initial commit / add all books

12 files changed, 288 insertions, 0 deletions

diff --git a/3843/CH11/EX11.1/Ex11_1.sce b/3843/CH11/EX11.1/Ex11_1.sce
new file mode 100644
index 000000000..c18be5462
--- /dev/null
+++ b/3843/CH11/EX11.1/Ex11_1.sce
@@ -0,0 +1,25 @@
+// Example 11_1

+clc;funcprot(0);

+// Given data

+N_1=78;// The number of moles for nitrogen in mol

+N_2=22;// The number of moles for oxygen in mol

+M_1=28;// The molecular weight of nirogen in kg/kmol

+M_2=32;// The molecular weight of oxygen in kg/kmol

+Rbar=8.314;// The universal gas constant kJ/kmol.K

+

+// Calculation

+// (a)

+N=N_1+N_2;// The total number of moles in mol

+y_1=N_1/N;// The mole fraction for nitrogen

+y_2=N_2/N;// The mole fraction for oxygen

+// (b)

+m_1=N_1*M_1;// The mass of nitrogen in kg

+m_2=N_2*M_2;// The mass of oxygen in kg

+m=m_1+m_2;// The total mass of the mixture in kg

+mf_1=m_1/m;// The mass fraction for nitrogen

+mf_2=m_2/m;// The mass fraction for oxygen

+// (c)

+M=m/N;// The molecular weight of the mixture in kg/k.mol

+// (d)

+R=Rbar/M;// The gas constant for air in kJ/kg.K

+printf("\n(a)The mole fraction for nitrogen,y_1=%0.2f \n   The mole fraction for oxygen,y_2=%0.2f \n(b)The mass fraction for nitrogen,mf_1=%0.3f \n   The mass fraction for oxygen,mf_2=%0.3f \n(c)The molecular weight of the mixture,M=%2.1f kg/k.mol \n(d)The gas constant for air,R=%0.3f kJ/kg.K",y_1,y_2,mf_1,mf_2,M,R);

diff --git a/3843/CH11/EX11.10/Ex11_10.sce b/3843/CH11/EX11.10/Ex11_10.sce
new file mode 100644
index 000000000..0179672ee
--- /dev/null
+++ b/3843/CH11/EX11.10/Ex11_10.sce
@@ -0,0 +1,33 @@
+// Example 11_10

+clc;funcprot(0);

+// Given data

+T_1=5;// °C

+T_2=25;// °C

+phi_1=40/100;// The relative humidity at state 1

+phi_2=40/100;// The relative humidity at state 2

+V=60;// m^3/min

+P=100;// kPa

+P_g1=0.872;// kPa

+R_a=0.287;// kJ/kg.K

+

+// Calculation

+// (a)

+P_a1=P-(phi_1*P_g1);// kPa

+rho_a1=P_a1/(R_a*(T_1+273));// kg/m^3

+mdot_a=(V/60)*rho_a1;// The mass flux of dry air in kg/s

+// Using psychrometric chart

+h_1=10;// kJ/kg air

+h_2=31;// kJ/kg air

+Q=mdot_a*(h_2-h_1);// The rate of heat transfer in kJ/s

+// (b)

+w_2=0.0021;// kgH2O/kg dry air

+w_3=0.008;// kgH2O/kg dry air

+mdot_s=(w_3-w_2)*mdot_a;// The rate of steam supplied in kg/s

+// (c)

+h_2=31;// kJ/kg

+h_3=45;// kJ/kg

+h_s=(mdot_a/mdot_s)*(h_3-h_2);// kJ/kg

+h_fs=604.7;// kJ/kg

+h_fgs=2133.8;// kJ/kg

+x_s=(h_s-h_fs)/h_fgs;// The state of the steam introduced

+printf("\n(a)The rate of heat transfer,Q=%2.1f kJ/s \n(b)The rate of steam supplied,mdot_s=%0.4f kg/s \n(c)The state of the steam introduced,x_s=%0.2f",Q,mdot_s,x_s);

diff --git a/3843/CH11/EX11.11/Ex11_11.sce b/3843/CH11/EX11.11/Ex11_11.sce
new file mode 100644
index 000000000..41b141294
--- /dev/null
+++ b/3843/CH11/EX11.11/Ex11_11.sce
@@ -0,0 +1,23 @@
+// Example 11_11

+clc;funcprot(0);

+// Given data

+T_1=80;// °F

+phi_1=90;// The relative humidity at state 1

+T_2=75;// °F

+phi_2=40;// The relative humidity at state 2

+

+// Calculation

+// (a)

+// From psychrometric chart

+w_2=0.0177;// lbm H2O/lbm dry air

+w_3=0.0075;// lbm H2O/lbm dry air

+dw=w_3-w_2;// The amount of moisture removed in lbm H2O/lbm dry air

+// (b)

