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Diffstat (limited to '926/CH5')
20 files changed, 444 insertions, 0 deletions
diff --git a/926/CH5/EX5.1/Chapter5_Example1.sce b/926/CH5/EX5.1/Chapter5_Example1.sce new file mode 100644 index 000000000..d2ca16bb0 --- /dev/null +++ b/926/CH5/EX5.1/Chapter5_Example1.sce @@ -0,0 +1,52 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 1, Page 111
+//Title: Calculation of composition of a saturated mixture
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = [442,745 760]; //Vapor pressure of ethyl ether, working pressure and standard pressure in mm Hg
+T1 = [20 0]; //Temperature of system and standard temperature in degree C
+V = 1; //cu ft of mixture(Basis of calculation for part(a) )
+MW = [74 28]; //Molecular weight of ethyl ether and nitrogen in lb/lb-mole
+
+//CALCULATIONS
+//Part(a)
+V1 = V*(P(1)/P(2)); //Pure component volume of vapor in cu ft
+v1 = V1*100; //Composition by volume of Ether vapor
+v2 = (1-V1)*100; //Composition by volume of Nitrogen
+//Part(b)
+W1 = V1*MW(1); //lb of ether vapor present
+W2 = (1-V1)*MW(2); //lb of Nitrogen present
+W = W1+W2; //Total weight of mixture in lb
+w1 = W1*100/W; //Composition by weight of Ether vapor
+w2 = W2*100/W; //Composition by weight of Nitrogen
+//Part(c)
+T = T1+273; //Converting temperature in K
+V2 = 359*(P(3)/P(2))*(T(1)/T(2)); //cu ft of mixture
+w3 = W1/V2; //lb of vapor per cu ft of mixture
+//Part(d)
+w4 = W1/W2; //lb of vapor per lb lb of nitrogen
+//Part(e)
+v3 = V1/(1-V1); //lb mole of vapor per lb moles of Nitrogen
+
+//OUTPUT
+// Console output
+mprintf('\n (a) Composition by Volume \n Ether Vapor %3.1f %% \n Nitrogen %3.1f %% ',v1,v2);
+mprintf('\n (b) Composition by Weight \n Ether Vapor %3.1f %% \n Nitrogen %3.1f %% ',w1,w2);
+mprintf('\n (c) Weight of ether per cu ft of mixture is %4.3f lb ',w3);
+mprintf('\n (d) Weight of vapor per lb Nitrogen is %3.2f lb ',w4);
+mprintf('\n (e) Moles of vapor per mole of nitrogen is %4.3f',v3);
+
+// File output
+fd= mopen('.\Chapter5_Example1_Output.txt','w');
+mfprintf(fd,'\n (a) Composition by Volume \n Ether Vapor %3.1f %% \n Nitrogen %3.1f %%',v1,v2);
+mfprintf(fd,'\n (b) Composition by Weight \n Ether Vapor %3.1f %% \n Nitrogen %3.1f %%',w1,w2);
+mfprintf(fd,'\n (c) Weight of ether per cu ft of mixture is %4.3f lb ',w3);
+mfprintf(fd,'\n (d) Weight of vapor per lb Nitrogen is %3.2f lb ',w4);
+mfprintf(fd,'\n (e) Moles of vapor per mole of nitrogen is %4.3f',v3);
+mclose(fd);
+
+//=============================END OF PROGRAM==================================
diff --git a/926/CH5/EX5.1/Chapter5_Example1_Output.txt b/926/CH5/EX5.1/Chapter5_Example1_Output.txt new file mode 100644 index 000000000..685f90ae6 --- /dev/null +++ b/926/CH5/EX5.1/Chapter5_Example1_Output.txt @@ -0,0 +1,10 @@ +
+ (a) Composition by Volume
+ Ether Vapor 59.3 %
+ Nitrogen 40.7 %
+ (b) Composition by Weight
+ Ether Vapor 79.4 %
+ Nitrogen 20.6 %
+ (c) Weight of ether per cu ft of mixture is 0.112 lb
+ (d) Weight of vapor per lb Nitrogen is 3.86 lb
+ (e) Moles of vapor per mole of nitrogen is 1.459
\ No newline at end of file diff --git a/926/CH5/EX5.11/Chapter5_Example11.sce b/926/CH5/EX5.11/Chapter5_Example11.sce new file mode 100644 index 000000000..2c2a839d3 --- /dev/null +++ b/926/CH5/EX5.11/Chapter5_Example11.sce @@ -0,0 +1,35 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 11, Page 127
