diff options
Diffstat (limited to 'Basic_Principles_And_Calculations_In_Chemical_Engineering/ch27.ipynb')
| -rw-r--r-- | Basic_Principles_And_Calculations_In_Chemical_Engineering/ch27.ipynb | 339 |
1 files changed, 339 insertions, 0 deletions
diff --git a/Basic_Principles_And_Calculations_In_Chemical_Engineering/ch27.ipynb b/Basic_Principles_And_Calculations_In_Chemical_Engineering/ch27.ipynb new file mode 100644 index 00000000..0515e5c2 --- /dev/null +++ b/Basic_Principles_And_Calculations_In_Chemical_Engineering/ch27.ipynb @@ -0,0 +1,339 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 27 : Ideal Processes Efficiency and the Mechanical Energy Balance" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 27.1 page no. 838\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Calculation of the Work done during Evaporation of a Liquid\n", + "# Solution E27.1\n", + "\n", + "# Variables\n", + "V_w = 1. ;\t\t\t# Volume of given water -[L]\n", + "P_atm = 100. ;\t\t\t# Atmospheric pressure - [kPa]\n", + "\n", + "#W = -p*del_V\n", + "V_H2O = 0.001043 ;\t\t\t# Specific volume of water from steam table according to book- [cubic metre] \n", + "V_vap = 1.694 ;\t\t\t# Specific volume of vapour from steam table according to book- [cubic metre] \n", + "V1 = 0 ;\t\t\t# Initial volume of H2O in bag-[cubic metre]\n", + "\n", + "# Calculations\n", + "V2 = (V_w*V_vap)/(1000*V_H2O) ;\t\t\t# Final volume of water vapour -[cubic metre] \n", + "W = -P_atm*(V2 -V1)* 1000 ;\t\t\t# Work done by saturated liquid water -[J]\n", + " \n", + "# Results \n", + "print ' Work done by saturated liquid water is %.3e J.'%W\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Work done by saturated liquid water is -1.624e+05 J.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + " Example 27.2 page no. 840\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Calculation of Work in a Batch Process\n", + "# Solution E27.2\n", + "\n", + "# Variables\n", + "m_N2 = 1. ;\t\t\t# Moles of N2 taken -[kg mol]\n", + "p = 1000.;\t\t\t# Pressure of cylinder-[kPa]\n", + "T = 20. + 273. ;\t\t\t# Temperature of cylinder -[K]\n", + "a_pis = 6. ;\t\t\t# Area of piston - [square centimetre]\n", + "m_pis = 2. ;\t\t\t# Mass of pston - [kg]\n", + "R = 8.31 ;\t\t\t# Ideal gas constant - [(kPa*cubic metre)/(K * kgmol)]\n", + "\n", + "# Calculations\n", + "V = (R*T)/p ;\t\t\t# Specific volue of gas at initial stage -[cubic metre/kg mol]\n", + "V1 = V * m_N2 ;\t\t\t# Initial volume of gas - [cubic metre]\n", + "V2 = 2.*V1 ;\t\t\t# Final volume of gas according to given condition -[cubic metre]\n", + "\n", + "# Assumed surrounding pressure constant = 1 atm\n", + "p_atm = 101.3 ;\t\t\t# Atmospheric pressure-[kPa]\n", + "del_Vsys = V2 -V1 ;\t\t\t# Change in volume of system -[cubic metre]\n", + "del_Vsurr = - del_Vsys ;\t\t\t# Change in volume of surrounding -[cubic metre]\n", + "W_surr = -p_atm*del_Vsurr ;\t\t\t# Work done by surrounding - [kJ]\n", + "W_sys = -W_surr ;\t\t\t# Work done by system - [kJ]\n", + "\n", + "# Results\n", + "print ' Work done by gas(actually gas + piston system) is %.0f kJ.'%W_sys\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Work done by gas(actually gas + piston system) is -247 kJ.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + " Example 27.3 page no. 845\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Efficiency of Power Generation by a Hydroelectric Plant\n", + "# Solution \n", + "\n", + "# Variables\n", + "p_plant = 20. ;\t\t\t# Power generated by plant-[MW]\n", + "h = 25. ;\t\t\t# Height of water level - [m]\n", + "V = 100. ;\t\t\t# Flow rate of water -[cubic metre/s]\n", + "d_water = 1000. ;\t\t\t# Density of water - [ 1000 kg / cubic metre]\n", + "g = 9.807 ;\t\t\t# Acceleration due to gravity-[m/square second]\n", + "\n", + "# Calculations\n", + "M_flow = V*d_water ;\t\t\t# Mass flow rate of water -[kg/s]\n", + "del_PE = M_flow*g*h ;\t\t\t# Potential energy change of water per second -[W]\n", + "eff = (p_plant*10**6) /(del_PE) ;\t\t\t# Efficiency of plant \n", + "\n", + "# Results\n", + "print ' Efficiency of plant is %.2f .'%eff\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Efficiency of plant is 0.82 .