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diff --git a/1919/CH9/EX9.1/Ex9_1.sce b/1919/CH9/EX9.1/Ex9_1.sce
new file mode 100755
index 000000000..80246982b
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@@ -0,0 +1,31 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 1

+

+clear ;clc;

+

+//Given data

+X1 = 0.3                // volume fraction of N2 

+X2 = 0.5                // volume fraction of He 

+X3 = 0.2                // volume fraction of CO2 

+m = 3000                // mass flowrate of oil in kg/hr

+Cp = 1.2                // molar heat capacity of oil in kJ/mol

+T1_o = 523.15           // entering temperature of oil in K

+T2_o = 323.15           // leaving temperature of oil in K

+T1_g = 303.15           // entering temperature of oil in K

+T2_g = 323.15           // leaving temperature of oil in K

+Cp1 = 29.1783         // molar heat capacity of N2 in kJ/mol

+Cp2 = 20.7860         // molar heat capacity of He in kJ/mol

+Cp3 = 37.1444        // molar heat capacity of CO2 in kJ/mol

+

+// Calculations

+Cp_m = X1*Cp1 + X2*Cp2 + X3*Cp3     // molar heat capacity of gas mixture

+

+N = m*Cp*(T1_o-T2_o)/(Cp_m*(T2_g-T1_g))

+

+

+// Output Results

+mprintf('Flow rate of gas mixture to cool oil = %4.3f kmol/h' ,N);

+

diff --git a/1919/CH9/EX9.10/Ex9_10.sce b/1919/CH9/EX9.10/Ex9_10.sce
new file mode 100755
index 000000000..96c962470
--- /dev/null
+++ b/1919/CH9/EX9.10/Ex9_10.sce
@@ -0,0 +1,39 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 10

+

+clear ;clc;

+

+//Given data

+RH = 0.6                // relative humidity moist air

+T1 = 313.15             // Temp of air entering cooling device in K

+T2 = 293.15             // temperature at air required to leave in K

+P = 101.325             // pressure of entering air in kPa

+V = 300                 // volumatric flow rate of air in m^3/min

+M = 28.97               // molecular weight of air

+R = 8.314               // gas constant

+

+// data from psychometric chart

+SH1 = 0.0284            // in kg H2O/kg air

+pw1 = 4.4               // in kPa

+h1 = 115                // in kJ/kg air

+SH2 = 0.0148            // in kg H2O/kg air

+h2 = 58                 // in kJ/kg air

+hw3 = 83.95             // in kJ/kg at T = 20`C

+

+// at TDB = 40 `C;  RH =  1 

+ma1 = (101.325-pw1)*1e3*V*M/(R*1e3*T1)

+

+// masss balance for H2O: ma1*SH1 = ma2*SH2 + mw3

+mw3 = ma1*(SH1-SH2)         // amount of water condensed in kg/min

+

+// Energy balance: ma1*h1 = ma2*h2 + mw3*hw3 + Qc

+Qc = ma1*(h1-h2) - mw3*hw3      // in kJ/min

+Qc = Qc*1e-3/60               // units conversion kJ/min to MW

+

+

+// Output Results

+mprintf('The amount of water condensed = %4.1f kg/min' , mw3);

+mprintf('\n The rate of cooling required = %4.1f MW' , Qc);

diff --git a/1919/CH9/EX9.11/Ex9_11.sce b/1919/CH9/EX9.11/Ex9_11.sce
new file mode 100755
index 000000000..0ecc3d561
--- /dev/null
+++ b/1919/CH9/EX9.11/Ex9_11.sce
@@ -0,0 +1,41 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 11

+

+clear ;clc;

