*ft^2",abs(Q));
+
+printf("Now in problem 11.5,\n");
+deltaT=R*Q //ohm's law (fourier's equation) //Change in temperature //fahrenheit
+//For Brick,
+deltaT=Q*R1; //Unit:fahrenheit //ohm's law (fourier's equation) //Change in temperature
+t1=deltaT;
+//For Concrete,
+deltaT=Q*R2; //Unit:fahrenheit //ohm's law (fourier's equation) //Change in temperature
+t2=deltaT;
+//For plaster,
+deltaT=Q*R3; //Unit:fahrenheit //ohm's law (fourier's equation) //Change in temperature
+t3=deltaT;
+
+deltaTo=t1+t2+t3; //Overall Change in temperature //fahrenheit
+printf("The overall change in temperature is %f F\n",abs(deltaTo));
+//The interface temperature are:
+printf("The interface temperature are:\n");
+printf("For brick-concrete : %f fahrenheit\n",abs(T2)+abs(t1));
+printf("For concrete-plaster : %f fahrenheit\n",abs(T2)+abs(t1)+abs(t2));
diff --git a/2417/CH11/EX11.6/Ex11_6.sce b/2417/CH11/EX11.6/Ex11_6.sce new file mode 100755 index 000000000..fdf0c4a66 --- /dev/null +++ b/2417/CH11/EX11.6/Ex11_6.sce @@ -0,0 +1,42 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.6\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.6 (page no. 559)
+// Solution
+
+//For Brick,
+deltaX=0.150; //Unit:m //150 mm = 0.150 m //deltaX=length //unit:meter
+A=1; //area //unit:m^2
+k=0.692; //Unit:W/(m*C) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:C/W
+printf("For brick,\n");
+printf("The resistance is %f Celcius/W\n\n",R);
+R1=R;
+
+//For Concrete,
+deltaX=0.012; //Unit:m //12 mm = 0.0120 m //deltaX=length //unit:meter
+A=1; //area //unit:m^2
+k=1.385; //Unit:W/(m*C) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:C/W
+printf("For Concrete,\n");
+printf("The resistance is %f Celcius/W\n\n",R);
+R2=R;
+
+//For plaster,
+deltaX=0.0120; //Unit:m //12 mm = 0.0120 m //deltaX=length //unit:meter
+A=1; //area //unit:m^2
+k=0.519; //Unit:W/(m*C) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:C/W
+printf("For plaster,\n");
+printf("The resistance is %f Celcius/W\n\n",R);
+R3=R;
+
+Ro=R1+R2+R3; //Rot=The overall resistance //unit:C/W
+printf("The overall resistance is %f Celcius/W\n",Ro);
+T1=0; //temperature maintained at one face //Celcius
+T2=20; //tempetature maintained at other face //Celcius
+deltaT=T2-T1; //Change in temperature //Celcius
+Q=deltaT/Ro; //Q=Heat transfer //Unit:W/m^2; //ohm's law (fourier's equation)
+printf("Heat transfer per square meter of wall is %f W/m^2",abs(Q));
diff --git a/2417/CH11/EX11.7/Ex11_7.sce b/2417/CH11/EX11.7/Ex11_7.sce new file mode 100755 index 000000000..2651ad4ac --- /dev/null +++ b/2417/CH11/EX11.7/Ex11_7.sce @@ -0,0 +1,63 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.7\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.7 (page no. 560)
+// Solution
+
+printf("In problem 11.6,\n");
+//For Brick,
+deltaX=0.150; //Unit:m //150 mm = 0.150 m //deltaX=length //unit:meter
+A=1; //area //unit:meter^2
+k=0.692; //Unit:W/(m*C) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:Celcius/W
+printf("For brick,\n");
+printf("The resistance is %f Celcius/W\n\n",R);
+R1=R;
+
+//For Concrete,
+deltaX=0.012; //Unit:m //12 mm = 0.0120 m //deltaX=length //unit:meter
+A=1; //area //unit:meter^2
+k=1.385; //Unit:W/(m*C) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:Celcius/W
+printf("For Concrete,\n");
+printf("The resistance is %f Celcius/W\n\n",R);
+R2=R;
+
+//For plaster,
+deltaX=0.0120; //Unit:m //12 mm = 0.0120 m //deltaX=length //unit:meter
+A=1; //area //unit:meter^2
+k=0.519; //Unit:W/(m*C) //k=proportionality constant //k=thermal conductivity //From the table
+R=deltaX/(k*A); //Thermal resistance //Unit:Celcius/W
+printf("For plaster,\n");
+printf("The resistance is %f Celcius/W\n\n",R);
+R3=R;
+
+Ro=R1+R2+R3; //Rot=The overall resistance Celcius/W
+printf("The overall resistance is %f Celcius/W\n",Ro);
+T1=0; //temperature maintained at one face //Celcius
+T2=20; //tempetature maintained at other face //Celcius
+deltaT=T2-T1; //Change in temperature //Celcius
+Q=deltaT/Ro; //Q=Heat transfer //Unit:W/m^2;
+printf("Heat transfer per square meter of wall is %f W/m^2\n\n",abs(Q));
+
+printf("Now in problem 11.5,\n");
+//deltaT=R*Q //ohm's law (fourier's equation)
+//For Brick,
+deltaT=Q*R1; //Unit:Celcius //Change in temperature
+t1=deltaT;
+//For Concrete,
+deltaT=Q*R2; //Unit:Celcius //Change in temperature
+t2=deltaT;
+//For plaster,
+deltaT=Q*R3; //Unit:Celcius //Change in temperature
+t3=deltaT;
+
+deltaTo=t1+t2+t3; //The overall Change in temperature //Celcius
+printf("The overall change in temperature is %f celcius\n",abs(deltaTo));
+//The interface temperature are:
+printf("The interface temperature are:\n");
+printf("%f Celcius\n",abs(deltaTo)-abs(t1));
+printf("%f Celcius\n",abs(deltaTo)-abs(t1)-abs(t2));
+printf("%f Celcius\n",abs(deltaTo)-abs(t1)-abs(t2)-abs(t3));
diff --git a/2417/CH11/EX11.8/Ex11_8.sce b/2417/CH11/EX11.8/Ex11_8.sce new file mode 100755 index 000000000..5ba6b9e9a --- /dev/null +++ b/2417/CH11/EX11.8/Ex11_8.sce @@ -0,0 +1,47 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.8\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.8 (page no. 561)
+// Solution
+
+deltaX=4/12; //4 inch = 6/12 feet //deltaX=length //unit:ft
+A=7*2; //area //area=hight*width //unit:ft^2
+k=0.090; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity for fir //From the table
+Rfir=deltaX/(k*A); //Resistance of fir //Unit:(hr*F)/Btu
+printf("For fir,\n");
+printf("The resistance is %f (hr*F)/Btu\n\n",Rfir);
+
+deltaX=4/12; //4 inch = 6/12 feet //deltaX=length //unit:ft
+A=7*2; //area //area=hight*width //unit:ft^2
+k=0.065; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity for pine //From the table
+Rpine=deltaX/(k*A); //Resistance of pine //Unit:(hr*F)/Btu
+printf("For pine,\n");
+printf("The resistance is %f (hr*F)/Btu\n\n",Rpine);
+
+deltaX=4/12; //4 inch = 6/12 feet //deltaX=length //unit:ft
+A=7*2; //area //area=hight*width //unit:ft^2
+k=0.025; //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity for corkboard //From the table
+Rcorkboard=deltaX/(k*A); //Resistance of corkboard //Unit:(hr*F)/Btu
+printf("For corkboard,\n");
+printf("The resistance is %f (hr*F)/Btu\n\n",Rcorkboard);
+
+Roverall=inv(inv(Rfir)+inv(Rpine)+inv(Rcorkboard));
+printf("The overall resistance is %f (hr*F)/Btu\n\n",Roverall);
+
+T1=60; //temperature maintained at one face //unit:fahrenheit
+T2=80; //tempetature maintained at other face //unit:fahrenheit
+deltaT=T2-T1; //Change in temperature //unit:fahrenheit
+Qtotal=deltaT/Roverall; //Q=Total Heat loss //Unit:Btu/hr; //ohm's law (fourier's equation)
+printf("Total Heat loss from the wall is %f Btu/hr\n",abs(Qtotal));
+
+//As a check,
+Qfir=deltaT/Rfir; //Q=Fir Heat loss //Unit:Btu/hr; //ohm's law (fourier's equation)
+printf("Heat loss from the wall made of fir is %f Btu/hr\n",abs(Qfir));
+Qpine=deltaT/Rpine; //Q=Pine Heat loss //Unit:Btu/hr; //ohm's law (fourier's equation)
+printf("Heat loss from the wall made of pine is %f Btu/hr\n",abs(Qpine));
+Qcorkboard=deltaT/Rcorkboard; //Q=corkboard Heat loss //Unit:Btu/hr; //ohm's law (fourier's equation)
+printf("Heat loss from the wall made of corkboard is %f Btu/hr\n",abs(Qcorkboard));
+Qtotal=Qfir+Qpine+Qcorkboard; //Total Heat loss from the wall //unit:Btu/hr
+printf("Total Heat loss from the wall is %f Btu/hr\n",abs(Qtotal));
diff --git a/2417/CH11/EX11.9/Ex11_9.sce b/2417/CH11/EX11.9/Ex11_9.sce new file mode 100755 index 000000000..1889703c6 --- /dev/null +++ b/2417/CH11/EX11.9/Ex11_9.sce @@ -0,0 +1,18 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 11.9\n\n\n");
+// Chapter 11 : Heat Transfer
+// Problem 11.9 (page no. 565)
+// Solution
+
+//A bare steel pipe
+ro=3.50; //Outside diameter //Unit:in.
+ri=3.00; //inside diameter //Unit:in.
+Ti=240; //Inside temperature //unit:fahrenheit
+To=120; //Outside temperature //unit:fahrenheit
+L=5; //Length //Unit:ft
+deltaT=Ti-To; //Change in temperature //unit:fahrenheit
+k=26 //Unit:Btu/(hr*ft*F) //k=proportionality constant //k=thermal conductivity
+Q=(2*%pi*k*L*deltaT)/log(ro/ri); //The heat loss from the pipe //unit:Btu/hr
+printf("The heat loss from the pipe is %f Btu/hr",Q);
diff --git a/2417/CH2/EX2.10/Ex2_10.sce b/2417/CH2/EX2.10/Ex2_10.sce new file mode 100755 index 000000000..99da26ba9 --- /dev/null +++ b/2417/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.10\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.10 (page no. 70)
+// Solution
+
+m=10 //Unit:kg //m=mass
+Z=10 //Unit:m //Z=The distance,the body is raised from its initial position when the force is applied
+g= 9.81 //Unit:m/s^2 //g=The local gravity
+//There are no losses in the system
+//So,initial potential energy plus initial kinetic energy equal to sum of final potential energy plus final kinetic energy
+//So, PE1+KE1=PE2+KE2
+//From the figure,KE1=0; PE2=0;
+//So,PE1=KE2;
+PE1=m*g*Z; //PE=Potential Energy //Unit:Joule
+//KE2=(m*v^2)/2
+v=(PE1*2)/m;
+V=sqrt(v); //Unit:m/s //velocity
+printf("Velocity = %f m/s",V);
+KE2=PE1; //kinetic energy //Unit:Joule
+printf("\nKinetic energy is %f N*m",PE1);
diff --git a/2417/CH2/EX2.11/Ex2_11.sce b/2417/CH2/EX2.11/Ex2_11.sce new file mode 100755 index 000000000..cd9a69ab8 --- /dev/null +++ b/2417/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.11\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.11 (page no. 74)
+// Solution
+
+printf("At the entrance of device,\n");
+p1=100; //pressure at the entance //Unit:psia,lbf/in^2
+Rho1=62.4; //Unit:lbm/ft^3 //Rho=The density
+v1=144*(1/Rho1) //Specific Volume at entrance or reciprocal of fluid density // 144 in^2=1 ft^2
+//1 Btu = 778 ft*lbf
+J=778; //Unit:ft*lbf/Btu //conversion factor
+FW1=(p1*v1)/J; //Flow work //Btu/lbm
+printf("Flow work = %f Btu/lbm\n",FW1);
+
+printf("At the exit of device,\n");
+p2=50; //pressure at the exit //Unit:psia,lbf/in^2
+Rho2=30; //Unit:lbm/ft^3 //Rho=The density
+v2=144*(1/Rho2) //Specific Volume at exit or reciprocal of fluid density // 144 in^2=1 ft^2
+//1 Btu = 778 ft*lbf
+J=778; //Unit:ft*lbf/Btu //conversion factor
+FW2=(p2*v2)/J; //Flow work //Btu/lbm
+printf("Flow work = %f Btu/lbm\n",FW2);
diff --git a/2417/CH2/EX2.12/Ex2_12.sce b/2417/CH2/EX2.12/Ex2_12.sce new file mode 100755 index 000000000..0b583dce7 --- /dev/null +++ b/2417/CH2/EX2.12/Ex2_12.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.12\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.12 (page no. 75)
+// Solution
+
+printf("At the entrance of device,\n");
+p1=200*1000; //200kPa*1000 Pa/kPa //pressure at the entrance //Unit:N/m^2
+Rho1=1000; //kg/m^3 //Fluid density at entrance
+v1=1/Rho1; //Specific Volume at entrance or reciprocal of fluid density
+FW1=p1*v1; //Flow work at entrance //Unit:N*m/kg
+printf("Flow work = %fN*m/kg\n",FW1);
+
+printf("At the exit of device,\n");
+p2=100*1000; //200kPa*1000 Pa/kPa //pressure at the exit //Unit:N/m^2
+Rho2=250; //kg/m^3 //Fluid density at exit
+v2=1/Rho2; //Specific Volume at entrance or reciprocal of fluid density
+FW2=p2*v2; //Flow work at exit//Unit:N*m/kg
+printf("Flow work = %f N*m/kg\n",FW2);
diff --git a/2417/CH2/EX2.14/Ex2_14.sce b/2417/CH2/EX2.14/Ex2_14.sce new file mode 100755 index 000000000..93bdb57a5 --- /dev/null +++ b/2417/CH2/EX2.14/Ex2_14.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.14\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.14 (page no. 78)
+// Solution
+
+//It is necessary that pressure be expressed as psfa when the volume is in cubic feet
+//100 psia = 100*144 psfa
+p1=100*144; //Unit:psfa //initial pressure
+v1=2; //Unit:ft^3/lb //Initial Specific Volume
+v2=1; //Unit:ft^3/lb //Final Specific Volume
+w=p1*v1*log(v2/v1); //work done on fluid //Unit:ft*lbf/lbm
+printf("Work done on fluid = %f ft*lbf/lb\n",w);
+//1 Btu = 778 ft*lbf
+printf("Work done on the fluid per pound of fluid is %f Btu/lbm",w/778);
diff --git a/2417/CH2/EX2.15/Ex2_15.sce b/2417/CH2/EX2.15/Ex2_15.sce new file mode 100755 index 000000000..b658e33b0 --- /dev/null +++ b/2417/CH2/EX2.15/Ex2_15.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.15\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.15 (page no. 79)
+// Solution
+
+//p1*v1=p2*v2
+p1=200*1000; //p1=Initial Pressure //Unit:Pa
+p2=800*1000; //p2=Final Pressure //Unit:Pa
+v1=0.1; //v1=Initial Special Volume //Unit:m^3/kg
+v2=(p1/p2)*v1; //v1=final Special Volume //Unit:m^3/kg
+w=p1*v1*log(v2/v1); //workdone //Unit:kJ/kg
+printf("Work done per kilogram of gas is %f kJ/kg (into the system)",w/1000);
diff --git a/2417/CH2/EX2.2/Ex2_2.sce b/2417/CH2/EX2.2/Ex2_2.sce new file mode 100755 index 000000000..7712c8105 --- /dev/null +++ b/2417/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,12 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.2\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.2 (page no. 62)
+// Solution
+
+//Given
+k=100; // Unit:lbf/in. //k=spring constant
+l=2; //Unit:inch //l= length of compression of string
+work=(1/2)*k*l^2; //force-displacement relation //Unit:in*lbf
+printf("Workdone is %f inch*lbf",work);
diff --git a/2417/CH2/EX2.3/Ex2_3.sce b/2417/CH2/EX2.3/Ex2_3.sce new file mode 100755 index 000000000..6d1652dbd --- /dev/null +++ b/2417/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,12 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.3\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.3 (page no. 62)
+// Solution
+
+//Given
+k=20*1000; // Unit:N/m //k=20kN //k=spring constant
+l=0.075; //Unit:meter //l=75 mm //l= length of compression of string
+work=(1/2)*k*l^2; //force-displacement relation //Unit:N*m
+printf("Workdone is %f Jule",work);
diff --git a/2417/CH2/EX2.4/Ex2_4.sce b/2417/CH2/EX2.4/Ex2_4.sce new file mode 100755 index 000000000..e47b333fa --- /dev/null +++ b/2417/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.4\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.4 (page no. 66)
+// Solution
+
+//Given
+Z=600; //Unit:ft //Z=The distance,the body is raised from its initial position when the force is applied
+gc=32.174; //Unit: (lbm*ft)/(lbf*s^2) //gc is constant of proportionality
+g=gc; //Unit:ft/s^2 //g=The local gravity
+m=1; //Unit:lbm //m=mass
+PE=(m*g*Z)/gc; //potential energy //Unit:ft*lbf
+printf("%f ft*lbf work is done lifting the water to elevation ",PE)
diff --git a/2417/CH2/EX2.5/Ex2_5.sce b/2417/CH2/EX2.5/Ex2_5.sce new file mode 100755 index 000000000..1028513ae --- /dev/null +++ b/2417/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.5\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.5 (page no. 66)
+// Solution
+
+
+m=1; //Unit:kg //m=mass
+g= 9.81 //Unit:m/s^2 //g=The local gravity
+Z=50 //Unit:m ////Z=The distance,the body is raised from its initial position when the force is applied //In this case Z=delivered water from well to pump
+PE=m*g*Z; //PE=Potential Energy //Unit:Joule
+printf("Change in potential energy per kg of water is %f J ",PE); //J=Joule=N*m=kg*m^2/s^2
diff --git a/2417/CH2/EX2.6/Ex2_6.sce b/2417/CH2/EX2.6/Ex2_6.sce new file mode 100755 index 000000000..c29387009 --- /dev/null +++ b/2417/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.6\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.6 (page no. 66)
+// Solution
+
+Rho=62.4; //Unit:lbm/ft^3 //Rho=The density of water
+A=10000; //Flow=10000; gal/min
+V=(231/1728); // 12 inch=1 ft //So,1 ft^3=1728 in^3 // One Gallon is a volumetric measure equal to 231 in^3
+//A*V //Unit:ft^3/min
+
+//In example, 2.4:
+printf("From example 2.4\n");
+Z=600; //Unit:ft //Z=The distance,the body is raised from its initial position when the force is applied
+gc=32.174; //Unit: (lbm*ft)/(lbf*s^2) //gc is constant of proportionality
+g=gc; //Unit:ft/s^2 //g=The local gravity
+m=1; //Unit:lbm //m=mass
+PE=(m*g*Z)/gc; //potential energy //Unit:ft*lbf
+printf("%f ft*lbf work is done lifting the water to elevation\n ",PE);
+
+//So,
+printf("In example 2.5 \n")
+M=Rho*A*V; //M=the mass flow
+Power=M*PE; //Unit:ft*lbf/lbm
+printf("Generated Power is %f ft*lbf/lbm \n",Power);
+// 1 horsepower = 33,000 ft*lbf/min
+printf("Power = %f hp\n",Power/33000);
diff --git a/2417/CH2/EX2.7/Ex2_7.sce b/2417/CH2/EX2.7/Ex2_7.sce new file mode 100755 index 000000000..380559a06 --- /dev/null +++ b/2417/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.7\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.7 (page no. 67)
+// Solution
+
+printf("In problem 2.5\n");
+m=1; //Unit:kg //m=mass
+g= 9.81 //Unit:m/s^2 //g=The local gravity
+Z=50 //Unit:m ////Z=The distance,the body is raised from its initial position when the force is applied //In this case Z=delivered water from well to pump
+PE=m*g*Z; //PE=Potential Energy //Unit:Joule
+printf("Change in potential energy per kg of water is %f J \n",PE); //J=Joule=N*m=kg*m^2/s^2
+//Given data in problem 2.7 is
+M=1000; //Unit;kg/min//M=Water density
+Power=PE*M*(1/60); //1 min=60 seconds //power //unit:Joule/s=W
+printf("Power is %f Watt\n",Power); //Watt=N*m/s = Joule/s =Watt
+//1 Hp=746 Watt
+printf("Power is %f Horsepower",Power/745);
diff --git a/2417/CH2/EX2.8/Ex2_8.sce b/2417/CH2/EX2.8/Ex2_8.sce new file mode 100755 index 000000000..680b7ba5e --- /dev/null +++ b/2417/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.8\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.8 (page no. 69)
+// Solution
+
+m=10; //Unit:lb //m=Mass
+V1=88; //Unit://ft/s V1=Velocity before it is slowed down
+V2=10; //Unit;ft/s //V2=Velocity after it is slowed down
+gc=32.174; //Unit: (lbm*ft)/(lbf*s^2) //gc is constant of proportionality
+
+KE1=m*V1^2/(2*gc); //The kinetic energy of the body before it is slowed down //Unit:ft*lbf
+printf("The kinetic energy of the body before it is slowed down is %f ft*lbf\n",KE1);
+
+KE2=m*V2^2/(2*gc); //The kinetic energy of the body before it is slowed down //Unit:ft*lbf
+printf("The kinetic energy of the body before it is slowed down is %f ft*lbf\n",KE2);
+
+KE=KE1-KE2; //KE=Change in kinetic energy //Unit:ft*lbf
+printf("Change in kinetic energy is %f ft*lbf",KE);
diff --git a/2417/CH2/EX2.9/Ex2_9.sce b/2417/CH2/EX2.9/Ex2_9.sce new file mode 100755 index 000000000..28a2e9796 --- /dev/null +++ b/2417/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 2.9\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.9 (page no. 70)
+// Solution
+
+m=1500; //Unit:kg //m=mass
+V1=50; //Km/hour V1=Velocity before it is slowed down
+//V1=(50*1000 m/hour)^2/(3600 s/hour)^2
+KE1=(m*(V1*1000)^2/3600^2)/2; //KE1=Initial kinetic energy //Unit:Joule
+
+//After slowing down
+V2=30; //Unit:KM/hour //V2=Velocity after it is slowed down
+//V2=(30*1000 m/hour)^2/(3600 s/hour)^2
+KE2=(m*(V2*1000)^2/3600^2)/2; //KE2=After slowing down, the kinetic energy //Unit:Joule
+
+KE=KE1-KE2; //KE=Change in kinetic energy //Unit:Joule
+printf("Change in kinetic energy is %f kJ",KE/1000);
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);
diff --git a/2417/CH4/EX4.1/Ex4_1.sce b/2417/CH4/EX4.1/Ex4_1.sce new file mode 100755 index 000000000..bf325932b --- /dev/null +++ b/2417/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,31 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.1\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.1 (page no. 148)
+// Solution
+
+//given data
+t1=1000; //(unit:fahrenheit) //Source temperature
+t2=80; //(unit:fahrenheit) //Sink temperature
+//solution
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+
+printf("Solution for (a)\n");
+ans=((T1-T2)/T1)*100;//(ans in %) //Efficiency of the engine
+printf("Efficiency of the engine is %f percentage\n\n",ans);
+
+printf("Solution for (b)\n");
+T1=2000+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+ans=((T1-T2)/T1)*100;//(ans in %) //Efficiency of the engine
+printf("When the upper tempretrature is increased upto certain ,Efficiency of the engine is %f percentage \n\n",ans);
+
+printf("Solution for (c)\n");
+T1=t1+460; //Source temperature //Unit:R
+T2=160+460; //Sink temperature //Unit:R
+ans=((T1-T2)/T1)*100;//(ans in %) //Efficiency of the engine
+printf("When the lower tempretrature is increased upto certain ,Efficiency of the engine is %f percentage \n\n",ans);
diff --git a/2417/CH4/EX4.10/Ex4_10.sce b/2417/CH4/EX4.10/Ex4_10.sce new file mode 100755 index 000000000..30ba0ca22 --- /dev/null +++ b/2417/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,12 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.10\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.10 (page no. 159)
+// Solution
+
+hfg=1959.7; //Unit:kJ/kg //Evaporative enthalpy
+T=195.07+273; //Converted into Kelvin //Temperature
+deltaS=hfg/T; //Change in entropy //kJ/kg*K
+printf("Change in entropy at 1.4MPa for the vaporization of 1 kg is %f kJ/kg*K",deltaS); //Values compares very closely to the Steam Tables value
diff --git a/2417/CH4/EX4.11/Ex4_11.sce b/2417/CH4/EX4.11/Ex4_11.sce new file mode 100755 index 000000000..639d3aa22 --- /dev/null +++ b/2417/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,21 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.11\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.11 (page no. 159)
+// Solution
+
+//Let is assume that a Carnot engine cycle operates between two temperatures in each case.