+//Title: Calculation of dry bulb and wet bulb temperature
+//=============================================================================
+clear
+clc
+
+//INPUT
+T = [190 90]; //Dry bulb and wet bulb temperature of air entering dryer in degree F
+n = 0.028; //lb-mole of water evaporated per lb mole of dry air entering
+
+//DATA FROM GRAPH
+mH = 0.011; //Molal humidity corresponding to DBT of 190 degree F and WBT of 90 degree F from Fig 19, Pg 120
+DBT = 116; //Dry bulb temperature in degree F corresponding to molal humidity of 0.039 from Fig 19, Pg 120
+PS = 35; //Percentage saturation corresponding to molal humidity of 0.039 and DBT of 116 degree F from Fig 19, Pg 120
+
+//CALCULATIONS
+mH1 = n+mH; //Molal humidity of leaving air
+WBT = T(2); //Wet bulb temperature of leaving air in degree F
+
+//OUTPPUT
+// Console output
+mprintf('\n Dry bulb temperature of air leaving the drier = %3.0f degree F',WBT);
+mprintf('\n wet bulb temperature of air leaving the drier = %2.0f degree F',WBT);
+mprintf('\n Percentage saturation of air leaving the drier is %2.0f percent',PS);
+
+// File output
+fd= mopen('.\Chapter5_Example11_Output.txt','w');
+mfprintf(fd,'\n Dry bulb temperature of air leaving the drier = %3.0f degree F',WBT);
+mfprintf(fd,'\n wet bulb temperature of air leaving the drier = %2.0f degree F',WBT);
+mfprintf(fd,'\n Percentage saturation of air leaving the drier is %2.0f percent',PS);
+mclose(fd);
+
+//=============================END OF PROGRAM==================================
diff --git a/926/CH5/EX5.11/Chapter5_Example11_Output.txt b/926/CH5/EX5.11/Chapter5_Example11_Output.txt new file mode 100644 index 000000000..8866d0175 --- /dev/null +++ b/926/CH5/EX5.11/Chapter5_Example11_Output.txt @@ -0,0 +1,4 @@ +
+ Dry bulb temperature of air leaving the drier = 90 degree F
+ wet bulb temperature of air leaving the drier = 90 degree F
+ Percentage saturation of air leaving the drier is 35 percent
\ No newline at end of file diff --git a/926/CH5/EX5.12/Chapter5_Example12.sce b/926/CH5/EX5.12/Chapter5_Example12.sce new file mode 100644 index 000000000..669dd7fa4 --- /dev/null +++ b/926/CH5/EX5.12/Chapter5_Example12.sce @@ -0,0 +1,26 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 12, Page 127
+//Title: Calculation of molal humidity
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = 1; //Pressure of entering gas in atm
+DBT = 120; //Temperature of entering gas in degree F
+
+//DATA FROM GRAPH
+WBT = 71; //Wet bulb temperature in degree F corresponding to DBT of 120 degree F from Fig 20 Page 122
+mH = 0.027; //Molal humidity corresponding to DBT of 120 degree F from Fig 20 Page 122
+
+//OUTPUT
+// Console output
+mprintf('\n The temperature and molal humidity of saturated carbon dioxide leaving the chamber is %2.0f degree F and %4.3f respectively',WBT,mH);
+
+// File output
+fd= mopen('.\Chapter5_Example12_Output.txt','w');
+mfprintf(fd,'\n The temperature and molal humidity of saturated carbon dioxide leaving the chamber is %2.0f degree F and %4.3f respectively',WBT,mH);
+mclose(fd);
+
+//=============================END OF PROGRMAM=================================
diff --git a/926/CH5/EX5.12/Chapter5_Example12_Output.txt b/926/CH5/EX5.12/Chapter5_Example12_Output.txt new file mode 100644 index 000000000..ead55fe14 --- /dev/null +++ b/926/CH5/EX5.12/Chapter5_Example12_Output.txt @@ -0,0 +1,2 @@ +