+

+//Given data

+RH1 = 0.8               // relative humidity moist air entering

+T1 = 308.15             // Temp of air entering cooling device in K

+P = 101.325             // pressure of entering air in kPa

+RH4 = 0.4               // relative humidity moist air leaving

+T4 = 293.15             // temperature at air required to leave in K

+m = 2                   // flowrate dry air in kg/s

+

+

+// data from psychometric chart

+SH1 = 0.029             // in kg H2O/kg air

+h1 = 110                // in kJ/kg air

+SH4 = 0.0055            // in kg H2O/kg air

+h4 = 34                 // in kJ/kg air

+SH2 = SH4

+RH2 = 1

+h2 = 19                 // in kJ/kg air

+T2 = 278.15             // in K

+hw3 = 20.97             // at 5 degree C in kJ/kg

+

+// masss balance for H2O: ma1*SH1 = ma2*SH2 + mw3

+mw3 = m*(SH1-SH2)         // amount of water condensed in kg/s

+

+// Energy balance: ma1*h1 = ma2*h2 + mw3*hw3 + Qc

+Qc = m*(h1-h2) - mw3*hw3      // in kJ/s

+

+Qh = m*(h4-h2)

+

+

+// Output Results

+mprintf('The amount of water condensed in refrigeration unit = %4.3f kg/s' , mw3);

+mprintf('\n The rate of cooling required in refrigeration unit  = %2.0f kW' , Qc);

+mprintf('\n The rate of heating required in refrigeration unit  = %2.0f kW' , Qh);

diff --git a/1919/CH9/EX9.12/Ex9_12.sce b/1919/CH9/EX9.12/Ex9_12.sce
new file mode 100755
index 000000000..f73e3cdc3
--- /dev/null
+++ b/1919/CH9/EX9.12/Ex9_12.sce
@@ -0,0 +1,35 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 12

+

+clear ;clc;

+

+//Given data

+RH1 = 0.3               // relative humidity of summer day atmospheric air

+T1 = 313.15             // Temperature of air in summer day in K

+V = 3                   // evaporative cooler rate in m^3/s

+t = 10                  // cooler operation time in hours

+M = 28.97               // molecular weight of air

+R = 8.314               // gas constant

+

+// data from psychometric chart for T = 40`C and RH = 0.3

+SH1 = 0.0135            // in kg H2O/kg air

+pw1 = 2.05              // in kPa

+// Follow the constant wet bulb temperature line(23`C) from that point 

+// till it intersects with RH = 0.8

+SH2 = 0.0184            // in kg H2O/kg air

+T2 = 27.3               // in degree C

+

+ma = (101.325-pw1)*1e3*V*M/(R*1e3*T1)

+

+mw = ma*(SH2-SH1)

+

+q_wat = mw*3600*10      // quantity of water required for 10 hours in kg

+

+

+// Output Results

+mprintf('The temperature of the air leaving the cooler = %4.1f degree C' , T2);

+mprintf('\n The specific humidity of air leaving cooler = %4.4f kg H2O/kg air' , SH2);

+mprintf('\n Quantity of water required to operate cooler for 10 hours = %4.1f kg',q_wat);

diff --git a/1919/CH9/EX9.13/Ex9_13.sce b/1919/CH9/EX9.13/Ex9_13.sce
new file mode 100755
index 000000000..4f14f6c5b
--- /dev/null
+++ b/1919/CH9/EX9.13/Ex9_13.sce
@@ -0,0 +1,51 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 13

+

+clear ;clc;

+

+//Given data

+mw3 = 1000              // cooling tower supply rate in kg/min

+T1 = 303.15             // Temp of air entering cooling tower in K

+RH1 = 0.3               // relative humidity of air entering cooler

+T2 = 308.15             // Temp of air leaving cooling tower in K

+RH2 = 0.8               // relative humidity of air leaving cooler

+T3 = 318.15             // Temp of water entering cooling tower in K

+T4 = 300.15             // Temp of water leaving cooling tower in K

+

+// subscript 1 and 2 denotes the state of air entering and leaving the cooling tower respectively

+// subscript 3 and 4 denotes the state of water entering and leaving the cooling tower respectively

+

+// data from psychometric chart for T = 30 degree C and RH = 0.3

+SH1 = 0.0078            // in kg H2O/kg air

+h1 = 51                 // in kJ/kg air

+// data from psychometric chart for T = 35 degree C and RH = 0.8

+SH2 = 0.029             // in kg H2O/kg air

+h2 = 110                // in kJ/kg air

+hw3 = 188.45            // in kJ/kg

+hw4 = 113.25            // in kJ/kg

+

+// mass balance for H2O: 