+t=1000; //(unit:fahrenheit)
+//converting temperatures to absolute temperatures;
+T1=t+460;
+//T1*deltaS=Qin;
+Qin=100; //Unit:Btu //heat added to the cycle
+deltaS=Qin/T1; //Change in entropy //Btu/R
+T2=50+460; //converting 50 F temperature to absolute temperature;
+Qr=T2*deltaS; //Heat rejected //Unit:Btu
+printf("%f Btu energy is unavailable with respect to a receiver at 50 fahrenheit \n",Qr);
+T2=0+460; //converting 0 F temperature to absolute temperature;
+Qr=T2*deltaS; //Heat rejected //unit:Btu
+printf("%f Btu energy is unavailable with respect to a receiver at 0 fahrenheit \n",Qr);
diff --git a/2417/CH4/EX4.12/Ex4_12.sce b/2417/CH4/EX4.12/Ex4_12.sce new file mode 100755 index 000000000..f9fb0921f --- /dev/null +++ b/2417/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,22 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.12\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.12 (page no. 160)
+// Solution
+
+Qin=1000; //Unit:Joule //heat entered to the system
+t=500; //(unit:Celcius) //temperature
+//converting temperature
+T1=t+273; //Unit:Kelvin
+deltaS=Qin/T1; //Change in entropy //Unit:J/K
+printf("Solution for (a),\n");
+T2=20+273; //converted 20 Celcius temperature to Kelvin;
+Qr=T2*deltaS; //Heat rejected at 20 celcius //Joule
+printf("%f Joule energy is unavailable with respect to a receiver at 20 Celcius\n\n",Qr);
+
+printf("Solution for (b),\n")
+T2=0+273; //converted 0 Celcius temperature to Kelvin
+Qr=T2*deltaS; //heat rejected at 0 celcius //Joule
+printf("%f Joule energy is unavailable with respect to a receiver at 0 Celcius\n",Qr);
diff --git a/2417/CH4/EX4.13/Ex4_13.sce b/2417/CH4/EX4.13/Ex4_13.sce new file mode 100755 index 000000000..3c3cf2370 --- /dev/null +++ b/2417/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,21 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.13\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.13 (page no. 161)
+// Solution
+
+//deltas=Cp*ln(T2/T1)
+//Multiplying both the sides of equation by the mass m,
+//DeltaS=m*Cp*ln(T2/T1)
+m=6; //mass //Unit:lbm
+Cp=0.361; //Btu/lbm*R //Specific heat constant
+DeltaS=-0.7062; //Unit:Btu/R //change in entropy
+t=1440; //(unit:fahrenheit)
+//converting temperatures to absolute temperatures;
+T1=t+460; //Unit:R
+//Rearranging the equation,
+T2=T1*exp(DeltaS/(m*Cp)); //final temperature //Unit:R
+printf("Final temperature is %f R",T2);
+printf("or %f fahrenheit",T2-460);
diff --git a/2417/CH4/EX4.14/Ex4_14.sce b/2417/CH4/EX4.14/Ex4_14.sce new file mode 100755 index 000000000..8fbdd0d09 --- /dev/null +++ b/2417/CH4/EX4.14/Ex4_14.sce @@ -0,0 +1,25 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.14\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.14 (page no. 162)
+// Solution
+
+//1 lbm of water at 500F is mixed with 1 lbm of water at 100F
+m1=1; //Unit:lbm //mass
+m2=1; //Unit:lbm //mass
+c1=1; //Specific heat constant
+c2=1; //Specific heat constant
+t1=500; //(unit:fahrenheit)
+t2=100; //(unit:fahrenheit)
+cmix=1; //Specific heat constant of mixture
+//now, m1*c1*t1 +m2*c2*t2 = (m1+m2)*cmix*t
+//So,
+t=((m1*c1*t1)+(m2*c2*t2))/((m1+m2)*cmix) //resulting temperature of the mixture
+printf("The resulting temperature of the mixture is %f fahrenheit\n",t);
+//For this problem,the hot steam is cooled
+deltas=cmix*log((t+460)/(t1+460)); //temperatures converted to absolute temperatures; //deltas=change in entropy //Unit:Btu/(lbm*R)
+//The cold steam is heated
+deltaS=cmix*log((t+460)/(t2+460)); //temperatures converted to absolute temperatures; //deltaS=change in entropy //Unit:Btu/(lbm*R)
+printf("The net change in entropy is %f Btu/(lbm*R)\n",deltaS+deltas);
diff --git a/2417/CH4/EX4.15/Ex4_15.sce b/2417/CH4/EX4.15/Ex4_15.sce new file mode 100755 index 000000000..82fca6503 --- /dev/null +++ b/2417/CH4/EX4.15/Ex4_15.sce @@ -0,0 +1,33 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.15\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.15 (page no. 163)
+// Solution
+
+//In problem 4.15,
+//1 lbm of water at 500F is mixed with 1 lbm of water at 100F
+m1=1; //Unit:lbm //mass
+m2=1; //Unit:lbm //mass
+c1=1; //Specific heat constant
+c2=1; //Specific heat constant
+t1=500; //(unit:fahrenheit)
+t2=100; //(unit:fahrenheit)
+cmix=1; //Specific heat constant of mixture
+//now, m1*c1*t1 +m2*c2*t2 = (m1+m2)*cmix*t //So,
+t=((m1*c1*t1)+(m2*c2*t2))/((m1+m2)*cmix) //resulting temperature of the mixture
+printf("In problem 4.14,The resulting temperature of the mixture is %f fahrenheit\n",t);
+
+//Now,in problem 4.15,taking 0F as a reference temperature,
+//For hot fluid,
+deltas=cmix*log((t1+460)/(0+460)); //temperatures converted to absolute temperatures; //deltas=change in entropy //Unit:Btu/(lbm*R)
+//For cold fluid,
+s=cmix*log((t2+460)/(0+460)); //temperatures converted to absolute temperatures; //s=change in entropy //Unit:Btu/(lbm*R)
+//At final mixture temperature of t F,the entropy of each system above 0F is,for the hot fluid
+s1=cmix*log((t+460)/(0+460)); //temperatures converted to absolute temperatures; //s1=change in entropy //Unit:Btu/(lbm*R)
+//and for the cold fluid,
+s2=cmix*log((t+460)/(0+460)); //temperatures converted to absolute temperatures; //s2=change in entropy //Unit:Btu/(lbm*R)
+printf("The change in the entropy for hot fluid is %f Btu/(lbm*R)\n",s1-deltas);
+printf("The change in the entropy for cold fluid is %f Btu/(lbm*R)\n",s2-s);
+printf("The total change in entropy if %f Btu/(lbm*R",s1-deltas+s2-s);
diff --git a/2417/CH4/EX4.2/Ex4_2.sce b/2417/CH4/EX4.2/Ex4_2.sce new file mode 100755 index 000000000..204476d2d --- /dev/null +++ b/2417/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,27 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.2\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.2 (page no. 149)
+// Solution
+
+//given data
+Qin=100; //heat added to the cycle
+
+printf("In problem 4.1,\n")
+//given data
+t1=1000; //(unit:fahrenheit) //Source temperature
+t2=80; //(unit:fahrenheit) //Sink temperature
+//solution
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+printf("Solution for (a)\n");
+printf("Efficiency of the engine is %f percentage\n\n",((T1-T2)/T1)*100);
+
+printf("Now in problem 4.2,\n")
+W=0.63*Qin; //W=W/J; //Efficiency in problem 4.1
+W=Qin*(W/Qin); //amount of work
+Qr=Qin-W; //Qin-Qr=W/J //Qr=heat rejected by the cycle
+printf("The heat removed from the reservoir %f units",Qr);
diff --git a/2417/CH4/EX4.3/Ex4_3.sce b/2417/CH4/EX4.3/Ex4_3.sce new file mode 100755 index 000000000..c57a6582e --- /dev/null +++ b/2417/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,22 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.3\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.3 (page no. 149)
+// Solution
+
+//given data
+t1=70; //(unit:fahrenheit) //Source temperature
+t2=15; //(unit:fahrenheit) //Sink temperature
+Qin=125000; //(unit=Btu/hr) //Qin=heat added to the cycle
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+Qr=Qin*(T2/T1); //Qr=heat rejected by the cycle
+printf("Qr is %f in Btu/hr\n",Qr);
+work=Qin-Qr; //reversed cycle requires atleast input //work //btu/hr
+printf("Work is %f in Btu/hr\n",work);
+// 1 hp = 33000 ft*LBf/min
+// 1 Btu = 778 ft*LBf //1 hr = 60 min
+printf("Minimum horsepower input required is %f hp",work*778/(60*33000));
diff --git a/2417/CH4/EX4.4/Ex4_4.sce b/2417/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 000000000..a6f672526 --- /dev/null +++ b/2417/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,24 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.4\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.4 (page no. 150)
+// Solution
+
+W=(50*33000)/778;//output //W=W/J
+// 1 hp = 33000 ft*LBf/min
+// 1 Btu = 778 ft*LBf
+printf("Output is %f in Btu/min\n",W);
+t1=1000; //Source temperature //(unit:fahrenheit)
+t2=100; //Sink temperature //(unit:fahrenheit)
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+n=(1-(T2/T1))*100; //efficiency
+printf("Efficiency is %f percentage\n",n);//(in %)
+//n=(W/J)/Qin
+Qin=W/(n/100);//(unit Btu/hr) //Qin=heat added to the cycle
+printf("Heat added to the cycle is %f in Btu/min\n",Qin);
+Qr=Qin*(1-(n/100));//(unit Btu/hr) //Qr=heat rejected by the cycle
+printf("Heat rejected by the cycle is %f in Btu/min \n",Qr);
diff --git a/2417/CH4/EX4.5/Ex4_5.sce b/2417/CH4/EX4.5/Ex4_5.sce new file mode 100755 index 000000000..43e9cc1dd --- /dev/null +++ b/2417/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,22 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.5\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.5 (page no. 151)
+// Solution
+
+t1=700; //Source temperature //Unit:Celcius
+t2=20; //Sink temperature //Unit:Celcius
+//converting in F
+T1=t1+273; //Source temperature //Unit:R
+T2=t2+273; //Sink temperature //Unit:R
+n=(T1-T2)/T1*100; //Efficiency
+printf("Efficiency is %f percentage\n",n);//(in %)
+output=65;//in hp //Given
+work=output*0.746;//(unit kJ/s) // 1 hp = 746 W
+printf("Work is %f kJ/s\n",work);
+Qin=work/(n/100);//(unit kJ/s) //Qin=heat added to the cycle
+printf("Heat added to the cycle is %f kJ/s \n",Qin);
+Qr=Qin*(1-(n/100));//(unit kJ/s) //Qr=heat rejected by the cycle
+printf("Heat rejected by the cycle is %f kJ/s \n",Qr);
diff --git a/2417/CH4/EX4.7/Ex4_7.sce b/2417/CH4/EX4.7/Ex4_7.sce new file mode 100755 index 000000000..60fde9f89 --- /dev/null +++ b/2417/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,20 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.7\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.7 (page no. 152)
+// Solution
+
+t1=700; //(unit:fahrenheit) //Source temperature
+t2=200; //(unit:fahrenheit) //Sink temperature
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+//n1=(T1-Ti)/T1 and n2=(Ti-T2)/Ti //n1 & n2 are efficiency
+//(T1-Ti)/T1=(Ti-T2)/Ti;
+Ti=sqrt(T1*T2); //Exhaust temperature //Unit:R
+printf("Exhaust temperature of first engine is %f in R\n",Ti);
+//converting absolute temperature to normal F temperature
+//Ti(fahrenheit)=Ti(R)-460;
+printf("Exhaust temperature of first engine is %f fahrenheit\n",Ti-460);
diff --git a/2417/CH4/EX4.8/Ex4_8.sce b/2417/CH4/EX4.8/Ex4_8.sce new file mode 100755 index 000000000..c83682752 --- /dev/null +++ b/2417/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,15 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.8\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.8 (page no. 157)
+// Solution
+
+//For reversible isothermal process,
+q=843.7; //Heat //Unit:Btu //at 200 psia
+t=381.86; //(unit:fahrenheit) //temperature
+////converting temperatures to absolute temperatures;
+T=t+460; //temperature //unit:R
+deltaS=(q/T); //Change in entropy //Unit:Btu/lbm*R
+printf("Change in entropy is %f Btu/lbm*R\n",deltaS); //1 LBm of saturated water
diff --git a/2417/CH4/EX4.9/Ex4_9.sce b/2417/CH4/EX4.9/Ex4_9.sce new file mode 100755 index 000000000..37c8350b6 --- /dev/null +++ b/2417/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,38 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 4.9\n\n\n");
+// Chapter 4 : The Second Law Of Thermodynamics
+// Problem 4.9 (page no. 158)
+// Solution
+
+//For reversible isothermal process,
+//In problem 4.8,
+q=843.7; //Heat //Unit:Btu //at 200 psia
+t=381.86; //(unit:fahrenheit)
+//converting temperatures to absolute temperatures;
+T=t+460; //Unit:R"
+deltaS=(q/T); //Change in entropy //Btu/lbm
+printf("Change in entropy is %f Btu/lbm*R\n",deltaS); //1 LBm of saturated water
+
+//In problem 4.9
+t1=381.86; //(unit:fahrenheit) //Source temperature
+t2=50; //(unit:fahrenheit) //Sink temperature
+//converting temperatures to absolute temperatures;
+T1=t1+460; //Source temperature //Unit:R
+T2=t2+460; //Sink temperature //Unit:R
+qin=q;//heat added to the cycle
+n=(1-(T2/T1))*100; //Efficiency
+printf("Efficiency is %f percentage\n",n);
+wbyJ=qin*n*0.01;//work output
+printf("Work output is %f Btu/lbm\n",wbyJ);
+Qr=qin-wbyJ; //heat rejected
+printf("Heat rejected is %f Btu/lbm\n\n",Qr);
+printf("As an alternative solution and refering to figure 4.12,\n")
+qin=T1*deltaS; //heat added //btu/lbm
+Qr=T2*deltaS; //Heat rejected //btu/lbm
+printf("Heat rejected is %f Btu/lbm\n",Qr);
+wbyJ=qin-Qr; //Work output //Btu/lbm
+printf("Work output is %f Btu/lbm\n",wbyJ);
+n=(wbyJ/qin)*100; //Efficiency
+printf("Efficiency is %f percentage\n",n);
diff --git a/2417/CH5/EX5.1/Ex5_1.sce b/2417/CH5/EX5.1/Ex5_1.sce new file mode 100755 index 000000000..3ef10e17b --- /dev/null +++ b/2417/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,16 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.1\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.1 (page no. 182)
+// Solution
+
+p=0.6988; //Unit:psia //absolute pressure
+vg=467.7; //Unit:ft^3/lbm //Saturated vapour specific volume
+ug=1040.2; //Unit:Btu/lbm //Saturated vapour internal energy
+J=778; //J=Conversion factor
+// 1 Btu = 778 ft*LBf
+//h=u+(p*v)/J
+hg=ug+((p*vg*144)/J); //The enthalpy of saturated steam //1 ft^2=144 in^2 //Btu/lbm
+printf("The enthalpy of saturated steam at 90 F is %f Btu/lbm",hg); //The value is matched with the value in table 1
diff --git a/2417/CH5/EX5.10/Ex5_10.sce b/2417/CH5/EX5.10/Ex5_10.sce new file mode 100755 index 000000000..8d627dee1 --- /dev/null +++ b/2417/CH5/EX5.10/Ex5_10.sce @@ -0,0 +1,15 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.10\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.10 (page no. 194)
+// Solution
+
+//For the wet mixture,hx=hf+(x*hfg),solving for x gives us
+//Using table 1,we have,
+hx=2000; //kJ/kg //Enthalpy of wet mixture at 30 C
+hf=125.79; //kJ/kg //saturated liquid enthalpy
+hfg=2430.5; // //Evap. Enthalpy //kJ/kg
+x=(hx-hf)/hfg; //quality
+printf("The quality is %f percentage of a wet steam at 30 C\n",x*100);
diff --git a/2417/CH5/EX5.11/Ex5_11.sce b/2417/CH5/EX5.11/Ex5_11.sce new file mode 100755 index 000000000..c3ce4cc6c --- /dev/null +++ b/2417/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.11\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.11 (page no. 197)
+// Solution
+
+//The values of temperature and pressure are listed in Table 3(Figure 5.10) and can be read directly.
+printf("Specific volume of superheated steam at 330 psia and 450F is v=1.4691 ft^3/lbm\n");
+printf("Internal Energy of superheated steam at 330 psia and 450F is u=1131.8 Btu/lbm\n");
+printf("Enthalpy of superheated steam at 330 psia and 450F is h=1221.5 Btu/lbm\n");
+printf("Entropy of superheated steam at 330 psia and 450F is s=1.5219 Btu/lbm*R\n");
diff --git a/2417/CH5/EX5.12/Ex5_12.sce b/2417/CH5/EX5.12/Ex5_12.sce new file mode 100755 index 000000000..d08305232 --- /dev/null +++ b/2417/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.12\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.12 (page no. 197)
+// Solution
+
+//The values of temperature and pressure are listed in Table 3(Figure 5.10) and can be read directly.
+printf("Specific volume of superheated steam at 2.0 MPa and 240 C is v=0.10845 m^3/lbm\n");
+printf("Internal Energy of superheated steam at 2.0 MPa and 240 C is u=2659.6 kJ/kg\n");
+printf("Enthalpy of superheated steam at 2.0 MPa and 240 C is h=2876.5 kJ/kg\n");
+printf("Entropy of superheated steam at 2.0 MPa and 240 C is s=6.4952 kJ/kg*K\n");
diff --git a/2417/CH5/EX5.13/Ex5_13.sce b/2417/CH5/EX5.13/Ex5_13.sce new file mode 100755 index 000000000..222b21b90 --- /dev/null +++ b/2417/CH5/EX5.13/Ex5_13.sce @@ -0,0 +1,37 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.13\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.13 (page no. 197)
+// Solution
+
+
+//The necessary interpolations(between 450F and 460F at 330 psia) are best done in tabular forms as shown:
+// t v
+// 460 1.4945
+// 455 1.4818
+// 450 1.4691
+v=1.4691+(1/2)*(1.4945-1.4691); //ft^3/lbm //specific volume
+printf("The specific volume of saturated steam at 330 psia & 455F is %f ft^3/lbm\n",v);
+
+// t u
+// 460 1137.0
+// 455 1134.4
+// 450 1131.8
+u=1131.8+(1/2)*(1137.0-1131.8); //Btu/lbm //internal energy
+printf("The internal energy of saturated steam at 330 psia & 455F is %f Btu/lbm\n",u);
+
+// t h
+// 460 1228.2
+// 455 1224.9
+// 450 1221.5
+h=1221.5+(1/2)*(1228.2-1221.5); //enthaply //Btu/lbm
+printf("The enthalpy of saturated steam at 330 psia & 455F is %f Btu/lbm\n",h);
+
+// t s
+// 460 1.5293
+// 455 1.5256
+// 450 1.5219
+s=1.5219+(1/2)*(1.5293-1.5219); //entropy //Btu/lbm*R
+printf("The entropy of saturated steam at 330 psia & 455F is %f Btu/lbm*R\n",s);
diff --git a/2417/CH5/EX5.14/Ex5_14.sce b/2417/CH5/EX5.14/Ex5_14.sce new file mode 100755 index 000000000..a46c78e47 --- /dev/null +++ b/2417/CH5/EX5.14/Ex5_14.sce @@ -0,0 +1,35 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.14\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.14 (page no. 198)
+// Solution
+
+//From Table3, we first obtain the properties at 337 psia and 460 F and then 337 psia and 470 F.
+//The necessary interpolations are best done in tabular forms as shown:
+//Proceeding with the calculation,at 460 F,
+// p v // p h
+// 340 1.4448 // 340 1226.7
+// 337 1.4595 // 337 1227.2
+// 335 1.4693 // 335 1227.5
+v=1.4696-(2/5)*(1.4693-1.4448); h=1227.5-(2/5)*(1227.5-1226.7);
+//ft^3/lbm //specific volume //Btu/lbm //enthaply
+
+//And at 470 F,
+// p v // p h
+// 340 1.4693 // 340 1233.4
+// 337 1.4841 // 337 1233.9
+// 335 1.4940 // 335 1234.2
+v=1.4640-(2/5)*(1.4640-1.4693); h=1234.2-(2/5)*(1234.2-1233.4);
+//ft^3/lbm //specific volume //Btu/lbm //enthaply
+
+//Therefore,at 337 psia and 465 F
+// t v // t h
+// 470 1.4841 // 470 1233.9
+// 465 1.4718 // 465 1230.7
+// 460 1.4595 // 460 1227.5
+v=1.4595+(1/2)*(1.4841-1.4595); h=1227.5+(1/2)*(1233.9-1227.5);
+//ft^3/lbm //specific volume //Btu/lbm //enthaply
+printf("At 465 F and 337 psia,specific volume=%f ft^3/lbm and enthalpy=%f Btu/lbm\n",v,h);
+
diff --git a/2417/CH5/EX5.15/Ex5_15.sce b/2417/CH5/EX5.15/Ex5_15.sce new file mode 100755 index 000000000..f48d4d002 --- /dev/null +++ b/2417/CH5/EX5.15/Ex5_15.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.15\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.15 (page no. 202)
+// Solution
+
+//The values of temperature and pressure are listed in Table 4(Figure 5.10) and can be read directly.
+printf("Specific volume of subcooled water at 1000 psia and 300F is v=0.017379 ft^3/lbm\n");
+printf("Internal Energy of subcooled water at 1000 psia and 300F is u=268.24 Btu/lbm\n");
+printf("Enthalpy of subcooled water at 1000 psia and 300F is h=271.46 Btu/lbm\n");
+printf("Entropy of subcooled water at 1000 psia and 300F is s=0.43552 Btu/lbm*R\n");
diff --git a/2417/CH5/EX5.16/Ex5_16.sce b/2417/CH5/EX5.16/Ex5_16.sce new file mode 100755 index 000000000..8e1632008 --- /dev/null +++ b/2417/CH5/EX5.16/Ex5_16.sce @@ -0,0 +1,21 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.16\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.16 (page no. 202)
+// Solution
+
+//It is necessary to ontain the saturation values corresponding to 300 F.This is done by reading Table A.1 in Appendix 3,which gives
+pf=66.98; //psia //pressure
+vf=0.017448; //ft^3/lbm //specific volume
+hf=269.73; //Btu/lbm //enthaply
+//Now,
+p=1000; //psia //pressure
+J=778; //Conversion factor //ft*lbf/Btu
+//From eq.5.5,
+h=hf+((p-pf)*vf*144)/J; //1ft^2=144 in^2 //The enthalpy of subcooled water //Btu/lbm
+printf("The enthalpy of subcooled water is %f Btu/lbm\n",h);
+//The difference between this value and the value found in problem 5.15,expressed as a percentage is
+percentoferror=(h-271.46)/271.46;
+printf("Percent of error is %f\n",percentoferror*100);
diff --git a/2417/CH5/EX5.2/Ex5_2.sce b/2417/CH5/EX5.2/Ex5_2.sce new file mode 100755 index 000000000..20255fa8e --- /dev/null +++ b/2417/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,16 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.2\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.2 (page no. 187)
+// Solution
+
+p=4.246; //Unit:kPa //absolute pressure
+vg=32.894; //Unit:m^3/kg //specific volume
+ug=2416.6; //Unit:kJ/kg //internal energy
+J=778; //J=Conversion factor
+// 1 Btu = 778 ft*LBf
+//h=u+(p*v)
+hg=ug+(p*vg); //The enthalpy of saturated steam //1 ft^2=144 in^2 //unit:kJ/kg
+printf("The enthalpy of saturated steam at 30 C is %f kJ/kg",hg); //The value is matched with the value in table 1
diff --git a/2417/CH5/EX5.25/Ex5_25.sce b/2417/CH5/EX5.25/Ex5_25.sce new file mode 100755 index 000000000..841f64fd1 --- /dev/null +++ b/2417/CH5/EX5.25/Ex5_25.sce @@ -0,0 +1,11 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.25\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.25 (page no. 211)
+// Solution
+
+//On a chart in Appendix 3,it is necessary to estimate the 90 F point on the saturation line.From the chart or the table in the upper left of the chart,we note that 90 F is between 1.4 and 1.5 in. of mercury.Estimating the intersection of this value with the saturation curve yields
+printf("Enthalpy of saturated steam hg=1100 Btu/lbm\n");
+//This is a good agreement with results of problem 5.1
diff --git a/2417/CH5/EX5.26/Ex5_26.sce b/2417/CH5/EX5.26/Ex5_26.sce new file mode 100755 index 000000000..95c134f61 --- /dev/null +++ b/2417/CH5/EX5.26/Ex5_26.sce @@ -0,0 +1,11 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.26\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.26 (page no. 212)
+// Solution
+
+//The Mollier chart has lines of constant moisture in the wet region which correspond to (1-x).Therefore,we read at 20% moisture(80% Quality) and 120 psia,
+printf("The enthalpy of a wet steam mixture at 120 psia having quality 80 percent is 1015 Btu/lbm\n");
+//Which also agrees well with the calculated value in problem 5.7
diff --git a/2417/CH5/EX5.27/Ex5_27.sce b/2417/CH5/EX5.27/Ex5_27.sce new file mode 100755 index 000000000..a58278063 --- /dev/null +++ b/2417/CH5/EX5.27/Ex5_27.sce @@ -0,0 +1,11 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.27\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.27 (page no. 213)
+// Solution
+
+//Entering the Mollier chart at 900 Btu/lbm and estimating 90 F(near the 1.5-in. Hg dashed line) yields a constant moisture percent of 19.2%.