+ The temperature and molal humidity of saturated carbon dioxide leaving the chamber is 71 degree F and 0.027 respectively
\ No newline at end of file diff --git a/926/CH5/EX5.2/Chapter5_Example2.sce b/926/CH5/EX5.2/Chapter5_Example2.sce new file mode 100644 index 000000000..426e5302a --- /dev/null +++ b/926/CH5/EX5.2/Chapter5_Example2.sce @@ -0,0 +1,34 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 2, Page 113
+//Title: Calculation of relative saturation and percentage saturation
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = [745 184.8] //Working pressure and vapor pressure of acetone at 20 degree C in mm Hg
+v = 14.8 //Composition by volume of acetone
+
+//CALCULATIONS
+P1 = P(1)*v/100; //Partial pressure of acetone in mm Hg
+RS = P1*100/P(2); //Relative saturation of mixture
+n1 = v/100; //lb mole of acetone
+n2 = 1-n1; //lb mole of nitrogen
+n3 = n1/n2; //lb moles of acetone per lb moles of nitrogen
+V1 = P(2)*100/P(1); //Percentage by volume of acetone
+n4 = V1/100; //lb moles of acetone
+n5 = 1-n4; //lb moles of nitrogen
+n6 = n4/n5; //Moles of acetone per moles of nitrogen
+PS = n3*100/n6; //Percentage saturation
+
+//OUTPUT
+//Console output
+mprintf('\n Relative saturation of the mixture at given conditions = %3.1f %% \n Percentage saturation = %3.1f %%',RS,PS);
+
+// File output
+fd= mopen('.\Chapter5_Example2_Output.txt','w');
+mfprintf(fd,'\n Relative saturation of the mixture at given conditions = %3.1f %% \n Percentage saturation = %3.1f %%',RS,PS);
+mclose(fd);
+
+//=============================END OF PROGRAM==================================
diff --git a/926/CH5/EX5.2/Chapter5_Example2_Output.txt b/926/CH5/EX5.2/Chapter5_Example2_Output.txt new file mode 100644 index 000000000..8cabf5efd --- /dev/null +++ b/926/CH5/EX5.2/Chapter5_Example2_Output.txt @@ -0,0 +1,3 @@ +
+ Relative saturation of the mixture at given conditions = 59.7 %
+ Percentage saturation = 52.7 %
\ No newline at end of file diff --git a/926/CH5/EX5.4/Chapter5_Example4.sce b/926/CH5/EX5.4/Chapter5_Example4.sce new file mode 100644 index 000000000..67097606e --- /dev/null +++ b/926/CH5/EX5.4/Chapter5_Example4.sce @@ -0,0 +1,36 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 4, Page 114
+//Title: Calculation of dew point
+//=============================================================================
+clear
+clc
+
+//INPUT
+v1 = 10.1; //Composition by volume of benzene
+P = [750 750 700]; //Various pressurea in mm Hg at which dew point is to be computed
+
+//CALCULATIONS
+P1 = P(1)*v1/100; //Partial pressure of benzene in mm Hg at 25 degree C and 750 mm Hg
+T1 = 20; //Temperature in degree C corresponding to pressure P1 obtained from vapor-pressure data of Benzene Fig. 15 Page 84
+P2 = P(2)*v1/100; //Partial pressure of benzene in mm Hg at 30 degree C and 750 mm Hg
+T2 = 20; //Temperature in degree C corresponding to pressure P2 obtained from vapor-pressure data of Benzene Fig. 15 Page 84
+P3 = P(3)*v1/100; //Partial pressure of benzene in mm Hg at 30 degree C and 700 mm Hg
+T3 = 18.7; //Temperature in degree C corresponding to pressure P3 obtained from vapor-pressure data of Benzene Fig. 15 Page 84
+
+//OUTPUT
+// Console output
+mprintf('\n Dew point of benzene vapor and air mixture at \n (a) 25 degree C and 750 mm Hg = %2.0f degee C ',T1);
+mprintf('\n (b) 30 degree C and 750 mm Hg = %2.0f degree C ',T3);