+h_3=20;// Btu/lbm dry air

+h_1=39.5;// Btu/lbm dry air

+q=h_3-h_1;// The heat removed in Btu/lbm dry air

+// (c)

+h_3=20;// Btu/lbm dry air

+h_4=26.5;// Btu/lbm dry air

+q_c=h_4-h_3;// The necessary added heat in Btu/lbm dry air

+printf("\n(a)The amount of moisture removed,dw=%0.3f lbm H2O/lbm dry air \n(b)The heat removed,q=%2.1f Btu/lbm dry air \n(c)The necessary added heat,q=%1.1f Btu/lbm dry air",dw,q,q_c);

diff --git a/3843/CH11/EX11.12/Ex11_12.sce b/3843/CH11/EX11.12/Ex11_12.sce
new file mode 100644
index 000000000..ea4fb18c8
--- /dev/null
+++ b/3843/CH11/EX11.12/Ex11_12.sce
@@ -0,0 +1,10 @@
+// Example 11_12

+clc;funcprot(0);

+// Given data

+w_1=0.0046;// kg H2O/kg dry air

+w_2=0.010;// kg H2O/kg dry air

+

+// Calculation

+// (b)

+dw=w_2-w_1;// The amount of water added in kg H2O/kg dry air

+printf("\n(b)The amount of water added,w_2-w_1=%0.4f kg H2O/kg dry air",dw);

diff --git a/3843/CH11/EX11.14/Ex11_14.sce b/3843/CH11/EX11.14/Ex11_14.sce
new file mode 100644
index 000000000..eecb7c8b8
--- /dev/null
+++ b/3843/CH11/EX11.14/Ex11_14.sce
@@ -0,0 +1,28 @@
+// Example 11_14

+clc;funcprot(0);

+// Given data

+m_w3=10000;// kg/min

+T_ain=20;// The temperature of air at inlet in °C

+phi_1=50;// Humidity in %

+T_aout=32;// The temperature of air at exit in °C

+phi_2=98;// Humidity in %

+T_win=40;// The temperature of water at inlet in °C

+T_wout=25;// The temperature of water at exit in °C

+

+// Calculation

+// (a)

+// From the psychrometric chart we find

+h_1=37;// kJ/kg of dry air

+h_2=110;// kJ/kg of dry air

+w_1=0.0073;// kgH2O/kg dry air

+w_2=0.0302;// kgH2O/kg dry air

+// From steam tables

+h_3=167.5;// kJ/kg

+h_4=104.9;// kJ/kg

+m_a=(m_w3*(h_4-h_3))/(h_1-h_2+((w_2-w_1)*h_4));// kg/min

+// From the psychrometric chart we find

+v_1=0.84;// m^3/ kg dry air

+Vdot=m_a*v_1;// m^3/min

+// (b)

+m_4=m_w3-((w_2-w_1)*m_a);// kg/min

+printf("\n(a)The volume flow rate of air into the cooling tower,Vdot=%4.0f m^3/min \n(b)The mass flux of water,m_4=%4.0f kg/min",Vdot,m_4);

diff --git a/3843/CH11/EX11.2/Ex11_2.sce b/3843/CH11/EX11.2/Ex11_2.sce
new file mode 100644
index 000000000..a3e760804
--- /dev/null
+++ b/3843/CH11/EX11.2/Ex11_2.sce
@@ -0,0 +1,22 @@
+// Example 11_2

+clc;funcprot(0);

+// Given data

+T=25;// °C

+P=2;// MPa

+m_1=2;// The mass of nitrogen in kg

+m_2=4;// The mass of CO_2 in kg

+M_1=28;// The molecular weight of the nitrogen in kg/k.mol

+M_2=44;// The molecular weight of the CO_2 in kg/k.mol

+Rbar=8.314;// The universal gas constant kJ/kmol.K

+

+// Calculation

+N_1=m_1/M_1;// The number of moles for nitrogen in mol

+N_2=m_2/M_2;// The number of moles for CO_2 in mol

+N=N_1+N_2;// The total number of moles in mol

+y_1=N_1/N;// The mole fraction for nitrogen

+y_2=N_2/N;// The mole fraction for CO_2

+P_1=y_1*P;// The partial pressure for nitrogen in MPa

+P_2=y_2*P;// The partial pressure for CO_2 in MPa

+M=(M_1*y_1)+(M_2*y_2);// The molecular weight of the mixture in kg/k.mol

+R=Rbar/M;// The gas constant of the mixture in kJ/kg.K

+printf("\nThe partial pressure for nitrogen,P_1=%0.2f MPa \nThe partial pressure for CO_2,P_2=%1.2f MPa \nThe gas constant of the mixture,R=%0.3f kJ/kg.K",P_1,P_2,R);

diff --git a/3843/CH11/EX11.3/Ex11_3.sce b/3843/CH11/EX11.3/Ex11_3.sce
new file mode 100644
index 000000000..f474e568e
--- /dev/null
+++ b/3843/CH11/EX11.3/Ex11_3.sce
@@ -0,0 +1,18 @@
+// Example 11_3