+printf("The quality is %f percent\n",(1-0.192)*100);
+//We show good agreement with the calculated value.
diff --git a/2417/CH5/EX5.28/Ex5_28.sce b/2417/CH5/EX5.28/Ex5_28.sce new file mode 100755 index 000000000..48c14d081 --- /dev/null +++ b/2417/CH5/EX5.28/Ex5_28.sce @@ -0,0 +1,11 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.28\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.28 (page no. 214)
+// Solution
+
+//From the chart,
+printf("The enthalpy of steam at 330 psia is h=1220 Btu/lbm\n");
+//Compared to 1221.5 Btu/lbm found in problem 5.11
diff --git a/2417/CH5/EX5.29/Ex5_29.sce b/2417/CH5/EX5.29/Ex5_29.sce new file mode 100755 index 000000000..4114b576e --- /dev/null +++ b/2417/CH5/EX5.29/Ex5_29.sce @@ -0,0 +1,11 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.29\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.29 (page no. 214)
+// Solution
+
+//We note that the steam is superheated.From the Mollier chart in SI units,
+printf("The enthalpy h=2876.5 kJ/kg and entropy s=6.4952 kJ/kg*K\n");
+//Values are matched with problem 5.12
diff --git a/2417/CH5/EX5.3/Ex5_3.sce b/2417/CH5/EX5.3/Ex5_3.sce new file mode 100755 index 000000000..a87476923 --- /dev/null +++ b/2417/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,38 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.3\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.3 (page no. 188)
+// Solution
+
+//The necessary interpolations are best done in tabular forms as shown:
+// p hg
+// 115 1190.4 table 2
+// 118 1190.8 (hg)118=1190.8
+// 120 1191.1
+hg=1190.4+(3/5)*(1191.1-1190.4); //Btu/lbm //enthaply
+printf("The enthalpy of saturated steam at 118 psia is %f Btu/lbm\n",hg);
+
+// p vg
+// 115 3.884 table 2
+// 118 3.792 (vg)118=3.790
+// 120 3.730
+vg=3.884-(3/5)*(3.884-3.730); //ft^3/lbm //specific volume
+printf("The specific volume of saturated steam at 118 psia is %f ft^3/lbm\n",vg);
+
+// p sg
+// 115 1.5921 table 2
+// 118 1.5900 (sg)118=1.5900
+// 120 1.5886
+sg=1.5921-(3/5)*(1.5921-1.5886); //entropy
+printf("The entropy of saturated steam at 118 psia is %f\n",sg);
+
+// p ug
+// 115 1107.7 table 2
+// 118 1108.06 (ug)118=1180.1
+// 120 1108.3
+ug=1107.7-(3/5)*(1108.3-1107.7); //internal energy
+printf("The internal energy of saturated steam at 118 psia is %f\n",ug);
+//The interpolation process that was done in tabular form for this problem can also be demonstated by refering to figure 5.8 for the specific volume.It will be
+//seen that the results of this problem and the tabulated values are essentially in exact agreement and that linear interpolation is satisfactory in these tables.
diff --git a/2417/CH5/EX5.30/Ex5_30.sce b/2417/CH5/EX5.30/Ex5_30.sce new file mode 100755 index 000000000..f1fe37fb6 --- /dev/null +++ b/2417/CH5/EX5.30/Ex5_30.sce @@ -0,0 +1,11 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.30\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.30 (page no. 215)
+// Solution
+
+//Because neither pressure nor temperature is shown directly,it is necessary to estimate to obtain the desired value.
+printf("The enthalpy of steam is h=1231 Btu/lbm\n");
+//In problem 5.14,h=1230.7 Btu/lbm,Which is matched here.
diff --git a/2417/CH5/EX5.31/Ex5_31.sce b/2417/CH5/EX5.31/Ex5_31.sce new file mode 100755 index 000000000..b8fdcb43f --- /dev/null +++ b/2417/CH5/EX5.31/Ex5_31.sce @@ -0,0 +1,11 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.31\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.31 (page no. 215)
+// Solution
+
+//Reading the chart at 30 C and saturation gives us,
+printf("The enthalpy of saturated steam is hg=2556 kJ/kg\n");
+//Which matches with value of problem 5.2
diff --git a/2417/CH5/EX5.32/Ex5_32.sce b/2417/CH5/EX5.32/Ex5_32.sce new file mode 100755 index 000000000..ec59533ac --- /dev/null +++ b/2417/CH5/EX5.32/Ex5_32.sce @@ -0,0 +1,11 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.32\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.32 (page no. 215)
+// Solution
+
+//Reading the chart in wet region at 1.0 MPa and x=0.85(moisture of 15%) gives us
+printf("hx=2476 kJ/kg and sx=5.92 kJ/kg*K\n");
+//The chart does not give ux or vx directly
diff --git a/2417/CH5/EX5.33/Ex5_33.sce b/2417/CH5/EX5.33/Ex5_33.sce new file mode 100755 index 000000000..4af1a7589 --- /dev/null +++ b/2417/CH5/EX5.33/Ex5_33.sce @@ -0,0 +1,10 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.33\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.33 (page no. 215)
+// Solution
+
+//Locate 30 C on the saturation line.Now follow a line of constant pressure,which is also a line of constant temperature in wet region,until an enthalpy of 2000kJ/kg is reached.
+printf("The moisture content is 23 percent or x=77 percent\n");
diff --git a/2417/CH5/EX5.34/Ex5_34.sce b/2417/CH5/EX5.34/Ex5_34.sce new file mode 100755 index 000000000..0f2870ab3 --- /dev/null +++ b/2417/CH5/EX5.34/Ex5_34.sce @@ -0,0 +1,11 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.34\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.34 (page no. 216)
+// Solution
+
+//We enter the chart in the superheat region at 2.0MPa and 240 C to read the enthalpy and entropy.This procedure gives
+printf("Enthalpy h=2877 kJ/kg and entropy s=6.495 kJ/kg*K\n");
+//The other properties cant be obtained directly from the chart
diff --git a/2417/CH5/EX5.35/Ex5_35.sce b/2417/CH5/EX5.35/Ex5_35.sce new file mode 100755 index 000000000..1d4f5037e --- /dev/null +++ b/2417/CH5/EX5.35/Ex5_35.sce @@ -0,0 +1,59 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.35\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.35 (page no. 218)
+// Solution
+
+//As already noted,h1=h2 for this process.On the Mollier chart,h2 is found to be 1170 Btu/lbm at 14.7 psia and 250 F.Proceeding to the left on the chart,the constant-enthalpy value of 1170 Btu/lbm to 150 psia yields a moisture of 3% or a quality of 97%.
+//If we use the tables to obtain the solution to this problem,we would first obtain h2 from the superheated vapor tables as 1168.8 Btu/lbm.Because hx=hf+(x*hfg),we obtain x as
+hx=1168.8; //Btu/lbm
+hf=330.75; //Btu/lbm //values of 150 psia
+hfg=864.2; //Btulbm //values of 150 psia
+x=(hx-hf)/hfg; //Quality
+printf("Moisture in the steam flowing in the pipe is %f percent\n",(1-x)*100);
+printf("or quality of the steam is %f percent\n",x*100);
+//very often,it is necessary to perform multiple interpolations if the tables are used,and the Mollier chart yields results within the rquired accuracy for most engineering problems and saves considerable time.
+//We can also use the computerised programs to solve this program.We first enter the 250F and 14.7 psia to obtain h of 1168.7 Btu/lbm.We then continue by entering h of 1168.7 Btu/lbm and p of 150 psia.The printout gives us x of 0.9699 or 97%.While the computer solution is quick and easy to use,you should still sketch out the problem on an h-s or T-s diagram to show the path of the process.
+
+// Saturation Properties
+//--------------------------
+// T=250.00 degF
+// P=29.814 psia
+// z z1 zg
+// v(ft^3/lbm) 0.01700 13.830
+// h(Btu/lbm) 218.62 1164.1
+// s(Btu/lbm*F) 0.3678 1.7001
+// u(Btu/lbm) 218.52 1087.8
+
+//Thermo Properties
+//------------------------
+// T= 250.00 degF
+// P= 14.700 psia
+// v= 28.417 ft^3/lbm
+// h= 1168.7 Btu/lbm
+// s= 1.7831 Btu/lbm*F
+// u= 1091.4 Btu/lbm
+
+// Saturation Properties
+//--------------------------
+// T=340.06 degF
+// P=118.00 psia
+// z z1 zg
+// v(ft^3/lbm) 0.01787 3.7891
+// h(Btu/lbm) 311.39 1190.7
+// s(Btu/lbm*F) 0.4904 1.5899
+// u(Btu/lbm) 311.00 1108.0
+
+//Thermo Properties
+//------------------------
+// T= 358.49 degF
+// P= 150.00 psia
+// v= 2.9248 ft^3/lbm
+// h= 1168.7 Btu/lbm
+// s= 1.5384 Btu/lbm*F
+// u= 1087.5 Btu/lbm
+// x= 0.9699
+
+//Region:Saturated
diff --git a/2417/CH5/EX5.36/Ex5_36.sce b/2417/CH5/EX5.36/Ex5_36.sce new file mode 100755 index 000000000..38f0ba3e6 --- /dev/null +++ b/2417/CH5/EX5.36/Ex5_36.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.36\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.36 (page no. 219)
+// Solution
+
+//Because the tank volume is 10 ft^3,the final specific volume of the steam is 10 ft^3/lbm.Interpolations in Table A.2 yield a final pressure of 42 psia.The heat added is simply difference in internal energy between the two states.
+u2=1093.0; //internal energy //Btu/lbm
+u1=117.95; //internal energy //Btu/lbm
+q=u2-u1; //heat added //Btu/lbm
+printf("The final pressure is 42 psia and the heat added is %f Btu/lbm\n",q);
diff --git a/2417/CH5/EX5.37/Ex5_37.sce b/2417/CH5/EX5.37/Ex5_37.sce new file mode 100755 index 000000000..82e08494a --- /dev/null +++ b/2417/CH5/EX5.37/Ex5_37.sce @@ -0,0 +1,43 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.37\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.37 (page no. 220)
+// Solution
+
+//The mass in the tank is constant,and the heat added will be the change in internal energy of the contents of the tank between the two states.The initial mass in the tank is found as follows:
+Vf=45; //volume of water //ft^2
+vf=0.016715;
+Vg=15; //Volume of steam //ft^2
+vg=26.80;
+mf=Vf/vf; //lbm
+mg=Vg/vg; //lbm
+total=mf+mg; //total mass
+//The internal energy is the sum of the internal energy of the liquid plus vapor:
+ug=1077.6;
+uf=180.1;
+Ug=mg*ug; //Btu
+Uf=mf*uf; //Btu
+Total=Ug+Uf; //total internal energy
+printf("The total internal energy is %f Btu\n",Total);
+//Because the mass in the tank is constant,the final specific volume must equal the initial specific volume,or
+vx=(Vf+Vg)/(mf+mg); //ft^3/lbm
+//But vx=vf+(x*vfg).Therefore using table A.2 at 800 psia,
+vx=0.022282;
+vf=0.02087;
+vfg=0.5691-0.02087;
+x=(vx-vf)/vfg;
+printf("The final amount of vapor is %f lbm\n",x*total); //x*total mass
+mg=x*total;
+printf("The final amount of liquid is %f lbm\n",total-(x*total)); //total mass minus final amount of vapor
+mf=total-(x*total);
+//The final internal energy is found as before:
+ug=1115.0;
+uf=506.6;
+Ug=mg*ug; //Btu
+Uf=mf*uf; //Btu
+Total1=Ug+Uf;
+difference=Total1-Total; //final internal energy-initial internal energy
+//per unit mass heat added is,
+printf("The heat added per unit is %f Btu/lbm\n",difference/total); //the difference of internal energy/total mass
diff --git a/2417/CH5/EX5.38/Ex5_38.sce b/2417/CH5/EX5.38/Ex5_38.sce new file mode 100755 index 000000000..c55dbc4ad --- /dev/null +++ b/2417/CH5/EX5.38/Ex5_38.sce @@ -0,0 +1,62 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.38\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.38 (page no. 222)
+// Solution
+
+//As shown in Fig. 5.21b,the process dscribed in this problem is a vertical line on the Mollier Chart.For 800 psia and 600F,the Mollier chart yeilds h1=1270 Btu/lbm and s1=1.485.Proceeding vertically down the chart at constant s to 200 psia yields a final enthalpy h2=1148 Btu/lbm.The change in enthaly using the process is 1270-1148=122 Btu/lbm.
+//We may also solve this problem using the steam tables in Appendix 3.Thus,the enthalpy at 800 psia and 600 F is 1270.4 Btu/lbm,and its entropy is 1.4861 Btu/lbm*R.
+//Because the process is isentropic,the final entropy at 200psia must be 1.4861.From the saaturation table,the entropy of saturated steam at 200 psia is 1.5464,which indicates the final steam condition must be wet because the entropy of the final steam is less than the entropy of saturation.Using the wet steam relation yields,
+//sx=sf+(x*sfg)
+h1=1270.4; sx=1.4861; sf=0.5440; sfg=1.0025 ;hf=355.6; hfg=843.7;
+x=(sx-sf)/sfg; //Quality
+//Therefore,the final enthalpy is
+hx=hf+(x*hfg); //Btu/lbm
+printf("The final enthalpy is %f Btu/lbm\n",hx);
+printf("The change in enthalpy is %f Btu/lbm\n",h1-hx); //Note the agreement with the Mollier chart solution
+//we can also use the computer program to solve this problem.For 600F and 800 psia, h=1270. Btu/lbm and s=1.4857 Btu/lbm*R.Now using p=200 psia and s=1.4857,we obtain
+//h=1148.1 Btu/lbm.The change in enthalpy is 1270.0-1148.1=121.9 Btu/lbm.Note the effort saved using either the Mollier chart or the computer program.
+
+// Saturation Properties
+//--------------------------
+// T=600.00 degF
+// P=1541.7 psia
+// z z1 zg
+// v(ft^3/lbm) 0.02362 0.2675
+// h(Btu/lbm) 616.59 1166.2
+// s(Btu/lbm*F) 0.8129 1.3316
+// u(Btu/lbm) 609.85 1089.9
+
+//Thermo Properties
+//------------------------
+// T= 600.00 degF
+// P= 800.00 psia
+// v= 0. ft^3/lbm
+// h= 1168.7 Btu/lbm
+// s= 1.5384 Btu/lbm*F
+// u= 1087.5 Btu/lbm
+//Region:Superheated
+
+// Saturation Properties
+//--------------------------
+// T=381.87 degF
+// P=200.00 psia
+// z z1 zg
+// v(ft^3/lbm) 0.01839 2.2883
+// h(Btu/lbm) 355.60 1199.0
+// s(Btu/lbm*F) 0.5440 1.5462
+// u(Btu/lbm) 354.92 1114.3
+
+//Thermo Properties
+//------------------------
+// T= 381.87 degF
+// P= 200.00 psia
+// v= 2.1512 ft^3/lbm
+// h= 1148.1 Btu/lbm
+// s= 1.4857 Btu/lbm*F
+// u= 1068.5 Btu/lbm
+// x= 0.9396
+
+//Region:Saturated
diff --git a/2417/CH5/EX5.39/Ex5_39.sce b/2417/CH5/EX5.39/Ex5_39.sce new file mode 100755 index 000000000..e0433db04 --- /dev/null +++ b/2417/CH5/EX5.39/Ex5_39.sce @@ -0,0 +1,17 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.39\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.39 (page no. 226)
+// Solution
+
+//As refering to figure 5.21,it will be seen that the final temperature and enthalpy will both be higher than for the isentropic case.
+//80% of the isentropic enthalpy difference
+deltah=0.8*122; //change in enthalpy //Btu/lbm
+h1=1270; //Btu/lbm //initial enthalpy
+h2=h1-deltah; //the final enthalpy //Btu/lbm
+printf("The final enthalpy is %f Btu/lbm\n",h2);
+printf("and the final pressure is 200 psia\n");
+printf("The Mollier chart indicates the final state to be in the wet region,\n");
+printf("with 3.1percent moisture content and an entropy of 1.514 Btu/lbm*R");
diff --git a/2417/CH5/EX5.4/Ex5_4.sce b/2417/CH5/EX5.4/Ex5_4.sce new file mode 100755 index 000000000..d04e0d7f6 --- /dev/null +++ b/2417/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,21 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.4\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.4 (page no. 189)
+// Solution
+
+//By defination,
+//hg=ug+(p*vg)/J
+//hf=uf+(p*vf)/J
+//hfg = hg-hf = (ug-uf) + p*(vg-vf)/J = ufg + p*(vg-vf)/J
+//From table 2 at 115 psia,
+p=115; //Unit:psia //absolute pressure
+ufg=798.8; //Unit:Btu/lbm //Evap. internal energy
+ug=3.884; //Unit:ft^3/lbm //Saturated vapour internal energy
+vf=0.017850; //Unit:ft^3/lbm //Saturated liquid specific volume
+J=778; //J=Conversion factor //Unit:ft*lbf/Btu
+//1 ft^2=144 in^2
+hfg=ufg+(p*144*(ug-vf))/J; //Evap. Enthalpy //Unit:Btu/lbm
+printf("hfg for saturated steam at 115 psia is %f Btu/lbm",hfg); //The tabulated values are matched
diff --git a/2417/CH5/EX5.40/Ex5_40.sce b/2417/CH5/EX5.40/Ex5_40.sce new file mode 100755 index 000000000..34ea1f73b --- /dev/null +++ b/2417/CH5/EX5.40/Ex5_40.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.40\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.40 (page no. 226)
+// Solution
+
+//Using the Mollier chart,
+h1=2942; //kJ/kg //initial enthalpy
+//Proceeding as shown in figure 5.21b,that is,vertically at constant entropy to a pressure of 0.1 MPa,gives us
+h2=2512; //kJ/kg //final enthalpy
+printf("Neglecting kinetic & potential energy,The change in enthalpy of the steam is %f kJ/kg",h1-h2);
diff --git a/2417/CH5/EX5.41/Ex5_41.sce b/2417/CH5/EX5.41/Ex5_41.sce new file mode 100755 index 000000000..faa5406e2 --- /dev/null +++ b/2417/CH5/EX5.41/Ex5_41.sce @@ -0,0 +1,15 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.41\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.41 (page no. 226)
+// Solution
+
+//From the conditions given in problem 5.38,the isentropic change in enthalpy is 122 Btu/lbm
+//So,
+h1minush2=122; //Btu/lbm //change in enthalpy
+J=778; //Conversion factor
+gc=32.17; //lbm*ft/lbf*s^2 //constant of proportionality
+V2=sqrt(2*gc*J*(h1minush2)); //final velocity //ft/s
+printf("As the steam leaves the nozzle,The final velocity is %f ft/s",V2);
diff --git a/2417/CH5/EX5.42/Ex5_42.sce b/2417/CH5/EX5.42/Ex5_42.sce new file mode 100755 index 000000000..139e32305 --- /dev/null +++ b/2417/CH5/EX5.42/Ex5_42.sce @@ -0,0 +1,16 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.42\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.42 (page no. 227)
+// Solution
+
+//Because the process is irreversible,we cannot show it on the Mollier diagram.However,the analysis of problem 3.22 for the nozzle is still valid,and all that is needed is the enthalpy at the beginning and the end of the expansion.From the problem 5.38,
+h1=1270; //Btu/lbm //initial enthalpy
+//For h2 we locate the state point on the Mollier diagram as being saturated vapor at 200 psia.This gives us
+h2=1199; //Btu/lbm //final enthalpy
+J=778; //Conversion factor
+gc=32.17; //lbm*ft/lbf*s^2 //constant of proportionality
+V2=sqrt(2*gc*J*(h1-h2)); //final velocity //Ft/s
+printf("As the steam leaves the nozzle,The final velocity is %f ft/s",V2);
diff --git a/2417/CH5/EX5.43/Ex5_43.sce b/2417/CH5/EX5.43/Ex5_43.sce new file mode 100755 index 000000000..304ce0b17 --- /dev/null +++ b/2417/CH5/EX5.43/Ex5_43.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.43\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.43 (page no. 229)
+// Solution
+
+//From the saturation table,500 psia corresponds to a temperature of 467.13F,and the saturated vapor has an enthalpy of 1205.3 Btu/lbm.At 500 psia and 800 F,the saturated vapor has an enthalpy of 1412.1 Btu/lbm.Because this process is a steady-flow process at constant pressure,the energy equation becomes q=h2-h1,assuming that differences in the kinetic energy and potential energy terms are negligible.Therefore,
+h2=1412.1; //Btu/lbm //final enthalpy
+h1=1205.3; //Btu/lbm //initial enthalpy
+q=h2-h1; //heat added //Btu/lbm
+printf("%f Btu/lbm heat per pound of steam was added\n",q);
diff --git a/2417/CH5/EX5.44/Ex5_44.sce b/2417/CH5/EX5.44/Ex5_44.sce new file mode 100755 index 000000000..8033facec --- /dev/null +++ b/2417/CH5/EX5.44/Ex5_44.sce @@ -0,0 +1,41 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.44\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.44 (page no. 229)
+// Solution
+
+//From the saturation table at 1 psia,
+hf=69.74; //Btu/lbm //saturated liquid enthalpy
+hfg=1036.0; //Btu/lbm //Evap. Enthalpy
+hg=1105.8; //Btu/lbm //The enthalpy of saturated steam
+x=0.97; //Quality
+//Because the condensation process is carried out at constant pressure,the energy equation is q=deltah.
+hx=hf+(x*hfg); //the initial enthalpy //Btu/lbm
+printf("The initial enthalpy is %f Btu/lbm\n",hx);
+//The final enthalpy is hf=69.74.So,
+deltah=hx-hf; //The enthalpy difference //Btu/lbm
+printf("At 1 psia,The enthalpy difference is %f Btu/lbm\n",deltah);
+printf("By the computer solution,the enthalpy difference is 1004.6 Btu/lbm");
+// Saturation Properties
+//--------------------------
+// T=101.71 degF
+// P=1.0000 psia
+// z z1 zg
+// v(ft^3/lbm) 0.01614 333.55
+// h(Btu/lbm) 69.725 1105.4
+// s(Btu/lbm*F) 0.1326 1.9774
+// u(Btu/lbm) 69.722 1043.6
+
+//Thermo Properties
+//------------------------
+// T= 101.71 degF
+// P= 1.0000 psia
+// v= 323.55 ft^3/lbm
+// h= 1074.3 Btu/lbm
+// s= 1.9221 Btu/lbm*F
+// u= 1014.4 Btu/lbm
+// x= 0.9700
+
+//Region:Saturated
diff --git a/2417/CH5/EX5.5/Ex5_5.sce b/2417/CH5/EX5.5/Ex5_5.sce new file mode 100755 index 000000000..781cd9fad --- /dev/null +++ b/2417/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,16 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.5\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.5 (page no. 190)
+// Solution
+
+//From table 2 at 1.0 MPa,
+p=1000; //Unit:kN/m^2 //absolute pressure
+ufg=1822.0; //Unit:kJ/kg //Evap. internal energy
+vf=0.0011273; //Unit:m^3/kg //Saturated liquid specific volume
+vg=0.19444; //Unit:m^3/kg //Saturated vapour specific volume
+vfg=vg-vf; //Evap. specific volume //m^3/kg
+hfg=ufg+(p*vfg); //Evap. Enthalpy //Unit:kJ/kg
+printf("hfg for saturated steam at 1.0 MPa is %f kJ/kg",hfg); //The tabulated values are matched
diff --git a/2417/CH5/EX5.6/Ex5_6.sce b/2417/CH5/EX5.6/Ex5_6.sce new file mode 100755 index 000000000..119dca3d0 --- /dev/null +++ b/2417/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,12 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.6\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.6 (page no. 190)
+// Solution
+
+//For constant-temperature,reversible vaporization, hfg=deltah=T*deltas=T*sfg
+hfg=(388.12+460)*(1.1042); //Evap. Enthalpy //Unit:Btu/lbm
+printf("By considering the process to be a reversible,constant-temperature,hfg for saturated steam at 115 psia is %f Btu/lbm",hfg); //ans is wrong in the book
+//Values are matched with tabulated values.Use of -459.67 F for absolute zero,which is the value used in table,gives almost exact agreement.