\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + " Example 27.4 page no. 845\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Calculation of Plant Efficiency\n", + "# Solution Fig.E27.4\n", + "\n", + "# Variables\n", + "LHV = 36654. ;\t\t\t# LHV value of fuel - [kJ/ cubic metre]\n", + "Q1 = 16. ;\t\t\t#- [kJ/ cubic metre]\n", + "Q2 = 0 ;\t\t\t#- [kJ/ cubic metre]\n", + "Q3 = 2432. ;\t\t\t#- [kJ/ cubic metre]\n", + "Q4 = 32114. ;\t\t\t#- [kJ/ cubic metre]\n", + "Q41 = 6988. ;\t\t\t#- [kJ/ cubic metre]\n", + "Q8 = 1948. ;\t\t\t#- [kJ/ cubic metre]\n", + "Q9 = 2643. ;\t\t\t#- [kJ/ cubic metre]\n", + "Q81 = 2352. - Q8 ;\t\t\t# - [kJ/ cubic metre]\n", + "Q567 = 9092. ;\t\t\t# Sum of Q5, Q6 and Q7- [kJ/ cubic metre]\n", + "\n", + "# Calculations and Results\n", + "#(a)\n", + "G_ef = (LHV+ Q1 +Q2 + Q3 - Q9)/(LHV) ;\t\t\t# Gross efficiency\n", + "print '(a) Gross efficiency is %.3f .'%G_ef\n", + "\n", + "#(b)\n", + "T_ef = (Q567+Q8)/(LHV+ Q1 +Q2 + Q3) ;\t\t\t#Thermal efficiency \n", + "print ' (b) Thermal efficiency is %.3f .'%T_ef\n", + "\n", + "#(c)\n", + "C_ef = Q4/(Q4 + Q41) ;\t\t\t# Combustion efficiency\n", + "print ' (c) Combustion efficiency is %.3f .'%C_ef\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) Gross efficiency is 0.995 .\n", + " (b) Thermal efficiency is 0.282 .\n", + " (c) Combustion efficiency is 0.821 .\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 27.5 page no. 850" + ] + }, + { + "cell_type": "code", + "collapsed": true, + "input": [ + "# Comparison of the reversible work for a batch process with that of a flow process operating under the same conditions.\n", + "\n", + "from scipy.integrate import quad\n", + "\n", + "# Variables\n", + "V1 = 5. ;\t\t\t# Volume of gas initially - [cubic feet]\n", + "P1 = 1. ;\t\t\t# Initial pressure - [atm]\n", + "P2 = 10. ;\t\t\t# Final pressure - [atm]\n", + "T1 = 100. + 460 ;\t\t\t# initial temperature - [degree Rankine]\n", + "R = 0.7302 ;\t\t\t# Ideal gas constant -[(cubic feet*atm)/(lb mol)*(R)]\n", + "\t\t\t#Equation of state pV**1.4 = constant\n", + "\n", + "\n", + "# Calculations and Results\n", + "V2 = V1*(P1/P2)**(1/1.4) ;\t\t\t# Final volume - [cubic feet] \n", + "\n", + "def f(V):\n", + " return -(P1)*(V1/V)**(1.4)\n", + " \n", + "W1_rev = quad(f,V1,V2)[0] ;\t\t\t# Reversible work done in compresion in a horizontal cylinder with piston -[cubic feet *atm]\n", + "W1 = W1_rev *1.987/.7302 ;\t\t\t# Conversion to Btu -[Btu]\n", + "print '(a)Reversible work done in compression in a horizontal cylinder with piston is %.1f Btu . '%W1\n", + "\n", + "#(b)\n", + "n1 = (P1*V1)/(R*T1) ;\t\t\t# Number of moles of gas\n", + "\n", + "def f1(P):\n", + " return (V1)*(P1/P)**(1/1.4)\n", + "W2_rev = quad(f1,P1,P2)[0]\t\t# Reversible work done in compresion in a rotary compressor -[cubic feet *atm]\n", + "W2 = W2_rev *1.987/.7302 ;\t\t\t# Conversion to Btu -[Btu]\n", + "\n", + "print '(b)Reversible work done in a rotary compressor is %.1f Btu . '%W2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Reversible work done in compression in a horizontal cylinder with piston is 31.7 Btu . \n", + "(b)Reversible work done in a rotary compressor is 44.3 Btu . \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 27.6 page no. 853" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Application of the mechanical energy balance to the pumping water\n", + "\n", + "from scipy.integrate import quad\n", + "# Variables\n", + "m_water = 1. ;\t\t\t# Mass flow rate of water -[lb/min]\n", + "P1 = 100. ;\t\t\t# Initial pressure - [psia]\n", + "P2 = 1000. ;\t\t\t# Final pressure - [psia]\n", + "T1 = 80. + 460 ;\t\t\t# initial temperature - [degree Rankine]\n", + "T2 = 100. + 460 ;\t\t\t# final temperature - [degree Rankine]\n", + "h = 10. ;\t\t\t# Difference in water level between entry and exit of stream-[ft]\n", + "g = 32.2 ;\t\t\t# Accleration due to gravity - [ft/ square second]\n", + "gc = 32.2 ;\t\t\t#[(ft*lbm)/(lbf*square second)]\n", + "\n", + "v1 = .01607 ;\t\t\t# specific volume of liquid water at 80 degree F -[cubic feet/lbm]\n", + "v2 = .01613 ;\t\t\t# specific volume of liquid water at 100 degree F -[cubic feet/lbm] \n", + "v= 0.0161 ;\t\t\t# -[cubic feet/lbm]\n", + "\n", + "# Calculations\n", + "del_PE = (h*g)/(gc*778) ;\t\t\t# Change in potential energy - [Btu/lbm]\n", + "\n", + "def f(P):\n", + " return (v)*(12**2/778.)\n", + " \n", + "PV_work = quad(f,P1,P2)[0]\t\t\t# PV work done -[Btu/lbm]\n", + "#From eqn. (A)\n", + "W = PV_work + del_PE ;\t\t\t# Work per minute required to pump 1 lb water per minute - [Btu/lbm]\n", + "\n", + "# Results\n", + "print ' Work per minute required to pump 1 lb water per minute is %.2f Btu/lbm . '%W\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Work per minute required to pump 1 lb water per minute is 2.69 Btu/lbm . \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file |