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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /1247/CH5 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '1247/CH5')
61 files changed, 1492 insertions, 0 deletions
diff --git a/1247/CH5/EX5.1/example5_1.sce b/1247/CH5/EX5.1/example5_1.sce new file mode 100755 index 000000000..110300aa6 --- /dev/null +++ b/1247/CH5/EX5.1/example5_1.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.1
+// Page 186
+printf("Example 5.1, Page 186 \n \n");
+
+// solution
+
+// basis pumping of 1 l/s of water
+Hadd = 52 // kW
+Hlost = 21 // kW
+fi = Hadd - Hlost // kW
+p1 = 101325 // Pa
+p2 = p1
+Z1 = -50 // m
+Z2 = 10 // m
+g = 9.80665 // m/s sq
+gc = 1 // kg.m/(N.s sq)
+row = 1 // kg/l
+W = 1.5*.55 // kW
+// energy balance b/w A and B
+// dE = E2-E1 = W + Q + (Z1-Z2)*(g/gc)*qm
+dE = 31.237 // kW
+printf("Increase in internal energy between the storage tank and the bottom of the well = "+string(dE)+" kW.")
diff --git a/1247/CH5/EX5.10/example5_10.sce b/1247/CH5/EX5.10/example5_10.sce new file mode 100755 index 000000000..cbe4a81ea --- /dev/null +++ b/1247/CH5/EX5.10/example5_10.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.10
+// Page 225
+printf("Example 5.10, Page 225 \n \n");
+
+// solution
+
+// (a)
+Pc = 3732 // kPa
+Tc = 630.3 // K
+Tb = 417.6 //K
+TBr = Tb/Tc
+lambdav = 8.314472*417.6*(1.092*(log(3732)-5.6182)/(.930-.6625))
+// (b)
+T1 = 298.15 //K
+lambdav1 = 36240*[(630.3-298.15)/(630.3-417.6)]^.38
+printf(" (a) \n \n Latent heat of vaporization at Tb using Riedel eq is "+string(lambdav)+" kJ/kmol. \n \n \n (b) \n \n Latent heat of vaporizaation at 298.15 K using Watson eq is "+string(lambdav1)+" kJ/kmol.")
diff --git a/1247/CH5/EX5.11/example5_11.sce b/1247/CH5/EX5.11/example5_11.sce new file mode 100755 index 000000000..4bee91619 --- /dev/null +++ b/1247/CH5/EX5.11/example5_11.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.11
+// Page 225
+printf("Example 5.11, Page 225 \n \n");
+
+// solution
+
+// (a)
+Pc = 61.37 // bar
+Tc = 514 //K
+Tb = 351.4
+P = 1 // atm
+TBr = Tb/Tc
+// Riedel eq
+lambdav1 = 8.314472*Tb*1.092*(log(6137)-5.6182)/(.930-TBr)
+// NIST eq
+lambdav2 = 50430*exp(-(-.4475*TBr))*(1-TBr)^.4989
+// (b)
+T1 = 298.15
+TBr1 = T1/Tc
+// Watson eq
+lambdav21 = 38563*[(514-298.15)/(514-351.4)]^.38
+// NIST eq
+lambdav22 = 50430*exp(-(-.4475*TBr1))*(1-TBr1)^.4969
+printf(" (a) \n \n Latent heat of vaporization at Tb using \n Riedel eq is "+string(lambdav1)+" kJ/kmol \n NIST eq is "+string(lambdav2)+" kJ/kmol \n \n \n (b) \n \n Latent heat of vaporization at 298.15 K using \n Watson eq is "+string(lambdav21)+" kJ/kmol \n NIST eq is "+string(lambdav22)+" kJ/kmol")
diff --git a/1247/CH5/EX5.12/example5_12.sce b/1247/CH5/EX5.12/example5_12.sce new file mode 100755 index 000000000..2fe4ea2fa --- /dev/null +++ b/1247/CH5/EX5.12/example5_12.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.12
+// Page 227
+printf("Example 5.12, Page 227 \n \n");
+
+// solution
+
+// using Appendix IV.2
+Ps1 = 75
+Ps2 = 80
+T1 = 563.65
+T2 = 568.12
+T = 565.15
+Ps = 75*exp((T2*(T-T1)*log(80/75)/(T*(T2-T1))))
+printf(" Saturation Pressure of steam at 565.15K is "+string(Ps)+" bar.")
diff --git a/1247/CH5/EX5.13/example5_13.sce b/1247/CH5/EX5.13/example5_13.sce new file mode 100755 index 000000000..804c1ee18 --- /dev/null +++ b/1247/CH5/EX5.13/example5_13.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.13
+// Page 236
+printf("Example 5.13, Page 236 \n \n");
+
+// solution
+
+// basis 1 kmol equimolar mix
+npent = .5 // kmol
+nhex = .5 // kmol
+P = 101.325 // kPa
+x1 = .5
+x2 = x1
+Ts1 = 309.2 // K
+Ts2 = 341.9 // K
+T1 = (Ts1+Ts2)/2
+// using these data, we get table 5.10 and 5.11
+Tbb = 321.6 //K
+Tdp = 329.9 //K
+printf(" Bubble point = "+string(Tbb)+" K and \n Dew point = "+string(Tdp)+" K.")
diff --git a/1247/CH5/EX5.14/example5_14.sce b/1247/CH5/EX5.14/example5_14.sce new file mode 100755 index 000000000..e214a5894 --- /dev/null +++ b/1247/CH5/EX5.14/example5_14.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.14
+// Page 237
+printf("Example 5.14, Page 237 \n \n");
+
+// solution
+
+// basis 1000 kg/h of condensate at the saturation temperature corresponding to 8 bar a
+// using Appendix IV.2
+H = 720.94 // kJ/kg
+Hm = 419.06 // kJ/kg
+x = poly(0,'x')
+condensate = 1000-x
+Hcondensate1 = 1000*H
+Hcondensate2 = condensate*419.06
+Ht = x*2676
+p = Hcondensate2+Ht-Hcondensate1
+printf(" The quqntity of flash steam produced = "+string(roots(p))+" kg/h.")
diff --git a/1247/CH5/EX5.15/example5_15.sce b/1247/CH5/EX5.15/example5_15.sce new file mode 100755 index 000000000..49416723c --- /dev/null +++ b/1247/CH5/EX5.15/example5_15.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.15
+// Page 238
+printf("Example 5.15, Page 238 \n \n");
+
+// solution
+
+qv1 = 50 // l/s
+qm = qv1*1.08 // kg/s
+fi = qm*3.08*(263.15-258.15) // kW
+lv = 384.19-168.7 // kJ/kg
+qm2 = fi/lv
+H = 256.35 // kJ/kg
+x = poly(0, 'x')
+p = H*(qm2+x) - 168.7*qm2-x*384.19
+a = qm2+roots(p)
+printf(" Flow of vapor from he chiller = "+string(a)+" kg/s.")