diff --git a/2417/CH5/EX5.7/Ex5_7.sce b/2417/CH5/EX5.7/Ex5_7.sce new file mode 100755 index 000000000..33510409d --- /dev/null +++ b/2417/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,34 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.7\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.7 (page no. 192)
+// Solution
+
+//Using Table 2 ans a quality of 80%(x=0.8),we have
+//at 120 psia
+x=0.8;
+sf=0.49201; //saturated liquid entropy //Unit:Btu/lbm*R
+sfg=1.0966; //Evap. Entropy //Unit:Btu/lbm*R
+hf=312.67; //saturated liquid enthalpy //Unit:Btu/lbm
+hfg=878.5; //Evap. Enthalpy //Unit:Btu/lbm
+uf=312.27; //saturated liquid internal energy //Unit:Btu/lbm
+ufg=796.0; //Unit:Btu/lbm //Evap. internal energy
+vf=0.017886; //Saturated liquid specific volume //Unit:ft^3/lbm
+vfg=(3.730-0.017886); //evap. specific volume //Unit:ft^3/lbm
+sx=sf+(x*sfg); //entropy //Btu/lbm*R
+printf("Entropy of a wet steam mixture at 120 psia is %f Btu/lbm*R\n",sx);
+hx=hf+(x*hfg); //enthalpy //Btu/lbm*R
+printf("Enthalpy of a wet steam mixture at 120 psia is %f Btu/lbm\n",hx);
+ux=uf+(x*ufg); //internal energy //Btu/lbm*R
+printf("Internal energy of a wet steam mixture at 120 psia is %f Btu/lbm\n",ux);
+vx=vf+(x*vfg); ///specific volume //ft^3/lbm
+printf("Specific Volume of a wet steam mixture at 120 psia is %f ft^3/lbm\n",vx);
+//As a check,
+J=778; //ft*lbf/Btu //Conversion factor
+px=120; //psia //pressure
+ux=hx-((px*vx*144)/J); //1 ft^2=144 in^2 //internal energy
+printf("As a check,\n")
+printf("Internal energy of a wet steam mixture at 120 psia is %f Btu/lbm\n",ux);
+printf("Which agrees with the values obtained above");
diff --git a/2417/CH5/EX5.8/Ex5_8.sce b/2417/CH5/EX5.8/Ex5_8.sce new file mode 100755 index 000000000..7f13d6769 --- /dev/null +++ b/2417/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,37 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.8\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.8 (page no. 193)
+// Solution
+
+//Using Table 2 ans a quality of 85%(x=0.85),we have
+//at 1.0 MPa
+x=0.85;
+sf=2.1387; //saturated liquid entropy //Unit:kJ/kg*K
+sfg=4.4487; //Evap. Entropy //Unit:kJ/kg*K
+hf=762.81; //saturated liquid enthalpy //Unit:kJ/kg
+hfg=2015.3; //Evap. Enthalpy //Unit:kJ/kg
+uf=761.68; //saturated liquid internal energy //Unit:kJ/kg
+ufg=1822.0; //Unit:kJ/kg //Evap. internal energy
+vf=1.1273; //Saturated liquid specific volume //Unit:m^3/kg
+vfg=(194.44-1.1273); //evap. specific volume //Unit:m^3/kg
+sx=sf+(x*sfg); //entropy //kJ/kg*K
+printf("Entropy of a wet steam mixture at 1.0 MPa is %f kJ/kg*K\n",sx);
+hx=hf+(x*hfg); //enthalpy //kJ/kg*K
+printf("Enthalpy of a wet steam mixture at 1.0 MPa is %f kJ/kg\n",hx);
+ux=uf+(x*ufg); //internal energy //kJ/kg*K
+printf("Internal energy of a wet steam mixture at 1.0 MPa is %f kJ/kg\n",ux);
+vx=(vf+(x*vfg))*(0.001); //specific volume //m^3/kg
+printf("Specific Volume of a wet steam mixture at 1.0 MPa is %f m^3/kg\n",vx);
+//As a check,
+px=10^6; //psia //pressure
+ux=hx-((px*vx)/10^3); //1 ft^2=144 in^2 //internal energy
+printf("As a check,\n")
+printf("Internal energy of a wet steam mixture at 120 psia is %f kJ/kg\n",ux);
+printf("Which agrees with the values obtained above");
+
+
+
+
diff --git a/2417/CH5/EX5.9/Ex5_9.sce b/2417/CH5/EX5.9/Ex5_9.sce new file mode 100755 index 000000000..0f0588c26 --- /dev/null +++ b/2417/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,15 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 5.9\n\n\n");
+// Chapter 5 : Properties Of Liquids And Gases
+// Problem 5.9 (page no. 193)
+// Solution
+
+//For the wet mixture,hx=hf+(x*hfg),solving for x gives us
+//Using table 1,we have,
+hx=900; //Btu/lbm //Enthalpy of wet mixture at 90F
+hf=58.07; //Btu/lbm //saturated liquid enthalpy
+hfg=1042.7; //Btu/lbm //Evap. Enthalpy
+x=(hx-hf)/hfg; //quality
+printf("The quality is %f percentage of a wet steam at 90F\n",x*100);
diff --git a/2417/CH6/EX6.1/Ex6_1.sce b/2417/CH6/EX6.1/Ex6_1.sce new file mode 100755 index 000000000..f4889bc95 --- /dev/null +++ b/2417/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.1\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.1 (page no. 241)
+// Solution
+
+P1=100; //Pressure at volume V1=100 ft^3 //Unit:psia
+V1=100; //Unit:ft^3 //V1=Volume at 100 psia
+P2=30 // Reduced Pressure //Unit:psia
+//Boyle's law,P1*V1=P2*V2
+V2=(P1*V1)/P2; //Volume occupied by the gas //ft^3
+printf("Volume occupied by the gas = %f ft^3",V2);
diff --git a/2417/CH6/EX6.10/Ex6_10.sce b/2417/CH6/EX6.10/Ex6_10.sce new file mode 100755 index 000000000..550c0bdb1 --- /dev/null +++ b/2417/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.10\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.10 (page no. 252)
+// Solution
+
+//The table in Appendix 3 does not give us the enthalpy data at 960R and 540R that we need.Interpolating yields
+// T hbar T hbar
+// 537 3729.5 900 6268.1
+// 540 3750.4 960 6694.0
+// 600 4167.9 1000 6977.9
+//So,
+hbar540=3729.5+(3/63)*(4167.9-3729.5); //enthalpy //unit:Btu/lbm
+hbar960=6268.1+(60/100)*(6977.9-6268.1); //enthalpy //unit:Btu/lbm
+//Note that hbar is given for a mass of 1 lb mole.To obtain the enthalpy per pound,it is necessary to divide the values og h by the molecular weight,28.
+h2=6694.0; //enthalpy //unit:Btu/lbm
+h1=3750.4; //enthalpy //unit:Btu/lbm
+T2=500+460; //absolute final temperature //unit:R
+T1=80+460; //absolute initial temperature //unit:R
+cbar=(h2-h1)/(28*(T2-T1)); //The mean specific heat at constant pressure //unit:Btu/lbm*R
+printf("The mean specific heat at constant pressure is %f Btu/lbm*R\n",cbar);
+//With the more extesive Gas tables,these interpolations are avoided.The Gas Tables provide a relatively easy and accurate method of obtaining average specific heats.Also,these tables have been computerized for ease of application.
diff --git a/2417/CH6/EX6.11/Ex6_11.sce b/2417/CH6/EX6.11/Ex6_11.sce new file mode 100755 index 000000000..8cc98e9b4 --- /dev/null +++ b/2417/CH6/EX6.11/Ex6_11.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.11\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.11 (page no. 253)
+// Solution
+
+T2=500+460; //absolute final temperature //unit:R
+T1=80+460; //absolute initial temperature //unit:R
+//cp=0.219 + (3.42*10^-5*T) - (2.93*10^-9*T^2); //Unit:Btu/lbm*R
+//Comparing with c=A+(B*T)+(D*T^2)
+A=0.219; //constant
+B=3.42*10^-5; //constant
+D=2.93*10^-9; //constant
+//Using these values and equation cbar=A+((B/2)(T2+T1))+((D/3)*(T2^2+(T2*T1)+T1^2))
+cpbar=A+((B/2)*(T2+T1))+((D/3)*(T2^2+(T2*T1)+T1^2)); //The mean specific heat //Btu/lbm*R
+printf("The mean specific heat at constant pressure for air between 80F and 500F is %f Btu/lbm*R\n",cpbar);
diff --git a/2417/CH6/EX6.12/Ex6_12.sce b/2417/CH6/EX6.12/Ex6_12.sce new file mode 100755 index 000000000..076503ae1 --- /dev/null +++ b/2417/CH6/EX6.12/Ex6_12.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.12\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.12 (page no. 255)
+// Solution
+
+//The molecular weight of oxygen is 32.Therefore,
+R=1545/32; //Unit:ft*lbf/lbm*R //constant of proportionality
+J=778; //conversion factor
+cp=0.24; //Unit:Btu/lbm*R //specific heat at constant pressure
+//cp-cv=R/J
+cv=cp-(R/J); //specific heat at constant volume //unit:Btu/lbm*R
+printf("Specific heat at constant volume is %f Btu/lbm*R\n",cv);
diff --git a/2417/CH6/EX6.13/Ex6_13.sce b/2417/CH6/EX6.13/Ex6_13.sce new file mode 100755 index 000000000..2726b2bd7 --- /dev/null +++ b/2417/CH6/EX6.13/Ex6_13.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.13\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.13 (page no. 255)
+// Solution
+
+//From equation,cv=R/(k-1) ,
+R=8.314/32; //constant of proportionality //kJ/kg*K //The molecular weight of oxygen is 32
+k=1.4 //for oxygen //given //k=cp/cv
+cv=R/(k-1); //Specific heat at constant volume //unit:kJ/kg*K
+printf("Specific heat at constant volume is %f kJ/kg*K\n",cv);
+cp=k*cv; //specific heat at constant pressure //Unit:kJ/kg*K
+printf("Specific heat at constant pressure is %f kJ/kg*K\n",cp);
diff --git a/2417/CH6/EX6.14/Ex6_14.sce b/2417/CH6/EX6.14/Ex6_14.sce new file mode 100755 index 000000000..62ea0a13c --- /dev/null +++ b/2417/CH6/EX6.14/Ex6_14.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.14\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.14 (page no. 255)
+// Solution
+
+R=60; //Unit:ft*lbf/lbm*R //constant of proportionality
+deltah=500; //Btu/lbm //change in enthalpy
+deltau=350; //Btu/lbm //change in internal energy
+J=778; //conversion factor
+//Because deltah-(cp*deltaT) and deltau=cv*deltaT
+// deltah/deltau=(cp*deltaT)/(cv*deltaT)=cp/cv=k
+k=deltah/deltau; //Ratio of specific heats
+printf("Ratio of specific heats k is %f\n",k);
+//From equation cv=R/(J*(k-1))
+cv=R/(J*(k-1)); //specific heat at constant volume //Btu/lbm*R
+printf("Specific heat at constant volume is %f Btu/lbm*R\n",cv);
+cp=k*cv; //Specific heat at constant pressure //Btu/lbm*R
+printf("Specific heat at constant pressure is %f Btu/lbm*R\n",cp);
diff --git a/2417/CH6/EX6.15/Ex6_15.sce b/2417/CH6/EX6.15/Ex6_15.sce new file mode 100755 index 000000000..8402b0a02 --- /dev/null +++ b/2417/CH6/EX6.15/Ex6_15.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.15\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.15 (page no. 256)
+// Solution
+
+//When solving this type of problem,it is necessary to note carefully the information given and to write the correct energy equation for this process.Because the process is carried out at constant volume,the heat added equals the change in inernal energy.Because the change in internal energy per pound for the ideal gas is cv*(T2-T1),the total change in internal energy for m pounds must equals the heat added.Thus,
+//data given
+Q=0.33; //heat
+//Initial conditions
+V=60; //in^3 //volume
+m=0.0116; //lbs //mass
+p1=90; //psia //pressure
+T1=460+40; //Fahrenheit temperature converted to absolute temperature
+//Final condition=Initial condition + heat
+V=60; //in^3 //volume
+m=0.0116; //lbs //mass
+p2=108; //psia //pressure
+T2=460+140; //Fahrenheit temperature converted to absolute temperature //unit:R
+//Q=m*(u2-u1)=m*cv*(T2-T1)
+cv=Q/(m*(T2-T1)); //specific heat at constant volume //Btu/lbm*R
+printf("Specific heat at constant volume is %f Btu/lbm*R\n",cv);
+//To obtain cp,it is first necessary to obtain R.Enough information was given in the initial conditions of the problem to apply eqn. p*V=m*R*T
+R=(144*p1*(V/1728))/(m*T1); //1 ft^2=144 in^2 //1 ft^3=1728 in^3 //Unit:ft*lbf/lbm*R //constant of proportionality
+printf("Constant of proportionality R is %f ft*lbf/lbm*R\n",R);
+//cp-cv=(R/J)
+J=778; //conversion factor
+cp=cv+(R/J); //Specific heat at constant pressure //Btu/lbm*R
+printf("Specific heat at constant pressure is %f Btu/lbm*R\n",cp);
diff --git a/2417/CH6/EX6.16/Ex6_16.sce b/2417/CH6/EX6.16/Ex6_16.sce new file mode 100755 index 000000000..624a5a796 --- /dev/null +++ b/2417/CH6/EX6.16/Ex6_16.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.16\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.16 (page no. 260)
+// Solution
+
+//data
+cp=0.24; //Specific heat at constant pressure //Btu/lbm*R
+p2=15; //psia //final pressure
+p1=100; //psia //initial pressure
+T2=460+0; //absolute final temperature //unit:R
+T1=460+100; //absolute initial temperature //unit:R
+J=778; //conversion factor
+R=1545/29; //moleculer weight=29 //Unit:ft*lbf/lbm*R //constant of proportionality
+ //On the basis of the data given,
+deltas=(cp*(log(T2/T1)))-((R/J)*(log(p2/p1))); //change in entropy //Btu/lbm*R
+printf("The change in enthalpy is %f Btu/lbm*R\n",deltas);
diff --git a/2417/CH6/EX6.17/Ex6_17.sce b/2417/CH6/EX6.17/Ex6_17.sce new file mode 100755 index 000000000..531651dc7 --- /dev/null +++ b/2417/CH6/EX6.17/Ex6_17.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.17\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.17 (page no. 261)
+// Solution
+
+//data of problem6.16
+cp=0.24; //Specific heat at constant pressure //Btu/lbm*R
+p2=15; //psia //final pressure
+p1=100; //psia //initial pressure
+T2=460+0; //absolute final temperature //unit:R
+T1=460+100; //absolute initial temperature //unit:R
+J=778; //conversion factor
+R=1545/29; //moleculer weight=29 //Unit:ft*lbf/lbm*R //constant of proportionality
+//Because cp and R are given,let us first solve for cv,
+//cp=(R*k)/(J*(k-1))
+k=(cp*J)/((cp*J)-R); //k=cp/cv //ratio of specific heats
+printf("Ratio of specific heats k is %f\n",k);
+//k=cp/cv
+cv=cp/k; //Specific heat at constant volume //Btu/lbm*R
+printf("Specific heat at constant volume is %f Btu/lbm*R\n",cv);
+//Now, deltas=(cv*log(p2/p1))+(cp*log(v2/v1));
+//But, v2/v1=(T2*p1)/(T1*p2)
+v2byv1=(T2*p1)/(T1*p2); // v2/v1 //unitless
+deltas=(cv*log(p2/p1))+(cp*log(v2byv1)); //The change in enthalpy //unit:Btu/lbm*R
+printf("The change in enthalpy is %f Btu/lbm*R\n",deltas);
+//The agreement is very good.
diff --git a/2417/CH6/EX6.18/Ex6_18.sce b/2417/CH6/EX6.18/Ex6_18.sce new file mode 100755 index 000000000..438eea95d --- /dev/null +++ b/2417/CH6/EX6.18/Ex6_18.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.18\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.18 (page no. 261)
+// Solution
+
+//data,
+cp=0.9093; //Specific heat at constant pressure //kJ/kg*R
+p2=150; //kPa //final pressure
+p1=500; //kPa //initial pressure
+T2=273+0; //final temperature //Celsius converted to kelvin
+T1=273+100; //initial temperature //Celsius converted to kelvin
+//J=778; //conversion factor
+R=8.314/32; //moleculer weight of oxygen=32 //Unit:ft*lbf/lbm*R //constant of proportionality
+//Using equation, and dropping J gives,
+deltas=(cp*(log(T2/T1)))-((R)*(log(p2/p1))); //change in entropy //kJ/kg*K
+//For 2 kg,
+deltaS=2*deltas; //The change in enthalpy in kJ/K
+printf("For 2 kg oxygen,The change in enthalpy is %f kJ/K\n",deltaS);
diff --git a/2417/CH6/EX6.19/Ex6_19.sce b/2417/CH6/EX6.19/Ex6_19.sce new file mode 100755 index 000000000..e6e8ee493 --- /dev/null +++ b/2417/CH6/EX6.19/Ex6_19.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.19\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.19 (page no. 262)
+// Solution
+
+//from the equation, deltas/cv = (k*log(v2/v1))+ log(p2/p1) //change in entropy
+k=1.4; //k=cp/cv //ratio of specific heats
+//deltas=(1/4)*cv //so,
+// 1/4= (k*log(v2/v1))+ log(p2/p1)
+v2=1/2; //Because,v2=(1/2)*v1 //initial specific volume
+v1=1; //final specific volume
+p2byp1=exp((1/4)-(k*log(v2/v1))); //increase in pressure
+printf("p2/p1=%f\n",p2byp1);
+printf("So,increase in pressure is %f ",p2byp1);
diff --git a/2417/CH6/EX6.2/Ex6_2.sce b/2417/CH6/EX6.2/Ex6_2.sce new file mode 100755 index 000000000..3db9344ad --- /dev/null +++ b/2417/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.2\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.2 (page no. 241)
+// Solution
+
+P1=10^6; //Pressure at volume V1=2 m^3 //Unit:Pa
+V1=2; //Unit:m^3 //V1=Volume at 10^6 Pa
+P2=8*10^6 // Increased Pressure //Unit:Pa
+//Boyle's law,P1*V1=P2*V2
+V2=(P1*V1)/P2; //Volume occupied by gas //unit:m^3
+printf("Volume occupied by gas = %f m^3",V2);
diff --git a/2417/CH6/EX6.20/Ex6_20.sce b/2417/CH6/EX6.20/Ex6_20.sce new file mode 100755 index 000000000..89d575833 --- /dev/null +++ b/2417/CH6/EX6.20/Ex6_20.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.20\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.20 (page no. 264)
+// Solution
+
+//data
+T2=460+270; //Fahrenheit temperature converted to absolute final temperature //unit:R
+T1=460+70; //Fahrenheit temperature converted to absolute initial temperature //unit:R
+cv=0.17; //specific heat at constant volume //Btu/lbm*R
+//Now,
+deltas=cv*log(T2/T1); //change in entropy //Unit:Btu/lbm*R
+//For 1/2 lb,
+deltaS=(1/2)*deltas; //The change in enthalpy in Btu/R
+printf("For 1/2 lb of gas,The change in enthalpy is %f Btu/R\n",deltaS);
diff --git a/2417/CH6/EX6.21/Ex6_21.sce b/2417/CH6/EX6.21/Ex6_21.sce new file mode 100755 index 000000000..61d26a3c2 --- /dev/null +++ b/2417/CH6/EX6.21/Ex6_21.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.21\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.21 (page no. 264)
+// Solution
+
+//data
+T2=100+273; //Celsius temperature converted to Kelvin //final temperature
+T1=20+273; //Celsius temperature converted to Kelvin //initial temperature
+cv=0.7186; //specific heat at constant volume //kJ/kg*K
+//Now,
+deltas=cv*log(T2/T1); //change in entropy //Unit:kJ/kg*K
+//For 0.2 kg,
+deltaS=(0.2)*deltas; //The change in enthalpy in kJ/K
+printf("For 0.2 kg of air,The change in enthalpy is %f kJ/K\n",deltaS);
diff --git a/2417/CH6/EX6.22/Ex6_22.sce b/2417/CH6/EX6.22/Ex6_22.sce new file mode 100755 index 000000000..22a74043e --- /dev/null +++ b/2417/CH6/EX6.22/Ex6_22.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.22\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.22 (page no. 264)
+// Solution
+
+//data
+deltas=0.0743; //change in entropy //Unit:Btu/lbm*R
+T1=460+100; //Fahrenheit temperature converted to absolute initial temperature
+cv=0.219; //specific heat at constant volume //Btu/lbm*R
+//Now,
+//deltas=cv*log(T2/T1);
+T2=T1*exp(deltas/cv); //higher temperature //absolute temperature //unit:R
+printf("The higher temperature is %f R\n",T2)
diff --git a/2417/CH6/EX6.23/Ex6_23.sce b/2417/CH6/EX6.23/Ex6_23.sce new file mode 100755 index 000000000..1761aa117 --- /dev/null +++ b/2417/CH6/EX6.23/Ex6_23.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.23\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.23 (page no. 265)
+// Solution
+
+//data
+deltaS=0.4386; //change in entropy //Unit:kJ/K
+T2=273+425; //Celsius temperature converted to kelvin //initial temperature
+cv=0.8216; //specific heat at constant volume //kJ/kg*K
+m=1.5; //mass //kg
+//Now,
+//deltas=m*cv*log(T2/T1);
+T1=T2/(exp(deltaS/(m*cv))) //initial temperature //unit:K
+printf("The initial temperature of the process is %f K or %f C\n",T1,T1-273)
diff --git a/2417/CH6/EX6.24/Ex6_24.sce b/2417/CH6/EX6.24/Ex6_24.sce new file mode 100755 index 000000000..f5e3ea2fc --- /dev/null +++ b/2417/CH6/EX6.24/Ex6_24.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.24\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.24 (page no. 267)
+// Solution
+
+//data given
+T2=460+400; //Fahrenheit temperature converted to absolute final temperature //unit:R
+T1=460+70; //Fahrenheit temperature converted to absolute initial temperature //unit:R
+cp=0.24; //specific heat at constant pressure //Btu/lbm*R
+J=778; //conversion factor
+R=1545/29; //moleculer weight=29 //Unit:ft*lbf/lbm*R //constant of proportionality
+//From the energy equation for the constant-pressure process,the heat transferred is deltah.Therefore,
+//q=deltah=cp*(T2-T1)
+deltah=cp*(T2-T1); //heat transferred //Btu/lb //into system
+printf("The heat transferred is %f Btu/lb(into system)\n",deltah);
+deltas=cp*log(T2/T1); //increase in entropy //Btu/lbm*R
+printf("The increase in entropy is %f Btu/lbm*R\n",deltas);
+//The flow work change is (p2*v2)/J - (p1*v1)/J = (R/J)*(T2-T1)
+flowworkchange=(R/J)*(T2-T1); //Btu/lbm //The flow work change per pound of air
+printf("The flow work change per pound of air is %f Btu/lbm\n",flowworkchange);
+//In addition to each of the assumptions made in all the process being considered,it has further been tacitly assumed that these processes are carried out quasi- statically and without friction.
diff --git a/2417/CH6/EX6.25/Ex6_25.sce b/2417/CH6/EX6.25/Ex6_25.sce new file mode 100755 index 000000000..805cfff20 --- /dev/null +++ b/2417/CH6/EX6.25/Ex6_25.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.25\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.25 (page no. 268)
+// Solution
+
+//data given
+T2=500+273; //Celsius temperature converted to Kelvin //final temperature
+T1=20+273; //Celsius temperature converted to Kelvin //initial temperature
+cp=1.0062; //specific heat at constant pressure //kJ/kg*K
+//From the energy equation for the constant-pressure process,the heat transferred is deltah.Therefore,
+//q=deltah=cp*(T2-T1)
+deltah=cp*(T2-T1); //heat transferred //kJ/kg //into system
+printf("The heat transferred is per kilogram of air %f kJ/kg\n",deltah);
+deltas=cp*log(T2/T1); //increase in entropy //kJ/kg*K
+printf("The increase in entropy per kilogram of air is %f kJ/kg*K\n",deltas);
diff --git a/2417/CH6/EX6.26/Ex6_26.sce b/2417/CH6/EX6.26/Ex6_26.sce new file mode 100755 index 000000000..7bf02f2fd --- /dev/null +++ b/2417/CH6/EX6.26/Ex6_26.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.26\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.26 (page no. 270)
+// Solution
+
+//data given
+v2=2; //Because,v2=(2)*v1 //volume increases to its twice its final volume
+v1=1; //initial volume
+T=460+200; //Fahrenheit temperature converted to absolute temperature
+J=778; //conversion factor
+R=1545/28; //moleculer weight of nitrogen=28 //Unit:ft*lbf/lbm*R //constant of proportionality
+//From the equation, w/J=q=T*deltas=((R*T)/J)*log(v2/v1)
+q=((R*T)/J)*log(v2/v1); //Btu/lbm //the heat added to system
+//For 0.1 lb,
+Q=0.1*q; //Btu //the heat added to system
+printf("The heat added to system is %f Btu\n",Q);
+//The work out of the system is equal to the heat added;thus,
+WbyJ=Q; //The work out of the system(out of the system) //unit:Btu
+printf("The work out of the system is %f Btu(out of the system)\n",WbyJ);
diff --git a/2417/CH6/EX6.27/Ex6_27.sce b/2417/CH6/EX6.27/Ex6_27.sce new file mode 100755 index 000000000..bbc038fbc --- /dev/null +++ b/2417/CH6/EX6.27/Ex6_27.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.27\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.27 (page no. 270)
+// Solution
+
+//data given
+T=50+273; //Celsius temperature converted to Kelvin //final temperature //unit:K
+v2=1/2; //Because,v2=(1/2)*v1 //volume increases to its half its final volume
+v1=1;
+R=8.314/32; //moleculer weight of oxygen=32 //Unit:kJ/kg*K //constant of proportionality
+//From the equation, q=((R*T))*log(v2/v1)
+q=R*T*log(v2/v1); //heat added //kJ/kg
+printf("The heat added to system is %f kJ/kg(heat out of system)\n",q);
+//The work out of the system is equal to the heat added;thus,
+W=q; //The work out of the system //unit:kJ/kg
+printf("The work out of the system is %f kJ/kg(into system)\n",W);
diff --git a/2417/CH6/EX6.28/Ex6_28.sce b/2417/CH6/EX6.28/Ex6_28.sce new file mode 100755 index 000000000..96cafe745 --- /dev/null +++ b/2417/CH6/EX6.28/Ex6_28.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.28\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.28 (page no. 271)
+// Solution
+
+//data given in problem 6.27
+T=50+273; //Celsius temperature converted to Kelvin //final temperature
+v2=1/2; //Because,v2=(1/2)*v1 //volume increases to its half its final volume
+v1=1;
+R=8.314/32; //moleculer weight of oxygen=32 //Unit:kJ/kg*K //constant of proportionality
+//From the equation, q=((R*T))*log(v2/v1)
+q=R*T*log(v2/v1); //heat added //kJ/kg
+printf("The heat added to system is %f kJ/kg(heat out of system)\n",q);
+//For a constant temperature,
+deltas=q/T; //Change in entropy //unit:kJ/kg*K
+printf("The change in entropy is %f kJ/kg*K\n",deltas);
diff --git a/2417/CH6/EX6.29/Ex6_29.sce b/2417/CH6/EX6.29/Ex6_29.sce new file mode 100755 index 000000000..64c261a66 --- /dev/null +++ b/2417/CH6/EX6.29/Ex6_29.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.29\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.29 (page no. 274)
+// Solution
+
+//data given
+T1=1000; //absolute initial temperature //unit:R
+p2=1; //unit:atm //absolute final pressure
+p1=5; //unit:atm //absolute initial pressure
+J=778; //conversion factor
+R=1545/29; //moleculer weight=29 //Unit:ft*lbf/lbm*R //constant of proportionality
+k=1.4; //k=cp/cv //ratio of specific heats
+//From the equation,
+T2=T1*((p2/p1)^((k-1)/k)); //Unit:R //The absolute final temperature
+printf("The absolute final temperature is %f R\n",T2);
+work=(R*(T2-T1))/(J*(1-k)); //Btu/lbm //The work done by air(out)
+printf("The work done by air is %f Btu/lbm(out)\n",work)
diff --git a/2417/CH6/EX6.3/Ex6_3.sce b/2417/CH6/EX6.3/Ex6_3.sce new file mode 100755 index 000000000..ab79bf7b9 --- /dev/null +++ b/2417/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.3\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.3 (page no. 242)
+// Solution
+
+T1=32+460; //Temperature at volume V1=150 ft^3 //Unit:R
+V1=150; //Unit:ft^3 //V1=Volume at 32 F
+T2=100+460 // Increased Temperature //Unit:R
+//Charles's law,V1/V2 = T1/T2
+V2=(T2*V1)/T1; //Volume occupied by gas //unit:m^3
+printf("Volume occupied by gas = %f m^3",V2);
diff --git a/2417/CH6/EX6.30/Ex6_30.sce b/2417/CH6/EX6.30/Ex6_30.sce new file mode 100755 index 000000000..782f13225 --- /dev/null +++ b/2417/CH6/EX6.30/Ex6_30.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.30\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.30 (page no. 274)
+// Solution
+
+//data given
+//mass of 1 kg
+T1=500+273; //Celsius temperature converted to Kelvin //final temperature
+p2=1; //atm //absolute final pressure
+p1=5; //atm //absolute initial pressure
+J=778; //conversion factor
+R=8.314/29; //moleculer weight=29 //Unit:kJ/kg*K //constant of proportionality
+k=1.4; //k=cp/cv //ratio of specific heat
+//From the equation,
+T2=T1*((p2/p1)^((k-1)/k)); //Unit:Kelvin //The absolute final temperature
+printf("The absolute final temperature is %f K or %f C\n",T2,T2-273);
+work=(R*(T2-T1))/((1-k)); //kJ/kg //The work done by air(out)
+printf("The work done by air is %f kJ/kg(out)\n",work)
diff --git a/2417/CH6/EX6.31/Ex6_31.sce b/2417/CH6/EX6.31/Ex6_31.sce new file mode 100755 index 000000000..3409ecb82 --- /dev/null +++ b/2417/CH6/EX6.31/Ex6_31.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.31\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.31 (page no. 275)
+// Solution
+
+//data given
+T1=800+273; //Celsius temperature converted to Kelvin //initial temperature
+T2=500+273; //Celsius temperature converted to Kelvin //final temperature
+p2=1; //atm //absolute final pressure
+p1=5; //atm //absolute initial pressure
+//A gas expands isentropically
+//From the equation,
+//T2/T1=((p2/p1)^((k-1)/k));
+//rearranging,
+k=inv(1-((log(T2/T1)/log(p2/p1)))); //k=cp/cv //Ratio of specific heats
+printf("Ratio of specific heats (k) is %f\n",k);
diff --git a/2417/CH6/EX6.32/Ex6_32.sce b/2417/CH6/EX6.32/Ex6_32.sce new file mode 100755 index 000000000..9020de4ab --- /dev/null +++ b/2417/CH6/EX6.32/Ex6_32.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.32\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.32 (page no. 279)
+// Solution
+
+//data given
+n=1.3; //p*v^1.3=constant
+k=1.4; //k=cp/cv Ratio of specific heats
+cp=0.24; //specific heat at constant pressure //Btu/lbm*R
+T2=600; //absolute final temperature //unit:R
+T1=1500; //absolute initial temperature //unit:R
+R=53.3; //Unit:ft*lbf/lbm*R //constant of proportionality
+J=778; //conversion factor
+cv=cp/k; //specific heat at constant volume //Btu/lbm*R
+//Therefore,
+cn=cv*((k-n)/(1-n)); //Polytropic specific heat //Btu/lbm*R
+printf("Polytropic specific heat(cn) is %f Btu/lbm*R\n",cn);
+//The negative sign of cn indicates that either the heat transfer for the process comes from the system or there is a negative temperature change while heat is transferred to the system.
