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INTRODUCTION:
This web page presents a practical mathematical model for characterizing the water seepage into a Deep Geologic Repository (DGR) that is located below the local water table as compared to similar seepage into a well, a deep mine and through a high hydroelectric dam.
It is shown that in order for the ongoing pumping cost at the Ontario Power Generation (OPG) proposed Bruce DGR to be acceptable the rock material constant Kr must be two orders of magnitude larger than the Kr value at the Con mine and must be several orders of magnitude larger than the minimum Kr value that is acceptable at a major hydroelectric dam.
Most soil mechanics is based on Darcy's Law, which assumes laminar flow in the capillary channels through soil. Darcy's Law is usually valid for water flow through gravel, sand or clay. However, Darcy's Law fails in hard rock close to a DGR where the pressure gradient and flow channel size are sufficient to make the water flow turbulent rather than laminar.
Real hard rock can be mathematically modeled as an ideal solid containing Np pores per m^3 that are interconnected by capillary channels and are randomly distributed through the solid.
The water in each capillary channel is accelerated along the channel by the pressure difference between the adjacent pores.
Due to surface tension water molecules fill and coat the inside walls of the capillary channels. At high pore densities (and hence small pore to pore pressure differences) the flow through the capillary channels is laminar, the viscous drag force is proportional to capillary axial flow velocity Vc and the Darcy Law applies.
However, the capillary channels are not smooth so that the axial flow becomes highly turbulent at relatively low axial velocities. In hard rock near an unlined DGR with a low pore density and hence a high pore to pore pressure differences the flow through the capillary channels is turbulent. In turbulent flow the drag force is proportional to Vc^2 and the Darcy Law is invalid.
Thus the Darcy Law, which may adequately describe the subsurface hydraulic pressure distribution prior to DGR formation, is invalid in the proximity of an unlined DGR. Failure to recognize this issue has led OPG into serious problems with its proposed Bruce DGR. These problems can in principle be fixed by fitting the Bruce DGR with a rigid water tight lining, but the cost of the required extra excavation plus the cost of the required lining will increase the overall cost of the Bruce DGR by several billion dollars. To avoid this extra cost OPG should choose an alternative high, dry and accessible DGR site.
REYNOLDS NUMBER:
The Reynolds Number is an indicator of the degree of turbulence in a flow channel.
The Reynolds number Re in a smooth capillary channel is given by:
Re = (Rho Vc Dc) / Mu
where:
Rho = 1000 kg / m^3 = density of water
Vc = water axial velocity along the channel
Mu = 1.3 X 10^-3 N s / m^2 = dynamic viscosity of water
Dc = channel diameter
The water flow in a smooth capillary channel is significantly turbulent at Reynolds numbers above about 10.
Assume Dc = 10^-5 m,
Assume Vc = 1 m / s.
Then:
Re = (10^3 kg / m^3 X 1 m /s X 10^-5 m) / (1.3 X 10^-3 N s / m^2)
= [(10 / 1.3) kg / m s] / (kg m s/ s^2 m^2)
= (10 / 1.3)
Under these circumstances the channel flow is starting to be non-laminar.
The pressure difference dP required to produce a 1 m / s axial flow is given by:
dP = (Rho) Vc^2
= (1000 kg / m^3) X (1 m / s)^2
= 1000 kg / m s^2
= 1000 Pa
Thus if the pore to pore pressure differences exceed 1000 Pa the flow can be turbulent.
The volumetric flow rate along this channel is:
Ac Vc = Pi Rc^2 Vc = 3.14 X (.5 X 10^-5 m)^2 X 1 m / s channel
= .785 X 10^-10 m^3 / s channel
= .0785 X 10^-9 m^3 / s channel
= .0785 mm^3 / s channel
Assume a seepage rate of:
7.42 X 10^-3 m / h (As in Con mine)
Seepage rate
= 7.42 X 10^-3 m / h X 1 h / 3600 s
= 2.06 X 10^-6 m / s
Number of channels per unit of exposed surface area = Seepage Rate / Flow rate per channel
= (2.06 X 10^-6 m / s) / .785 X 10^-10 m^3 / s channel
= 2.624 X 10^4 channels / m^2
Thus a seepage rate of 7.42 mm / h is consistent with there being a 10 micron diameter capillary channel every 0.7 cm along the exposed surface.