diff --git a/1247/CH5/EX5.16/example5_16.sce b/1247/CH5/EX5.16/example5_16.sce new file mode 100755 index 000000000..b80a942bb --- /dev/null +++ b/1247/CH5/EX5.16/example5_16.sce @@ -0,0 +1,44 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.16
+// Page 238
+printf("Example 5.16, Page 238 \n \n");
+
+// solution
+
+// basis liquifaction capacity = 0.116 kg/s
+p1 = 101 // kPa
+Ts1 = 239.15
+lv1 = 288.13 // kJ/kg
+p2 = 530 // kPa
+Ts2 = 290.75 // K
+lv2 = 252.93 // kJ/kg
+// referring to table 5.3 and using eq 5.21
+H1 = -39.246*(Ts2-Ts1)+1401.223*10^-3*(Ts2^2-Ts1^2)/2-6047.226*10^-6*(Ts2^3-Ts1^3)/3+8591.4*10^-9*(Ts2^4-Ts1^4)/4 // kJ/kmol
+T3 = 313.15
+H2 = [28.5463*(T3-Ts1)+23.8795*10^-3*(T3^2-Ts1^2)/2-21.3631*10^-6*(T3^3-Ts1^3)/3+6.4726*10^-9*(T3^4-Ts1^4)/4]/70.903 // kJ/kg
+fi2 = .116*H2
+Cl2evp = fi2/lv1 // kg/s
+Cl2recy = Cl2evp/(1-.185)
+R = Cl2recy/.116 // kg/kg fresh feed
+// T4/T1 = (p2/p1)^[(gamma-1)/gamma]
+gm = 1.355
+p22 = 326.3
+p21 = 101
+T4 = Ts1*(p2/p1)^[(gm-1)/gm]
+T5 = 313.15
+fi3 = 1.88*10^-3*(343.1+91.6-26.2+2.5) // kW
+Fwater1 = fi3/(8*4.1868) // kg/s
+// similarly
+T6 = 379.9
+fi4 = 1.88*10^-3*[28.5463*(T6-T5)+23.8795*10^-3*(T6^2-T5^2)/2-21.3631*10^-6*(T6^3-T5^3)/3+6.4726*10^-9*(T6^4-T5^4)/4] // kW
+Fwater2 = fi4/(8*4.1868) // kg/s
+Wreq = Fwater1+Fwater2
+fi5 = 1.88*10^-3*[28.5463*(T5-Ts2)+23.8795*10^-3*(T5^2-Ts2^2)/2-21.3631*10^-6*(T5^3-Ts2^3)/3+6.4726*10^-9*(T5^4-Ts2^4)/4] +.1333*252.93 // kW
+printf(" (a) \n \n Recycle ratio = "+string(R)+" kg Cl2/kg fresh feed \n \n \n (b) \n \n Cooling water required at \n interface = "+string(Fwater1)+" kg/s \n after cooler = "+string(Wreq)+" kg/s \n \n \n (c) \n \n Refrigiration load of chiller = "+string(fi5)+" kW.")
diff --git a/1247/CH5/EX5.17/example5_17.sce b/1247/CH5/EX5.17/example5_17.sce new file mode 100755 index 000000000..aea19d4ad --- /dev/null +++ b/1247/CH5/EX5.17/example5_17.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.17
+// Page 242
+printf("Example 5.17, Page 242 \n \n");
+
+// solution
+
+// basis 100 kg of tin
+T1 = 303.15
+T2 = 505.15
+n = 100/118.7 // kmol
+// Q1 = n*[intgr from T1 to T2 (Cms dT)]
+Q1 = 4973.3 // kJ
+lf = 7201
+Q2 = n*lf // kJ
+Q = Q1+Q2
+lv = 278 // kJ/kg
+vp = Q/lv // kg
+printf(" Quantity of eutectic mixture condensed = "+string(vp)+" kg per 100 kg of tin melted at its melting point.")
diff --git a/1247/CH5/EX5.18/example5_18.sce b/1247/CH5/EX5.18/example5_18.sce new file mode 100755 index 000000000..4dcd9a76f --- /dev/null +++ b/1247/CH5/EX5.18/example5_18.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.18
+// Page 243
+printf("Example 5.18, Page 243 \n \n");
+
+// solution
+
+Ts1 = (438.2+436)/2
+Ta = 300
+fi1 = .045*(Ts1-Ta)*3600
+theta1 = 307293/fi1 //h
+Ts2 = (436+434)/2
+fi2 = .045*(Ts2-Ta)*3600
+theta2 = 302415/fi2
+Ts3 = (434+432.1)/2
+fi3 = .045*(Ts3-Ta)*3600
+theta3 = 313859/fi3
+theta = theta1+theta2+theta3
+printf(" total time required = "+string(theta)+" hrs.")
diff --git a/1247/CH5/EX5.19/example5_19.sce b/1247/CH5/EX5.19/example5_19.sce new file mode 100755 index 000000000..6c4438619 --- /dev/null +++ b/1247/CH5/EX5.19/example5_19.sce @@ -0,0 +1,39 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.19
+// Page 245
+printf("Example 5.19, Page 245 \n \n");
+
+// solution
+
+H1 = 482.9 // kJ/kg
+H2 = 273.4
+fi1 = 100*(H1-H2) // kJ/h
+T1 = 313.15
+T2 = 403.15
+fi11 = 21.3655*(T2-T1)+64.2841*10^-3*(T2^2-T1^2)/2-41.0506*10^-6*(T2^3-T1^3)/3+9.7999*10^-9*(T2^4-T1^4)/4 // kJ/h
+// at 20 MPa
+h1 = 211.1
+Ts = 277.6
+H11 = 427.8
+x = poly(0, 'x')
+p = x*h1+(100-x)*H11-100*H2
+a = roots(p)
+fi2 = (100-a)*(H11-h1) // kJ/h
+h2 = -148.39
+H3 = 422.61
+y = poly(0, 'y')
+p1 = 100*176.18-(100-y)*H3+h2*y
+b = roots(p1)
+fi3 = 100*(h1-176.8)
+H = fi3+24021
+H4 = H/(100-43.16)
+// from ref 23
+T = 262.15
+printf(" (a) \n \n Yield of dry ice = "+string(b)+" kg. \n \n \n (b) \n \n Percent liquifaction = "+string(a)+". \n \n \n (c) \n \n Temp of vented gas = "+string(T)+" K.")
diff --git a/1247/CH5/EX5.2/example5_2.sce b/1247/CH5/EX5.2/example5_2.sce new file mode 100755 index 000000000..106cbda67 --- /dev/null +++ b/1247/CH5/EX5.2/example5_2.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.2
+// Page 197
+printf("Example 5.2, Page 197 \n \n");
+
+// solution
+
+// using table 5.1
+// basis 1 kmol of methane
+T1 = 303.15 // K
+T2 = 523.15 // K
+// using eq 5.17
+H = 19.2494*(T2-T1) + 52.1135*10^-3*(T2^2-T1^2)/2 + 11.973*10^-6*(T2^3-T1^3)/3 - 11.3173*(T2^4-T1^4)*10^-9/4 // kJ
+printf(" Heat added = "+string(H)+" kJ/kmol methane.")
diff --git a/1247/CH5/EX5.20/example5_20.sce b/1247/CH5/EX5.20/example5_20.sce new file mode 100755 index 000000000..6b1f9a67c --- /dev/null +++ b/1247/CH5/EX5.20/example5_20.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.20
+// Page 247
+printf("Example 5.20, Page 247 \n \n");
+
+// solution
+
+// basis 200 kg/h of Sulphur firing
+F = 200/32 // kmol/h
+O2req = 6.25*1.1
+airin = O2req/.21
+N2in = airin-O2req
+T1 = 1144.15
+T2 = 463.15
+fi = 788852.2 // kJ/h
+H = 15*4.1868+1945.2
+qm = fi*.9/2008 // kg/h
+printf(" Amount of steam produced = "+string(qm)+" kg/h.")
