ultra large energy storage system
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Large Energy Storage SystemEduard Heindl
Hochschule Furtwangen University, Germany
Renewable Energy needs storage
wind 60%solar 40%
supply change
source: Lueder von Bremen, EWEC 2009
storage for up to seven days
required!
Electicity: Daily Production
Base load: coal, nuclear daily production: 1600GWh
Problem: how to use this energy
70GW
280GW
Hydro Storage capacity in Germany: 40GWh
0h midnight 24h time
„priceless energie“
solar power,wind power
Consumption in Germany
12h high noon
Energy demand: Joe Public
• 21kWh electricity a day
• 147kWh electricity a week
• Battery pricing– Car lead battery 150$/kWh – Lithium 1000$/kWh
• Storage per person for one week
25.000$ (2,2t) ... 130.000$
1km
Think Big!
1km
500m 1700GWh
capacitypower converter h
Basic Principle
•Water is pumped into a subsurface cavity, using cheap electrical power•The rock above is lifted by hydraulic forces•The storage is discharged when the energy price is high, using a power converter
power gridc
on
ne
c
t
Construction
1km
mine shaft
base tunnel, water intake
blast hole
1. tunnel
1km
construction road
deep well drilling plant
blast hole
Construction work
1km
1. tunnel
lower saw mill
construction road
wire saw
upper saw mill
Diamond wire sawing
Surface
1. Tunnel
traction
cut surface Diamondwire saw
rock
drilling holes
r
Diamond wire sawing
surface
1. tunnel
traction
Diamond wire saw
rock
drilling holes
r
cut surface
Baseplate seperation
base-tunnel
2. tunnel
traktion
sawing string
rock
drilling holes
r
cutted rock
remaining rock
top view!
Rock Bottom Area SeparationThe remaining rock granite: • Tensile strength: Ts = 11,3N/mm²= 11,3MN/m²• Burst strength : Bs = 214 N/mm² = 214MN/m²• Mass above : ma = ρ h a0 • Pressure by ma : pm= Fm/a0 = g ρ h• Neccesary remaining rock: Fm = Bsar
• Sepparating Force: Fs = Fm + Ts ar
Sepparating pressure ps
ps = Fs /(a0 –ar)ps = (Fm + Ts ar)/(a0 –ar)ps = (g ρ h a0 + Ts ar)/(a0 –ar)ps = (g ρ h a0 + g ρ h a0 Ts/Bs)/(a0 – g ρ h a0/Bs)ps = g ρ h a0 (1 + Ts /Bs)/(a0 – g ρ h a0/Bs)ps = g ρ h (1 + Ts/Bs)/(1 – g ρ h/Bs)
ma
Fm
ar
a0=ar+as
hρ
Fs
ps Bs;Ts
ps=304barh=1000m:
Cutting completed
1km
1. tunnel
2. tunnel
construction road
the wall and the floorof the cylinder are separated from thesurrounding rock
h=r
D=2r
l=2r
Energy Storage CapacityGenerated surface:M = 2 * π * r * l = 4 * π * r² (1)
Real densityρ2 = ρ1 – ρ3 (2)
Potential energy in gravitation fieldE = g * m * h (3)
Real mass of cylinder m = π * r² * 2 h * ρ2 (4)
Eq. 4 and eq. 3 using h = r results in:E = g * π * r² * 2 * r * ρ2 * r (5)
Eq. 5 condensed: E = 2 π g ρ2 * r4 the power
V=4πr³
Energiespeichervolumen als Funktion des Radius
1
10
100
1.000
10.000
100.000
1.000.000
10.000.000
100.000.000
10 100 1000 10000
Radius r [m]
En
erg
ie [
GW
h]
500
1.700
Beispiel
Daily electrical power production in Germany
Energy storage capacity depends on the Radius r
Example
Source: www.eia.doe.gov/cneaf/electricity/epm/table1_1.html
Daily electrical power production in USA
800
Sealing
• The piston has to be sealed against the wall
• The stone walls have to be free of cracks (or the cracks must be closed)
• The sealing needs to function during the movement of the piston
• The sealing has to be positioned above the centre of gravity (half of piston height)
Sealing the Walls
rock mass piston
1. tunnel
platform
sealed wall
support column
sealing
water under pressure
Saw line
Closing the Cracks
rock mass piston
piston1. tunnel
water under pressure
wall with sealant
ad sealant
seal
crack
Complete the Sealing
rock mass piston
piston 1. tunnel
water under pressure
wall with sealant
ad sealing
seal
crack
Sealing
rock mass
piston
pin joint with sensor
seal
steel
podest
water pressure
Water Volume
• Amount of water in the case of 500m radius: 390 Mio m³
• Inlet with one month filling time: 152m³/sFor comparison:
– Schluchseekraftwerk: 80m³/s– Rhine at Speyer: 1 200m³/s mean
• Inlet with one week filling time: 650m³/s