+//The heat transferred is cn*(T2-T1).Therefore,
+q=cn*(T2-T1); //heat transferred //Btu/lbm(to the system)
+printf("The heat transferred is %f Btu/lbm(to the system)\n",q);
+//The work done can be found using equation,
+w=(R*(T2-T1))/(J*(1-n)); //Btu/lbm //the workdone(from the system)
+printf("The work done is %f Btu/lbm(from the system)\n",w);
+deltas=cn*log(T2/T1)' //change in entropy //Btu/lbm*R
+printf("The change in enthalpy is %f Btu/lbm*R\n",deltas);
diff --git a/2417/CH6/EX6.33/Ex6_33.sce b/2417/CH6/EX6.33/Ex6_33.sce new file mode 100755 index 000000000..1a173ea77 --- /dev/null +++ b/2417/CH6/EX6.33/Ex6_33.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.33\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.33 (page no. 279)
+// Solution
+
+//data given in problem 6.32,
+n=1.3; //p*v^1.3=constant
+k=1.4; //k=cp/cv //ratio of specific heats
+cp=0.24; //specific heat at constant pressure //Btu/lbm*R
+T2=600; //absolute final temperature //unit:R
+T1=1500; //absolute initial temperature //unir:R
+R=53.3; //Unit:ft*lbf/lbm*R //constant of proportionality
+J=778; //conversion factor
+//Equation,
+// T1/T2=((p1/p2)^((n-1)/n));
+//rearranging,
+p1byp2=exp(log(T1/T2)/((n-1)/n)); //The ratio of inlet to outlet pressure
+printf("The ratio of inlet to outlet pressure is %f\n",p1byp2);
diff --git a/2417/CH6/EX6.34/Ex6_34.sce b/2417/CH6/EX6.34/Ex6_34.sce new file mode 100755 index 000000000..02d39fc24 --- /dev/null +++ b/2417/CH6/EX6.34/Ex6_34.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.34\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.34 (page no. 284)
+// Solution
+
+//From the table at 1000 R: //From the table at 500 R:
+h2=240.98; h1=119.48;
+//Btu/lbm //enthalpy //Btu/lbm //enthalpy
+u2=172.43; u1=85.20;
+//Btu/lbm //internal energy //Btu/lbm //internal energy
+fy2=0.75042; fy1=0.58233;
+//Btu/lbm*R //Btu/lbm*R
+
+//The change in enthalpy is
+deltah=h2-h1; //Btu/lbm
+//The change in internal energy is
+deltau=u2-u1; //Btu/lbm
+printf("The change in enthalpy is %f Btu/lbm & the change in internal energy is %f Btu/lbm\n",deltah,deltau);
+//Because in the constant-pressure process -R*log(p2/p1) is zero,
+deltas=fy2-fy1; //Btu/lbm*R //The entropy when air is heated at constant pressure
+printf("The entropy when air is heated at constant pressure is %f Btu/lbm/R\n",deltas);
diff --git a/2417/CH6/EX6.35/Ex6_35.sce b/2417/CH6/EX6.35/Ex6_35.sce new file mode 100755 index 000000000..1317109d6 --- /dev/null +++ b/2417/CH6/EX6.35/Ex6_35.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.35\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.35 (page no. 285)
+// Solution
+
+//In this problem,the air expands from 5 atm absolute to 1 atm absolute from an initial temperature of 1000R,
+pr=12.298; //relative pressure //unit:atm
+h=240.98; //Btu/lbm //enthalpy
+pr=12.298/5; //The value of the final relative pressure //unit:atm
+//Interpolation in the air table yields the following:
+// T pr
+// 620 2.249
+// 2.4596
+// 640 2.514
+T=620+(((2.4596-2.249)/(2.514-2.249))*20); //the final temperature //unit:R
+printf("The absolute final temperature is %f R\n",T);
+u1=172.43; //initial internal energy //Btu/lbm
+u2=108.51; //final internal energy //Btu/lbm
+work=u1-u2; //Btu/lbm The work done by air in an isentropic nonflow expansion //where the value of u2 is obtained by interpolation at T temperature and the value of u1 is read from the air table at 1000 R.
+printf("The work done by air in an isentropic nonflow expansion is %f Btu/lbm(out)\n",work)
diff --git a/2417/CH6/EX6.36/Ex6_36.sce b/2417/CH6/EX6.36/Ex6_36.sce new file mode 100755 index 000000000..acc6f34b1 --- /dev/null +++ b/2417/CH6/EX6.36/Ex6_36.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.36\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.36 (page no. 288)
+// Solution
+
+T=1000+460; //Fahrenheit temperature converted to absolute temperature
+//The velocity of sound in air at 1000 F is
+Va=49.0*sqrt(T); //velocity //ft/s
+printf("The velocity of sound air at 1000 F is %f ft/s\n",Va);
+//Hydrogen has a specific heat ratio of 1.41 and R=766.53.Therefore,
+khydrogen=1.41; //specific heats ratio for air
+kair=1.40; //specific heats ratio for air
+Rhydrogen=766.53; //gas constant //ft*lbf/lbm*R
+Rair=53.36; //gas constant //ft*lbf/lbm*R
+// Vahydrogen/Vaair = sqrt((Rhydrogen*khydrogen)/(Rair*kair))
+//rearranging,
+Vahydrogen=Va*sqrt((Rhydrogen*khydrogen)/(Rair*kair)); //The velocity of sound in hydrogen at 1000 F //unit:ft/s
+printf("The velocity of sound in hydrogen at 1000 F is %f ft/s\n",Vahydrogen);
diff --git a/2417/CH6/EX6.37/Ex6_37.sce b/2417/CH6/EX6.37/Ex6_37.sce new file mode 100755 index 000000000..5cc863aab --- /dev/null +++ b/2417/CH6/EX6.37/Ex6_37.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.37\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.37 (page no. 288)
+// Solution
+
+T=200+460; //Fahrenheit temperature converted to absolute temperature //unit:R
+V=1500; //ft/s //the local velocity
+Va=49.0*sqrt(T); //velocity of sound air at 200 F //unit:ft/s
+printf("The velocity of sound air at 200 F is %f ft/s\n",Va);
+M=V/Va; //The Mach number=the local velocity/velocity of sound //unitless
+printf("The Mach number is %f\n",M);
diff --git a/2417/CH6/EX6.38/Ex6_38.sce b/2417/CH6/EX6.38/Ex6_38.sce new file mode 100755 index 000000000..002f2731d --- /dev/null +++ b/2417/CH6/EX6.38/Ex6_38.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.38\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.38 (page no. 290)
+// Solution
+
+//data given
+V=1000; //ft/s //the fluid velocity
+gc=32.17; //Unit:(LBm*ft)/(LBf*s^2) //gc is constant of proportionality
+J=778; //conversion factor
+h=1204.4; //Btu/lbm //enthalpy of saturated steam
+//h0-h=V^2/(2*gc*J)
+h0=h+((V^2)/(2*gc*J)); //Btu/lbm //h0=stagnation enthalpy
+printf("The total enthalpy is %f Btu/lbm\n",h0);
+//It will be noted for this problem that if the initial velocity had been 100ft/s,deltah would have been 0.2 Btu/lbm,and for most practical purpposes,the total properties and those of the flowing fluid would have been essentially the same.Thus,for low-velocity fluids,the difference in total and steam properties can be neglected.
diff --git a/2417/CH6/EX6.39/Ex6_39.sce b/2417/CH6/EX6.39/Ex6_39.sce new file mode 100755 index 000000000..bd3081df7 --- /dev/null +++ b/2417/CH6/EX6.39/Ex6_39.sce @@ -0,0 +1,46 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.39\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.39 (page no. 297)
+// Solution
+
+k=1.4; //the specific heats ratio //k=cp/cv
+M=1; //(table 6.5) //The Mach number=the local velocity/velocity of sound
+T0=800; //absolute temperature //unit:R
+gc=32.17; //Unit:(LBm*ft)/(LBf*s^2) //gc is constant of proportionality
+R=53.35; //gas constant //ft*lbf/lbm*R
+p0=300; //psia //pressure
+
+// * or "star" subscripts to conditions in which M=1;
+// "0" subscript refers to isentropic stagnation
+//Refer to figure 6.26,
+//Tstar/T0=0.8333
+Tstar=T0*0.8333; //temperature when M=1 //unit:R
+printf("If the mach number at the outlet is unity,temperature is %f R\n",Tstar);
+Vat=sqrt(gc*R*Tstar*k); //ft/s //Vat=V2 //local velocity of sound
+printf("If the mach number at the outlet is unity,velocity is %f ft/s\n\n",Vat)
+
+//For A/Astar=2.035
+//The table yields
+M1=0.3; //mach number at inlet
+printf("At inlet,The mach number is %f\n",M1)
+//pstar/p0=0.52828
+pstar=p0*0.52828; //pressure when M=1 //psia
+//also,
+//T1/T0=0.98232 and p1/p0=0.93947
+//Therefore,
+T1=T0*0.982332; //unit:R //T1=temperature at inlet
+printf("At inlet,The temperature is %f R\n",T1);
+p1=p0*0.93947; //psia //p1=pressure at inlet
+printf("At inlet,The pressure is %f psia\n",p1);
+//From the inlet conditions derived,
+Va1=sqrt(gc*k*R*T1); //ft/s //V1=velocity at inlet
+V1=M1*Va1; //ft/s //velocity
+printf("At inlet,The velocity is %f ft/s\n",V1);
+//The specific volume at inlet is found from the equation of state for an ideal gas:
+v=(R*T1)/(p1*144); //ft^3/lbm //1 ft^2=144 in^2(for conversion of unit) //specific volume
+rho=inv(v); //inverse of specific volume //density
+A=2.035; //area //ft^2
+m=rho*A*V1; //mass flow //unit:lbm/s
+printf("At inlet,The mass flow is %f lbm/s\n",m);
diff --git a/2417/CH6/EX6.4/Ex6_4.sce b/2417/CH6/EX6.4/Ex6_4.sce new file mode 100755 index 000000000..ce7f1f880 --- /dev/null +++ b/2417/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.4\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.4 (page no. 242)
+// Solution
+
+//If for this process T2=1.25*T1,
+// T2/T1 = 1.25
+//Therefore,
+// p2/p1 = T2/T1 //Charles's law(volume constant)
+//Thus,
+printf("The absolute gas pressure increases by 25 percent\n");
diff --git a/2417/CH6/EX6.40/Ex6_40.sce b/2417/CH6/EX6.40/Ex6_40.sce new file mode 100755 index 000000000..d16672d41 --- /dev/null +++ b/2417/CH6/EX6.40/Ex6_40.sce @@ -0,0 +1,59 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.40\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.40 (page no. 299)
+// Solution
+
+// * or "star" subscripts to conditions in which M=1;
+// "0" subscript refers to isentropic stagnation
+//This problem will be solved by two methods(A and B)
+printf("Method A\n"); //By equations:
+k=1.4; //the specific heat ratio //k=cp/cv
+R=53.3; //gas constant //ft*lbf/lbm*R
+M=2.5; //mach number=the local velocity/velocity of sound
+printf("Solution for (a)\n");
+// T/Tstar = (k+1)/(2*(1+((1/2)*(k-1)*M^2)))
+// Tstar/T0=2/(k+1)
+//Therefore,
+// (Tstar/T0)*(T/Tstar) = (T/T0)=1/(1+((1/2)*(k-1)*M^2))
+T0=560; //absolute temperature or stagnation temperature //unit:R
+T=T0/(1+((1/2)*(k-1)*M^2)); //temperature at M=2.5
+printf("The temperature is %f R\n\n",T);
+printf("Solution for (b)\n");
+p=0.5; //static pressure //unit:psia
+// p0/p = (T0/T)^(k/(k-1))
+p0=p*14.7*((T0/T)^(k/(k-1))); //pressure at M=2.5 //unit:psia
+printf("The pressure is %f psia\n\n",p0);
+printf("Solution for (c)\n");
+gc=32.17; //Unit:(LBm*ft)/(LBf*s^2) //gc is constant of proportionality
+Va=sqrt(gc*k*R*T); //ft/s //local velocity of sound
+V=M*Va; //valocity at M=2.5 //unit:ft/s
+printf("The velocity is %f ft/s\n\n",V);
+printf("Solution for (d)\n");
+v=(R*T)/(p*14.7*144); //ft^3/lbm //1 ft^2=144 in^2 //specific volume at M=2.5
+printf("The specific volume is %f ft^3/lbm\n\n",v);
+printf("Solution for (e)\n");
+//Mass velocity is definrd as the mass flow per unit area
+// m/A=(A*V)/(v*A)=V/v
+printf("The mass velocity is %f lbm/(s*ft^2)\n\n\n",V/v); //mass velocity at M=2.5
+
+
+printf("Method B\n"); //By the gas tables: //table 6.5 gives
+M=2.5; //mach number=the local velocity/velocity of sound
+printf("Solution for (a)\n");
+T0=560; //absolute temperature or stagnation temperature
+//T/T0=0.44444
+T=T0*0.44444; //temperature at M=2.5
+printf("The temperature is %f R\n\n",T)
+printf("Solution for (b)\n");
+p=0.5; //static pressure
+//p/p0=0.05853
+p0=(p*14.7)/0.05853; //pressure at M=2.5
+printf("The pressure is %f psia\n\n",p0);
+printf("Solution for (c)\n");
+printf("As before %f ft/s\n\n",V)
+printf("Solution for (d)\n");
+printf("As before %f ft^3/lbm\n\n",v)
+printf("Solution for (e)\n");
+printf("As before %f lbm/(s*ft^1)\n",V/v)
diff --git a/2417/CH6/EX6.41/Ex6_41.sce b/2417/CH6/EX6.41/Ex6_41.sce new file mode 100755 index 000000000..0b89e2b0a --- /dev/null +++ b/2417/CH6/EX6.41/Ex6_41.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.41\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.41 (page no. 304)
+// Solution
+
+//For Methane(CH4,MW=16)
+p=500; //evaluate specific volume at p pressure //Unit:psia
+pc=674; //critical temperature //Unit:psia
+T=50+460; //evaluate specific volume at T temperature //Unit:R
+Tc=343; //critical temperature //Unit:R
+R=1545/16; //gas constant R = 1545/Molecular Weight //ft*lbf/lbm*R
+pr=p/pc; //reduced pressure //unit:psia
+Tr=T/Tc; //reduced temperature //unit:R
+//Reading figure 6.28 at these values gives
+Z=0.93; //compressibility factor
+//Z=(p*v)/(R*T)
+v=Z*((R*T)/(p*144)); //ft^3/lbm //1 ft^2=144 in^2(for conversion of unit) //specific volume
+printf("Using the value of Z=0.93,the specific volume is %f ft^3/lbm\n",v);
+//For ideal gas,
+v=(R*T)/(p*144); //ft^3/lbm //1 ft^2=144 in^2(for conversion of unit) //specific volume
+printf("For the ideal gas,the specific volume is %f ft^3/lbm\n",v);
diff --git a/2417/CH6/EX6.5/Ex6_5.sce b/2417/CH6/EX6.5/Ex6_5.sce new file mode 100755 index 000000000..f0fb4bc81 --- /dev/null +++ b/2417/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,12 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.5\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.5 (page no. 242)
+// Solution
+
+V1=4; //m^3 //initial volume
+T2=0+273; //celsius converted to kelvin //gas is cooled to 0 C //final temperature
+T1=100+273; //celsius converted to kelvin //initial temperature
+V2=V1*(T2/T1); //final volume //Charles's law(pressure constant) //unit:m^3
+printf("The final volume is %f m^3",V2);
diff --git a/2417/CH6/EX6.6/Ex6_6.sce b/2417/CH6/EX6.6/Ex6_6.sce new file mode 100755 index 000000000..b6d4bdb0d --- /dev/null +++ b/2417/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.6\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.6 (page no. 245)
+// Solution
+
+//Let us first put each of the given variables into a consistent set of units:
+p=(200+14.7)*(144); //Unit:psfa*(lbf/ft^2) //1 ft^2=144 in^2 //pressure
+T=(460+73); //Fahrenheit temperature converted to absolute temperature //unit:R
+V=120/1728; //1 ft^3=1728 in^3 //total volume //unit:ft^3
+R=1545/28; //Unit:ft*lbf/lbm*R //because the molecular weight of nitrogen is 28 //constant of proportionality
+//Applying, p*v=R*T, //ideal gas law
+v=(R*T)/p; //Unit:ft^3/lbm //specific volume
+printf("The specific volume is %f ft^3/lbm\n",v);
+//The mass of gas is the total volume divided by the specific volume
+printf("The gas in the container is %f lbm\n",V/v);
+//The same result is obtained by direct use of eq. p*V=m*R*T
+m=(p*V)/(R*T); //The gas in the container //unit:lbm //ideal gas law
+printf("The gas in the container is %f lbm\n",m);
diff --git a/2417/CH6/EX6.7/Ex6_7.sce b/2417/CH6/EX6.7/Ex6_7.sce new file mode 100755 index 000000000..7f3e5358d --- /dev/null +++ b/2417/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,14 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.7\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.7 (page no. 245)
+// Solution
+
+//Applying , (p1*V1)/T1 = (p2*V2)/T2
+//and p2=p1*(T2/T1) because V1=V2
+p1=200+14.7; //Unit:psia //initial pressure
+T2=460+200; //final temperature is 200 F //Fahrenheit temperature converted to absolute temperature //unit:R
+T1=460+73; //Fahrenheit temperature converted to absolute temperature //unit:R
+p2=p1*(T2/T1); //final pressure //Unit:psia //Charles's law(volume constant)
+printf("The final pressure is %f psia",p2);
diff --git a/2417/CH6/EX6.8/Ex6_8.sce b/2417/CH6/EX6.8/Ex6_8.sce new file mode 100755 index 000000000..0ed5ce1c4 --- /dev/null +++ b/2417/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.8\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.8 (page no. 246)
+// Solution
+
+//For CO2,
+R=8.314/44; //Unit:kJ/kg*K //constant of proportionality //Molecular weight of CO2=44
+p=500; //Unit:kPa //pressure
+V=0.5; //Unit:m^3 //volume
+T=(100+273); //Unit:K //Celsius converted to kelvin
+//Applying p*V=m*R*T ,
+m=(p*V)/(R*T); //mass //kg //ideal gas law
+printf("The mass of gas in the tank is %f kg\n",m);
diff --git a/2417/CH6/EX6.9/Ex6_9.sce b/2417/CH6/EX6.9/Ex6_9.sce new file mode 100755 index 000000000..1c54aa4eb --- /dev/null +++ b/2417/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+printf("\t\t\tProblem Number 6.9\n\n\n");
+// Chapter 6: The Ideal Gas
+// Problem 6.9 (page no. 252)
+// Solution
+
+T2=500+460; //absolute final temperature //unit:R
+T1=80+460; //absolute initial temperature //unit:R
+//The equation cpbar= 0.338-(1.24*10^2/T)+(4.15*10^4)/T^2 has a form , cbar= Adash+(Bdash/T)+(Ddash/T^2)
+//So,
+Adash=0.338; //constant
+Bdash=-1.24*10^2; //constant
+Ddash=4.15*10^4; //constant
+//Therefore,from equation,cbar=Adash+((Bdash*log(T2/T1))/(T2-T1))+(Ddash/(T2*T1))
+cpbar=Adash+((Bdash*log(T2/T1))/(T2-T1))+(Ddash/(T2*T1)); //The mean specific heat //Btu/lbm*R
+printf("The mean specific heat at constant pressure between 80F and 500F is %f Btu/lbm*R\n",cpbar);
diff --git a/2417/CH7/EX7.1/Ex7_1.sce b/2417/CH7/EX7.1/Ex7_1.sce new file mode 100755 index 000000000..009180163 --- /dev/null +++ b/2417/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,36 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.1\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.1 (page no. 322)
+// Solution
+
+//As the basis of the calculation,assume that we have 1 lbm of mixture.Also,take the molecular weight of oxygen to be 32.00 and nitrogen to be 28.02.(from table7.1)
+printf("Solution for (a)\n");
+nO2=0.2315/32; //no of moles of oxygen=ratio of mass and molecular weight //0.2315 lb of oxygen per pound
+printf("The moles of oxygen is %f mole/lbm of mixture\n",nO2);
+nN2=0.7685/28.02; //no of moles of nitrogen=ratio of mass and molecular weight //0.7685 lb of nitrogen per pound
+printf("The moles of nitrogen is %f mole/lbm of mixture\n",nN2);
+nm=nO2+nN2; //Unit:Mole/lbm //number of moles of gas mixture is sum of the moles of its constituent gases
+printf("The total number of moles is %f mole/lbm\n",nm);
+xO2=nO2/nm; //mole fraction of oxygen=ratio of no of moles of oxygen and total moles in mixture
+xN2=nN2/nm; //mole fraction of nitrogen=ratio of no of moles of oxygen and total moles in mixture
+printf("The mole fraction of oxygen is %f and the mole fraction of nitrogen is %f\n",xO2,xN2);
+//(Check:xO2+xN2=1)
+printf("xO2+xN2=%f\n\n",xO2+xN2);
+
+printf("Solution for (b)\n");
+// the air is at 14.7 psia
+pO2=xO2*14.7; //the partial pressure of oxygen=pressure of air * the mole fraction of oxygen //psia
+printf("The partial pressure of oxygen is %f psia\n",pO2);
+pN2=xN2*14.7; //the partial pressure of nitrogen=pressure of air * the mole fraction of nitrogen //psia
+printf("The partial pressure of nitrogen is %f psia\n\n",pN2);
+
+printf("Solution for (c)\n");
+MWm=(xO2*32) + (xN2*28.02); //the molecular weight of air=sum of products of mole fraction of each gas component
+printf("The molecular weight of air is %f\n\n",MWm);
+
+printf("Solution for (d)\n");
+Rm=1545/MWm; //the gas constant of air
+printf("The gas constant of air is %f\n\n",Rm);
diff --git a/2417/CH7/EX7.10/Ex7_10.sce b/2417/CH7/EX7.10/Ex7_10.sce new file mode 100755 index 000000000..788142d22 --- /dev/null +++ b/2417/CH7/EX7.10/Ex7_10.sce @@ -0,0 +1,20 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.10\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.10 (page no. 332)
+// Solution
+
+//Problem 7.9 is carried out as a nonflow mixing process.