+//mw3-mw4 = ma*(SH2-SH1)

+

+// energy balance gives:

+// mw3*hw3 - mw4*hw4 = ma*(h2-h1)

+

+// x(1) = ma; x(2)= mw4;

+function[f] =F(x)

+    f(1) = mw3-x(2)-x(1)*(SH2-SH1);

+    f(2) = mw3*hw3 - x(2)*hw4 -x(1)*(h2-h1);

+endfunction

+x = [10 10];

+y = fsolve(x,F)

+

+ma = y(1);                      // air flow rate in kg/min

+mw4 = y(2);

+

+wat_mak = mw3-mw4;              // make up water required

+

+// Output Results

+mprintf('Make up water required = %4.2f kg/min' , wat_mak);

+mprintf('\n Air flow rate = %4.1f kg/min' , ma);

diff --git a/1919/CH9/EX9.2/Ex9_2.sce b/1919/CH9/EX9.2/Ex9_2.sce
new file mode 100755
index 000000000..ebcc7603b
--- /dev/null
+++ b/1919/CH9/EX9.2/Ex9_2.sce
@@ -0,0 +1,35 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 2

+

+clear ;clc;

+

+//Given data

+X1 = 0.5                // volume fraction of propane 

+X2 = 0.5                // volume fraction of oxygen 

+P1 = 0.1                // initial pressure of mixture in MPa 

+T1 = 300                // initial temperature of mixture in K

+P2 = 1.0                // final pressure of mixture in MPa

+N = 100                 // number moles of gas mixture

+Cp1 = 74.0565           // molar heat capacity of propone in kJ/mol K

+Cv1 = 65.7442           // molar heat capacity of propone in kJ/mol K

+Cp2 = 29.4912           // molar heat capacity of oxygen  in kJ/mol K

+Cv2 = 21.1776           // molar heat capacity of oxygen  in kJ/mol K

+

+// Calculations

+Cp_m = X1*Cp1 + X2*Cp2  // molar heat capacity of gas mixture

+Cv_m = X1*Cv1 + X2*Cv2  // molar heat capacity of gas mixture

+

+gam = Cp_m/Cv_m         // specific heat ratio

+

+T2 = T1*((P2/P1)^((gam-1)/gam))  // final temperature of mixture

+

+W = -N*Cv_m*(T2-T1)              // Work  required to compress gas mixture

+

+W = W*1e-3                       // units conversion J to kJ

+

+// Output Results

+mprintf('Final temperature of gas mixture = %4.1f K' ,T2);

+mprintf('\n Work  required to compress 100 mol of gas mixture = %4.2f kJ' ,W);

diff --git a/1919/CH9/EX9.3/Ex9_3.sce b/1919/CH9/EX9.3/Ex9_3.sce
new file mode 100755
index 000000000..ee7d849ee
--- /dev/null
+++ b/1919/CH9/EX9.3/Ex9_3.sce
@@ -0,0 +1,30 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 3

+

+clear ;clc;

+

+//Given data

+gam = 1.5               // specific heat ratio

+Cp1 = 20.7860           // molar heat capacity of Helium in kJ/mol K

+Cv1 = 12.4717           // molar heat capacity of Helium in kJ/mol K

+Cp2 = 28.6455           // molar heat capacity of Hydrogen  in kJ/mol K

+Cv2 = 20.3311           // molar heat capacity of Hydrogen in kJ/mol K

+

+// Calculations

+// X1 be the molefraction of He in the mixture

+// CP_m = X1*Cp1 + (1-X1)*Cp2

+// Cv_m = X1*Cv1 + (1-X1)*Cv2

+// gam = (X1*Cp1 + (1-X1)*Cp2)/(X1*Cv1 + (1-X1)*Cv2)

+deff('y=mol_frac(X1)', 'y = gam-(X1*Cp1 + (1-X1)*Cp2)/(X1*Cv1 + (1-X1)*Cv2)') 

+X1 = fsolve(0.1,mol_frac)

+

+

+X2 = (1-X1)*100         // volume composition of H2

+X1 = X1 * 100           // volume composition of He

+

+// Output Results

+mprintf('Volume percent composition of Helium = %4.2f ' ,X1);