• The amount of water is manageable, especially if the system is for long term storage
Speed of movement
• Depending on the speed of filling and emptying, the piston will move with different speeds
• If we fill the cavity within a month
v = s/tvm = 500m / 2 592 000s
vm = 0.19mm/s
(filling within one day: vd = 5.8mm/s)
• The speed seems to be very low, an advantage for the sealing system
Piston position fine tuning• using a four sector pressure control
piston
low pressure sealing about
1 bar
x x
Seal low pressureseal low
pressure
seal high pressure
exchange pump
Thermal effects
• Due to the water that is pumped into the system, the stone will cool down
• This will reduce the radius of the piston• Calculation: The temperature at the bottom of
the system will be about 50°C* and the temperature will be reduced to 10°C due to the cold water, ΔT=40K, 50m depth of penetration, resulting in 6mm contraction of the granite piston
• This seems to be no problem, in addition, the sign of the change is with the proper direction
*Quelle: Regierungspräsidiums Freiburg www.lgrb.uni-freiburg.de
Price of the system
• Tunnel 50 000€/m
• Sawing 10€/m²
• Coverage 100€/m²
• Sealing 1 000€/m
• Turbines
• Generators not included
• Pumps
Kosten pro kWh Speichervolumen(ohne Konverter)
0,00
0,01
0,10
1,00
10,00
100,00
1.000,00
10.000,00
10 100 1.000 10.000
Anlagenradius r [m]
Pre
is [
€/k
Wh
]
pump storage power station
0,39
500100 1.000
example system
Price per kWh storage volume
(excluding the energy converter)
pri
ce
system radius [m]
Kosten pro kWh Speichervolumen(Verschiedene Konverter)
0,00
0,01
0,10
1,00
10,00
100,00
1.000,00
10.000,00
10 100 1.000 10.000
Anlagenradius r [m]
Pre
is [
€/k
Wh
]
0,39
500100 1.000
day
week
month
year
200€/kW
pump storage power station
example system
system radius [m]
pri
ce
Price per kWh storage volume
(using different converters, 200€/kW)
Calculation
Building cost per unit Mio €
Tunnel 50 000€/m 464
Drilling 1 000€/m 157
Sawing 10€/m² 39
Seal shield 100€/m² 157
Sealing 1 000€/m 3
Total 821
Calculation
Cost Mio €
Storage system
821
Pumps and
Generators
200€/kWh
(T=7d)
2 040
Total 2 861
Return on Investment
Turnover Mio €
Power buy per year
20€/MWh
(T=100d)
490
Power sell
Per year
70€/MWh
(T=100d)
1 200
(Q=70%)
Total 710DRAFT!
Assumption: there are 100 days (2,400h) where the power could be bought with a mean price of 20€/MWh and there are 2400h where it could be sold for 70€/MWh, the other times of the year, there is no activity. The conversion efficiency is 70% (worst case)
Return on Investment
• Under the presented assumptions, the system will be completely depreciated within 4 years
• In the future, a growing market share of regenerative energy sources, like wind and solar, will widen the spread of the price
Requirements: Storage Systems
☺Energy and power density 2MWh/m²☺Lifetime unlimited☺High current acceptance yes☺Fast response to demand for power yes☺Simple request to storage current yes☺Wide temperature range yes☺High safety standards yes☺High efficiency 70-80%☺Low service cost yes☺Simple load measurement yes☺Simple and cost efficient recycling yes
source Dirk Uwe Sauer
Maintenance Steps
• Empty system with pumps
• Open supply tunnel
• Assess sealing
• Replace sealing if necessary
• Assess wall surface
• Replace wall surfacing if necessary
Worst case scenario
• Complete broken sealing– Water exits the system along the sawing slit– Using simple assumption, about 600m³/s leek– This amount of water could be managed by
the hydro system
• Preparation– Drainage along the sawing slit– Back up sealing ring – Usage of a high viscose cover liquid
Research Topics
• Search for good places• Understanding of rock properties• Drilling deep, aligned holes• Configuration of long sawing strings• Coverage of rock for sealing• Sealing ring• Removal of ground plate• Temperature effects• High viscose material sealing• Optimal system size• High pressure pump
Thank you for your interest!Questions?
eduard-heindl.de/energy-storage
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