+clc;funcprot(0);

+// Given data

+m=20;// The mass of the mixture in lbm

+T_1=80;// °F

+T_2=300;// °F

+c_v1=0.177;// Btu/lbm-°R

+c_v2=0.158;// Btu/lbm-°R

+c_v3=0.157;// Btu/lbm-°R

+mf_1=20/100;// The mole fraction for nitrogen

+mf_2=40/100;// The mole fraction for CO_2

+mf_3=40/100;// The mole fraction for oxygen

+

+// Calculation

+c_v=(mf_1*c_v1)+(mf_2*c_v2)+(mf_3*c_v3);// // Btu/lbm-°R

+delT=T_2-T_1;// °F

+Q=m*c_v*delT;// The heat transfer in Btu

+printf("\nThe heat transfer,Q=%3.0f Btu",Q);

diff --git a/3843/CH11/EX11.4/Ex11_4.sce b/3843/CH11/EX11.4/Ex11_4.sce
new file mode 100644
index 000000000..e7e5aa7f4
--- /dev/null
+++ b/3843/CH11/EX11.4/Ex11_4.sce
@@ -0,0 +1,33 @@
+// Example 11_4

+clc;funcprot(0);

+// Given data

+N_1=2;// The number of moles for CO_2 in mol

+N_2=4;// The number of moles for nitrogen in mol

+M_1=44;// The molecular weight of the CO_2 in kg/k.mol

+M_2=28;// The molecular weight of nirogen in kg/kmol

+P_1=100;// kPa

+T_1=20+273;// K

+P_2=2000;// kPa

+c_v1=0.653;// kJ/kg.K

+c_v2=0.745;// kJ/kg.K

+c_p1=0.842;// kJ/kg.K

+c_p2=1.042;// kJ/kg.K

+Rbar=8.314;// The universal gas constant kJ/kgmol.K

+

+// Calculation

+// (a)

+N=N_1+N_2;// The total number of moles in mol

+m_1=N_1*M_1;// The mass of CO_2 in kg

+m_2=N_2*M_2;// The mass of nitrogen in kg

+m=m_1+m_2;// The mass of the mixture in kg

+m_f1=m_1/m;// The mole fraction for CO_2

+m_f2=m_2/m;// The mole fraction for nitrogen

+c_v=(m_f1*c_v1)+(m_f2*c_v2);// kJ/kg.K

+c_p=(m_f1*c_p1)+(m_f2*c_p2);// kJ/kg.K

+k=c_p/c_v;// The ratio of specific heats

+T_2=T_1*(P_2/P_1)^((k-1)/(k));// K

+// (b)

+W=(-m*c_v*(T_2-T_1))/10^3;// MJ

+// (c)

+dels=(c_p*log(T_2/T_1))-((Rbar/(((N_1/N)*M_1)+((N_2/N)*M_2)))*log(P_2/P_1));// The entropy change in kJ/kg.K

+printf("\n(a)The final temperature,T_2=%3.0f K or %3.0f°C \n(b)The work required,W=%2.1f MJ \n(c)The change in entropy,dels=%0.5f kJ/kg.K",T_2,T_2-273,W,dels);

diff --git a/3843/CH11/EX11.5/Ex11_5.sce b/3843/CH11/EX11.5/Ex11_5.sce
new file mode 100644
index 000000000..bb8c67bc2
--- /dev/null
+++ b/3843/CH11/EX11.5/Ex11_5.sce
@@ -0,0 +1,27 @@
+// Example 11_5

+clc;funcprot(0);

+// Given data

+T=25;// °C

+P=100;// kPa

+V=150;// m^3

+phi=60/100;// The relative humidity at state 1

+P_g=3.169;// kPa

+M_v=18;// kg/k.mol

+M_a=28.97;// kg/k.mol

+R_a=0.287;// kJ/kg.K

+

+// Calculation

+// (a)

+P_v=P_g*phi;// kPa

+P_a=P-P_v;// The partial pressure of air in kPa

+w=0.622*(P_v/P_a);// The humidity ratio in kg H2O/kg dry air

+// (b)

+// From psychrometric chart

+T_dp=16.6;// The dew point temperature in °C

+// (c)

+m_v=w*((P_a*V)/(R_a*(T+273)));// The mass of water vapor in kg

+// (d)