+mprintf('\n Volume percent composition of Hydrogen = %4.2f ' ,X2);

diff --git a/1919/CH9/EX9.4/Ex9_4.sce b/1919/CH9/EX9.4/Ex9_4.sce
new file mode 100755
index 000000000..906276247
--- /dev/null
+++ b/1919/CH9/EX9.4/Ex9_4.sce
@@ -0,0 +1,42 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 4

+

+clear ;clc;

+

+//Given data

+V = 5                  // volume of tank in m^3

+X1 = 0.3               // volume fraction of Hydrogen

+X2 = 0.3               // volume fraction of Argon 

+X3 = 0.4               // volume fraction of Methane 

+Xf = 0.6               // final mixture composition of methane

+T = 300                // initial temperature of mixture in K

+P = 1                  // initial pressure of mixture in MPa

+R = 8.314              // gas constant

+

+// Calculations

+p1 = X1*P              // pressure of Hydrogen

+p2 = X2*P              // pressure of Argon

+p3 = X3*P              // pressure of Methane

+

+N1 = p1*1e3*V/(R*T)        // pressure of Hydrogen

+N2 = p2*1e3*V/(R*T)        // pressure of Argon

+N3 = p3*1e3*V/(R*T)        // pressure of Methane

+

+

+

+// to determine the number of moles of methane in final mixture

+deff('y=moles(Nf)', 'y = Xf-(Nf/(N1+N2+Nf))') 

+Nf = fsolve(0.1,moles)

+

+CH4_add = Nf-N3         // CH4 moles to be added

+

+N = N1 + N2 + Nf        // total number of moles

+

+Pf = N*R*T/V * 1e-3     // Final pressure in MPa

+

+// Output Results

+mprintf('Amount of methane to be added = %4.4f kmol' , CH4_add);

+mprintf('\n Final pressure of mixture in Tank = %4.1f MPa' , Pf);

diff --git a/1919/CH9/EX9.5/Ex9_5.sce b/1919/CH9/EX9.5/Ex9_5.sce
new file mode 100755
index 000000000..82aa4fbd8
--- /dev/null
+++ b/1919/CH9/EX9.5/Ex9_5.sce
@@ -0,0 +1,48 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 5

+

+clear ;clc;

+

+//Given data

+V = 1                  // volume of tank in m^3

+P1 = 0.1               // pressure of tank with He in MPa

+T1 = 300               // temperature of tank with He in K

+P2 = 0.4               // pressure of tank with CO2 in MPa

+T2 = 600               // temperature of tank with CO2 in K

+Pf = 0.4               // pressure of tank after Co2 enters in MPa

+Cv1 = 12.4717          // molar heat capacity of He in kJ/kmol K

+Cv2 = 28.8354          // molar heat capacity of CO2 in kJ/kmol K

+Cp2 = 37.1444          // molar heat capacity of CO2 in kJ/kmol K

+R = 8.314              // gas constant

+

+// Calculations

+// N2*h = Uf-U0         (A)// First law of thermodynamics of transient flow

+// U0 = N1*Cv1*T1

+U0 = P1*1e3*V*Cv1/(R)       // in kJ

+

+// Uf = (N1*Cv1 + N2*Cv2) * Tf

+//Tf = Pf*V/((N1+N2)*R)

+N1 = P1*1e3*V/(R*T1)

+

+// Uf = (N1*Cv1 + N2*Cv2) * Pf*V/((N1+N2)*R)

+

+// N2*h = (N1*Cv1 + N2*Cv2) * Pf*V/((N1+N2)*R) - U0

+

+// to determine the number of moles of CO2

+deff('y=moles(N2)', 'y = N2*Cp2*T2 - ((N1*Cv1 + N2*Cv2) * Pf*1e3*V/((N1+N2)*R)) + U0') 

+N2 = fsolve(0.1,moles)

+

+Tf = Pf*1e3*V/((N1+N2)*R)       // Final temperature of mixture in tank

+

+X2 = N2/(N1+N2)             // molefraction of CO2

+X1 = 1 - X2                 // molefraction of He

+

+// Output Results

+mprintf('Final temperature of mixture in tank = %4.2f K' , Tf);

