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"source": [
"# Chapter 12 : Flywheel Energy storage"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex: 12.1 Pg: 849"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
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"outputs": [
{
"name": "stdout",
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"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)"
]
}
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