+N_v=m_v/M_v;// mol

+N_a=((P_a*V)/(R_a*(T+273)))/M_a;// mol

+y_v=N_v/(N_a+N_v);// The mole fraction of the water vapor 

+printf("\n(a)The humidity ratio,w=%0.5f kg H2O/kg dry air \n(b)The dew point temperature,T_dp=%2.1f°C \n(c)The mass of water vapor,m_v=%1.2f kg \n(d)The mole fraction of the water vapor,y=%0.4f",w,T_dp,m_v,y_v);

diff --git a/3843/CH11/EX11.6/Ex11_6.sce b/3843/CH11/EX11.6/Ex11_6.sce
new file mode 100644
index 000000000..04b51eb91
--- /dev/null
+++ b/3843/CH11/EX11.6/Ex11_6.sce
@@ -0,0 +1,23 @@
+// Example 11_6

+clc;funcprot(0);

+// Given data

+T=25;// °C

+T_dp=10;// The dew point temperature in °C

+P=100;// kPa

+V=150;// m^3

+P_g=3.169;// kPa

+M_v=18;// kg/k.mol

+M_a=28.97;// kg/k.mol

+R_a=0.287;// kJ/kg.K

+

+// Calculation

+// (a)

+P_v=1.228;// kPa

+P_a=P-P_v;// The partial pressure of air in kPa

+w_1=0.622*(P_v/P_a);// The humidity ratio in kg H2O/kg dry air

+w_2=0.01205;// kg H2O/kg dry air

+dw=w_2-w_1;// The difference in humidity ratio in kg H2O/kg dry air

+dm_v=dw*((P_a*V)/(R_a*(T+273)));// kg H2O

+// (b)

+phi=1.608*((w_1*P_a)/(P_g));// The relative humidity in %

+printf("\n(a)The amount of water vapor that will condense,delm_v=%0.3f kg H2O \n(b)The relative humidity,phi=%0.3f or %2.1f percentage.",dm_v,phi,phi*100);

diff --git a/3843/CH11/EX11.7/Ex11_7.sce b/3843/CH11/EX11.7/Ex11_7.sce
new file mode 100644
index 000000000..83cf7dea2
--- /dev/null
+++ b/3843/CH11/EX11.7/Ex11_7.sce
@@ -0,0 +1,23 @@
+// Example 11_7

+clc;funcprot(0);

+// Given data

+T_1=100;// °F

+T_2=80;// °F

+P=14.7;// psia

+P_g1=0.9503;// psia

+P_g2=0.5073;// psia

+c_p=0.24;// Btu/lbm-°R

+h_fg2=1048;// Btu/lbm

+h_g1=1105;// Btu/lbm

+h_f2=48.09;// Btu/lbm

+

+// Calculation

+// (a)

+w_2=0.622*(P_g2/(P-P_g2));// lbm H2O/lbm dry air

+w_1=((w_2*h_fg2)+(c_p*(T_2-T_1)))/(h_g1-h_f2);// lbm H2O/lbm dry air

+// (b)

+P_v1=(w_1*P)/(0.622*(1+(w_1)));// psia

+phi=P_v1/P_g1;// The relative humidity in %

+// (c)

+h=(c_p*T_1)+(w_1*h_g1);// Btu/lbm dry air

+printf("\n(a)The humidity ratio,w_1=%0.5f lbm H2O/lbm dry air \n(b)The relative humidity,phi=%0.3f or %2.1f percentage. \n(c)The specific enthalpy of the air,h=%2.1f Btu/lbm dry air",w_1,phi,phi*100,h);

diff --git a/3843/CH11/EX11.9/Ex11_9.sce b/3843/CH11/EX11.9/Ex11_9.sce
new file mode 100644
index 000000000..37ecc0a5e
--- /dev/null
+++ b/3843/CH11/EX11.9/Ex11_9.sce
@@ -0,0 +1,23 @@
+// Example 11_9

+clc;funcprot(0);

+// Given data

+T_1=5;// °C

+T_2=25;// °C

+phi_1=70/100;// The relative humidity at state 1

+V=50;// m^3/min

+P=100;// kPa

+P_g1=0.872;// kPa

+R_a=0.287;// kJ/kg.K

+

+// Calculation

+P_a1=P-(phi_1*P_g1);// kPa

+rho_a1=P_a1/(R_a*(T_1+273));// kg/m^3

+mdot_a=(V/60)*rho_a1;// The mass flux of dry air in kg/s

+// Using psychrometric chart

+h_1=14;// kJ/kg air

+h_2=35;// kJ/kg air

+Q=mdot_a*(h_2-h_1);// The rate of heat transfer in kJ/s

+// From the chart 

+phi_2=19;// The relative humidity at state 2

+printf("\nThe rate of heat transfer,Q=%2.1f kJ/s \nThe final relative humidity,phi_2=%2.0f percentage.",Q,phi_2);

+// The answer provided in the textbook is wrong