+//Given in problem 7.9,: cp of oxygen is 0.23 Btu/lbm*R.cp of nitrogen is 0.25 Btu/lbm*R. 160 lbm/hr of oxygen and 196 lbm/hr of nitrogen are mixed.oxygen is at 500F and nitrogen is at 200 F. //cp=specific heat at constant pressure
+//Given in problem 7.10,: cv of oxygen is 0.164 Btu/lbm*R.cv of nitrogen is 0.178 Btu/lbm*R. //cv=specific heat at constant volume
+
+//Because this is a nonflow process,the energy equation for this process requires the internal energy of the mixture to equal to the sum of the internal energy of its components.
+//Alternatively,the decrease in internal energy of the oxygen must equal the increase in internal energy of the nitrogen.Using latter statement gives us,
+// (160*0.164*(500-tm)) = (196*0.178*(tm-200))
+//where m*cv*deltat has been used for deltau. //Unit for cp & cv is Btu/lbm*R
+//rearranging the above equation,
+tm=((500*160*0.164)+(196*0.178*200))/((196*0.178)+(160*0.164)); //tm=mixture temperature //Unit:fahrenheit
+printf("The final temperature of the mixture is %f F\n",tm);
+
diff --git a/2417/CH7/EX7.12/Ex7_12.sce b/2417/CH7/EX7.12/Ex7_12.sce new file mode 100755 index 000000000..9e6906994 --- /dev/null +++ b/2417/CH7/EX7.12/Ex7_12.sce @@ -0,0 +1,69 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.12\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.12 (page no. 334)
+// Solution
+
+//The change in entropy of the mixture is the sum of the changes in entropy of each component.
+//Given in problem 7.9,: cp of oxygen is 0.23 Btu/lbm*R.cp of nitrogen is 0.25 Btu/lbm*R. 160 lbm/hr of oxygen and 196 lbm/hr of nitrogen are mixed.oxygen is at 500F and nitrogen is at 200 F. //cp=specific heat at constant pressure
+//In 7.9,for the oxygen,the temperature starts at 500F(960 R) and decreases to 328.7 F.For the nitrogen,the temperature starts at 200F(660 R) and increase to 328.7 F.
+//deltas = (cp*log(T2/T1)); //Unit:Btu/lbm*R //change in entropy
+
+//For the oxygen,
+cp=0.23; //specific heat at constant pressure //Unit:Btu/lbm*R
+T2=328.7+460; //Unit:R //final temperature
+T1=500+460; //Unit:R //starting temperature
+deltas=(cp*log(T2/T1)); //Unit:Btu/lbm*R //change in entropy for oxygen
+DeltaS=160*deltas; //Btu/R //The total change in entropy of the oxygen
+printf("The total change in entropy of the oxygen is %f Btu/R\n",DeltaS);
+
+//For the nitrogen,
+cp=0.25; //specific heat at constant pressure //Unit:Btu/lbm*R
+T2=328.7+460; //Unit:R //final temperature
+T1=200+460; //Unit:R //starting temperature
+deltas=(cp*log(T2/T1)); //Unit:Btu/lbm*R //change in entropy for nitrogen
+deltaS=196*deltas; //Btu/R //The total change in entropy of the nitrogen
+printf("The total change in entropy of the nitrogen is %f Btu/R\n",deltaS);
+deltaS=deltaS+DeltaS; //the total change in entropy for the mixture //Btu/lbm*R
+printf("The total change in entropy for the mixture is %f Btu/R\n",deltaS);
+
+//Per pound of mixture,
+deltasm=deltaS/(196+160); //increase in entropy per pound mass of mixture
+printf("Increase in entropy per pound mass of mixture is %f Btu/lbm*R\n\n",deltasm);
+
+
+printf("An alternative solution:\n");
+//As an alternative solution,assume an arbitrary datum of 0 F(460 R).
+cp=0.23; //specific heat at constant pressure //Unit:Btu/lbm*R
+//For initial entropy of oxygen,
+T2=500+460; //Unit:R //final temperature
+T1=0+460; //Unit:R //starting temperature
+deltas=cp*log(T2/T1); //the initial change in entropy for oxygen // Btu/lbm*R
+printf("The initial change in entropy for oxygen is %f Btu/lbm*R\n",deltas);
+//For final entropy of oxygen,
+T2=328.7+460; //Unit:R //final temperature
+T1=0+460; //Unit:R //starting temperature
+Deltas=cp*log(T2/T1); //the final change in entropy for oxygen // Btu/lbm*R
+printf("The final change in entropy for oxygen is %f Btu/lbm*R\n",Deltas);
+deltaS=Deltas-deltas; //The entropy change of the oxygen //Btu/lbm*R
+printf("The entropy change of the oxygen is %f Btu/lbm*R\n",deltaS);
+
+//For nitrogen,
+cp=0.25; //specific heat at constant pressure //Unit:Btu/lbm*R
+//For initial entropy of nitrogen,
+T2=200+460; //Unit:R //final temperature
+T1=0+460; //Unit:R //starting temperature
+deltas=cp*log(T2/T1); //the initial change in entropy for nitrogen // Btu/lbm*R
+printf("The initial change in entropy for nitrogen is %f Btu/lbm*R\n",deltas);
+//For final entropy of nitrogen,
+T2=328.7+460; //Unit:R //final temperature
+T1=0+460; //Unit:R //starting temperature
+Deltas=cp*log(T2/T1); //the final change in entropy for nitrogen // Btu/lbm*R
+printf("The final change in entropy for nitrogen is %f Btu/lbm*R\n",Deltas);
+deltaS=Deltas-deltas; //The entropy change of the nitrogen //Btu/lbm*R
+printf("The entropy change of the nitrogen is %f Btu/lbm*R\n",deltaS);
+
+//The remainder of the problem is as before.The advantage of using this alternative method is the negative logarithms are avoided by choosing a reference temperature lower than any other temperature in the system
+
diff --git a/2417/CH7/EX7.13/Ex7_13.sce b/2417/CH7/EX7.13/Ex7_13.sce new file mode 100755 index 000000000..b284677cc --- /dev/null +++ b/2417/CH7/EX7.13/Ex7_13.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.13\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.13 (page no. 338)
+// Solution
+
+//Referring to figure 7.6,it will be seen that the cooling of an air-water vapor mixture from B to A proceeds at constant pressure until the saturation curve is reached.
+//At 80 F(the mixture temperature),the Steam Tables give us a saturation pressure of a 0.5073 psia,and because the relative humidity is 50%,the vapor pressure of the water is 0.5*0.5073=0.2537 psia.
+//Using the steam tables,the saturation temperature corresponding to 0.2537 psia is 60 F.
+//So,
+printf("The dew point temperature of the air is 60 F\n")
diff --git a/2417/CH7/EX7.14/Ex7_14.sce b/2417/CH7/EX7.14/Ex7_14.sce new file mode 100755 index 000000000..da8c561dc --- /dev/null +++ b/2417/CH7/EX7.14/Ex7_14.sce @@ -0,0 +1,31 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.14\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.14 (page no. 338)
+// Solution
+
+//To solve this probelm,it is necessary to determine the properties of the saturated mixture 90 F.If the air is saturated at 90 F,the partial pressure of the water vapor is found directly from the Steam Tables as 0.6988 psia,and the specific volume of the water vapor is 467.7 ft^3/lbm of vapor.
+printf("The partial pressure of the dry air is %f psia\n",14.7-0.6988); //the mixture is at 14.7 psia
+R=1545/28.966; //gas constant of dry air=1545/Molecular weight
+T=90+460; //temperature of dry air //Unit:R
+pdryair=14.0; //psia //pressure of dry air
+//Applying the ideal gas equation to the air,
+vdryair=(R*T)/(pdryair*144); //volume of dry air //ft^3/lbm //1 in^2=144 Ft^2
+//the mass of dry air in the 467.7 ft^3 container
+printf("The mass of dry air in the 467.7 ft^3 container is %f lbm\n",467.7/vdryair);
+//To obtain relative humidity(phy),it is necessary to determine the mole fraction of water vapor for both the saturated mixture and the mixture in question.
+//The saturated mixture contains 1 lbm of water vapor or 1/18.016 moles =0.055 mole of water vapor and (467.7/vdryair)/28.966=1.109 moles of dry air.
+//For the saturated mixture, the ratio of moles of water vapor to moles of mixture is 0.055/(0.055+1.109)=0.0477
+//For the actual mixture,the moles of water vapor per pound of dry air is 0.005/18.016=0.000278 and 1 lbm of dry air is 1/28.966=0.0345 mole.So,the mole of water vapor per mole of mixture at the conditions of the mixture is 0.000278/(0.0345+0.000278)=0.00799
+//From the defination of relative humidity,
+printf("The relative humidity of the mixture is %f \n",(0.00799/0.0477)*100);
+
+//Because the mole ratio is also the ratio of the partial pressures for the ideal gas,phy can be expressed as the ratio of the partial pressure of the water vapor in the mixture to the partial pressure of the water vapor at saturation.Therefore,
+printf("The partial pressure of the vapor at saturation is %f psia\n",(0.00799/0.0477)*0.6988);
+printf("And the partial pressure of the dry air in the mixture is %f psia\n",14.7-((0.00799/0.0477)*0.6988)); //14.7-The partial pressure of the vapor at saturation
+//The dew point temperature is the saturation temperature corresponding to the partial pressure of the water vapor in the mixture.So,
+printf("The dew point temperature corresponding to %f psia is 39F\n",(0.00799/0.0477)*0.6988);
+
+
diff --git a/2417/CH7/EX7.15/Ex7_15.sce b/2417/CH7/EX7.15/Ex7_15.sce new file mode 100755 index 000000000..2a3bdef18 --- /dev/null +++ b/2417/CH7/EX7.15/Ex7_15.sce @@ -0,0 +1,26 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.15\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.15 (page no. 343)
+// Solution
+
+//Problem 7.14 using equations, Rm=((ma/(ma+mv))*Ra)+((mv/(ma+mv))*Rv) and phy*pvs=pv
+W=0.005; //Humidity ratio
+pm=14.7; //mixture is at 14.7 psia
+//W=0.622*(pv/(pm-pv))
+//Rearranging,
+pv=(W*pm)/(0.622+W); //the partial pressure of the water vapor
+printf("The partial pressure of the water vapor is %f psia\n",pv);
+pa=pm-pv; //pa=the partial pressure of the dry air in the mixture
+printf("The partial pressure of dry air is %f psia\n",pa);
+//It is necessary to obtain pvs from the Steam Tables at 90 F.This is 0.6988 psia.
+pvs=0.6988; //saturation pressure of water vapor at the temperature of mixture
+printf("The partial pressure of the water vapor at saturation is %f psia\n",pvs);
+//Therefore,
+phy=pv/pvs; //relative humidity
+printf("The relative humidity is %f percent\n",phy*100);
+//The dew point temperature is the saturation temperature corresponding to 0.117 psia,which is found from the Steam Tables to be 39 F.
+printf("The dew point temperature of the mixture is 39 F\n");
+//The results of this problem and problem 7.14 are in good agreement
diff --git a/2417/CH7/EX7.16/Ex7_16.sce b/2417/CH7/EX7.16/Ex7_16.sce new file mode 100755 index 000000000..9a8bab28b --- /dev/null +++ b/2417/CH7/EX7.16/Ex7_16.sce @@ -0,0 +1,25 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.16\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.16 (page no. 343)
+// Solution
+
+pm=14.7; //the barometer is at 14.7 psia //mixture is at 14.7 psia
+//The amount of water vapor removed (per pound of dry air) is the difference between the humidity ratio (specific humidity) at inlet and outlet of the conditioning unit.We shall therefore evalute W for both specified conditions.Because phy=pv/pvs,
+//At 90F:
+phy=0.7; //relative humidity
+pvs=0.6988; //psia //saturation pressure of water vapor at the temperature of mixture
+pv=phy*pvs; //psia //the partial pressure of the water vapor
+pa=pm-pv; //psia //pa=the partial pressure of the dry air in the mixture
+W=0.622*(pv/pa); //Humidity ratio
+
+//At 80F:
+phy=0.4; //relative humidity
+pvs=0.5073; //psia //saturation pressure of water vapor at the temperature of mixture
+pv=phy*pvs; //psia //the partial pressure of the water vapor
+pa=pm-pv; //psia //pa=the partial pressure of the dry air in the mixture
+w=0.622*(pv/pa); //Humidity ratio
+
+printf("The amount of water removed per pound of dry air is %f\n",W-w);
diff --git a/2417/CH7/EX7.17/Ex7_17.sce b/2417/CH7/EX7.17/Ex7_17.sce new file mode 100755 index 000000000..0c40b079e --- /dev/null +++ b/2417/CH7/EX7.17/Ex7_17.sce @@ -0,0 +1,11 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.17\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.17 (page no. 347)
+// Solution
+
+//Problem 7.13 using the psychrometric chart
+//Entering figure 7.11 at a dry-bulb temperature of 80 F,we proceed vertically until we reach 50% humidity curve.At this intersection,we proceed horizontally and read the dew-point temperature as approximately 60 F.
+printf("The dew point temperature of air is 60 F\n");
diff --git a/2417/CH7/EX7.18/Ex7_18.sce b/2417/CH7/EX7.18/Ex7_18.sce new file mode 100755 index 000000000..7b3b6804a --- /dev/null +++ b/2417/CH7/EX7.18/Ex7_18.sce @@ -0,0 +1,17 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.18\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.18 (page no. 347)
+// Solution
+
+//Problem 7.14 using the psychrometric chart
+//In this problem,we are given the moisture content of the air to be 0.005 lb per pound of dry air.
+//This corresponds to 0.005*7000=35 grains per pound of dry air.
+//Entering the chart at 90F and proceeding verticaly to 35 grains per pound of dry air,we find the dew point to be 39F by proceeding horizontally to the intersection with the saturation curve.
+printf("The dew-point temperature of the mixture is 39 F\n");
+printf("The relative humidity is approximately 17 percent\n");
+//From the leftmost scale,we read the pressure of water vapor to be 0.12 psia.
+printf("The partial pressure of the air is %f psia\n",14.7-0.12);
+//Comparing these results to problem 7.14,indicated good agreement between the results obtained by chart and by calculation
diff --git a/2417/CH7/EX7.19/Ex7_19.sce b/2417/CH7/EX7.19/Ex7_19.sce new file mode 100755 index 000000000..2d6e825ed --- /dev/null +++ b/2417/CH7/EX7.19/Ex7_19.sce @@ -0,0 +1,14 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.19\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.19 (page no. 348)
+// Solution
+
+//Problem 7.16 using the psychrometric chart
+//The initial conditions are 90 F and 70% relative humidity
+//Entering the chart at 90 F dry bulb temperature and proceeding vertically to 70% relative humidity,we find the air to have 150 grains water vapor per pound of dry air.At the final condition of 80F and 40% relative humidity,we read 61 grains of water/lb of dry air.
+//So,
+printf("The water removed is %f grains per pound of dry air\n",150-61);
+printf("Or %f lb of water per pound of dry air is removed\n",(150-61)/7000);
diff --git a/2417/CH7/EX7.2/Ex7_2.sce b/2417/CH7/EX7.2/Ex7_2.sce new file mode 100755 index 000000000..fb58e89ee --- /dev/null +++ b/2417/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,20 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.2\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.2 (page no. 323)
+// Solution
+
+//For Gaseous Freon-12 (CCl2F2)
+//MW of air=29 & MW of freon-12=120.9
+//initial pressure in tank is atmospheric pressure that is 14.7 psia
+//final pressure of tank is 1000 psia
+//The partial pressure of the Freon-12 is 1000-14.7
+printf("The partial pressure of the Freon-12 is %f\n",1000-14.7)
+//the mole fraction of air=the initial pressure / final pressure
+printf("The mole fraction of air is %f\n",14.7/1000)
+//the mole fraction of freon=the partial pressure of freon / the final pressure
+printf("The mole fraction of Freon-12 is %f\n",(1000-14.7)/1000)
+MWm=((14.7/1000)*29) + (((1000-14.7)/1000)*120.9);//the molecular weight of mixture=sum of products of mole fraction of each gas component
+printf("The molecular weight of the mixture is %f",MWm);
diff --git a/2417/CH7/EX7.20/Ex7_20.sce b/2417/CH7/EX7.20/Ex7_20.sce new file mode 100755 index 000000000..47d8dfbf1 --- /dev/null +++ b/2417/CH7/EX7.20/Ex7_20.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.22\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.22 (page no. 349)
+// Solution
+
+//dry bulb temperature is 50 F
+//relative humidity is 50 percent
+//We first locate 50 F and 50 percent relative humidity on figure 7.11.At this state,we read 26 grains of water per pound of dry air and a total heat of 16.1 Btu per pound of a dry air.
+//We now proceed horizontally to 80 F at a constant value of 26 grains of water per pound of dry air and read a total heat of 23.4 Btu per pound of dry air.
+printf("The heat required is %f Btu per pound of dry air",23.4-16.1)
diff --git a/2417/CH7/EX7.21/Ex7_21.sce b/2417/CH7/EX7.21/Ex7_21.sce new file mode 100755 index 000000000..597f1e36e --- /dev/null +++ b/2417/CH7/EX7.21/Ex7_21.sce @@ -0,0 +1,12 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.21\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.21 (page no. 352)
+// Solution
+
+//An evaporative cooling process
+//Because the exit air is saturated,we find the exit condition on the curve corresponding to a wet-bulb temperature of 50 F.The process is carried out at constant total enthalpy,which is along a line of constant wet-bulb temperature.
+//Proceeding along the 50 F wet-bulb temperature line of figure 7.11 diagonally to the right until it intersects with the vertical 80 F dry-bulb temperature line yields a relative humidity of approximately 4 %
+printf("For An evaporative cooling process,The relative humidity of the entering air is 4 percent");
diff --git a/2417/CH7/EX7.22/Ex7_22.sce b/2417/CH7/EX7.22/Ex7_22.sce new file mode 100755 index 000000000..ec92e822a --- /dev/null +++ b/2417/CH7/EX7.22/Ex7_22.sce @@ -0,0 +1,11 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.22\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.22 (page no. 356)
+// Solution
+
+//As noted from figure 7.27, 1 lb of mixture,4/5 lb of indoor air,and 1/5 lb of outdoor air are mixed per pound of mixture.
+//We now locate the two end states on the psychrometric chart and connect them with a straight line.The line connecting the end states is divided into 5 equal parts. Using the results of equation, (ha-ha2)/(ha-ha1) = (W2-W)/(W-W1) = ma1/ma2 = l1/l2 ,we now proceed from the 75 F indoor air state 1 part toward the 90F outdoor air state.This Locates
+printf("The final mixture,which is found to be a dry-bulb temperature of approximately 78 F,a wet-bulb temperature of 66 F and relative humidity of 54 percent\n");
diff --git a/2417/CH7/EX7.23/Ex7_23.sce b/2417/CH7/EX7.23/Ex7_23.sce new file mode 100755 index 000000000..39cf6ce8a --- /dev/null +++ b/2417/CH7/EX7.23/Ex7_23.sce @@ -0,0 +1,32 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.23\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.23 (page no. 358)
+// Solution
+
+//The cooling tower
+//From the Steam tables,
+//For water:
+h100F=68.05; //Btu/lbm //enthalpy at 100 F
+h70F=38.09; //Btu/lbm //enthalpy at 70 F
+//For air:
+h=20.4; //Unit:Btu/lb //at inlet,total heat/lb dry air
+w=38.2; //Unit:grains/lb //at inlet,moisture pickup/lb dry air (at 60F D.B. and 50% R.H.)
+H=52.1; //Unit:Btu/lb //at outlet,total heat/lb dry air
+W=194.0; //Unit:grains/lb //at outlet,moisture pickup/lb dry air (at 90F D.B. and 90% R.H.)
+
+//Per pound of dry air,the heat interchange is H-h Btu per pound of dry air.
+//Per pound of dry air,the moisture increase is (W-w)/7000 lb per pound of dry air.
+//From the equation, ma*(H-h) = 200000*h100F - mwout*h70F //ma=mass of air mwout=mass of cooled water
+//and ma*((W-w)/7000) = 200000 - mwout
+//Solving the latter equation for mwout,we have mwout=200000-(ma*((W-w)/7000))
+//Substituting this into the heat balance yields,
+// ma*(H-h) = 200000*h100F - 200000*h70F + ma*h70F*((W-w)/7000)
+//Solving gives us,
+ma=(200000*(h100F-h70F))/((H-h)-(h70F*((W-w)/7000))); //The amount of air required per hour //Unit:lbm/hr of dry air
+printf("The amount of air required per hour is %f lbm/hr of dry air\n",ma);
+printf("The amount of water lost per hour due to evaporation is %f lbm/hr\n",ma*((W-w)/7000));
+//note that the water evaporated is slightly over 2% of the incoming water,and this is the makeup that has to be furnished to the tower.
+//answer are slightly differ because of value of (W-w)/7000 is given 0.0233 instead of 0.0225
diff --git a/2417/CH7/EX7.3/Ex7_3.sce b/2417/CH7/EX7.3/Ex7_3.sce new file mode 100755 index 000000000..e5b5b56e4 --- /dev/null +++ b/2417/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,39 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.3\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.3 (page no. 323)
+// Solution
+
+//Ten pounds of air,1 lb of carbon dioxide,and 5 lb of nitrogen are mixed at constant temperature until the mixture pressure is constant
+nair=10/29; //no of moles of air=ratio of mass and molecular weight //10 lb of nitrogen per pound //molecular weight of air=29
+printf("The moles of air is %f mole/lbm of mixture\n",nair);
+nCO2=1/44; //no of moles of carbon dioxide=ratio of mass and molecular weight //1 lb of per pound //molecular weight of CO2=44
+printf("The moles of carbon dioxide is %f mole/lbm of mixture\n",nCO2);
+nN2=5/28; //no of moles of nitrogen=ratio of mass and molecular weight //5 lb of nitrogen per pound //molecular weight of N2=28
+printf("The moles of nitrogen is %f mole/lbm of mixture\n",nN2);
+nm=nair+nCO2+nN2; //Unit:Mole/lbm //number of moles of gas mixture is sum of the moles of its constituent gases
+printf("The total number of moles is %f mole/lbm\n\n",nm);
+
+xair=nair/nm //mole fraction of air=ratio of no of moles of air and total moles in mixture
+xCO2=nCO2/nm; //mole fraction of carbon dioxide=ratio of no of moles of carbon dioxide and total moles in mixture
+xN2=nN2/nm; //mole fraction of nitrogen=ratio of no of moles of oxygen and total moles in mixture
+printf("The mole fraction of air is %f \n",xair);
+printf("The mole fraction of carbon dioxide is %f\n",xCO2)
+printf("The mole fraction of nitrogen is %f\n\n",xN2);
+
+//final pressure of is 100 psia
+pair=xair*100; //the partial pressure of air= final pressure * the mole fraction of air //psia
+printf("The partial pressure of air is %f psia\n",pair);
+pCO2=xCO2*100; //the partial pressure of carbon dioxide= final pressure * the mole fraction of CO2 //psia
+printf("The partial pressure of carbon dioxide is %f psia\n",pCO2);
+pN2=xN2*100; //the partial pressure of nitrogen=final pressure * the mole fraction of nitrogen //psia
+printf("The partial pressure of nitrogen is %f psia\n\n",pN2);
+
+//the molecular weight of mixture=sum of products of mole fraction of each gas component
+MWm=(xair*29) + (xCO2*44) + (xN2*28); //The molecular weight of air
+printf("The molecular weight of air is %f\n\n",MWm);
+
+Rm=1545/MWm; //the gas constant of air
+printf("The gas constant of air is %f\n\n",Rm);
diff --git a/2417/CH7/EX7.4/Ex7_4.sce b/2417/CH7/EX7.4/Ex7_4.sce new file mode 100755 index 000000000..43a56a10a --- /dev/null +++ b/2417/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,52 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.4\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.4 (page no. 325)
+// Solution
+
+//five moles of oxygen and 10 moles of hydrogen are mixed
+//The total number of moles is 10+5=15.Therefore,mole fraction of each constituent is
+xO2=5/15; //The mole fraction of oxygen
+xH2=10/15; //The mole fraction of hydrogen
+printf("The mole fraction of oxygen is %f and of hydrogen is %f\n",xO2,xH2);
+//the molecular weight of mixture=sum of products of mole fraction of each gas component(MW of O2=32 and MW of H2=2.016)
+printf("The molecular weight of the final mixture is %f\n",((5/15)*32)+((10/15)*2.016));
+R=1545/32; //the gas constant of oxygen
+T=460+70; //absolute temperature //Unit:R
+p=14.7; //pressure //psia
+//The partial volume of the oxygen can be found as follows:per pound of oxygen,
+//p*vO2=R*T;
+vO2=(R*T)/(p*144); //ft^3/lbm //1 in^2=144 ft^2
+//Because there are 5 moles of oxygen,each containing 32 lbm,
+VO2=vO2*5*32; //ft^3 //partial volume of oxygen
+printf("The partial volume of oxygen is %f ft^3\n",VO2);
+//For the hydrogen,we can simplify the procedure by noting that the fraction of the total volume occupied by the oxygen is the same as its mole fraction.Therefore,
+Vm=3*VO2; //total volume occupied //ft^3
+printf("The mixture volume is %f ft^3\n",Vm);
+//and the hydrogen volume
+VH2=Vm-VO2; //Ft^2 //partial volume of hydrogen
+printf("From simplified procedure,The partial volume of hydrogen is %f ft^3\n",VH2);
+
+//We could obtain the partial volume of hydrogen by proceeding as we did for the oxygen.Thus,
+//p*vH2=R*T;
+R=1545/2.016; //the gas constant of hydrogen
+vH2=(R*T)/(p*144); //ft^3/lbm //1 in^2=144 ft^2
+//Because there are 10 moles of hydrogen,each containing 2.016 lbm,
+VH2=vH2*10*2.016; //ft^3 //partial volume of hydrogen
+printf("The partial volume of hydrogen is %f ft^3\n\n",VH2);
+//Which checks our previous values.