diff --git a/1247/CH5/EX5.21/example5_21.sce b/1247/CH5/EX5.21/example5_21.sce new file mode 100755 index 000000000..a0b6f22c0 --- /dev/null +++ b/1247/CH5/EX5.21/example5_21.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.21
+// Page 248
+printf("Example 5.21, Page 248 \n \n");
+
+// solution
+
+// enthalpy at Tbb
+Tbb = 321.6
+T1 = 298.15
+H1 = 65.4961*(Tbb-T1)+628.628*10^-3*(Tbb^2-T1^2)/2-1898.8*10^-6*(Tbb^3-T1^3)/3+3186.51*10^-9*(Tbb^4-T1^4)/4 // kJ/kmol
+H2 = 31.421*(Tbb-T1)+976.058*10^-3*(Tbb^2-T1^2)/2-2353.68*10^-6*(Tbb^3-T1^3)/3+3092.73*10^-9*(Tbb^4-T1^4)/4 // kJ/kmol
+Hsol = (H1+H2)/2 // kJ/kmol
+// enthalpy at Tdp
+lv1 = 25790*((469.7-329.9)/(469.7-309.2))^.38
+lv2 = 28850*((507.6-329.9)/(507.6-341.9))^.38
+Tdp = 329.9
+H21ig = 65.4961*(Tdp-T1)+628.628*10^-3*(Tdp^2-T1^2)/2-1898.8*10^-6*(Tdp^3-T1^3)/3+3186.51*10^-9*(Tdp^4-T1^4)/4 + lv1 // kJ/kmol
+H22ig = 31.421*(Tdp-T1)+976.058*10^-3*(Tdp^2-T1^2)/2-2353.68*10^-6*(Tdp^3-T1^3)/3+3092.73*10^-9*(Tdp^4-T1^4)/4 +lv2 // kJ/kmol
+Hmixig = (H21ig+H22ig)/2
+printf(" (a) \n \n H = "+string(Hsol)+" kJ/kmol \n \n \n (b) \n \n H = "+string(Hmixig)+" kJ/kmol")
diff --git a/1247/CH5/EX5.22/example5_22.sce b/1247/CH5/EX5.22/example5_22.sce new file mode 100755 index 000000000..3aa53fe72 --- /dev/null +++ b/1247/CH5/EX5.22/example5_22.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.22
+// Page 252
+printf("Example 5.22, Page 252 \n \n");
+
+// solution
+
+H1 = 23549 //kJ/kmol
+H2 = 16325
+H3 = 28332
+H4 = .4*H2+.6*H3
+printf("Enthalpy of vapor-liquid mixture after flashing = "+string(H4)+" kJ/mol.")
diff --git a/1247/CH5/EX5.23/example5_23.sce b/1247/CH5/EX5.23/example5_23.sce new file mode 100755 index 000000000..9c1869830 --- /dev/null +++ b/1247/CH5/EX5.23/example5_23.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.23
+// Page 253
+printf("Example 5.23, Page 253 \n \n");
+
+// solution
+
+// basis feed gas = 12000 Nm^3 = 535.4 kmol/h
+T1 = 147.65 // K
+n1 = 535.4*.3156 // kmol/h HP tail gas stream
+T = 118.5 // K
+n2 = (535.4-n1)*.0602 // kmol/h LP tail stream
+n3 = 535.4-n2-n1 // kmol/h product H2 stream
+p = 315.35*100/n3
+printf(" Purity of product H2 stream = "+string(p)+" percent.")
diff --git a/1247/CH5/EX5.24/example5_24.sce b/1247/CH5/EX5.24/example5_24.sce new file mode 100755 index 000000000..0109d3133 --- /dev/null +++ b/1247/CH5/EX5.24/example5_24.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.24
+// Page 256
+printf("Example 5.24, Page 256 \n \n");
+
+// solution
+
+// fi1 = integr (from 304.15 to 313.15) {11831.6+24997.4*10T^-3-5979.8*10^-6T^2-31.7*10^-9T3}dt
+fi1 = 170787.7 // kJ/h
+fi2 = 535.4*12086 - [344.36*8743.2+168.97*18036+22.07*15892] // kJ/h
+printf(" (a) \n \n Refrigiration requirement = "+string(fi1)+" kJ/h \n \n \n (b) \n \n Refrigiration requirement based on real enthalpies = "+string(fi2)+" kJ/h.")
diff --git a/1247/CH5/EX5.25/example5_25.sce b/1247/CH5/EX5.25/example5_25.sce new file mode 100755 index 000000000..e5d73797b --- /dev/null +++ b/1247/CH5/EX5.25/example5_25.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.25
+// Page 257
+printf("Example 5.25, Page 257 \n \n");
+
+// solution
+
+// basis 100 kmol/h of benzene feed rate
+Cl2 = .4*100
+HClp = 40
+Benzenecon = 37
+MCBp = 100*.37*.9189
+DCBp = Benzenecon-MCBp
+unreactBenzene = 100-Benzenecon
+Nt = HClp + MCBp + DCBp + unreactBenzene
+// using eq xi = Ni/(L(1-K1)+NtKi) and sigma xi = 1
+L = 89.669 // kmol/h
+V = Nt - L
+printf(" Liquid product stream = "+string(L)+" kmol/h \n Vapor product stream = "+string(V)+" kmol/h")
diff --git a/1247/CH5/EX5.26/example5_26.sce b/1247/CH5/EX5.26/example5_26.sce new file mode 100755 index 000000000..bf008b879 --- /dev/null +++ b/1247/CH5/EX5.26/example5_26.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.26
+// Page 260
+printf("Example 5.26, Page 260 \n \n");
+
+// solution
+
+// 2C + 2O2 = 2CO2 A
+// 2H2 + O2 = 2H2O B
+// C2H4 + 3O2 = 2CO2 + 2H2O C
+// A+B-C gives
+// 2C(g) + 2H2 = C2H4(g) D
+H = -2*393.51-2*241.82+1323.1 // kJ/mol
+printf(" Heat of formation of Ethylene is "+string(H)+" kJ/mol.")
diff --git a/1247/CH5/EX5.27/example5_27.sce b/1247/CH5/EX5.27/example5_27.sce new file mode 100755 index 000000000..2ed8303e8 --- /dev/null +++ b/1247/CH5/EX5.27/example5_27.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.27
+// Page 260
+printf("Example 5.27, Page 260 \n \n");
+
+// solution
+
+Hc = 2*(-393.51)-887.811+2*(-285.83)-(-73.6+0) //kJ/mol
+printf(" Heat of combustion of ethyl mercaptan = "+string(Hc)+" kJ/mol.")
diff --git a/1247/CH5/EX5.28/example5_28.sce b/1247/CH5/EX5.28/example5_28.sce new file mode 100755 index 000000000..f865e90aa --- /dev/null +++ b/1247/CH5/EX5.28/example5_28.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.28
+// Page 261
+printf("Example 5.28, Page 261 \n \n");
+
+// solution
+
+lv1 = 26694 // kj/kmol
+Tc = 466.74
+lv2 = lv1*((Tc-298.15)/(Tc-307.7))^.38/1000 // kJ/mol
+Hf = -252 // kJ/mol
+Hf1 = Hf-lv2 // kJ/kmol
+printf("Heat of formation of liquid di ethyl ether = "+string(Hf1)+" kJ/mol.")
diff --git a/1247/CH5/EX5.29/example5_29.sce b/1247/CH5/EX5.29/example5_29.sce new file mode 100755 index 000000000..a8811696e --- /dev/null +++ b/1247/CH5/EX5.29/example5_29.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.29
+// Page 261
+printf("Example 5.29, Page 261 \n \n");
+
+// solution
+
+// basis 1 kg motor spirit
+G = 141.5/(131.5+64)
+// r = C/H
+r = (74+15*G)/(26-15*G)
+C = r/6.605 // C content of motor spirit
+H2 = 1-C
+O2req = C+H2
+Hf = 44050-27829-18306 // kJ/kg
+printf(" Heat of formation of motor spirit = "+string(Hf)+" kJ/kg.")