+mprintf('\n (c) 30 degree C and 700 mm Hg = %3.1f degree C',T3);
+mprintf('\n Above results shows that the dew point does not depend on temperature but vary with the total pressure');
+
+// File output
+fd= mopen('.\Chapter5_Example4_Output.txt','w');
+mfprintf(fd,'\n Dew point of benzene vapor and air mixture at \n (a) 25 degree C and 750 mm Hg = %2.0f degee C ',T1);
+mfprintf(fd,'\n (b) 30 degree C and 750 mm Hg = %2.0f degree C ',T3);
+mfprintf(fd,'\n (c) 30 degree C and 700 mm Hg = %3.1f degree C',T3);
+mfprintf(fd,'\n Above results shows that the dew point does not depend on temperature but vary with the total pressure');
+mclose(fd);
+
+//=========================END OF PROGRAM======================================
diff --git a/926/CH5/EX5.4/Chapter5_Example4_Output.txt b/926/CH5/EX5.4/Chapter5_Example4_Output.txt new file mode 100644 index 000000000..a57b61d69 --- /dev/null +++ b/926/CH5/EX5.4/Chapter5_Example4_Output.txt @@ -0,0 +1,6 @@ +
+ Dew point of benzene vapor and air mixture at
+ (a) 25 degree C and 750 mm Hg = 20 degee C
+ (b) 30 degree C and 750 mm Hg = 19 degree C
+ (c) 30 degree C and 700 mm Hg = 18.7 degree C
+ Above results shows that the dew point does not depend on temperature but vary with the total pressure
\ No newline at end of file diff --git a/926/CH5/EX5.5/Chapter5_Example5.sce b/926/CH5/EX5.5/Chapter5_Example5.sce new file mode 100644 index 000000000..29cb289c8 --- /dev/null +++ b/926/CH5/EX5.5/Chapter5_Example5.sce @@ -0,0 +1,54 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 5, Page 116
+//Title: Calculation of different concentrations in vaporization process
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = [750 760]; //Barometric pressure and standard pressure in mm Hg
+PS = [116 185]; //Vapor pressure of acetone at 10 degree C and 20 degree C in mm Hg
+T = [30 25 0]; //Temperature of nitrogen entering evaporator and standard temperature in degree C
+n = 1; //lb mole of nitrogen (Basis of calculation in part(c),part(d) and part(e))
+MW = 58; //Molecualar weight of acetone in lb/lb mole
+
+//CALCULATIONS
+//Part(a)
+//Entering gases
+PP1 = PS(1); //Partial pressure of acetone in mm Hg
+PP2 = P(1)-PP1; //Partial pressure of nitrogen in mm Hg
+n1 = PP1/PP2; //Moles of acetone per mole of nitrogen
+//leaving gases
+PP3 = PS(2); //Partial pressure of acetone in mm Hg
+PP4 = P(1)-PS(2); //Partial pressure of nitrogen in mm Hg
+n2 = PP3/PP4; //Moles of acetone per mole of nitrogen
+//Part(b)
+n3 = n2-n1; //lb mole of acetone evaporated
+//Part(c)
+Tkelvin = T+273; //Converting temperature fron degree C to kelvin
+n4 = n+n1; //lb mole of total gas entering the process
+v1 = n4*359*(P(2)/P(1))*(Tkelvin(1)/Tkelvin(3)); //Volume of gas entering in cu ft
+m1 = MW*n3; //Weight of acetone evaporated
+m2 = m1*1000/v1; //lb of acetone evaporated per 1000 cu ft of gas entering
+//Part(d)
+n5 = n+n2; //lb mole of total gas leaving the process
+v2 = n5*359*(P(2)/P(1))*(Tkelvin(2)/Tkelvin(3)); //Volume of gas leaving in cu ft
+v3 = v2*1000/v1; //cu ft of gas leaving per 1000 cu ft of gas entering the process
+
+//OUTPUT
+// Console output
+mprintf('\n (a) Vapor concentration of gases entering and leaving = %4.3f and %4.3f respectively',n1,n2);
+mprintf('\n (b) Moles of acetone evaporated = %4.3f lb moles',n3);
+mprintf('\n (c) Weight of acetone evaporated per 1000 cu ft of gas entering = %3.1f lb',m2);
