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{
"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": {}
}
]
}
|