diff --git a/1247/CH5/EX5.3/example5_3.sce b/1247/CH5/EX5.3/example5_3.sce new file mode 100755 index 000000000..30a5da985 --- /dev/null +++ b/1247/CH5/EX5.3/example5_3.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.3
+// Page 198
+printf("Example 5.3, Page 198 \n \n");
+
+// solution
+
+// basis 1 kmol methane at 25 bar
+Pc = 46.04 // bar
+Tc = 190.5 // K
+Pr = 25/Pc
+// H-Ho = intgr(from303.15 to 523.15){CmpR dT}
+// solving it by simpson's rule
+HE = 255.2 // kJ/kmol
+H = 9175.1+HE
+printf(" Heat added = "+string(H)+" kJ/kmol of methane.")
diff --git a/1247/CH5/EX5.30/example5_30.sce b/1247/CH5/EX5.30/example5_30.sce new file mode 100755 index 000000000..081bd59e6 --- /dev/null +++ b/1247/CH5/EX5.30/example5_30.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.30
+// Page 267
+printf("Example 5.30, Page 267 \n \n");
+
+// solution
+
+// basis 1 kmol of styrene
+dH = 241749-189398 // kJ/mol
+Cmpn = dH/(600-298.15) // kJ/kmol K
+printf(" Mean heat capacity between 600K and 298.15 K is "+string(Cmpn)+" kJ/kmol K.")
diff --git a/1247/CH5/EX5.31/example5_31.sce b/1247/CH5/EX5.31/example5_31.sce new file mode 100755 index 000000000..f06ec4d99 --- /dev/null +++ b/1247/CH5/EX5.31/example5_31.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.31
+// Page 269
+printf("Example 5.31, Page 269 \n \n");
+
+// solution
+
+// basis 1 mol of SiO2 reacted
+Hf = [-2879+3*(-296.81)+3*0/2]-[3*(-1432.7)+1*(-903.5)] // kJ/mol SiO2
+printf(" Heat of reaction = "+string(Hf)+" kJ/mol SiO2.")
diff --git a/1247/CH5/EX5.32/example5_32.sce b/1247/CH5/EX5.32/example5_32.sce new file mode 100755 index 000000000..2630eff8d --- /dev/null +++ b/1247/CH5/EX5.32/example5_32.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.32
+// Page 269
+printf("Example 5.32, Page 269 \n \n");
+
+// solution
+
+// basis 100 kg of 2% ammonia solution
+NH3 = 2 // kg
+H2O = 98 // kg
+Hr = -361.2-(-45.94-285.83) // kJ/mol NH3 dissolved
+Hd = -(Hr*2*1000/17.0305) // kJ/100 kg sol.
+printf(" heat of reaction = "+string(Hd)+" kJ/100 kg solution.")
diff --git a/1247/CH5/EX5.33/example5_33.sce b/1247/CH5/EX5.33/example5_33.sce new file mode 100755 index 000000000..28ba0b737 --- /dev/null +++ b/1247/CH5/EX5.33/example5_33.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.33
+// Page 272
+printf("Example 5.33, Page 272 \n \n");
+
+// solution
+
+// basis 1 kmol of SO2 reacted
+a = 22.036-24.771-.5*(26.026)
+b = (121.624-62.948-.5*11.755)
+c = (-91.876+44.258-.5*(-2.343))
+d = (24.369-11.122-.5*(-.562))
+Hr = -395720+296810 // kJ/kmol
+Hro = Hr-a*298.15-b*10^-3*298.15^2/2-c*10^-6*298.15^3/3-d*10^-9*298.15^4/4
+T = 778.15
+Hrt = -Hro-15.748*T+26.4*10^-3*T^2-15.48*10^-6*T^3+3.382*10^-9*T^4
+printf(" Heat of reaction at 775K is "+string(Hrt)+" kJ/kmol.")
diff --git a/1247/CH5/EX5.34/example5_34.sce b/1247/CH5/EX5.34/example5_34.sce new file mode 100755 index 000000000..ad6446e69 --- /dev/null +++ b/1247/CH5/EX5.34/example5_34.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.34
+// Page 272
+printf("Example 5.34, Page 272 \n \n");
+
+// solution
+
+Hr = -480-285.83+277.2+484.2 // kJ/mol
+Hrt1 = Hr*1000 + [146.89+75.76-119.55-129.70]*75 // kJ/kmol
+a = 4.2905+50.845-100.92-155.48
+b = 934.378+213.08+111.8386+326.5951
+c = -2640-631.398-498.54-744.199
+d = 3342.58+648.746
+Hro = Hr*1000+a*(-298.15)+b*10^-3*(-298.15^2)/2+c*10^-6*(-298.15^3)/3+d*10^-9*(-298.15^4)/4
+T = 373.15
+Hrt = Hro+a*T+792.949*10^-3*T^2-1504.712*10^-6*T^3+997.832*10^-9*T^4
+printf(" Heat of reaction at 373 K is "+string(Hrt)+" kJ/kmol reactant.")
diff --git a/1247/CH5/EX5.35/example5_35.sce b/1247/CH5/EX5.35/example5_35.sce new file mode 100755 index 000000000..20a43daed --- /dev/null +++ b/1247/CH5/EX5.35/example5_35.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.35
+// Page 273
+printf("Example 5.35, Page 273 \n \n");
+
+// solution
+
+T2 = 800
+T1 = 298.15
+fi1 = 3614.577*(T2-T1)+305.561*10^-3*(T2^2-T2^2)/2+836.881*10^-6*(T2^3-T1^3)/3-393.707*10^-9*(T2^4-T1^4)/4 // kW
+T3 = 875
+fi2 = 3480.737*(T3-T1)+754.347*10^-3*(T3^2-T2^2)/2+442.159*10^-6*(T3^3-T1^3)/3-278.735*10^-9*(T3^4-T1^4)/4 // kW
+Hr = -98910 // kJ/kmol SO2 reacted by eg 5.33
+fi3 = (8.8511-.351)*Hr/3600 // kW
+dH = fi2/3600+fi3-fi1/3600
+printf(" Net enthalpy change = "+string(dH)+" kW.")
diff --git a/1247/CH5/EX5.36/example5_36.sce b/1247/CH5/EX5.36/example5_36.sce new file mode 100755 index 000000000..9d886fb20 --- /dev/null +++ b/1247/CH5/EX5.36/example5_36.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.36
+// Page 275
+printf("Example 5.36, Page 275 \n \n");
+
+// solution
+
+// basis 100 kmol outgoing gas mixture from scrubber
+moistin = 3127.7*.015/18 // kmol
+waterin = 40.2+moistin // kmol
+// using tables 5.29 and 5.30
+Hr = -27002658-(-26853359)
+Hr1 = Hr/246.4493 // kJ/kmol total reactants
+printf(" Heat of reaction = "+string(Hr1)+" kJ/kmol total reactants.")
diff --git a/1247/CH5/EX5.37/example5_37.sce b/1247/CH5/EX5.37/example5_37.sce new file mode 100755 index 000000000..7e05146d9 --- /dev/null +++ b/1247/CH5/EX5.37/example5_37.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.37
+// Page 276
+printf("Example 5.37, Page 276 \n \n");
+
+// solution
+
+fi3 = 15505407 // kJ/h
+lv = 296.2 // from table 5.6
+Ht = 17131551 // kJ/h
+r = Ht/lv // kg/h
+printf(" Downtherm circulation rate = "+string(r)+" kg/h.")
