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diff --git a/sample_notebooks/KAVANA B/chapter3.ipynb b/sample_notebooks/KAVANA B/chapter3.ipynb new file mode 100644 index 00000000..a66c91da --- /dev/null +++ b/sample_notebooks/KAVANA B/chapter3.ipynb @@ -0,0 +1,145 @@ +{
+ "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": {}
+ }
+ ]
+}
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