{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "CHAPTER 3: Heating and Cooling of Electrical Machines" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.1, Page number 37" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "d = 12.0 #Diameter of copper bar(mm)\n", "t = 1.5 #Thickness of micanite tube(mm)\n", "rho = 8.0 #Thermal resistivity(ohm-m)\n", "theta_diff = 25.0 #Temperature difference(\u00b0C)\n", "l = 0.2 #Length of bar(m)\n", "\n", "#Calculation\n", "S = math.pi*(d+t)*10**-3*l #Area of insulation in the path of heat flow(m^2)\n", "R_s = rho*t*10**-3/S #Thermal resistance in the micanite tube(ohm)\n", "Q_con = theta_diff/R_s #Loss that will pass from copper bar to iron(W)\n", "\n", "#Result\n", "print('Loss that will pass from copper bar to iron, Q_con = %.2f W' %Q_con)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Loss that will pass from copper bar to iron, Q_con = 17.67 W\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.2, Page number 37" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "ratio = 20.0 #Ratio of resistivity across lamination to along lamination\n", "t = 40.0 #Thickness of lamination stack(mm)\n", "S = 6000.0 #Cross-section of lamination(mm^2)\n", "theta = 20.0 #Temperature difference across laminations(\u00b0C)\n", "theta_1 = 5.0 #Temperature difference among laminations(\u00b0C)\n", "Q_con_1 = 25.0 #Heat conducted along laminations(W)\n", "S_1 = 2500.0 #Cross-section of lamination(mm^2)\n", "t_1 = 20.0 #Thickness of lamination stack(mm)\n", "\n", "#Calculation\n", "rho_1 = S_1*theta_1*10**-6/(Q_con_1*t_1*10**-3) #Thermal resistivity along direction of lamination(ohm-m)\n", "rho = ratio*rho_1 #Thermal resistivity across laminations(ohm-m)\n", "Q_con = S*10**-6*theta/(rho*t*10**-3) #Heat conducted across laminations(W)\n", "\n", "#Result\n", "print('Loss that will be conducted across laminations, Q_con = %.f W' %Q_con)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Loss that will be conducted across laminations, Q_con = 6 W\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.3, Page number 39" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "e = 0.8 #Co-efficient of emissivity\n", "temp_body = 60.0 #Temperature of body(\u00b0C)\n", "temp_walls = 20.0 #Temperature of walls of room(\u00b0C)\n", "\n", "#Calculation\n", "T_1 = 273+temp_body #Temperature of body(K)\n", "T_0 = 273+temp_walls #Temperature of walls of room(K)\n", "q_rad = 5.7*10**-8*e*(T_1**4-T_0**4) #Heat radiated from body(W/m^2)\n", "\n", "#Result\n", "print('Heat radiated from body, q_rad = %.1f W/m^2' %q_rad)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat radiated from body, q_rad = 224.6 W/m^2\n" ] } ], "prompt_number": 1 } ], "metadata": {} } ] }