{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 1 : Measured thermodynamic Properties and Other Basic Concepts" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.3 Page No : 28" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "P = 1.4 ; \t\t\t# [MPa]\n", "T = 333. \t\t\t#[K]\n", "T1 = 320. \t\t\t#[K]\n", "T2 = 360. \t\t\t#[K]\n", "P_low = 1. \t\t\t#[MPa]\n", "P_high = 1.5\n", "V_cap_T1_P1 = 0.2678\n", "V_cap_T2_P1 = 0.2873\n", "V_cap_T1_P1_5 = 0.1765\n", "V_cap_T2_P1_5 = 0.1899\n", "\n", "# Calculations\n", "#At P = 1 MPa\n", "V_cap_T333_P1 = V_cap_T1_P1 + (V_cap_T2_P1 - V_cap_T1_P1)*((T - T1)/(T2- T1))\n", "\n", "#Similarly at P=1.5 MPa\n", "V_cap_T333_P1_5 = V_cap_T1_P1_5 + (V_cap_T2_P1_5 - V_cap_T1_P1_5)*((T - T1)/(T2 - T1))\n", "\n", "#At T=333*C\n", "V_cap_P1_5 = V_cap_T333_P1_5 ;\n", "V_cap_P1 = V_cap_T333_P1 ;\n", "V_cap_P1_4 = V_cap_P1 + (V_cap_P1_5 - V_cap_P1)*((P - P_low)/(P_high - P_low)) ; \t\t\t#[m**3/kg]\n", "\n", "# Results\n", "print 'Required specific volume = %.5f m**3/kg'%(V_cap_P1_4);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Required specific volume = 0.19951 m**3/kg\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.4 Page No : 29" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Variables\n", "P = 1.4 ; \t\t\t#[MPa]\n", "P_low = 1 ;\t\t\t#[MPa]\n", "P_high = 1.5;\t\t\t#[MPa]\n", "\n", "#At T=333*C from interpolation we have\n", "v_cap_P1_5 = 0.18086 ;\t\t\t#[m**3/kg]\n", "v_cap_P1 = 0.27414 ;\t\t\t#[m**3/kg]\n", "\n", "# Calculations\n", "#Molar volume is inversely proportional to pressure\n", "v_cap_P1_4 = v_cap_P1 +(v_cap_P1_5 - v_cap_P1)*((1/P - 1/P_low)/(1/P_high - 1/P_low));\n", "x=(0.19951-v_cap_P1_4)/v_cap_P1_4*100 ;\n", "\n", "# Results\n", "print 'Specific volume m**3/kg) = %g m**3/kg'%(v_cap_P1_4);\n", "print 'Percentage difference = %.1f %%'%(x);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Specific volume m**3/kg) = 0.194186 m**3/kg\n", "Percentage difference = 2.7 %\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.5 page no : 29" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# variables\n", "V1 = 1 # volume\n", "m1 = 2.5 # water\n", "Vv = 5.042 # m**3/kg\n", "Vt = .001023 # m**3/kg\n", "T_44632 = 145 # C\n", "T_39278 = 150 # C\n", "\n", "# Calculations\n", "v1 = V1/m1 \n", "x1 = (v1 - Vt)/(Vv - Vt)\n", "\n", "T2 = T_44632 + ( T_39278 - T_44632) * (( 0.4 - 0.44632)/(0.39278 - 0.44632))\n", "\n", "# Restuls\n", "print \"Approximately %.f %% of the mass of water in the vapor\"%(round(x1,2)*100)\n", "print \"Temperature T2 = %.1f C\"%T2\n", "\n", "# Note : Answer in book is wrong. Please calculate manually.\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Approximately 8 % of the mass of water in the vapor\n", "Temperature T2 = 149.3 C\n" ] } ], "prompt_number": 8 } ], "metadata": {} } ] }