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{
"metadata": {
"name": "",
"signature": "sha256:179d1aa791f84982595aec77701ac00e5f4718cb4b0fb539bad5702fddb2a8a1"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 5:Chemical and Heating Effects of Electric Current"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.1: Page 74:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"t=200;# time in sec\n",
"M=111.83;# silver in mg\n",
"I=0.5;# current in A\n",
"\n",
"#calculations:\n",
"Z=(M/(I*t*1000))*1000# electro-chemical-equivalent\n",
"\n",
"#Results\n",
"print \"E.C.E,Z(mg/C) = \",Z"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"E.C.E,Z(mg/C) = 1.1183\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.2: page 74:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data :\n",
"Z=0.329*10**-3# IN g/C\n",
"I=1 # in amperes\n",
"t=90*60# in seconds\n",
"\n",
"#calculation:\n",
"M=Z*I*t# in grams \n",
"A=200#area in centimete square\n",
"S=8.9#density in g/cc\n",
"T=(M)/(2*A*S)#thickness in cm\n",
"\n",
"#Results\n",
"print \"thickness of copper in cm is\", round(T,6)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"thickness of copper in cm is 0.000499\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.3: Page 76:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"w=15 # in kg\n",
"t1=15# in degree celsius\n",
"t2=100#in degree celsius\n",
"t=25 # time in minutes\n",
"I=10 # in ampere\n",
"n=85 #efficiency of conversion in percentage\n",
"\n",
"#calculations:\n",
"ho=w*(t2-t1)#output heat required in kcal\n",
"R=((ho*4187*100)/(I**2*t*60*n))# resistance in ohms\n",
"\n",
"#Results\n",
"print \"resistance in ohms\",R"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"resistance in ohms 41.87\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 5.4: page 76:"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#given data:\n",
"w=20 # in kg\n",
"t1=10# in degree celsius\n",
"t2=90#in degree celsius\n",
"t=2*3600+19*60+34# time in seconds\n",
"I=4 # in ampere\n",
"n=80 #efficiency of conversion in percentage\n",
"\n",
"#calculations:\n",
"ho=w*(t2-t1)#output heat required in kcal\n",
"V=((ho*4187*100)/(I*t*n))# POTENTIAL DROP IN VOLTS\n",
"\n",
"#Results\n",
"print \"potential drop across heater element in volts is\", V"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"potential drop across heater element in volts is 250.0\n"
]
}
],
"prompt_number": 4
}
],
"metadata": {}
}
]
}
|