+mprintf('\n The amount of CO2 enters the tank = %4.4f kmol' , N2);

+mprintf('\n Composition of CO2 in mixture = %4.4f ' , X2);

+mprintf('\n Composition of He in mixture  = %4.4f ' , X1);

+

diff --git a/1919/CH9/EX9.6/Ex9_6.sce b/1919/CH9/EX9.6/Ex9_6.sce
new file mode 100755
index 000000000..252a04e9c
--- /dev/null
+++ b/1919/CH9/EX9.6/Ex9_6.sce
@@ -0,0 +1,34 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 6

+

+clear ;clc;

+

+//Given data

+V = 20*40*6             // dimensions of auditorium in cubic meters

+RH = 0.6                // relative humidy

+P = 101.325             // pressure of auditorium in kPa

+T = 303.15              // Temperature auditorium in K

+Ps_20 = 2.339           // saturation pressure of water at 20 degree C

+Ps_25 = 3.169           // saturation pressure of water at 25 degree C

+Ps_30 = 4.246           // saturation pressure of water at 30 degree C

+M = 18                  // molecular weight of water

+R = 8.314               // gas constant

+

+// Calculations

+pw = RH*Ps_30           // partial pressure of water at 30 degree C

+

+SH = 0.622*pw/(P-pw)    // specific humidity of air

+

+mw = pw*V*M/(R*T)       // mass of water vapor in the auditorium

+

+ps = pw                 // at dew point from steam table T = 21.3 degree C

+

+// Output Results

+mprintf('(a) Partial pressure of water at 30 degree C = %4.4f kPa' , pw);

+mprintf('\n (b) Specific humidity of air = %4.3f kg H20/kg air' , SH);

+mprintf('\n (c) Mass of water vapor in the auditorium = %4.2f kg' , mw);

+mprintf('\n (d) Dew point  = %4.4f kPa' , ps);

+

diff --git a/1919/CH9/EX9.7/Ex9_7.sce b/1919/CH9/EX9.7/Ex9_7.sce
new file mode 100755
index 000000000..50a2febf4
--- /dev/null
+++ b/1919/CH9/EX9.7/Ex9_7.sce
@@ -0,0 +1,43 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 7

+

+clear ;clc;

+

+//Given data

+T1 = 313.15             // Temp of air entering adiabatic saturator in K

+P1 = 101.325            // pressure air entering saturator in kPa

+T3 = 298.15             // Temp of air leaving adiabatic saturator in K

+P3 = 101.325            // pressure air entering saturator in kPa

+

+ps = 3.169              // from steam tables at T = 313.15K (in kPa)

+RH = 1                  // RH of steam leaving adiabatic saturator is 1

+pw = ps                 // from steam tables at 25 degree C

+

+SH_3 = 0.622*pw/(P3-pw)  // specific humidity of air

+

+hfg_3 = 2442.3           // in kJ/kg

+hg_1  = 2574.3          // in kJ/kg

+hf_2  = 104.89          // in kJ/kg

+Cp_air = 1.0045         // in kJ/kg K

+

+SH1 = (Cp_air*(T3-T1) + SH_3*hfg_3) / (hg_1-hf_2)

+

+//SH1 = 0.622*pw_2/(P1-pw_2)

+deff('y=dew(pw_2)', 'y = SH1 - 0.622*pw_2/(P1-pw_2)') 

+pw_2 = fsolve(0.01,dew)

+

+ps_40 = 7.384           // from steam table at 40 degree C

+

+RH1 = pw_2/ps_40

+

+ps_2 = pw_2         // at dew point ps = pw

+// Hence TDP = 18.8 degree C (from steam table)

+

+// Output Results

+mprintf('Specific humidity of entering air at 40 degree C = %4.4f kg H20/kg air' , SH1);

+mprintf('\n Relative humidity of air at 40 degree C = %4.3f ' , RH1);

+mprintf('\n Dew point  = %4.3f kPa' , ps_2);