+
+
+printf("From another method,\n");
+//As an alternative to the foregoing,we could also use the fact that at 14.7 psia and 32F a mole of any gas occupies a volume of 358 ft^3.
+printf("At 70F and 14.7 psia,a mole occupies %f ft^3\n",358*((460+70)/(460+32)));
+//Therefore, 5 moles of oxygen occupies
+VO2=5*358*((460+70)/(460+32)); //The partial volume of oxygen //ft^3
+printf("The partial volume of oxygen is %f ft^3\n",VO2);
+//and 10 moles of hydrogen occupies
+VH2=10*358*((460+70)/(460+32)); //The partial volume of hydrogen //ft^3
+printf("The partial volume of hydrogen is %f ft^3\n",VH2);
+//Both values are in good agreement with the previous calculations.
+
diff --git a/2417/CH7/EX7.5/Ex7_5.sce b/2417/CH7/EX7.5/Ex7_5.sce new file mode 100755 index 000000000..ea7380015 --- /dev/null +++ b/2417/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,40 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.5\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.5 (page no. 326)
+// Solution
+
+//Referring to figure 7.3,we have for CO2,
+nCO2=10/44; //mole //no of moles of carbon dioxide=ratio of mass and molecular weight //10 lb of per pound //molecular weight of CO2=44
+//and for N2,
+nN2=5/28.02; //mole //no of moles of nitrogen=ratio of mass and molecular weight //5 lb of nitrogen per pound
+printf("The total number of moles in the mixture is %f mole\n",nCO2+nN2);
+//Therefore,
+xCO2=nCO2/(nCO2+nN2); //mole fraction of carbon dioxide=ratio of no of moles of carbon dioxide and total moles in mixture
+xN2=nN2/(nCO2+nN2); //mole fraction of nitrogen=ratio of no of moles of oxygen and total moles in mixture
+printf("The mole fraction of carbon dioxide is %f and the mole fraction of nitrogen is %f\n",xCO2,xN2);
+//the molecular weight of mixture=sum of products of mole fraction of each gas component
+MWm=(xCO2*44) + (xN2*28.02); //the molecular weight of mixture
+printf("The molecular weight of air is %f\n",MWm);
+//Because the mixture is 15 lbm (10CO2 + 5N2),the volume of the mixture is found from pm*Vm=mm*Rm*Tm
+pm=100; //mixture pressure //psia
+Tm=460+70; //mixture temperature //R(absolute temperature)
+Rm=1545/37.0; //gas constant of mixture
+mm=15; //mass of mixture //Unit:lb
+//So,rearranging the equation,gives
+Vm=(mm*Rm*Tm)/(pm*144); //mixture volume //ft^3 //1 in^2= 144 ft^2
+printf("The mixture volume is %f ft^3\n",Vm);
+//the partial volume of carbon dioxide is the total volume multiplied by the mole fraction.Thus,
+VCO2=Vm*xCO2; //the partial volume of CO2 //ft^3
+printf("The partial volume of carbon dioxide is %f ft^3\n",VCO2);
+VN2=Vm*xN2; //the partial volume of N2 //ft^3
+printf("The partial volume of nitrogen is %f ft^3\n",VN2);
+//The partial pressure of each constituent is proportional to its mole fraction,for these conditions,
+pCO2=pm*xCO2; //the partial pressure of carbon dioxide= final pressure * the mole fraction of CO2 //psia
+printf("The partial pressure of carbon dioxide is %f psia\n",pCO2);
+pN2=pm*xN2; //the partial pressure of nitrogen=final pressure * the mole fraction of nitrogen //psia
+printf("The partial pressure of nitrogen is %f psia\n\n",pN2);
+
+
diff --git a/2417/CH7/EX7.6/Ex7_6.sce b/2417/CH7/EX7.6/Ex7_6.sce new file mode 100755 index 000000000..b8f2c9820 --- /dev/null +++ b/2417/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,25 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.6\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.6 (page no. 327)
+// Solution
+
+//we will assume that we have 100 volumes of gas mixture and set up table 7.2.In first coloumn,we tabulate the gas,and in the second coloumn,we tabulate the given volume fractions.Because the mole fraction equals to volume fraction,the values in coloumn 3 are the same as those in coloumn 2.
+//The molecular weight is obtained from table 7.1.Because the MW of the mixture is the sum of the individual mole fraction multiplied by the respective molecular weights,the next coloumn tabulates the product of the mole fraction multiplied by molecular weight(3*4).The sum of these entries is the molecular weight of the mixture,which for this case is 33.4.
+printf("Basis:100 volumes of gas mixture\n\n")
+printf("gas Volume Mole Molecular mass \n")
+printf(" fraction fraction x weight MW (x)MW fraction\n")
+printf("CO2 0.40 0.40 44.0 %f %f\n",(0.40*44.0),(0.40*44.0)/33.4)
+printf("N2 0.10 0.10 28.02 %f %f \n",(28.02*0.10),(28.02*0.10)/33.4)
+printf("H2 0.10 0.10 2.016 %f %f \n",(0.10*2.016),(0.10*2.016)/33.4)
+printf("O2 0.40 0.40 32.0 %f %f \n",(0.40*32.0),(0.40*32.0)/33.4)
+printf(" 1.00 1.00 33.4=MWm = 1.000 \n ")
+
+
+
+
+
+
+
diff --git a/2417/CH7/EX7.7/Ex7_7.sce b/2417/CH7/EX7.7/Ex7_7.sce new file mode 100755 index 000000000..62aee95e9 --- /dev/null +++ b/2417/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,22 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.7\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.7 (page no. 328)
+// Solution
+
+//We will take as a basis 100 lbm of mixture.
+//Dividing colomn 2 by 3 gives us mass/molecular weight or moles of each constituents.The total number of moles in the mixture is the sum of coloumn 4,and the molecular weight of the mixture is the mass of the mixture(100 lbm) divided by the number of moles
+//In coloumn 5,mole fraction is given by moles/total mole
+
+printf("Basis:100 pounds of gas mixture\n\n")
+printf("gas Mass Molecular Moles Mole Percent \n")
+printf(" lbm weight MW fraction Volume \n")
+printf("CO2 52.7 44.0 1.2 %f %f \n",(1.2/3),(1.2/3)*100)
+printf("N2 8.4 28.02 0.3 %f %f \n",(0.3/3),(0.3/3)*100)
+printf("H2 0.6 2.016 0.3 %f %f \n",(0.3/3),(0.3/3)*100)
+printf("O2 38.3 32.0 1.2 %f %f \n",(1.2/3),(1.2/3)*100)
+printf(" =100.0 =3.0 =1.00 = 100 \n ")
+printf(" MWm=100/3=33.3 ")
+
diff --git a/2417/CH7/EX7.8/Ex7_8.sce b/2417/CH7/EX7.8/Ex7_8.sce new file mode 100755 index 000000000..63e85e0e2 --- /dev/null +++ b/2417/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,21 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.8\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.8 (page no. 329)
+// Solution
+
+//We will take as a basis 100 lbm of mixture.
+//Dividing colomn 2 by 3 gives us mass/molecular weight or moles of each constituents.The total number of moles in the mixture is the sum of coloumn 4,and the molecular weight of the mixture is the mass of the mixture(100 lbm) divided by the number of moles
+//In coloumn 5,mole fraction is given by moles/total mole
+
+printf("Basis:100 pounds of gas mixture\n\n")
+printf("gas Mass Molecular Moles Mole Percent \n")
+printf(" lbm weight MW fraction Volume \n")
+printf("O2 23.18 32.00 0.724 %f %f \n",(0.724/3.45),(0.724/3.45)*100)
+printf("N2 75.47 28.02 2.693 %f %f \n",(2.692/3.45),(2.692/3.45)*100)
+printf("A 1.30 39.90 0.033 %f %f \n",(0.033/3.45),(0.033/3.45)*100)
+printf("CO2 0.05 44.00 - - - \n")
+printf(" =100.00 =3.45 =1.00 = 100 \n ")
+printf(" MWm=100/3.45=28.99 ")
diff --git a/2417/CH7/EX7.9/Ex7_9.sce b/2417/CH7/EX7.9/Ex7_9.sce new file mode 100755 index 000000000..cdc248f3d --- /dev/null +++ b/2417/CH7/EX7.9/Ex7_9.sce @@ -0,0 +1,25 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 7.9\n\n\n");
+// Chapter 7 : Mixtures Of Ideal Gases
+// Problem 7.9 (page no. 331)
+// Solution
+
+//Given: cp of oxygen is 0.23 Btu/lbm*R.cp of nitrogen is 0.25 Btu/lbm*R. 160 lbm/hr of oxygen and 196 lbm/hr of nitrogen are mixed.oxygen is at 500 F and nitrogen is at 200 F.
+
+//The energy equation for the steady-flow,adiaatic mixing process gives us the requirement that the enthalpy of the mixture must equal to the enthalpies of the components,because deltah=q=0.An alternative statement of this requirement is that the gain in enthalpy of the nitrogen must equal the decrease in enthalpy of the oxygen.Using the latter statement,that the change in enthalpy of nitrogen,yields
+// (160*0.23*(500-tm)) = (196*0.25*(tm-200)) where tm=mixture temperature
+//where m*cp*deltat has been used for deltah. //cp=specific heat at constant pressure //Unit for cp is Btu/lbm*R
+//rearranging the above equation,
+tm=((500*160*0.23)+(196*0.25*200))/((196*0.25)+(160*0.23)); //tm=mixture temperature //Unit:fahrenheit
+printf("The final temperature of the mixture is %f F\n",tm);
+//Using the requirement that the enthalpy of the mixture must equal to the sum of the enthalpies of the components yields an alternative solution to this problem.Let us assume that at 0 F,the enthalpy of each gas and of the mixture is zero.The enthalpy of the entering oxygen is (160*0.23*(500-0)),and the enthalpy of the entering nitrogen is (196*0.25*(200-0)).The enthalpy of the mixture is ((160+196)*cpm*(tm-0))
+//Therefore, (160*0.23*500)+(196*0.25*200) = ((160+196)*cpm*tm)
+cpm=((160/(160+196))*0.23)+((196/(160+196))*0.25); //specific heat at constant pressure for gas mixture //Btu/lbm*R
+printf("For mixture,Specific heat at constant pressure is %f Btu/lbm*R\n",cpm);
+//therefore,
+tm=((160*0.23*500)+(196*0.25*200))/(cpm*(160+196)); //tm=mixture temperature //Unit:fahrenheit
+printf("By using value of cpm,The final temperature of the mixture is %f F\n",tm);
+//The use of 0 F as a base was arbitrary but convenient.Any base would yield the same results.
+//The answer of cpm is wrong in the book.
diff --git a/2417/CH8/EX8.1/Ex8_1.sce b/2417/CH8/EX8.1/Ex8_1.sce new file mode 100755 index 000000000..991fdefef --- /dev/null +++ b/2417/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,27 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.1\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.1 (page no. 380)
+// Solution
+
+//From the Steam Tables or Mollier chart in Appendix 3,we find that
+hf=340.49; //Unit:kJ/kg //at 50kPa //enthalpy
+h1=hf; //at 50kPa //hf=enthalpy of saturated liquid //Unit:kJ/kg
+h4=3230.9; //Unit:kJ/kg //enthalpy
+h5=2407.4; //Unit:kJ/kg ////enthalpy
+//Here,point 5 is in the wet steam region.
+printf("Solution for (a)\n");
+//Neglecting pump work (h2=h1) gives
+nR=(h4-h5)/(h4-h1); //Thermal efficiency of the cycle
+printf("The thermal efficiency of the cycle is %f percentage\n\n",nR*100);
+
+printf("Solution for (b)\n");
+p2=3000; //Unit:kPa //Upper pressure
+p1=50; //Unit:kPa //Lower pressure
+vf=0.001030; //Specific volume of saturated liquid //m^3/kg
+Pumpwork=(p2-p1)*vf; //Unit:kJ/kg //pump work
+//The efficiency of the cycle including pump work is
+nR=((h4-h5)-Pumpwork)/((h4-h1)-Pumpwork); //Thermal efficiency of the cycle
+printf("The thermal efficiency of the cycle including pump work is %f percentage\n\n",nR*100);
diff --git a/2417/CH8/EX8.10/Ex8_10.sce b/2417/CH8/EX8.10/Ex8_10.sce new file mode 100755 index 000000000..73833e163 --- /dev/null +++ b/2417/CH8/EX8.10/Ex8_10.sce @@ -0,0 +1,19 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.10\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.10 (page no. 389)
+// Solution
+
+//Some of the property data required was found in problem8.4.In addition we have,
+//at 200 psia,s=1.7566 Btu/lbm*R,
+h7=1413.6; //Unit:Btu/lbm //Enthalpy
+//at 200 psia,s=1.8320 Btu/lbm*R,
+h4=1514.0; //Unit:Btu/lbm //Enthalpy
+//at 14.696 psia,s=1.8320 Btu/lbm*R,
+h5=1205.2; //Unit:Btu/lbm //Enthalpy
+h1=180.17; //Unit:Btu/lbm //enthalpy
+//Using these data,
+nreheat=((h4-h5)+(h4-h7))/((h4-h1)+(h4-h7)); //The efficiency of the reheat cycle
+printf("The efficiency of the reheat cycle is %f percentage",nreheat*100);
diff --git a/2417/CH8/EX8.11/Ex8_11.sce b/2417/CH8/EX8.11/Ex8_11.sce new file mode 100755 index 000000000..a1958f4a0 --- /dev/null +++ b/2417/CH8/EX8.11/Ex8_11.sce @@ -0,0 +1,41 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.11\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.11 (page no. 394)
+// Solution
+
+printf("Solution for (a)\n");
+//For the rankine cycle,the Mollier chart gives
+h4=1505; //Enthalpy //Unit:Btu/lbm
+h5=922; //Enthalpy //Unit:Btu/lbm
+h6=h5; //Enthalpy //Unit:Btu/lbm
+//and at the condenser,
+h1=69.74; //enthalpy //Unit:Btu/lbm
+nR=(h4-h5)/(h4-h1); //efficiency of rankine cycle
+printf("The efficiency of rankine cycle is %f percentage\n\n",nR*100);
+
+printf("Solution for (b)\n");
+//Figure 8.16 shows the regenerative cycle.After doing work(isentropically),W lbs of steam are bled from the turbine at 50 psia for each lbm of steam leaving the steam generator,and (1-W) pound goes through the turbine and is condensed in the condenser to saturated liquid at 1 psia.This condensate is pumped to the heater,where it mixes with the extraced steam and leaves as saturated liquid at 50 psia.The required enthalpies are:
+//Leaving turbine:
+h5=1168; //Btu/lbm at 50 psia
+//Leaving condenser:
+h7=69.74; //Btu/lbm at 1 psia // is equal to h8 if pump work is neglected
+//Leaving heater:
+h1=250.24; //Btu/lbm at 50 psia //is equal to h2 if pump work is neglected(saturated liquid)
+//A Heat balance around the heater gives
+//W*h5 + (1-W)*h7 = 1*h1
+W=((1*h1)-h7)/(h5-h7); //Unit:lbm //W lb of steam
+printf("W=%f lbm\n",W);
+work=(1-W)*(h4-922) + W*(h4-h5); //h5=922 from the mollier chart //Unit:Btu/lbm //The work output
+printf("The work output is %f Btu/lbm\n",work);
+//Heat into steam generator equals the enthalpy leaving minus the enthalpy of the saturated liquid entering at 50 psia:
+qin=h4-h1; //Unit:Btu/lbm //Heat in
+n=work/qin; //Efficiency of regenerative cycle
+printf("The efficiency of regenerative cycle is %f percentage\n",n*100);
+//The efficiency of a regenerative cycle with one open heater is given by
+n=1-(((h5-h1)*(h6-h7))/((h4-h1)*(h5-h7))); //efficiency of a regenerative cycle
+W=(h1-h7)/(h5-h7); //Unit:lbm //W lb of steam
+printf("When the rankine cycle is compared with regenerative cycle,\n");
+printf("W=%f lbm and the efficiency of a regenerative cycle with one open heater is given by %f percentage\n",W,n*100);
diff --git a/2417/CH8/EX8.12/Ex8_12.sce b/2417/CH8/EX8.12/Ex8_12.sce new file mode 100755 index 000000000..363cad59b --- /dev/null +++ b/2417/CH8/EX8.12/Ex8_12.sce @@ -0,0 +1,43 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.12\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.12 (page no. 396)
+// Solution
+
+//Figure 8.16(a) shows the cycle.For this cycle,W2 pounds are extracted at 100 psia,and W1 pounds are extracted at 50 psia for each pound produced by the steam generator.The enthalpies that are required are:
+//Leaving turbine: 922 //Btu/lbm at 1 psia
+//Leaving condenser: 69.74 //Btu/lbm at 1 psia (saturated liquid)
+//Leaving low pressure heater: 250.24 //Btu/lbm at 50 psia (saturated liquid)
+//Leaving high pressure heater: 298.61 //Btu/lbm at 100 psia
+//At low pressure extraction: 1168 //Btu/lbm at 50 psia
+//At high pressure extraction: 1228.6 //Btulbm at 100 psia
+//Entering turbine: 1505 //Btu/lbm
+//The heat balance around the high pressure heater gives us
+//W2*1228.6 + (1-W2)*250.24 = 1*298.61
+W2=((1*298.61)-250.24)/(1228.6-250.24); //lbm //W2 pounds are extracted at 100 psia
+printf("W2=%f lbm\n",W2);
+//A heat balance around the low pressure heater yields
+//W1*1168 + (1-W1-W2)*69.74 = (1-W2)*250.24
+W1=(((1-W2)*250.24)-69.74+(W2*69.74))/(1168-69.74); //lbm //W1 pounds are extracted at 50 psia
+printf("W1=%f lbm\n",W1);
+work=((1505-1228.6)*1)+((1-W2)*(1228.6-1168))+((1-W1-W2)*(1168-922)); //The work output //Btu/lbm
+printf("The work output is %f Btu/lbm\n",work);
+//Heat into the steam generator equals the enthalpy leaving minus the enthalpy of saturated liquid at 100 psia:
+qin=1505-298.61; //Btu/lbm //Heat in
+printf("Heat in = %f Btu/lbm\n",qin);
+n=work/qin; //The efficiency
+printf("The efficiency is %f percentage\n",n*100);
+//In terms of figure 8.16a,
+//W2=(h1-h11)/(h5-h11)
+//W1=(h5-h1/h6-h9)*(h10-h9/h5-h10) neglecting the pump work
+//n=1-(h7-h8/h4-h1)*(h5-h1/h5-h10)*(h6-h10/h6-h8)
+//For this problem , h8=h9 , h10=h11 and h1=h2.Thus
+W2=(298.61-250.24)/(1228.6-250.24); //lbm //W2 pounds are extracted at 100 psia
+printf("Comparing the results,\n");
+printf("W2=%f lbm\n",W2);
+W1=((1228.6-298.61)*(250.24-69.74))/((1168-69.74)*(1228.6-250.24)); //lbm //W1 pounds are extracted at 50 psia
+printf("W1=%f lbm\n",W1);
+n=1-(((922-69.74)*(1228.6-298.61)*(1168-250.24))/((1505-298.61)*(1228.6-250.24)*(1168-69.74))); //Efficiency
+printf("The efficiency is %f percentage\n",n*100);
diff --git a/2417/CH8/EX8.13/Ex8_13.sce b/2417/CH8/EX8.13/Ex8_13.sce new file mode 100755 index 000000000..7f0ae5d6d --- /dev/null +++ b/2417/CH8/EX8.13/Ex8_13.sce @@ -0,0 +1,36 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.13\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.13 (page no. 398)
+// Solution
+
+//Regenerative cycle
+//Assume that 1 lbm of steam leaves the steam generator and that W1 lbm is bled off to the closed heater at 100psia and that W2 lbm is bled off to the open heater at 50 psia.Alos,assume that the feedwater leaving the closed heater at 310F,18F less than the saturation temperature corresponding to 100 psia.For calculation purposes,we will use hf at 310 F for this enthalpy.Using the Mollier diagram and the steam tables,we find the following values of enthalpy:
+
+//h to turbine=1505 Btu/lbm(at 1000 psia and 1000F)
+//h at first extraction=1228 Btu/lbm(isentropically to 100 psia)
+//h at second extraction=1168 Btu/lbm(isentropically to 100 psia)
+//h at turbine exit=922 Btu/lbm (isentropically to 1 psia)
+//hf=298.61 Btu/lbm(at 100 psia)
+//hf=250.24 Btu/lbm(at 50 psia)
+//hf=280.06 Btu/lbm(at 310 F)
+//hf=69.74 Btu/lbm (at 1 psia)
+//A heat balance around the high pressure heater gives us
+//W1*(1228-298.61) = 1*(280.06-250.24)
+W1=((1*(280.06-250.24)))/(1228-298.61); //lbm //W1 lbm is extracted at 100 psia
+printf("W1=%f lbm\n",W1);
+//A heat balance around the open heater gives us
+//W2*1168 +(1-W1-W2)*69.74 + W1*268.61 = 1*250.24
+W2=((1*250.24)-(W1*268.61)-69.74+(W1*69.74))/(1168-69.74); //lbm //W2 lbm is extracted at 50 psia
+printf("W2=%f lbm\n",W2);
+//The work output of the cycle consists of the work that 1 lbm does in expanding isentropically to 100 psia,plus the work done by (1-W1)lbm expanding isentropicaly from 100 to 50 psia,plus the work done by (1-W1-W2)lbm expanding isentropically from 50 to 1 psia.
+//Numerically,the work is
+workoutput=(1*(1505-1228))+((1-W1)*(1228-1168))+((1-W1-W2)*(1168-922)); //Btu/lbm //the work output
+printf("The work output is %f Btu/lbm\n",workoutput);
+heatinput=1505-280.06; //Btu/lbm //the heat input
+printf("The heat input is %f Btu/lbm\n",heatinput);
+n=workoutput/heatinput; //Efficiency
+printf("The efficiency is %f percentage\n",n*100);
+//When compared to 8.11,we conclude that the addition of additional closed heater raises the efficiency.
diff --git a/2417/CH8/EX8.14/Ex8_14.sce b/2417/CH8/EX8.14/Ex8_14.sce new file mode 100755 index 000000000..7835efa4e --- /dev/null +++ b/2417/CH8/EX8.14/Ex8_14.sce @@ -0,0 +1,37 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.14\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.14 (page no. 426)
+// Solution
+
+//From problem 8.11,
+//Leaving turbine:
+h5=1168; //Btu/lbm at 50 psia
+//For the rankine cycle,the Mollier chart gives
+h4=1505; //Enthalpy //Unit:Btu/lbm
+h6=922; //Enthalpy //Unit:Btu/lbm //h6=h5;
+//and at the condenser,
+h1=69.74; //enthalpy //Unit:Btu/lbm
+//Leaving condenser:
+h7=69.74; //Btu/lbm at 1 psia // is equal to h8 if pump work is neglected
+//Leaving heater:
+h2=250.24; //Btu/lbm at 50 psia //is equal to h1 if pump work is neglected(saturated liquid)
+//A Heat balance around the heater gives
+//W*h5 + (1-W)*h7 = 1*h1
+W=((1*h2)-h7)/(h5-h7); //Unit:lbm
+liquidleaving=(W*h2)+(1-W)*h1; //Btu/lbm //liquid leaving the heatexchange
+
+//Using these data,,
+heatin=h4-liquidleaving; //Btu/lbm //heat in the boiler
+printf("Heat in at boiler is %f Btu/lbm\n",heatin);
+workout=((1-W)*(h4-h6))+(W*(h4-h5)); //Btu/lbm //The work out of turbine
+printf("The work out of turbine is %f Btu/lbm\n",workout);
+n=workout/heatin; //efficiency //The conventional thermal efficiency
+printf("The conventional thermal efficiency is %f percentage\n",n*100);
+//If at this time we have define the efficiency of energy utilization to be the ratio of the work out plus the useful heat out divided by the heat input to the cycle, nenergyutilization=((w+qoutuseful)/qin)*100
+qout=W*(h5-h2); //heat out //Btu/lbm
+n=(workout+qout)/heatin; //efficiency of energy utilization
+printf("Efficiency of energy utilization is %f percentage\n",n*100);
+//Comparing with 8.11, we see that conventional thermal efficiency is decreased and efficiency of energy utilization is increased
diff --git a/2417/CH8/EX8.2/Ex8_2.sce b/2417/CH8/EX8.2/Ex8_2.sce new file mode 100755 index 000000000..a9b1049dc --- /dev/null +++ b/2417/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,28 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.2\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.2 (page no. 381)
+// Solution
+
+//Using the computer disk to obtain the neccesary properties
+printf("Solution for (a)\n");
+//For the conditions given in problem8.1,the properties are found to be
+hf=340.49; //Unit:kJ/kg //at 50kPa //enthalpy
+h1=hf; //at 50kPa //hf=enthalpy of saturated liquid
+h2=h1; //Enthalpy //Unit:kJ/kg
+h4=3230.9; //Unit:kJ/kg //enthalpy
+h5=2407.4; //Unit:kJ/kg //enthalpy
+//Neglecting pump work
+nR=(h4-h5)/(h4-h2); //Thermal efficiency of the cycle
+printf("The thermal efficiency of the cycle is %f percentage\n\n",nR*100);
+
+printf("Solution for (b)\n");
+//For the pump work,we do not need the approximation,because the computerized tables give us the necessary values directly.
