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Diffstat (limited to 'Introduction_To_Chemical_Engineering/ch6.ipynb')
| -rw-r--r-- | Introduction_To_Chemical_Engineering/ch6.ipynb | 65 |
1 files changed, 0 insertions, 65 deletions
diff --git a/Introduction_To_Chemical_Engineering/ch6.ipynb b/Introduction_To_Chemical_Engineering/ch6.ipynb index dc10eb48..1a7b64a4 100644 --- a/Introduction_To_Chemical_Engineering/ch6.ipynb +++ b/Introduction_To_Chemical_Engineering/ch6.ipynb @@ -27,10 +27,8 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to find the flux and pressure difference\n", "\n", "import math \n", - "# Variables\n", "D_AB=6.75*10**-5 #in m2/s\n", "Z=0.03 #in m\n", "R=8314\n", @@ -38,11 +36,9 @@ "p_A2=1.5*10**4 #in Pa\n", "T=298 #in K\n", "\n", - "# Calculations and Results\n", "N_A=D_AB*(p_A1-p_A2)/(R*T*Z);\n", "print \"flux = %f kmol/sq m s\"%(N_A)\n", "\n", - "#for partial pressure\n", "Z=0.02; #in m\n", "p_A2=p_A1-((N_A*R*T*Z)/D_AB);\n", "print \"pressure = %f Pa\"%(p_A2)\n", @@ -74,17 +70,14 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to find the flux of NH3 and equimolar counter diffusion flux\n", "\n", "import math \n", "\n", - "# Variables\n", "Z=0.15 #in m\n", "P=1.013*10**5 #in Pa\n", "p_A1=1.5*10**4 #in Pa\n", "p_A2=5*10**3 #in Pa\n", "\n", - "# Calculations and Results\n", "p_B1=P-p_A1;\n", "p_B2=P-p_A2;\n", "\n", @@ -92,13 +85,11 @@ "R=8314.\n", "T=298. #in K\n", "\n", - "#for non diffusing N2\n", "p_BM=(p_B2-p_B1)/math.log (p_B2/p_B1);\n", "print p_B1, p_B2\n", "N_A=D_AB*(p_A1-p_A2)*P/(R*T*Z*p_BM);\n", "print \"flux = %.4e kmol/sq m s\"%(N_A)\n", "\n", - "#for diffusing N2\n", "N_A=D_AB*(p_A1-p_A2)/(R*T*Z);\n", "print \"flux = %.4e kmol/sq m s\"%(N_A)\n" ], @@ -130,13 +121,11 @@ "collapsed": false, "input": [ "import math \n", - "# Variables\n", "M_A=36.5 #molar mass of HCl\n", "M_B=18. #molar masss of water\n", "w_A1=12.; #weight % of HCL\n", "w_A2=4. #weight % of HCL\n", "\n", - "# Calculations and Results\n", "x_A1=(w_A1/M_A)/((w_A1/M_A)+((100-w_A1)/M_B));\n", "print 'x_A1 =%f'%(x_A1)\n", "\n", @@ -144,7 +133,6 @@ "M1=100./((w_A1/M_A)+((100-w_A1)/M_B));\n", "print \"molar mass at point 1 = %f kg/kmol\"%(M1)\n", "\n", - "#at point 2\n", "x_A2=(w_A2/M_A)/((w_A2/M_A)+((100-w_A2)/M_B));\n", "x_B2=1-x_A2;\n", "M2=100/((w_A2/M_A)+((100-w_A2)/M_B)); #avg molecular weight at point 2\n", @@ -190,10 +178,8 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to find the mean driving force and mass transfer area\n", "\n", "import math \n", - "# Variables\n", "Gs=700/22.4 #in kmol of dry air/hr\n", "Ls=1500./18 #in kmol of dry air/hr\n", "y1=0.05\n", @@ -203,14 +189,11 @@ "X1=(Gs/Ls)*(Y1-Y2);\n", "m=Gs*(Y1-Y2);\n", "\n", - "# Calculations and Results\n", - "#driving force\n", "delta_Y1=Y1-1.68*X1;\n", "delta_Y2=Y2-1.68*X2;\n", "delta_Y=(delta_Y1-delta_Y2)/(math.log (delta_Y1/delta_Y2));\n", "print \"driving force = %f kmol acetone/kmol dry air\"%(delta_Y)\n", "\n", - "#mass transfer area\n", "K_G=0.4 #in kmol acetone/kmol dry air\n", "A=m/(K_G*delta_Y);\n", "print \"area = %f sq m\"%(A)\n" @@ -241,17 +224,13 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to