+

diff --git a/1919/CH9/EX9.8/Ex9_8.sce b/1919/CH9/EX9.8/Ex9_8.sce
new file mode 100755
index 000000000..85dd8fe60
--- /dev/null
+++ b/1919/CH9/EX9.8/Ex9_8.sce
@@ -0,0 +1,48 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 8

+

+clear ;clc;

+

+//Given data

+T1 = 313.15             // Temp of air entering adiabatic saturator in K

+P1 = 101.325            // pressure air entering saturator in kPa

+T3 = 298.15             // Temp of air leaving adiabatic saturator in K

+P3 = 101.325            // pressure air entering saturator in kPa

+

+ps = 3.169              // from steam tables at T = 313.15K (in kPa)

+RH = 1                  // RH of steam leaving adiabatic saturator is 1

+pw = ps                 // from steam tables at 25 degree C

+

+SH_3 = 0.622*pw/(P3-pw)  // specific humidity of air

+

+hfg_3 = 2442.3           // in kJ/kg

+hg_1  = 2574.3          // in kJ/kg

+hf_2  = 104.89          // in kJ/kg

+Cp_air = 1.0045         // in kJ/kg K

+

+SH1 = (Cp_air*(T3-T1) + SH_3*hfg_3) / (hg_1-hf_2)

+

+//SH1 = 0.622*pw_2/(P1-pw_2)

+deff('y=dew(pw_2)', 'y = SH1 - 0.622*pw_2/(P1-pw_2)') 

+pw_2 = fsolve(0.01,dew)

+

+ps_40 = 7.384           // from steam table at 40 degree C

+

+RH1 = pw_2/ps_40

+

+ps_2 = pw_2         // at dew point ps = pw

+// Hence TDP = 18.8 degree C (from steam table)

+

+// enthalphy of mixture

+//h1 = h_a1 + SH1*h_w1

+h1 = Cp_air*(T1-273.15) + SH1*hg_1

+

+

+// Output Results

+mprintf('Specific humidity of entering air at 40 degre C = %4.3f kg H20/kg air' , SH1);

+mprintf('\n Relative humidity of air at 40 degree C = %3.2f ' , RH1);

+mprintf('\n Dew point  = %2.1f kPa' , ps_2);

+mprintf('\n Enthalphy of mixture  = %2.0f kJ/kg air' , h1);

diff --git a/1919/CH9/EX9.9/Ex9_9.sce b/1919/CH9/EX9.9/Ex9_9.sce
new file mode 100755
index 000000000..0f089681a
--- /dev/null
+++ b/1919/CH9/EX9.9/Ex9_9.sce
@@ -0,0 +1,38 @@
+

+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 9

+

+clear ;clc;

+

+//Given data

+V1 = 2                  // volume of air conditioner delivers in cubic meter

+T1 = 278.15             // Temp of air completely saturates in K

+V2 = 5                  // volume of air mixed in cubic meter

+T2 = 308.15             // Temp of air mixed in K

+RH = 0.4                // relative humidity in room

+R = 8.314               // gas constant

+M = 28.97               // molecular weight of air

+

+// data from psychometric chart

+SH1 = 0.0056            // in kg H2O/kg air

+pw1 = 0.8               // in kPa

+// at TDB = 5 degree C;  RH =  1 

+SH2 = 0.0138            // in kg H2O/kg air

+pw2 = 2.2               // in kPa

+

+// at TDB = 5 `C;  RH =  1 

+ma1 = (101.325-pw1)*1e3*V1*M/(R*1e3*T1)

+ma2 = (101.325-pw2)*1e3*V2*M/(R*1e3*T2)

+

+//ma1/ma2 = (SH2-SH3)/(SH3-SH1)

+deff('y=SH(SH3)', 'y = ma1/ma2 - (SH2-SH3)/(SH3-SH1)') 

+SH3 = fsolve(0.01,SH)

+

+// Join the states 1 and 2 by a straight line and read TDB = 26 degree C and

+// RH3=0.55 at the intersection of the line with SH3=0.0113 kg H20/kg air

+

+

+// Output Results

+mprintf('Specific humidity of air after mixing = %4.4f kg H20/kg air' , SH3);