+mprintf('\n (d) Volume of gas leaving per 1000 cu ft of gas entering = %4.0f cu ft',v3);
+
+// File output
+fd= mopen('.\Chapter5_Example5_Output.txt','w');
+mfprintf(fd,'\n (a) Vapor concentration of gases entering and leaving = %4.3f and %4.3f respectively',n1,n2);
+mfprintf(fd,'\n (b) Moles of acetone evaporated = %4.3f lb moles',n3);
+mfprintf(fd,'\n (c) Weight of acetone evaporated per 1000 cu ft of gas entering = %3.1f lb',m2);
+mfprintf(fd,'\n (d) Volume of gas leaving per 1000 cu ft of gas entering = %4.0f cu ft',v3);
+mclose(fd);
+
+//===========================END OF PROGRAM====================================
diff --git a/926/CH5/EX5.5/Chapter5_Example5_Output.txt b/926/CH5/EX5.5/Chapter5_Example5_Output.txt new file mode 100644 index 000000000..a1b23ddc3 --- /dev/null +++ b/926/CH5/EX5.5/Chapter5_Example5_Output.txt @@ -0,0 +1,5 @@ +
+ (a) Vapor concentration of gases entering and leaving = 0.183 and 0.327 respectively
+ (b) Moles of acetone evaporated = 0.144 lb moles
+ (c) Weight of acetone evaporated per 1000 cu ft of gas entering = 17.5 lb
+ (d) Volume of gas leaving per 1000 cu ft of gas entering = 1104 cu ft
\ No newline at end of file diff --git a/926/CH5/EX5.6/Chapter5_Example6.sce b/926/CH5/EX5.6/Chapter5_Example6.sce new file mode 100644 index 000000000..e32eab562 --- /dev/null +++ b/926/CH5/EX5.6/Chapter5_Example6.sce @@ -0,0 +1,56 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 6, Page 118
+//Title: Calculation of different concentrations in vaporization process
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = [750 760]; //Given pressure and standard pressure in mm Hg
+RH = 0.8; //Relative humidity
+PS = [17.5 9.2]; //Vapor pressure of water in mm Hg at 20 degree C and 10 degree C
+P_new = 35; //Increased pressure by condensing out some water in psi
+V = 1000; //Volume of original wet water in cu ft (basis for calculation in part (c))
+T = [20 0]; //Given temperature and standard temperature in degree C
+T_new = 10; //Reduced temperature by condensing out some water in degree C
+MW = 18; //Molecular weight of water in lb/lb-mole
+
+//CALCULATIONS
+//Part(a)
+PP1 = RH*PS(1); //initial partial pressure of water in mm Hg
+MH1 = PP1/(P(1)-PP1); //Initial molal humidity
+//Part(b)
+PP2 = PS(2); //Final partial pressure of water in mm Hg
+P_new1 = 35*(P(2)/14.7); //Final total pressure in mm Hg
+MH2 = PP2/(P_new1-PP2); //Final molal humidity
+//Part(c)
+T1 = T+273; //Given temperature and standard temperature in K
+PP3 = P(1)-PP1; //Partial pressure of dry air in mm Hg
+PV = V*(PP3/P(2))*(T1(2)/T1(1)); //partial volume of dry air at standard condition in cu ft
+n1 = PV/359; //Moles of dry air in lb-moles
+n2 = n1*MH1; //lb-mole of water originally present
+n3 = n1*MH2; //lb-mole of water finally present
+n4 = n2-n3; //lb-mole of water condensed
+w1 = n4*MW; //lb of water condensed
+//Part(d)
+T_new1 = T_new+273; //Final temperature in K
+n5 = n1+n3; //Total wet air finally present in lb-moles
+v1 = n5*359*(P(2)/P_new1)*(T_new1/T1(2)); //Final volume of wet air in cu ft
+
+//OUTPUT
+// Console output
+mprintf('\n (a) Initial molal humidity of air = %5.4f',MH1);
+mprintf('\n (b) Final molal humidity of air = %5.4f',MH2);
+mprintf('\n (c) Amount of water condensed = %4.3f lb',w1);
+mprintf('\n (d) Final volume of wet air = %3.0f cu ft',v1);
+
+// File output
