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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 3: HYDRO ELECTRIC STATIONS"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.1: Firm_capacity_and_Yearly_gross_output.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"// A Texbook on POWER SYSTEM ENGINEERING\n",
"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n",
"// DHANPAT RAI & Co.\n",
"// SECOND EDITION \n",
"\n",
"// PART I : GENERATION\n",
"// CHAPTER 3: HYDRO-ELECTRIC STATIONS\n",
"\n",
"// EXAMPLE : 3.1 :\n",
"// Page number 41\n",
"clear ; clc ; close ; // Clear the work space and console\n",
"\n",
"// Given data\n",
"Q = 95.0 // Minimum run-off(m^3/sec)\n",
"h = 40.0 // Head(m)\n",
"\n",
"// Calculations\n",
"w = 1000.0 // Density of water(kg/m^3)\n",
"weight = Q*w // Weight of water per sec(kg)\n",
"work_done = weight*h // Work done in one second(kg-mt)\n",
"kW_1 = 75.0/0.746 // 1 kW(kg-mt/sec)\n",
"power = work_done/kW_1 // Power production(kW)\n",
"hours_year = 365.0*24 // Total hours in a year\n",
"output = power*365*24.0 // Yearly gross output(kWhr)\n",
"\n",
"// Results\n",
"disp('PART I - EXAMPLE : 3.1 : SOLUTION :-')\n",
"printf('\nFirm capacity = %.f kW', power)\n",
"printf('\nYearly gross output = %.2e kWhr.', output)"
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.3: Available_continuous_power.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"// A Texbook on POWER SYSTEM ENGINEERING\n",
"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n",
"// DHANPAT RAI & Co.\n",
"// SECOND EDITION \n",
"\n",
"// PART I : GENERATION\n",
"// CHAPTER 3: HYDRO-ELECTRIC STATIONS\n",
"\n",
"// EXAMPLE : 3.3 :\n",
"// Page number 41\n",
"clear ; clc ; close ; // Clear the work space and console\n",
"\n",
"// Given data\n",
"A = 200.0 // Catchment area(Sq.km)\n",
"F = 1000.0 // Annual rainfall(mm)\n",
"H = 200.0 // Effective head(m)\n",
"K = 0.5 // Yield factor\n",
"n = 0.8 // Plant efficiency\n",
"\n",
"// Calculations\n",
"P = 3.14*n*K*A*F*H*10**-4 // Available continuous power(kW)\n",
"\n",
"// Results\n",
"disp('PART I - EXAMPLE : 3.3 : SOLUTION :-')\n",
"printf('\nAvailable continuous power of hydro-electric station , P = %.f kW', P)"
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3.4: Minimum_flow_of_river_water_to_operate_the_plant.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"// A Texbook on POWER SYSTEM ENGINEERING\n",
"// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar\n",
"// DHANPAT RAI & Co.\n",
"// SECOND EDITION \n",
"\n",
"// PART I : GENERATION\n",
"// CHAPTER 3: HYDRO-ELECTRIC STATIONS\n",
"\n",
"// EXAMPLE : 3.4 :\n",
"// Page number 41-42\n",
"clear ; clc ; close ; // Clear the work space and console\n",
"\n",
"// Given data\n",
"load_factor = 0.15 // Load factor\n",
"P = 10.0*10**3 // Rated installed capacity(kW)\n",
"H = 50.0 // Head of plant(m)\n",
"n = 0.8 // Efficiency of plant\n",
"\n",
"//Calculation\n",
"units_day = P*load_factor // Total units generated daily on basis of load factor(kWhr)\n",
"units_week = units_day*24.0*7 // Total units generated for one week(kWhr)\n",
"Q = units_week/(9.81*H*n*24*7) // Minimum flow of water(cubic mt/sec)\n",
"\n",
"//Result\n",
"disp('PART I - EXAMPLE : 3.4 : SOLUTION :-')\n",
"printf('\nMinimum flow of river water to operate the plant, Q = %.3f cubic mt/sec', Q)"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Scilab",
"language": "scilab",
"name": "scilab"
},
"language_info": {
"file_extension": ".sce",
"help_links": [
{
"text": "MetaKernel Magics",
"url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
}
],
"mimetype": "text/x-octave",
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"version": "0.7.1"
}
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"nbformat": 4,
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|