diff --git a/1247/CH5/EX5.38/example5_38.sce b/1247/CH5/EX5.38/example5_38.sce new file mode 100755 index 000000000..92c23b38e --- /dev/null +++ b/1247/CH5/EX5.38/example5_38.sce @@ -0,0 +1,50 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.38
+// Page 279
+printf("Example 5.38, Page 279 \n \n");
+
+// solution
+
+F = 100 // kmol/h feed rate of ethylene
+Econ = .99*F
+Econ1 = Econ*.998
+Econ2 = Econ-Econ1
+Cl2con = Econ1+2*Econ2
+Cl2in = F*1.1
+Cl2s3 = Cl2in-Cl2con
+HCls3 = Econ2
+TCEp = Econ2
+EDCp = Econ1
+nC2H4 = 1
+T = 328.15
+pv1 = exp(4.58518-1521.789/(T-24.67)) // bar
+pv2 = exp(4.06974-1310.297/(T-64.41)) // bar
+xEDC = Econ1/(Econ1+Econ2)
+xTEC = 1-xEDC
+pEDC = 37.2*xEDC
+pTEC = 12.64*xTEC
+pCl2HClC2H4 = 1.6*100-pEDC-pTEC
+yEDC = pEDC/160
+yTEC = pTEC/160
+nt = (Cl2s3+Econ2+1)*160/pCl2HClC2H4
+nEDC = yEDC*nt
+nTEC = yTEC*nt
+printf(" Compositions of gas streams : \n \n Component Stream 3 Stream 5 Stream 4 Stream 6 \n Cl2 "+string(Cl2s3)+" "+string(Cl2s3)+" \n HCl "+string(HCls3)+" "+string(HCls3)+" \n C2H4 "+string(nC2H4)+" "+string(nC2H4)+" \n EDC "+string(nEDC)+" 0.2355 3.3947 98.5665 \n TEC "+string(nTEC)+" Nil "+string(nTEC)+" "+string(TCEp)+" \n \n ")
+fi1 = (10.802*33.9+.198*29.1+1*43.6+3.6302*17.4+.0025*85.3)*(328.15-273.15)
+fi2 = 35.053*1000*3.3947+39.58*1000*.0025
+fi3 = (3.3947*129.4+.0025*144.4)*55/2
+fi = fi1+ fi2+ fi3 // kJ/h
+printf(" Heavy duty of Overhead condenser = "+string(fi)+" kJ/h. \n \n ")
+fi5 = (100*43.6+110*33.9)*(328.15-273.15)
+fi6 = 3.6302*1000*33.6+.0025*1000*38.166
+fi7 = (98.5665*129.4+.1988*144.4)*(328.15-273.15)
+fi8 = 216845.5*98.802+392394.5*.198
+ficol = fi5+fi8-fi1-fi6-fi7
+printf(" Heavy duty of external cooler = "+string(ficol)+" kJ/h.")
diff --git a/1247/CH5/EX5.39/example5_39.sce b/1247/CH5/EX5.39/example5_39.sce new file mode 100755 index 000000000..484f0c918 --- /dev/null +++ b/1247/CH5/EX5.39/example5_39.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.39
+// Page 284
+printf("Example 5.39, Page 284 \n \n");
+
+// solution
+
+To = 298.15
+T1 = 483.15
+// fi1 = intgr(from To to T1){12199.5+2241.4*10^-3*T+1557.7*10^-6*T^2-671.3*10^-9*T^3}dT
+fi1 = 2455874.6 // kJ/h
+dHr = 2*(-45.94) // kJ/mol N2 reacted
+fi2 = 91.88*1000*23.168
+fi3 = fi1+fi2
+// fi3 = intgr(from To to T2){10713.9+3841*10^-3*T+1278.8*10^-6*T^2-752.6*10^-9*T^3}dT
+// solving it
+T2 = 657.41 // K
+printf("Temperature of the gas mixture leaving the reactor = "+string(T2)+" K.")
diff --git a/1247/CH5/EX5.4/example5_4.sce b/1247/CH5/EX5.4/example5_4.sce new file mode 100755 index 000000000..6c76343ea --- /dev/null +++ b/1247/CH5/EX5.4/example5_4.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.4
+// Page 206
+printf("Example 5.4, Page 206 \n \n");
+
+// solution
+
+// using table 5.3
+// .25 kg/s toulene heated from 290.15K to 350.15K
+qm = .25/92 // kmol/s
+// reference 7
+fi = 2.717*10^-3*[1.8083*(350.15-290.15) + 812.223*10^-3*(350.15^2-290.15^2)/2 - 1512.67*10^-6*(350.15^3-290.15^3)/3 + 1630.01*10^-9*(350.15^4-290.15^4)/4]
+printf(" Heat required to be added to toulene = "+string(fi)+" kW.")
diff --git a/1247/CH5/EX5.40/example5_40.sce b/1247/CH5/EX5.40/example5_40.sce new file mode 100755 index 000000000..5665dc644 --- /dev/null +++ b/1247/CH5/EX5.40/example5_40.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.40
+// Page 292
+printf("Example 5.40, Page 292 \n \n");
+
+// solution
+
+// basis 4 kmol of HCl gas
+O2req = 1 // kmol
+O2spply = 1.35*1
+N2 = 1.35*79/21
+air = O2spply+N2
+HClbrnt = .8*4
+HCl = 4-HClbrnt
+O2 = O2spply-.8
+Cl2 = .8*2
+H2O = .8*2
+printf(" (a) \n \n Composition of dry product gas stream : \n Component Dry product gas stream,kmol \n HCl "+string(HCl)+" \n O2 "+string(O2)+" \n Cl2 "+string(Cl2)+" \n H2O "+string(H2O)+" \n N2 "+string(N2)+" \n \n \n (b) \n \n ")
+H2 = 114.4*1000*.8
+// H2 = intgr(from 298.15 to T){286.554+12.596*10^-3*T+63.246*10^-6*T^2-25.933*10^-9*T^3}dT
+// solving it
+T = 599.5 // K
+printf(" Adiabatic reaction temperature of product gas stream = "+string(T)+" K.")
diff --git a/1247/CH5/EX5.41/example5_41.sce b/1247/CH5/EX5.41/example5_41.sce new file mode 100755 index 000000000..d922efb9a --- /dev/null +++ b/1247/CH5/EX5.41/example5_41.sce @@ -0,0 +1,36 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.41
+// Page 294
+printf("Example 5.41, Page 294 \n \n");
+
+// solution
+
+// 1 kmol of EB vapors entering the reactor at 811.15 K
+// (from 811.15 to T1)intgr{-36.72+671.12*10^-3*T-422.02*10^-6*T^2+101.15*10^-9*T^3}dT = (from T1 to 978.15)intgr{487.38+1.19*10^-3*T+198.16*10^-6*T^2-68.21*10^-9*T^3}dT
+// we get
+T1 = 929.72 // K
+To = 298.15
+H1 = 493405 // kJ
+EBr = .35
+Styrenep = EBr*.9
+Benzeneb = EBr*.03
+Ethyleneb = Benzeneb
+Cb = EBr*.01
+Toulened = EBr*.06
+Hr1 = 147.36-29.92 // kJ/mol EB
+Hr2 = 82.93+52.5-29.92
+Hr3 = -29.92
+Hr4 = 50.17-74.52-147.36 // kJ/mol styrene
+dHr = 1000*(Hr1*(Styrenep+Toulened)+Hr2*Benzeneb+Hr3*Cb+Hr4*Toulened)
+H2 = H1-dHr
+// H2 = (from To t0 T2)intgr{Comp2dT
+// we get
+T2 = 798.79 // K
+printf(" Adiabatic reaction T at the outlet of the reactor is "+string(T2)+" K.")
diff --git a/1247/CH5/EX5.42/example5_42.sce b/1247/CH5/EX5.42/example5_42.sce new file mode 100755 index 000000000..7bf787630 --- /dev/null +++ b/1247/CH5/EX5.42/example5_42.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.42
+// Page 297
+printf("Example 5.42, Page 297 \n \n");
+
+// solution
+
+Hsol = 62.86 // kJ/mol solute
+Mcrystal = 286.1414
+Hcry = Hsol*1000/Mcrystal // kJ/kg solute
+printf(" Heat of crystallization of 1 kg crystal is "+string(Hcry)+" kJ.")