+fd= mopen('.\Chapter5_Example6_Output.txt','w');
+mfprintf(fd,'\n (a) Initial molal humidity of air = %5.4f',MH1);
+mfprintf(fd,'\n (b) Final molal humidity of air = %5.4f',MH2);
+mfprintf(fd,'\n (c) Amount of water condensed = %4.3f lb',w1);
+mfprintf(fd,'\n (d) Final volume of wet air = %3.0f cu ft',v1);
+mclose(fd);
+
+//===============================END OF PROGRAM=================================
diff --git a/926/CH5/EX5.6/Chapter5_Example6_Output.txt b/926/CH5/EX5.6/Chapter5_Example6_Output.txt new file mode 100644 index 000000000..10f9c5210 --- /dev/null +++ b/926/CH5/EX5.6/Chapter5_Example6_Output.txt @@ -0,0 +1,5 @@ +
+ (a) Initial molal humidity of air = 0.0190
+ (b) Final molal humidity of air = 0.0051
+ (c) Amount of water condensed = 0.629 lb
+ (d) Final volume of wet air = 395 cu ft
\ No newline at end of file diff --git a/926/CH5/EX5.7/Chapter5_Example7.sce b/926/CH5/EX5.7/Chapter5_Example7.sce new file mode 100644 index 000000000..dcc8a2c22 --- /dev/null +++ b/926/CH5/EX5.7/Chapter5_Example7.sce @@ -0,0 +1,35 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 7, Page 119
+//Title: Calculation of condensation pressure
+//=============================================================================
+clear
+clc
+
+//INPUT
+VP = [229.2 89.1]; //Vapour pressure of acetone at 25 degree C and 5 degree C
+P = 750; //Pressure of given mixture in mm Hg
+w = 90; //Amount of acetone to be condensed;
+
+
+//CALCULATION
+n1 = VP(1)/P; //lb moles of acetone present
+n2 = 1-n1; //lb moles of flue gases present
+n3 = (1-(w/100))*n1; //lb moles of acetone in final mixture
+n4 = n2+n3; //lb moles of final mixture of gas
+PP = VP(2); //Partial pressure of acetone in final mixture
+m = n3*100/n4; //Mole percent of acetone in final mixture
+Pf = PP/(m/100); //Final pressure in mm Hg
+
+//OUTPUT
+// Console output
+mprintf('\n Final pressure after condensation = %.0f mm Hg',Pf);
+
+// File output
+fd= mopen('.\Chapter5_Example7_Output.txt','w');
+mfprintf(fd,'\n Final pressure after condensation = %.0f mm Hg',Pf);
+mclose(fd);
+
+//=========================END OF PROGRAM======================================
+
+// Remark: Difference between the nswer given in the textbook (2110 mm Hg) and that computed using scilab code (2114 mm Hg) is due to round off error
diff --git a/926/CH5/EX5.7/Chapter5_Example7_Output.txt b/926/CH5/EX5.7/Chapter5_Example7_Output.txt new file mode 100644 index 000000000..5389ba5df --- /dev/null +++ b/926/CH5/EX5.7/Chapter5_Example7_Output.txt @@ -0,0 +1,2 @@ +
+ Final pressure after condensation = 2114 mm Hg
\ No newline at end of file diff --git a/926/CH5/EX5.8/Chapter5_Example8.sce b/926/CH5/EX5.8/Chapter5_Example8.sce new file mode 100644 index 000000000..e36a859ad --- /dev/null +++ b/926/CH5/EX5.8/Chapter5_Example8.sce @@ -0,0 +1,43 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 8, Page 123
+//Title: Estimation of humidity and saturation
+//=============================================================================
+clear
+clc
+
+//INPUT
+T = [100 85]; //Initial and wet bulb temperature of entering air in degree F
+P = 1; //Pressure of entering air in atm
+n = 1; //lb-moles of moisture free air
+T1 = 492; //Temperature corresponsing to STP in Rankine
+T2 = [120 115.3]; //Initial and wet bulb temperature of air in degree F in part(b)
+MW = 18; //Molecular weight of water
+
+//DATA FROM GRAPH
