{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 12 : Flywheel Energy storage" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex: 12.1 Pg: 849" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ " (a) the compressed sir temperature is 1162 degree C \n", " (b) the storage time is 7.71 hour \n", " (c) the total energy storage is 2426 MWh \n", " (d) the total energy delivered by the peaking turbine is 1455 MWh\n" ] } ], "source": [ "from __future__ import division\n", "#Input data\n", "V=64000#Volume in m**3\n", "Q=8300#Discharge in m**3/hr\n", "p1=1#Pressure in bar\n", "T1=20+273#Temperature in K\n", "p2=100#Pressure in bar\n", "pn=70#Polytropic efficiency in percent\n", "pt=60#Peaking turbine efficiency in percent\n", "g=1.4#Ratio of specific heats\n", "cp=1.005#Specific heat in kJ/kg.K\n", "R=0.287#Gas constant in kJ/kg.K\n", "\n", "#Calculations\n", "T2sT1=(p2/p1)**((g-1)/g)#Temperature ratio \n", "T2s=(T1*T2sT1)#Temperature in K\n", "T21=(T2s-T1)/(pn/100)#Difference in Temperatures in K\n", "T2=(T21+T1)-273#Temperature in degree C\n", "v=(R*T1)/(p2*100)#Specific volume in m**3/kg\n", "mf=(Q/(v*3600))#Mass flow rate in kg/s\n", "E=(mf*cp*T21)/1000#Rate of energy storage in MW\n", "t=(V/Q)#Storage time in hour\n", "tE=(E*t)#Total energy storage in MWh\n", "Ed=(tE*(pt/100))#Total energy delivered by the peaking turbine in MWh\n", "\n", "#Output\n", "print \" (a) the compressed sir temperature is %3.0f degree C \\n (b) the storage time is %3.2f hour \\n (c) the total energy storage is %3.0f MWh \\n (d) the total energy delivered by the peaking turbine is %3.0f MWh\"%(T2,t,tE,Ed)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex: 12.2 Pg: 850" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ " (a) the storage time is 13.969 h \n", " (b) the total energy stored in the accumulator is 16678.8 MWh \n", " (c) the total energy that can be delivered by the peaking turbine is 4002.92 MWh\n" ] } ], "source": [ "from __future__ import division\n", "from math import log\n", "#Input data\n", "V=175000#Volume in m**3\n", "d=4#Diameter in m\n", "U=1.5#Overall heat transfer coefficient in W/m**2.K\n", "p2=2#Pressure in bar\n", "p1=20#Pressure in bar\n", "Ta=20#Ambient temperature in degree C\n", "cp=4.35#Specific heta of water in kJ/kg.K\n", "e=96#Efficiency in percent\n", "ppe=25#Peaking plant efficiency in percent\n", "\n", "#Calculations\n", "#At 20 bar\n", "T1=212.37#Saturation temperature in degree C\n", "hf1=908.5#Enthalpy in kJ/kg\n", "vf1=0.0011766#Specific volume in m**3/kg\n", "#At 2 bar\n", "T2=120.23#Saturation temperature in degree C\n", "hf2=504.8#Enthalpy in kJ/kg\n", "vf2=0.0010605#Specific volume in m**3/kg\n", "ad=(1/2)*((1/vf1)+(1/vf2))#Average density of water in kg/m**3\n", "tc=(d*ad*cp*1000)/(4*U*3600)#Time constant in h\n", "ts=(log(1/(1-((1-((e/100)))/((T1-Ta)/(T1-T2))))))*tc#Storage time in h\n", "m=(V/vf1)#Mass of water needed in kg\n", "E=(m*(hf1-hf2))/(3600*10**3)#Total energy stored in MWh\n", "Ed=(E*(e/100)*(ppe/100))#Energy delivered in MWh\n", "\n", "#Output\n", "print \" (a) the storage time is %3.3f h \\n (b) the total energy stored in the accumulator is %3.1f MWh \\n (c) the total energy that can be delivered by the peaking turbine is %3.2f MWh\"%(ts,E,Ed)" ] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.9" } }, "nbformat": 4, "nbformat_minor": 0 }