diff --git a/1247/CH5/EX5.43/example5_43.sce b/1247/CH5/EX5.43/example5_43.sce new file mode 100755 index 000000000..56bd7adfa --- /dev/null +++ b/1247/CH5/EX5.43/example5_43.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.43
+// Page 297
+printf("Example 5.43, Page 297 \n \n");
+
+// solution
+
+Hf = -285.82 // kJ/mol of H2O
+Hcryst = -4327.26-(-1387.08+10*Hf)
+printf(" Heat of crystallization = "+string(Hcryst)+" kJ/mol.")
diff --git a/1247/CH5/EX5.44/example5_44.sce b/1247/CH5/EX5.44/example5_44.sce new file mode 100755 index 000000000..100ed34a8 --- /dev/null +++ b/1247/CH5/EX5.44/example5_44.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.44
+// Page 297
+printf("Example 5.44, Page 297 \n \n");
+
+// solution
+
+Hfs = -1094.33
+Hfao = -1072.32
+Hsol = Hfao-Hfs
+printf(" Heat of solution of Boric acid = "+string(Hsol)+" kJ/mol.")
diff --git a/1247/CH5/EX5.45/example5_45.sce b/1247/CH5/EX5.45/example5_45.sce new file mode 100755 index 000000000..25ef007dd --- /dev/null +++ b/1247/CH5/EX5.45/example5_45.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.45
+// Page 297
+printf("Example 5.45, Page 297 \n \n");
+
+// solution
+
+// (a)
+Hf = -982.8
+Hfcryst = -1053.904
+Hdis = Hfcryst-Hf
+// (b)
+Hfcr = -3077.75
+Hsol = Hfcryst+7*(-285.83)-(-3077.75)
+printf(" (a) \n \n Hdissolulition = "+string(Hdis)+" kJ/mol ZnSO4. \n \n \n (b) \n \n Hsolution = "+string(Hsol)+" kJ/kmol.")
diff --git a/1247/CH5/EX5.46/example5_46.sce b/1247/CH5/EX5.46/example5_46.sce new file mode 100755 index 000000000..600b5b3e9 --- /dev/null +++ b/1247/CH5/EX5.46/example5_46.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.46
+// Page 300
+printf("Example 5.46, Page 300 \n \n");
+
+// solution
+
+// using chart 5.16 we get
+T = 329.5 // K
+printf(" T = "+string(T)+" K.")
diff --git a/1247/CH5/EX5.47/example5_47.sce b/1247/CH5/EX5.47/example5_47.sce new file mode 100755 index 000000000..7dd98070f --- /dev/null +++ b/1247/CH5/EX5.47/example5_47.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.47
+// Page 300
+printf("Example 5.47, Page 300 \n \n");
+
+// solution
+// basis 100(m1) kg 46% sol
+NaOH = 46 // kg
+H2O = 54 // kg
+m2 = NaOH/.25
+NaOHo = 25 // kg
+H2Oo = 75 // kg
+Hf1 = -453.138 // kJ/mol
+Hf2 = -467.678 // kJ/mol
+Hs = Hf2-Hf1
+Hg = -Hs*1000*1.501
+// using Appendix IV.1
+Hw1 = 146.65
+Hw2 = 104.9
+Hadd = 84*(Hw1-Hw2)
+H = Hg+Hadd
+C1 = 3.55
+T2 = 298.15+H/(184*C1) // K
+printf(" Final sol T = "+string(T2)+" K.")
diff --git a/1247/CH5/EX5.48/example5_48.sce b/1247/CH5/EX5.48/example5_48.sce new file mode 100755 index 000000000..3cd7c1223 --- /dev/null +++ b/1247/CH5/EX5.48/example5_48.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.48
+// Page 301
+printf("Example 5.48, Page 301 \n \n");
+
+// solution
+// basis 100 kg of sol with 32% N
+MNH4NO3 = 80.0434
+MNH2CONO2 = 60.0553
+MN2 = 28.0134
+na = 32/(60.9516)
+Ureadis = 1.1758*na*MNH2CONO2 // kg
+water = 100-(na*MNH4NO3+Ureadis)
+ndis = 525
+m = ndis/water
+HE1 = 40.3044-2.5962*m+.1582*m^2-3.4782*10^-3*m^3
+HE = HE1*ndis
+printf("Heat effect of the sol = "+string(HE)+" kJ.")
diff --git a/1247/CH5/EX5.49/example5_49.sce b/1247/CH5/EX5.49/example5_49.sce new file mode 100755 index 000000000..4ae3d2973 --- /dev/null +++ b/1247/CH5/EX5.49/example5_49.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.49
+// Page 302
+printf("Example 5.49, Page 302 \n \n");
+
+// solution
+Hmix = 896
+M1 = 88 // molar mass of n-amyl alcohol
+M2 = 78 // molar mass of benzene
+B = .473*M2
+A = .527*M1
+Ha = Hmix/A
+Hb = Hmix/B
+printf(" Integral heat of sol of n-amyl alcohol = "+string(Ha)+" kJ/kg n-amyl alcohol and of benzene = "+string(Hb)+" kJ/kg benzene.")
diff --git a/1247/CH5/EX5.5/example5_5.sce b/1247/CH5/EX5.5/example5_5.sce new file mode 100755 index 000000000..f9a8c0061 --- /dev/null +++ b/1247/CH5/EX5.5/example5_5.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.5
+// Page 206
+printf("Example 5.5, Page 206 \n \n");
+
+// solution
+
+// basis 1kg of 20% NaOH sol
+// referring to fig 5.4
+C11 = 3.56 // kJ/kg.K at 280.15K
+C12 = 3.71 // kJ/kg.K at 360.15K
+C1m = (C11+C12)/2
+H = 1*C1m*(360.15-280.15) // kJ
+printf(" Heat required to be added = "+string(H)+"kJ.")
diff --git a/1247/CH5/EX5.50/example5_50.sce b/1247/CH5/EX5.50/example5_50.sce new file mode 100755 index 000000000..c927ee4fb --- /dev/null +++ b/1247/CH5/EX5.50/example5_50.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.50
+// Page 302
+printf("Example 5.50, Page 302 \n \n");
+
+// solution
+// from fig 5.18
+Ta = 379.5 // K
+dH = -274-(-106.5) // kJ/kg sol
+Cm = 2.05 // kJ/kg K
+dHc = Cm*(Ta-298.15)
+// basis 100 kg of 93 % acid
+// acid balance
+x = poly(0, 'x')
+p = .93*100+x*.15-(100+x)*.77
+y = roots(p)
+//from fig
+y1 = 25.3
+printf(" (a) \n \n Resultant T of 77 percent sol = "+string(Ta)+" K. \n \n \n (b) \n \n Heat to be removed to cool it to 298.15 K = "+string(dH)+" kJ/kg sol \n \n \n (c) \n \n By mean heat capacity method : "+string(dHc)+" kJ/kg sol \n \n \n (d) \n \n Quantity of 15 percent acid to be mixed = "+string(y)+" kg. \n \n \n (e) \n \n from fig : "+string(y1)+" kg.")