calculate minimum oil circulation rate\n", "\n", "import math \n", - "# Variables\n", "G1=(855/22.4)*(106.6/101.3)*(273/299.7);\n", "y1=0.02;\n", "Y1=y1/(1-y1);\n", "Gs=G1*(1-y1);\n", "\n", - "# Calculations\n", - "#for 95% removal\n", "Y2=0.05*Y1;\n", "x2=0.005;\n", "X2=x2/(1-x2);\n", @@ -261,7 +240,6 @@ "Ls_molar=(Gs*(Y1-Y2))/(X1-X2);\n", "Ls=Ls_molar*260;\n", "\n", - "# Results\n", "print \"minimum oil circulation rate = %f kg/hr\"%(Ls)\n" ], "language": "python", @@ -289,15 +267,12 @@ "cell_type": "code", "collapsed": false, "input": [ - "# to find the equilibrium composition\n", "\n", "import math \n", - "# Variables\n", "P_M = 53.32 #kPa\n", "P_W = 12.33 #in kpA\n", "P = 40 #IN K pA\n", "\n", - "# Calculations and Results\n", "x = (P - P_W)/(P_M-P_W);\n", "\n", "print \"liquid phase composition = %f\"%(x)\n", @@ -331,7 +306,6 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to find the top and bottom composition\n", "\n", "import math \n", "from matplotlib.pyplot import *\n", @@ -339,13 +313,9 @@ "\n", "%pylab inline\n", "\n", - "# Variables\n", "x = [1,0.69,0.40,0.192,0.045,0];\n", "y = [1,0.932,0.78,0.538,0.1775,0];\n", "plot(x,y)\n", - "#xlabel(\"x\")\n", - "#ylabel(\"y\")\n", - "#title(\"distillation curve\")\n", "x = linspace(0,1,10)\n", "y = linspace(0,1,10)\n", "plot(x,y)\n", @@ -362,7 +332,6 @@ "\n", "show()\n", "\n", - "# Results\n", "print \"composition of top product = %f mole percent of hexane\"%(y_D*100)\n", "print \"composition of bottom product = %f mole percent of hexane\"%(x_W*100)\n" ], @@ -415,19 +384,16 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to find the composite distillate and residue\n", "\n", "%pylab inline\n", "import math \n", "from numpy import *\n", "from matplotlib.pyplot import *\n", - "# Variables\n", "F = 100. #moles\n", "xf = 0.4;\n", "D = 60. #moles\n", "W = 40. #moles\n", "\n", - "# Calculations\n", "x = linspace(0.2,0.45,6)\n", "y = zeros(6)\n", "z = zeros(6)\n", @@ -443,7 +409,6 @@ "yd = (F*xf-W*xw)/D;\n", "show()\n", "\n", - "# Results\n", "print \"composition of distillate = %f\"%(yd)\n", "print \"composition of residue = %f\"%(xw)\n" ], @@ -488,22 +453,17 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to find the top and bottom product composition\n", "\n", "import math \n", "\n", - "# Variables\n", - "#part 1\n", "x=0.4;\n", "y=0.8;\n", "x_D=y;\n", "x_W=0.135; #bottom concentration\n", "\n", - "# Calculations and Results\n", "D=(100*x-100*x_W)/(y-x_W); #distillate amount\n", "print \"amount of distillate =%f moles/h\"%(D)\n", "\n", - "#part 2\n", "alpha=6; #relative volatility\n", "x_R=y/(y+(alpha*(1-y))); #liquid leaving partial condensor\n", "print \"liquid leaving partial condenser = %f\"%(x_R)\n", @@ -543,15 +503,12 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to find the percentage extraction of nicotine\n", "\n", "import math \n", - "# Variables\n", "x=0.01; #% of nicotine\n", "X0 = x/(1-x);\n", "w=150. #weight of nicotine water solution\n", "\n", - "# Calculations and Results\n", "A0=w*(1-X0);\n", "B0=250.; #kg keroscene\n", "X1 = A0*X0/(A0+B0*0.798);\n", @@ -590,15 +547,12 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to find the number of stages\n", "\n", "import math \n", - "# Variables\n", "x=0.01 #mole fraction of nicotine\n", "yN = 0.0006; #mole fraction in solvent\n", "xN = 0.001; #final mole fraction in water\n", "\n", - "# Calculations and Results\n", "X0=x/(1.