+PS = [52 84]; //Percentage saturation at WBT of 85 and 115.3 degree C respectively obtained from Fig 20 Page 122
+MH = [0.037 0.11 ]; //Molal humidity at WBT of 85 and 115.3 degree C respectively obtained from Fig 20 Page 122
+DP = 80.5; //Dew point corresponding to WBT of 85 degree F obtained from Fig 20 Page 122
+
+
+//CALCULATION
+//Part(b)
+n1 = n+MH(1); //lb-moles of wet air entering
+T3 = T+460; //initial temperature of entering air in Rankine
+v = n1*359*(T3(1)/T1); //Volume of wet air entering in cu ft
+n2 = MH(2)-MH(1); //lb-moles of water evaporated
+w1 = n2*MW; //lb of water evaporated
+w2 = w1*1000/v; //lb of water evaporated per 1000 cu ft of entering wet air
+
+//OUTPUT
+// Console output
+mprintf('\n (a) Molal humidity of the air = %4.3f \n Percentage saturation = %2.0f %%\n Dew point = %3.2f degree F',MH(1),PS(1),DP);
+mprintf('\n (b) Percentage saturation of the air leaving the evaporator = %2.0f %% \n weight of the water evaporated = %2.1f lb',PS(2),w2);
+
+// File output
+fd= mopen('.\Chapter5_Example8_Output.txt','w');
+mfprintf(fd,'\n (a) Molal humidity of the air = %4.3f \n Percentage saturation = %2.0f %%\n Dew point = %3.2f degree F',MH(1),PS(1),DP);
+mfprintf(fd,'\n (b) Percentage saturation of the air leaving the evaporator = %2.0f %% \n weight of the water evaporated = %2.1f lb',PS(2),w2);
+mclose(fd);
+
+//=============================END OF PROGRMAM=================================
diff --git a/926/CH5/EX5.8/Chapter5_Example8_Output.txt b/926/CH5/EX5.8/Chapter5_Example8_Output.txt new file mode 100644 index 000000000..02693e6fb --- /dev/null +++ b/926/CH5/EX5.8/Chapter5_Example8_Output.txt @@ -0,0 +1,6 @@ +
+ (a) Molal humidity of the air = 0.037
+ Percentage saturation = 52 %
+ Dew point = 80.50 degree F
+ (b) Percentage saturation of the air leaving the evaporator = 84 %
+ weight of the water evaporated = 3.1 lb
\ No newline at end of file diff --git a/926/CH5/EX5.9/Chapter5_Example9.sce b/926/CH5/EX5.9/Chapter5_Example9.sce new file mode 100644 index 000000000..b6ddb4c9d --- /dev/null +++ b/926/CH5/EX5.9/Chapter5_Example9.sce @@ -0,0 +1,27 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 9, Page 124
+//Title: Estimation of dry bulb and wet bulb temperature
+//=============================================================================
+clear
+clc
+
+//INPUT
+m = [12.1 0.1 7.6 80.2]; //Composition of CO2, CO, O2, N2 in percentage
+
+//DATA FROM GRAPH
+//Part(a)
+T1 = 87; //Wet bulb temperature obtained from Fig 19, Page 120 & Fig 20, 122
+//Part(b)
+T2 = 140; //Dry bulb temperature obtained from Fig 19, Page 120 & Fig 20, 122
+
+//OUTPUT
+// Console output
+mprintf('\n (a) Wet bulb temperature = %2.0f degree F \n (b) Dry bulb temperature = %3.0f degree F',T1,T2);
+
+// File output
+fd= mopen('.\Chapter5_Example9_Output.txt','w');
+mfprintf(fd,'\n (a) Wet bulb temperature = %2.0f degree F \n (b) Dry bulb temperature = %3.0f degree F',T1,T2);
+mclose(fd);
+
+//=========================END OF PROGRAM======================================
diff --git a/926/CH5/EX5.9/Chapter5_Example9_Output.txt b/926/CH5/EX5.9/Chapter5_Example9_Output.txt new file mode 100644 index 000000000..b72f90091 --- /dev/null +++ b/926/CH5/EX5.9/Chapter5_Example9_Output.txt @@ -0,0 +1,3 @@ +
+ (a) Wet bulb temperature = 87 degree F
+ (b) Dry bulb temperature = 140 degree F
\ No newline at end of file |