diff --git a/1247/CH5/EX5.51/example5_51.sce b/1247/CH5/EX5.51/example5_51.sce new file mode 100755 index 000000000..ef2277ba9 --- /dev/null +++ b/1247/CH5/EX5.51/example5_51.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.51
+// Page 304
+printf("Example 5.51, Page 304 \n \n");
+
+// solution
+// basis 100 kg of 93% acid and 25.8 kg of 15% acid
+Hfp = -814
+Hf1 = -830
+HE1 = Hf1-Hfp
+Hf2 = -886.2
+HE2 = Hf2-Hfp
+Hf3 = -851
+HE3 = Hf3-Hfp
+Hsol = .9876*1000*(-37)-[.9482*1000*(-16)+.0394*1000*(-72.2)]
+Hev = 100*(30-25)*1.6
+Hcon = 25.8*25*3.7
+netHev = -Hsol-Hcon+Hev
+T = 298.15+netHev/(125.8*2.05)
+printf(" Temp of sol = "+string(T)+" K.")
diff --git a/1247/CH5/EX5.52/example5_52.sce b/1247/CH5/EX5.52/example5_52.sce new file mode 100755 index 000000000..7521fe4bc --- /dev/null +++ b/1247/CH5/EX5.52/example5_52.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.52
+// Page 306
+printf("Example 5.52, Page 306 \n \n");
+
+// solution
+
+// basis 1000 kg of mixed acid
+C11 = 2.45
+H1 = -296.7+C11*(308.15-273.15)
+C12 = 2.2
+H2 = -87.8+C12*(308.15-273.15)
+C13 = 1.45
+H3 = -35.5+C13*(308.15-273.15)
+C14 = 1.8
+H4 = -148.9+C14*(308.15-273.15)
+Hmix = 1000*H4-[76.3*H1+345.9*H2+577.7*H3]
+printf(" Heat of mixing = "+string(Hmix)+" kJ.")
diff --git a/1247/CH5/EX5.53/example5_53.sce b/1247/CH5/EX5.53/example5_53.sce new file mode 100755 index 000000000..6de823af8 --- /dev/null +++ b/1247/CH5/EX5.53/example5_53.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.53
+// Page 308
+printf("Example 5.53, Page 308 \n \n");
+
+// solution
+
+F = 1135
+Benzenef = 400*.993
+HNO3con = Benzenef*63/78
+H1 = -186.5
+C11 = 1.88
+H11 = H1+C11*(298.15-273.15)
+H2 = -288.9
+C12 = 1.96
+H22 = H2+C12*(298.15-273.15)
+H3 = 0
+C13 = 1.98
+H33 = C13*(298.15-273.15)
+Hr = -285.83+12.5-(-174.1+49.08)
+Benzener = Benzenef/78.1118
+fi = 903.84*H22+HNO3con*H33-F*H11+Benzener*Hr*1000 // kJ/h
+printf(" Total heat exchanged = "+string(fi)+" kJ/h.")
diff --git a/1247/CH5/EX5.54/example5_54.sce b/1247/CH5/EX5.54/example5_54.sce new file mode 100755 index 000000000..f26effacb --- /dev/null +++ b/1247/CH5/EX5.54/example5_54.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.54
+// Page 311
+printf("Example 5.54, Page 311 \n \n");
+
+// solution
+
+// from ref 24
+H = 1600.83
+To = 273.15
+h = 200
+Hf1 = -79.3 // table 5.59
+Hf2 = -46.11
+Hsol = Hf1-Hf2
+Hg = Hsol*1000*140/17.0305
+Raq = 140/.15 // kg/h
+dT = Hg/(4.145*Raq)
+T = -dT+303
+printf(" Temp of resultant sol = "+string(T)+" K.")
diff --git a/1247/CH5/EX5.55/example5_55.sce b/1247/CH5/EX5.55/example5_55.sce new file mode 100755 index 000000000..96817a05e --- /dev/null +++ b/1247/CH5/EX5.55/example5_55.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.55
+// Page 311
+printf("Example 5.55, Page 311 \n \n");
+
+// solution
+
+Hf1 = -80.14
+Hf2 = -46.11
+Hsol = Hf1-Hf2
+Hg = Hsol*1000*2/17.0305
+printf(" Heat generated for making 2 percent solution = "+string(Hg)+" kJ/100 kg sol.")
diff --git a/1247/CH5/EX5.56/example5_56.sce b/1247/CH5/EX5.56/example5_56.sce new file mode 100755 index 000000000..dd0a9c850 --- /dev/null +++ b/1247/CH5/EX5.56/example5_56.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.56
+// Page 312
+printf("Example 5.56, Page 312 \n \n");
+
+// solution
+
+fi3 = 15505407
+fi4 = 11395056
+fi5 = fi3-fi4 // kJ/h
+fi6 = 111.375*62.75*1000
+fi7 = 1063379
+fi8 = 5532.15*4.1868*(303.15-298.15)
+fi9 = 9030.4*3.45*(323.15-298.15)
+fi = fi5+fi6+fi8-fi7-fi9
+printf(" Heat removal in the cooler = "+string(fi)+" kJ/h.")
diff --git a/1247/CH5/EX5.57/example5_57.sce b/1247/CH5/EX5.57/example5_57.sce new file mode 100755 index 000000000..d97aa779f --- /dev/null +++ b/1247/CH5/EX5.57/example5_57.sce @@ -0,0 +1,46 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.57
+// Page 314
+printf("Example 5.57, Page 314 \n \n");
+
+// solution
+
+To = 273.15
+T1 = 308.15
+H1 = 124.8*(T1-To) // kJ/kmol
+H2 = 134.9*(T1-To) // kJ/kmol
+HE1 = .1*.9*[542.4+55.4*(.9-.1)-132.8*(.9-.1)^2-168.9*(.9-.1)^3] // kJ/kmol of mix
+Ha = HE1+H1*.1+H2*.9
+HE2 = .2*.8*[542.4+55.4*(.8-.2)-132.8*(.8-.2)^2-168.9*(.8-.2)^3] // kJ/kmol of mix
+Hb = HE2+H1*.2+H2*.8
+HE3 = .3*.7*[542.4+55.4*(.7-.3)-132.8*(.7-.3)^2-168.9*(.7-.3)^3] // kJ/kmol of mix
+Hc = HE3+H1*.3+H2*.7
+HE4 = .4*.6*[542.4+55.4*(.6-.4)-132.8*(.6-.4)^2-168.9*(.6-.4)^3] // kJ/kmol of mix
+Hd = HE4+H1*.4+H2*.6
+HE5 = .5*.5*[542.4+55.4*(.5-.5)-132.8*(.5-.5)^2-168.9*(.5-.5)^3] // kJ/kmol of mix
+He = HE5+H1*.5+H2*.5
+HE6 = .6*.4*[542.4+55.4*(.4-.6)-132.8*(.4-.6)^2-168.9*(.4-.6)^3] // kJ/kmol of mix
+Hf = HE6+H1*.6+H2*.4
+HE7 = .7*.3*[542.4+55.4*(.3-.7)-132.8*(.3-.7)^2-168.9*(.3-.7)^3] // kJ/kmol of mix
+Hg = HE7+H1*.7+H2*.3
+HE8 = .8*.2*[542.4+55.4*(.2-.8)-132.8*(.2-.8)^2-168.9*(.2-.8)^3] // kJ/kmol of mix
+Hh = HE8+H1*.8+H2*.2
+HE9 = .9*.1*[542.4+55.4*(.1-.9)-132.8*(.1-.9)^2-168.9*(.1-.9)^3] // kJ/kmol of mix
+Hi = HE9+H1*.9+H2*.1
+HE10 = .0*1.*[542.4+55.4*(.0-1.)-132.8*(.0-1.)^2-168.9*(.0-1.)^3] // kJ/kmol of mix
+Hj = HE10+H1+H2*0
+x = linspace(0,1,100)
+y = linspace(4300,5000,100)
+y = 4721.5-57.4*x+1137.7*x^2-3993.6*x^3+3909.2*x^4-1351.2*x^5
+plot(x,y)
+title("H vs x1")
+xlabel("x1")
+ylabel("H (kJ/kg sol.)")