+//Assuming that the condensate leaving the condenser is saturated liquid gives us an enthalpy of 340.54 kJ/kg and an entropy of 1.0912 kJ/kg*K for an isentropic compression, the final cond-ition is the boiler pressure of 3Mpa and an entropy of 1.0912 kJ/kg*K. For these values,the program yields an enthalpy of 343.59 kJ/kg*K.The isentropic pump work is equal to
+Pumpwork=343.59-340.54; //Unit:kJ/kg //pumpwork
+//The efficiency of the cycle including pump work is
+nR=((h4-h5)-Pumpwork)/((h4-h1)-Pumpwork); //Thermal efficiency of the cycle
+printf("The thermal efficiency of the cycle including pump work is %f percentage\n\n",nR*100);
+//Final results in this problem agree with the result in problem8.1
diff --git a/2417/CH8/EX8.3/Ex8_3.sce b/2417/CH8/EX8.3/Ex8_3.sce new file mode 100755 index 000000000..a2d802863 --- /dev/null +++ b/2417/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,34 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.3\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.3 (page no. 382)
+// Solution
+
+//Solution for (a)
+//Figurre 8.3 with the cycle extending into the superheat region and expanding along 4->5 is the appropriate diagram for this process.
+
+printf("Solution for (b)\n");
+//This problem can be solved either by use of the Mollier chart or the Steam Tables.If the chart is used,14.696 psia is first located on the saturated vapor line.Because the expansion,4->5,is isentropic,a vertical line on the chart is the path of the process.The point corresponding to 4 in figure 8.3 is found where this vertical line intersects 400 psia.At this point,the ent-halpy is 1515 Btu/lbm,and the corresponding temperature is approximatelty 980F.Saturated vapor at 14.696 psia has an enthalpy of 1150.5 Btu/lbm(from the Mollier chart).The Steam Tables sh-ow that saturated liquid at 14.696 psia has an enthalpy of 180.15 Btu/lbm.In terms of figure 8.3,and neglecting pump work,we have
+h1=180.15; //Unit:Btu/lbm //enthalpy
+h2=h1; //Enthalpy //Unit:Btu/lbm
+h4=1515; //Unit:Btu/lbm //enthalpy
+h5=1150.5; //Unit:kJ/kg //enthalpy
+//Neglecting pump work yields
+nR=(h4-h5)/(h4-h2); //Thermal efficiency of the cycle
+printf("Neglecting the pump work,The thermal efficiency of the cycle is %f percentage\n\n",nR*100);
+p2=400; //Unit:Psia //Upper pressure
+p1=14.696; //Unit:Psia //Lower pressure
+vf=0.01167; //Specific volume of saturated liquid //ft^3/lbm
+J=778; //Conversion factor
+Pumpwork=((p2-p1)*vf*144)/J; //Unit:Btu/lbm //1ft^2=144 in^2 //pumpwork
+//The efficiency of the cycle including pump work is
+nR=((h4-h5)-Pumpwork)/((h4-h1)-Pumpwork); //Thermal efficiency of the cycle
+printf("The thermal efficiency of the cycle including pump work is %f percentage\n\n",nR*100);
+//where the denominator is h4-h2=h4-h1-(h2-h1).Neglecting pump work is obviously justified in this case.An alternative solution is obtained by using the Steam Tables:at 14.696 psia ans sat-uration,sg=1.7567 ; at 400 psia,s= 1.7567.From Table 3(at 400 psia)
+// s h t
+//1.7632 1523.6 1000
+//1.7567 1514.2 982.4
+//1.7558 1512.9 980
+
diff --git a/2417/CH8/EX8.4/Ex8_4.sce b/2417/CH8/EX8.4/Ex8_4.sce new file mode 100755 index 000000000..14e52fed0 --- /dev/null +++ b/2417/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,27 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.4\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.4 (page no. 383)
+// Solution
+
+//Refer to figure8.3.The desired quantities are obtained as follows:
+//at 14.696 psia,saturated vapor (x=1),s=1.7566 Btu/lbm*R
+h5=1150.4; //Unit:Btu/lbm //enthaply
+//at 14.696 psia,saturated liquid (x=0),s=0.3122 Btu/lbm*R
+h2=180.17; //Unit:Btu/lbm //enthaply
+h1=h2;
+//at 400 psia,s=1.7566 Btu/lbm*R,
+h4=1514.0; //Unit:Btu/lbm //Enthalpy
+t=982.07; //Unit:F //tempearature
+//at 400 psia,s=0.3122 Btu/lbm*R, //s=entropy
+h=181.39; //Unit:Btu/lbm //Enthalpy
+//Note the agreement of these values with the ones obtained for problem8.4.Alos,note the temperature of 982.07F compared to 982.4F.Continuing,
+//Neglecting pump work
+nR=(h4-h5)/(h4-h2); //Thermal efficiency of the cycle
+printf("Neglecting the pump work,The thermal efficiency of the cycle is %f percentage\n\n",nR*100);
+Pumpwork=h-h2; //Unit:kJ/kg ///pumpwork
+//The efficiency of the cycle including pump work is
+nR=((h4-h5)-Pumpwork)/((h4-h2)-Pumpwork); //Thermal efficiency of the cycle
+printf("The thermal efficiency of the cycle including pump work is %f percentage\n\n",nR*100);
diff --git a/2417/CH8/EX8.5/Ex8_5.sce b/2417/CH8/EX8.5/Ex8_5.sce new file mode 100755 index 000000000..418e8e422 --- /dev/null +++ b/2417/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,17 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.5\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.5 (page no. 385)
+// Solution
+
+//The Carnot cycle would operate between 982.4F and 212F.
+T1=982.4+460; //temperature converted to absolute temperature //Unit:R
+T2=212+460; //temperature converted to absolute temperature //Unit:R
+nc=((T1-T2)/T1)*100; //Efficiency of carnot cycle
+printf("The efficiency is %f percentage\n",nc);
+//In problem 8.3,
+nR=27.3; //Thermal efficiency of the cycle neglecting the pump work
+typen=(nR/nc)*100; //Type efficiency=ideal thermal efficiency/efficiency of carnot cycle operating between min and max temperature limits
+printf("The type efficiency of the ideal Rankine cycle is %f percentage\n",typen);
diff --git a/2417/CH8/EX8.6/Ex8_6.sce b/2417/CH8/EX8.6/Ex8_6.sce new file mode 100755 index 000000000..679e02294 --- /dev/null +++ b/2417/CH8/EX8.6/Ex8_6.sce @@ -0,0 +1,17 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.6\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.6 (page no. 385)
+// Solution
+
+//For the upper temperature of the cycle,we have 400C,and for 50kPa,the steam tables give us a saturation temperature of 81.33C.The efficiency of a Carnot cycle operating between the limits would be
+T1=400+273; //Celcius temperature converted to fahrenheit temperature
+T2=81.33+273; //temperature converted to fahrenheit temperature
+nc=((T1-T2)/T1)*100; //Efficiency of carnot cycle
+printf("The efficiency is %f percentage\n",nc);
+//In problem 8.1,
+nR=28.5; //Thermal efficiency of the cycle neglecting the pump work
+typen=(nR/nc)*100; //Type efficiency=ideal thermal efficiency/efficiency of carnot cycle operating between min and max temperature limits
+printf("The type efficiency of the ideal Rankine cycle is %f percentage\n",typen);
diff --git a/2417/CH8/EX8.7/Ex8_7.sce b/2417/CH8/EX8.7/Ex8_7.sce new file mode 100755 index 000000000..e4baf7669 --- /dev/null +++ b/2417/CH8/EX8.7/Ex8_7.sce @@ -0,0 +1,14 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.7\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.7 (page no. 386)
+// Solution
+
+//From problem 8.3,
+work=1515-1150.5; //Unit:Btu/lbm of steam //pump work is neglected //Useful ideal work
+//Because of the heat losses, 50 Btu/lbm of the 364.5 Btu/lbm becomes unavailable.
+available=364.5-50; //Unit:Btu/lbm
+n=available/(1515-180.15); //Thermal efficiency of the cycle neglecting pump work h4=1515; //Unit:Btu/lbm //enthalpy & h1=180.15; //Unit:Btu/lbm //enthalpy
+printf("The thermal efficiency of the cycle neglecting pump work is %f percentage\n\n",n*100);
diff --git a/2417/CH8/EX8.8/Ex8_8.sce b/2417/CH8/EX8.8/Ex8_8.sce new file mode 100755 index 000000000..78ca3ae67 --- /dev/null +++ b/2417/CH8/EX8.8/Ex8_8.sce @@ -0,0 +1,14 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.8\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.8 (page no. 387)
+// Solution
+
+//Neglecting the pump work,we have
+heatrate=3413/0.273; //Unit:Btu/kWh //0.273=efficiency //1 kWh=3413 //heat rate
+printf("The heat rate is %f Btu/kWh\n",heatrate);
+//Per pound of steam,1515-1150.5=364.5 Btu is delivered.
+//Because 1 kWh=3413
+printf("The steam rate is %f lbm of steam per kilowatt-hour\n",3413/(1515-1150.5));
diff --git a/2417/CH8/EX8.9/Ex8_9.sce b/2417/CH8/EX8.9/Ex8_9.sce new file mode 100755 index 000000000..ee72269c0 --- /dev/null +++ b/2417/CH8/EX8.9/Ex8_9.sce @@ -0,0 +1,17 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 8.9\n\n\n");
+// Chapter 8 : Vapor Power Cycles
+// Problem 8.9 (page no. 388)
+// Solution
+
+//The Mollier chart provides a convenient way of solving this problem.Expanding from 980F,400 psia,s=1.7567 to 200 psia yields a final enthalpy of 1413 Btu/lbm.Expanding from 200 psia ans an enthalpy of 1515 Btu/lbm to 14.696 psia yields a final enthaply of 1205 Btu/lbm.
+h4=1515; //Unit:Btu/lbm //enthalpy
+h5=1205; //Unit:Btu/lbm //enthalpy
+h7=1413; //Unit:Btu/lbm //enthalpy
+h1=180.15; //Unit:Btu/lbm //enthalpy
+nreheat=((h4-h5)+(h4-h7))/((h4-h1)+(h4-h7)); //The efficiency of the reheat cycle
+printf("The efficiency of the reheat cycle is %f percentage",nreheat*100);
+//It is apparent that for the conditions of this problem,the increase in efficiency is not very large.The final condition of the fluid after the second expansion is superheated steam at
+//14.696 psia.By condensing at this relatively high pressure condition,a large amount of heat is rejected to the condenser cooling water.7
diff --git a/2417/CH9/EX9.1/Ex9_1.sce b/2417/CH9/EX9.1/Ex9_1.sce new file mode 100755 index 000000000..729837c6d --- /dev/null +++ b/2417/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,30 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.1\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.1 (page no. 462)
+// Solution
+
+Rc=7; //Compression Ratio Rc=v2/v3
+k=1.4; //It is apparent incerease in compression ratio yields an increased cycle efficiency
+notto=(1-(1/Rc)^(k-1))*100; //Efficiency of an otto engine
+printf("The efficiency of the otto cycle is %f percentage\n",notto);
+//For the carnot cycle,
+//Nc=1-(T2/T4) //efficiency for the carnot cycle //T2=lowest temperature //T4=Highest temperature
+
+T2=70+460; //for converting to R //Conversion of unit
+//At 700 F
+T4=700+460; //temperatures converted to absolute temperatures;
+nc=(1-(T2/T4))*100; //efficiency of the carnot cycle
+printf("When peak temperature is 700 fahrenheit,efficiency of the carnot cycle is %f percentage\n",nc);
+
+//At 1000 F
+T4=1000+460; //temperatures converted to absolute temperatures;
+nc=(1-(T2/T4))*100; //efficiency of the carnot cycle
+printf("When peak temperature is 1000 fahrenheit,efficiency of the carnot cycle is %f percentage\n",nc);
+
+//At 3000 F
+T4=3000+460; //temperatures converted to absolute temperatures;
+nc=(1-(T2/T4))*100; //efficiency of the carnot cycle
+printf("When peak temperature is 3000 fahrenheit,efficiency of the carnot cycle is %f percentage\n",nc);
diff --git a/2417/CH9/EX9.10/Ex9_10.sce b/2417/CH9/EX9.10/Ex9_10.sce new file mode 100755 index 000000000..778440a3e --- /dev/null +++ b/2417/CH9/EX9.10/Ex9_10.sce @@ -0,0 +1,17 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.10\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.10 (page no. 470)
+// Solution
+
+//For four cycle,six cylinder engine,
+//Using the results of problem 9.5,
+hp=100; //Horsepower //Unit:hp
+L=4/12; //Unit:ft //stroke is 4 in.
+A=(%pi/4)*(3)^2*6; //Cylinder bore is 3 in.
+N=4000/2; //Power strokes per minute //2L engine //Unit:rpm
+//hp=(pm*LA*N)/33000;
+pm=(hp*33000)/(L*A*N); //The mean effective pressure //psia
+printf("The mean effective pressure is %f psia",pm);
diff --git a/2417/CH9/EX9.11/Ex9_11.sce b/2417/CH9/EX9.11/Ex9_11.sce new file mode 100755 index 000000000..a59066c56 --- /dev/null +++ b/2417/CH9/EX9.11/Ex9_11.sce @@ -0,0 +1,17 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.11\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.11 (page no. 470)
+// Solution
+
+//six cylinder engine,with displacement 3.3L
+//Using the results of problem 9.5,
+hp=230; //Horsepower //Unit:hp
+//3.3L*1000 cm^3/L*(in/2.54 cm)^3
+LA=3.3*1000*(1/2.54)^3; //mean //in^3
+N=5500/2; //Power strokes per minute //2L engine //Unit:rpm
+//hp=(pm*LA*N)/33000;
+pm=(hp*33000*12)/(LA*N); //1ft=12inch //The mean effective pressure //psia
+printf("The mean effective pressure is %f psia",pm);
diff --git a/2417/CH9/EX9.12/Ex9_12.sce b/2417/CH9/EX9.12/Ex9_12.sce new file mode 100755 index 000000000..7633c1daf --- /dev/null +++ b/2417/CH9/EX9.12/Ex9_12.sce @@ -0,0 +1,20 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.12\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.12 (page no. 478)
+// Solution
+
+//An air-standard Diesel engine
+rc=16; //Compression Ratio Rc=v2/v3
+v4byv3=2; //Cutoff ratio=v4/v3
+k=1.4; //with the cycle starting at 14 psia and 100 F //It is apparent incerease in compression ratio yields an increased cycle efficiency
+T2=100+460; //temperatures converted to absolute temperatures;
+ndiesel=1-((inv(rc))^(k-1)*(((v4byv3)^k-1)/(k*(v4byv3-1)))); //The efficiency of the diesel engine
+printf("The efficiency of the diesel engine is %f percentage\n",ndiesel*100);
+// T3/T2=rc^k-1 and T5/T4=(1/re^k-1) //re=expansion ratio=v5/v4
+//But T4/T3=v4/v3=rc/re
+//So,
+T5=T2*(v4byv3)^k; //The temperature of the exhaust of the cycle //Unit:R
+printf("The temperature of the exhaust of the cycle is %f R i.e. %f F",T5,T5-460);
diff --git a/2417/CH9/EX9.13/Ex9_13.sce b/2417/CH9/EX9.13/Ex9_13.sce new file mode 100755 index 000000000..699a7aa23 --- /dev/null +++ b/2417/CH9/EX9.13/Ex9_13.sce @@ -0,0 +1,29 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.13\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.13 (page no. 479)
+// Solution
+
+//Now,in problem 9.12,
+//An air-standard Diesel engine
+rc=16; //Compression Ratio Rc=v2/v3
+v4byv3=2; //Cutoff ratio=v4/v3
+k=1.4; //with the cycle starting at 14 psia and 100 F //It is apparent incerease in compression ratio yields an increased cycle efficiency
+T2=100; //Unit:F //temperature
+T5=1018; //Unit:F //Found in 9.12 //The temperature of the exhaust of the cycle //Unit:R
+ndiesel=0.614 //Efficiency of the diesel engine //Found in 9.12
+//Now,in problem 9.13,
+cp=0.24; //Unit:Btu/(lbm*R) //Specific heat constant for constant pressure process
+cv=0.172; //Unit:Btu/(lbm*R) //Specific heat constant for constant volume process
+
+Qr=cv*(T5-T2); //Heat rejected //Unit:Btu/lbm
+//ndeisel=1-(Qr/qin); //Efficiency=ndeisel //qin=heat added
+qin=Qr/(1-ndiesel); //Unit:Btu/lbm
+J=778; //J=Conversion factor
+networkout=J*(qin-Qr); //(ft*lbf)/lbm //Net work out per pound of gas
+printf("Net work out per pound of gas is %f (ft*lbf)/lbm\n",networkout);
+//The mean effective pressure is net work divided by (v2-v3):
+mep=networkout/((16-1)*144); //1ft^2=144 in^2 //Unit:psia //The mean effective pressure
+printf("The mean effective pressure is %f psia",mep);
diff --git a/2417/CH9/EX9.14/Ex9_14.sce b/2417/CH9/EX9.14/Ex9_14.sce new file mode 100755 index 000000000..5e2c4933f --- /dev/null +++ b/2417/CH9/EX9.14/Ex9_14.sce @@ -0,0 +1,31 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.14\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.14 (page no. 489)
+// Solution
+
+//A Brayton cycle
+rc=7; //Compression Ratio Rc=v2/v3
+k=1.4; //It is apparent incerease in compression ratio yields an increased cycle efficiency
+cp=0.24; //Unit:Btu/(lbm*R) //Specific heat constant for constant pressure process
+T3=1500; //(unit:fahrenheit) //peak tempeature
+p1=14.7; //Unit:psia //Initial condition
+T1=70+460; //temperatures converted to absolute temperatures; //Initial condition
+R=53.3; //Unit:ft*lbf/lbm*R //constant of proportionality
+nBrayton=1-((inv(rc))^(k-1)); //A Brayton cycle efficiency
+printf("A Brayton cycle efficiency is %f percentage\n",nBrayton*100);
+//If we base our calculation on 1 lbm of gas and use subscripts that corresponds to points (1),(2),(3) and (4) of fig.9.22,we have
+v1=(R*T1)/p1; //Unit:ft^3/lbm //specific volume at point 1
+//Because rc=7 then,
+v2=v1/rc; //Unit:ft^3/lbm //specific volume at point 2
+//After the isentropic compression, T2*v2^k-1 = T1*v1^k-1
+T2=T1*(v1/v2)^(k-1); //Unit:R //temperature at point 2
+T2=T2-460; //Unit:fahrenheit //temperature at point 2
+qin=cp*(T3-T2); //Heat in //Unit:Btu/lbm
+printf("The heat in is %f Btu/lbm\n",qin);
+//Because efficiency can be stated to be work out divided by heat in,
+wbyJ=nBrayton*qin; //The work out //Unit:Btu/lbm
+printf("The work out is %f Btu/lbm\n",wbyJ); //Answer is wrong in the book.cause they have taken efficiency value wrong
+printf("The heat rejected is %f Btu/lbm\n",qin-wbyJ); //Anser is affected because of value of wbyJ
diff --git a/2417/CH9/EX9.2/Ex9_2.sce b/2417/CH9/EX9.2/Ex9_2.sce new file mode 100755 index 000000000..0bcbb207b --- /dev/null +++ b/2417/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,25 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.2\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.2 (page no. 463)
+// Solution
+
+cv=0.172; //Unit:Btu/(lbm*R) //Specific heat constant
+Rc=7; //Compression Ratio Rc=v2/v3
+k=1.4; //It is apparent incerease in compression ratio yields an increased cycle efficiency
+T2=70+460; //for converting to R //Conversion of unit
+//For 1000 F
+T4=1000+460; //temperatures converted to absolute temperatures;
+T3byT2=Rc^(k-1); //Unit less
+T3=T3byT2*T2;
+qin=cv*(T4-T3); //Unit:Btu/lbm //Heat added
+//Qr=cv*(T5-T2)*(T5/T4)=(v2/v3)^(k-1)
+Qr=(inv(Rc))^(k-1); //Unit:Btu/lbm //Heat rejected
+T5=T4*Qr;
+Qr=cv*(T5-T2); //Unit:Btu/lbm //Heat rejected
+printf("The net work out is %f Btu/lbm\n",qin-Qr);
+notto=((qin-Qr)/qin)*100; //The efficiency of otto cycle
+printf("The efficiency of otto cycle is %f percentage",notto);
+//The value agrees with the results of problem 9.1
diff --git a/2417/CH9/EX9.3/Ex9_3.sce b/2417/CH9/EX9.3/Ex9_3.sce new file mode 100755 index 000000000..0e1605659 --- /dev/null +++ b/2417/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,18 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.3\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.3 (page no. 464)
+// Solution
+
+cv=0.7186; //Unit:kJ/(kg*K) //Specific heat constant for constant volume process
+Rc=8; //Compression Ratio Rc=v2/v3
+k=1.4; //It is apparent incerease in compression ratio yields an increased cycle efficiency
+T2=20+273; //20 C converted to its kelvin value
+qin=50; //Heat added //Unit:kJ
+T3byT2=Rc^(k-1);
+T3=T3byT2*T2; //Unit:K
+//qin=cv*(T4-T3) //heat added //Unit:kJ
+T4=(qin/cv)+T3; //The peak temperature of the cycle //Unit:K
+printf("The peak temperature of the cycle is %f Kelvin i.e. %f Celcius",T4,T4-273);
diff --git a/2417/CH9/EX9.4/Ex9_4.sce b/2417/CH9/EX9.4/Ex9_4.sce new file mode 100755 index 000000000..85018f7d8 --- /dev/null +++ b/2417/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,52 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.4\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.4 (page no. 465)
+// Solution
+
+//For an Otto cycle,
+rc=7; //Compression Ratio Rc=v2/v3
+q=50; //Unit:Btu/lbm //Heat added
+p2=14.7; //Unit:psia //pressure at point 2
+T2=60+460; //temperatures converted to absolute temperatures; //Unit:R
+cp=0.24; //Unit:Btu/(lbm*R) //Specific heat constant for constant pressure process
+cv=0.171; //Unit:Btu/(lbm*R) //Specific heat constant for constant volume process
+R=53.3; //Unit:ft*lbf/lbm*R //constant of proportionality
+k=1.4; //It is apparent incerease in compression ratio yields an increased cycle efficiency
+//Refering to figure 9.9,
+//At (2),we need v2.
+//p2*v2=R*T2
+v2=(R*T2)/(p2*144); //Unit:ft^3/lbm //1ft^2=144 in^2 //specific volume at point 2
+printf("At point (2),\nspecific volume v2=%f ft^3/lbm\n\n",v2);
+//For The isentropic path (2)&(3),p3*v3^k=p2*v2^k,so
+//So,p3=p2*(v2/v3)^k;
+p3=p2*rc^k; //Unit:psia //pressure at point 3
+printf("At path(2)&(3)\n");
+printf("pressure p3=%f psia\n",p3);
+v3=v2/rc; //Unit:ft^3/lbm //specific volume at point 3
+printf("specific volume v3=%f ft^3/lbm\n",v3);
+T3=(p3*v3*144)/R; //Unit:R //1ft^2=144 in^2 //temperature at point 3
+printf("temperature T3=%f R\n\n",T3);
+printf("At point(4),\n");
+//To obtain the values at (4),we note
+v4=v3; //Unit:ft^3/lbm //specific volume at point 4
+printf("specific volume v4=%f ft^3/lbm\n",v4);
+//qin=cv*(T4-T3)
+T4=T3+(q/cv); //Unit:R //temperature at point 4
+printf("temperature T4=%f R\n",T4);
+//For p4,
+p4=(R*T4)/(144*v4); //Unit:psia //1ft^2=144 in^2 //pressure at point 4
+printf("pressure p4=%f psia\n\n",p4);
+//The last point has the same specific volume as (2),giving
+printf("At last point,\n");
+v5=v2; //Unit:ft^3/lbm //specific volume at point 5
+printf("specific volume v5=%f ft^3/lbm\n",v5);
+//The isentropic path equation,p5*v5^k=p4*v4^k,so
+p5=p4*(v4/v5)^k; //Unit:psia //pressure at point 5
+printf("pressure p5=%f psia\n",p5);
+T5=(p5*v5*144)/(R); //Unit:R //1ft^2=144 in^2 temperature at point 5
+printf("temperature T5=%f R\n\n",T5);
+n=(((T4-T3)-(T5-T2))/(T4-T3))*100; //The efficiency of the cycle
+printf("The efficiency of the cycle is %f percentage",n);
diff --git a/2417/CH9/EX9.7/Ex9_7.sce b/2417/CH9/EX9.7/Ex9_7.sce new file mode 100755 index 000000000..172c0f2ab --- /dev/null +++ b/2417/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,15 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.7\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.7 (page no. 468)
+// Solution
+
+//For four cycle engine,
+//Using the results of problem 9.6,
+pm=1000; //Unit:kPa //mean effective pressure //Unit:psia
+N=4000/2; //Power strokes per minute //2L engine //Unit:rpm
+LA=2 //Mean //Unit:liters
+hp=(pm*LA*N)/44760; //The horsepower //Unit:hp
+printf("The horsepower is %f hp",hp);
diff --git a/2417/CH9/EX9.9/Ex9_9.sce b/2417/CH9/EX9.9/Ex9_9.sce new file mode 100755 index 000000000..6b96cbe38 --- /dev/null +++ b/2417/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,13 @@ +//scilab 5.4.1
+clear;
+clc;
+printf("\t\t\tProblem Number 9.9\n\n\n");
+// Chapter 9 : Gas Power Cycles
+// Problem 9.9 (page no. 469)
+// Solution
+
+//An otto engine
+c=0.2; //clearance equal to 20% of its displacement
+//Using results of problem 9.8,
+rc=(1+c)/c; //The compression ratio
+printf("The compression ratio is %f",rc);
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