{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 01 : Fundamentals of power System" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.1, Page No 11" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "Sb=100.0\t\t# base value of power(MVA)\n", "Vb=33.0\t\t\t# base value of voltage (Kv)\n", "\n", "#Calculations\n", "Vbl=Vb*110.0/32\n", "Vbm=Vbl*32.0/110\n", "Zp_ut=0.08*100*32*32/(110*33*33)\n", "Zp_u_l=50*100/(Vbl**2)\n", "Zp_um1=0.2*100*30*30.0/(30*33*33)\n", "Zp_um2=0.2*100*30*30.0/(20*33*33)\n", "Zp_um3=0.2*100*30*30.0/(50*33*33)\n", "\n", "#Results\n", "print(\"Base value of voltage in line = %.2f kV\" %Vbl)\n", "print(\"Base value of voltage in motor circuit=%.0f kV\" %Vbm)\n", "print(\"p.u value of reactance transformer =%.5f p.u\" %Zp_ut)\n", "print(\"p.u value of impedence of line=%.4f p.u\" %Zp_u_l)\n", "print(\"p.u value of reactance of motor 1 =%.4f p.u\" %Zp_um1)\n", "print(\"p.u value of reactance of motor 2 =%.3f p.u\" %Zp_um2)\n", "print(\"p.u value of reactance of motor 3 =%.4f p.u\" %Zp_um3)\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Base value of voltage in line = 113.44 kV\n", "Base value of voltage in motor circuit=33 kV\n", "p.u value of reactance transformer =0.06839 p.u\n", "p.u value of impedence of line=0.3886 p.u\n", "p.u value of reactance of motor 1 =0.5510 p.u\n", "p.u value of reactance of motor 2 =0.826 p.u\n", "p.u value of reactance of motor 3 =0.3306 p.u\n" ] } ], "prompt_number": 1 } ], "metadata": {} } ] }