+printf(" Enthalpy, kJ/kmol mix \n x1 HE H \n 0 0 "+string(H2)+" \n 0.1 "+string(HE1)+" "+string(Ha)+" \n 0.2 "+string(HE2)+" "+string(Hb)+" \n 0.3 "+string(HE3)+" "+string(Hc)+" \n 0.4 "+string(HE4)+" "+string(Hd)+" \n 0.5 "+string(HE5)+" "+string(He)+" \n 0.6 "+string(HE6)+" "+string(Hf)+" \n 0.7 "+string(HE7)+" "+string(Hg)+" \n 0.8 "+string(HE8)+" "+string(Hh)+" \n 0.9 "+string(HE9)+" "+string(Hi)+" \n 1.0 "+string(HE10)+" "+string(Hj)+"")
diff --git a/1247/CH5/EX5.58/example5_58.sce b/1247/CH5/EX5.58/example5_58.sce new file mode 100755 index 000000000..3a9ae28cd --- /dev/null +++ b/1247/CH5/EX5.58/example5_58.sce @@ -0,0 +1,16 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.58
+// Page 316
+printf("Example 5.58, Page 316 \n \n");
+
+// solution
+
+// see eg 5.57
+printf(" refer to eg 5.57")
diff --git a/1247/CH5/EX5.59/example5_59.sce b/1247/CH5/EX5.59/example5_59.sce new file mode 100755 index 000000000..c78bf8d5f --- /dev/null +++ b/1247/CH5/EX5.59/example5_59.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.59
+// Page 318
+printf("Example 5.59, Page 318 \n \n");
+
+// solution
+
+// from graph drawn in 5.57 we can see
+H1E1 = 300
+H1E2 = 63
+H2E1 = 30
+H2E2 = 214
+printf(" H1 at x1=0.3 is "+string(H1E1)+" kJ/kg sol \n H2 at x1=0.3 is "+string(H2E1)+" kJ/kg sol \n H1 at x1=0.6 is "+string(H1E2)+" kJ/kg sol \n H2 at x1=0.6 is "+string(H2E2)+" kJ/kg sol.")
diff --git a/1247/CH5/EX5.6/example5_6.sce b/1247/CH5/EX5.6/example5_6.sce new file mode 100755 index 000000000..2c06006ce --- /dev/null +++ b/1247/CH5/EX5.6/example5_6.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.6
+// Page 207
+printf("Example 5.6, Page 207 \n \n");
+
+// solution
+
+// basis 1kg Diphyl A-30
+Q = .7511*(553.15-313.15) + 1.465*10^-3*(553.15^2-313.15^2)/2 // kJ/kg
+fi = Q*4000 // kJ/h for mass flowrate 4000 kg/h
+Clm = (1.1807+1.5198)/2
+fi1 = Clm*(553.15-313.15)*4000/3600 // kJ/h
+err = (fi1-Q)*100/Q
+printf(" Heat to be supplied = "+string(fi1)+" kW \n Percent error = "+string(err)+".")
diff --git a/1247/CH5/EX5.60/example5_60.sce b/1247/CH5/EX5.60/example5_60.sce new file mode 100755 index 000000000..c24309120 --- /dev/null +++ b/1247/CH5/EX5.60/example5_60.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.60
+// Page 320
+printf("Example 5.60, Page 320 \n \n");
+
+// solution
+
+// basis 100 kg 96.1% H2SO4
+// from table 5.64
+m1SO3 = 78.4 // kg
+m1H2O = 21.6
+n1SO3 = m1SO3/80.063
+n1H2O = m1H2O/18.015
+// resultant sol has 23.2% H2SO4
+m2SO3 = 19
+m2H2O = 81
+Mrsol = m1SO3*100/m2SO3
+Mw = Mrsol-100
+w = Mrsol-m1SO3/18.015 // kmol
+HEosol = n1SO3*(-56940)+n1H2O*(-32657) // kJ
+HErsol = n1SO3*(-156168)+w-(-335)
+HE = HErsol-HEosol // kJ/kg original acid
+C = 3.43 // kJ/kg K
+dT = -HE/(Mrsol*C)
+T = 291.15+dT // K
+printf(" Heat of dilution = "+string(HE)+" kJ/kg original solution \n \n Final T of resultant solution = "+string(T)+" K.")
diff --git a/1247/CH5/EX5.61/example5_61.sce b/1247/CH5/EX5.61/example5_61.sce new file mode 100755 index 000000000..4d3d7906c --- /dev/null +++ b/1247/CH5/EX5.61/example5_61.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.61
+// Page 321
+printf("Example 5.61, Page 321 \n \n");
+
+// solution
+
+// basis 100 kg of original acid
+lv = 333.7 // kJ/kg
+H = -lv-18*4.1868
+HE = (-64277-H*312.63)/100 // kJ/kg
+printf(" Heat of dilution = "+string(HE)+" kJ/kg.")
diff --git a/1247/CH5/EX5.7/example5_7.sce b/1247/CH5/EX5.7/example5_7.sce new file mode 100755 index 000000000..3ee017a64 --- /dev/null +++ b/1247/CH5/EX5.7/example5_7.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.7
+// Page 208
+printf("Example 5.7, Page 208 \n \n");
+
+// solution
+
+T1 = 298.15 // K
+T2 = 775.15 //K
+// using eq 5.17
+Q = 28.839*(T2-T1)+2.0395*10^-3*(T2^2-T1^2)/2 + 6.9907*10^-6*(T2^3-T1^3)/3 - 3.2304*10^-9*(T2^4-T1^4)/4 // kJ/kmol
+printf(" Heat content of 1 kmol of gas mixture at 298K = "+string(Q)+" kJ/kmol.")
diff --git a/1247/CH5/EX5.8/example5_8.sce b/1247/CH5/EX5.8/example5_8.sce new file mode 100755 index 000000000..e8aad735d --- /dev/null +++ b/1247/CH5/EX5.8/example5_8.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.8
+// Page 210
+printf("Example 5.8, Page 210 \n \n");
+
+// solution
+
+// basis 8000 kg/h mixture is to be cooled
+qn1m = .118*8000 // kg/h
+qn1 = qn1m/93.1242 // kmol/h
+qn2m = 8000-qn1m // kg/h
+qn2 = qn2m/18 // kmol/h
+T1 = 373.15 //K
+T2 = 313.15 //K
+fi = qn1*[206.27*(T1-T2)-211.5065*10^-3*(T1^2-T2^2)/2+564.2902*10^-6*(T1^3-T2^3)/3] + qn2*[50.845*(T1-T2)+213.08*10^-3*(T1^2-T2^2)/2-631.398*10^-6*(T1^3-T2^3)/3+648.746*10^-9*(T1^4-T2^4)/4] // kJ/h
+printf(" Heat removal rate of subcooling zone of the condenser = "+string(fi)+" kJ/h.")
diff --git a/1247/CH5/EX5.9/example5_9.sce b/1247/CH5/EX5.9/example5_9.sce new file mode 100755 index 000000000..97c9166cb --- /dev/null +++ b/1247/CH5/EX5.9/example5_9.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Stoichiometry
+// Chapter 5
+// Energy Balances
+
+
+// Example 5.9
+// Page 220
+printf("Example 5.9, Page 220 \n \n");
+
+// solution
+
+// (a)
+T = 305.15 //K
+Pv1 = 10^(4.0026-(1171.530/(305.15-48.784))) // bar
+// (b)
+T = 395.15
+Pv2 = 10^(3.559-(643.748/(395.15-198.043))) // bar
+printf(" (a) \n \n V.P. of n-hexane at 305.15K = "+string(Pv1)+" bar. \n \n \n (b) \n \n V.P. of water at 395.15K = "+string(Pv2)+" bar.")
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