-x); #in kg nicotine/kg water\n", "YN =yN/(1.-yN); #in kg nicotine/kg keroscene\n", "XN = xN/(1.-xN);\n", @@ -608,7 +562,6 @@ "Y1=((A0*(X0-XN))/B0)+YN; #in kg nicotine/kg kerosene\n", "print \"Y1 = %f kg nicotine/kg kerosene\"%(Y1)\n", "\n", - "#for graph refer to the book\n", "number_of_stages = 8.4;\n", "print \"numnber of stages = %f\"%(number_of_stages)\n" ], @@ -638,16 +591,13 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to calculate the humidity\n", "\n", "import math \n", "\n", - "# Variables\n", "P = 101.3 #in kPa\n", "pA = 3.74 #in kPa\n", "p_AS = 7.415 #in kPa\n", "\n", - "# Calculations and Results\n", "H = (18.02/28.97)*(pA/(P-pA));\n", "print \"humidity = %f kg H2O/kg air\"%(H)\n", "\n", @@ -688,12 +638,10 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to find the air flow rate and outlet humidity\n", "\n", "import math \n", "from numpy import *\n", "\n", - "# Variables\n", "S=425.6 #in kg/h\n", "X1 = 0.035 #in kgwater/kg dry solid\n", "t_s1=25. #in degree C\n", @@ -705,7 +653,6 @@ "C_pS = 1.465 #in kJ/kg dry solid\n", "C_pA = 4.187 #in kg/ kg H2O K\n", "\n", - "# Calculations and Results\n", "H_G2=(1.005+1.88*H2)*(t_G2-0)+H2*2501;\n", "H_S1 = C_pS*(t_s1-0)+X1*C_pA*(t_s1-0); #in kJ/kg\n", "H_S2 = C_pS*(t_s2-0)+X2*C_pA*(t_s2-0); #in kJ/kg\n", @@ -715,8 +662,6 @@ "print \"Enthalpy of entering solid HS1 = %f kJ/kg dryair\"%(H_S1)\n", "print \"Enthalpy of exit solid HS2 = %f kJ/kg dryair\"%(H_S2)\n", "\n", - "#applying GHg2 + SHs1 = GHg1 +SHs2 +Q, we get two linear equations\n", - "#0.0175G+14.17248 = GH1 and 98.194G-29745.398 = 2562.664GH1\n", "A = array([[0.0175, -1],[98.194, -2562.664]]);\n", "b = array([[-14.17248],[29745.398]]);\n", "x = linalg.solve(A,b)\n", @@ -754,22 +699,16 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to find the crystal yield\n", "\n", "import math \n", "from numpy import *\n", - "# Variables\n", "M_Na2CO3 = 106\n", "M_10H2O = 180.2\n", "M_Na2CO3_10H2O = 286.2;\n", "w_Na2CO3 = 5000. #in kg\n", "water = 0.05 #% of water evaporated\n", "\n", - "# Calculations\n", "W = water*w_Na2CO3;\n", - "#solving material balance, we have two equations\n", - "#equation 1 -> 0.8230L +0.6296C = 3500\n", - "#equation 2 -> 0.1769L + 0.3703C = 1250\n", "\n", "A = array([[0.8230, 0.6296],[0.1769, 0.3703]])\n", "b = array([[3500],[1250]])\n", @@ -777,7 +716,6 @@ "L = x[0]\n", "C = x[1];\n", "\n", - "# Results\n", "print \"L = %f kg solution\"%(L)\n", "print \"C = %f kg of Na2CO3.10H2O crystals\"%(C)\n" ], @@ -807,15 +745,12 @@ "cell_type": "code", "collapsed": false, "input": [ - "#to find the crystal yield\n", "\n", "import math \n", "from numpy import *\n", - "# Variables\n", "A = array([[0.7380, 0.5117],[0.2619, 0.4882]])\n", "b = array([[1400],[600]])\n", "\n", - "# Calculations and Results\n", "x = linalg.solve(A,b)\n", "L = x[0]\n", "C = x[1];\n", |