Thus for hard rock with connected pore densities below 1 pore / cm^3 in a high pressure gradient the liquid flow is turbulent and Darcy's Law is invalid.
The fraction of rock volume occupied by fluid in the connected channels can be estimated by:
[(3 axis) X (2.624 X 10^4 channels / m^2) X (1 m^2 / axis) X ((1 m X Ac) / channel)] / 1 m^3
= (3 X 2.624 X 10^4 / m^2) X (Pi X (0.5 X 10^-5 m)^2)
= 6.18 X 10^-6
which is consistent with the "tight rock" reported by Mark Jensen of the NWMO. (DGR Review Panel Transcript October 29, 2013 Volume 24 Page 282 Line 2 to Page 283 Line 5.)
Thus the tests done by the NWMO on the "tight rock" are consistent with turbulent rather than laminar water flow through the rock. These tests are also consistent with a higher seepage rate than has been assumed by the NWMO.
DERIVATION:
Let Lc = the length of a typical capillary channel.
Let Ac = cross sectional area of a typical capillary channel
Consider the water flow along capillary channels that are parallel to the pressure gradient.
dP = change in pressure between two adjacent pores
Vc = axial flow velocity along the capillary channel
Rho = density of brine or water
At steady state in a channel:
Pressure force = inertial force + viscous force
Let Kv = constant of proportionality of viscous force.
Conservation of energy gives:
[dP Ac Lc] = (Rho / 2) Ac Lc Vc^2 + Kv Vc^2 2 Pi Rc Lc
or
dP = (Rho / 2) Vc^2 + Kv Vc^2 2 Pi Rc / Ac
or
dP = (Rho / 2) Vc^2 + Kv Vc^2 2 / Rc
or
dP = [(Rho / 2) + (2 Kv / Rc)] Vc^2
dX = (1 / Np)^.333 = distance between pores
Hence:
dP / dX = [(Rho / 2) + (2 Kv / Rc)] Vc^2 Np^.333
The number of pores per m^2 with capillary channels that link to one outside surface is:
(Np)^.6666
The open area per m^2 of these surface linked capillary channels is:
(Np)^.6666 Ac
Let Fv = total volumetric flow per unit time
Let A = perpendicular surface area through which flow Fv moves.
Then:
Fv / A = Np^.666 Ac Vc
or
Vc = (Fv / A) [1 / (Np^.666 Ac)]
or
Vc^2 = (Fv / A)^2 [1 / (Np^.666 Ac)]^2
Hence:
dP / dX = [ (Rho / 2) + (2 Kv / Rc)] Vc^2 Np^.333
= [(Rho / 2) + (2 Kv / Rc)]] Np^.333 (Fv / A)^2 [1 / (Np^.666 Ac)]^2
= [(Rho / 2) + (2 Kv / Rc)] (Fv / A)^2 [ 1 / (Np Ac^2)]
= (Fv / A)^2 Kr
where:
Kr = [(Rho / 2) + (2 Kv / Rc)][ 1 / (Np Ac^2)]
= characteristic constant indicating the quality of the rock.
Thus:
dP / dX = (Fv / A)^2 Kr
= Fs^2 Kr
where:
Fs = Fv / A
= Seepage Rate
Clearly if either Ac or Np decrease then Kr increases.
Now consider seepage water flow into a cylindrical cavity with length L which is long compared to its radius.
Under steady state conditions:
Fv = constant independent of radius R into the rock.
A = 2 Pi R L
where:
R = radius from cylindrical cavity axis
and
L = length of cylindrical cavity
Then:
dP / dR = (Fv / A)^2 Kr
= (Fv / (2 Pi R L))^2 Kr
or
dP = (Fv / (2 Pi L))^2 Kr dR / R^2
At the cavity cylindrical surface:
P = Ps
R = Rs
At large R:
P = Pr
R = infinity
Hence:
Pr – Ps = (Fv / (2 Pi L))^2 Kr (1 / Rs)
= [Fv / (2 Pi L Rs)]^2 Kr Rs
Seepage rate Fs through a cylindrical surface is given by:
Fs = [Fv / (2 Pi L Rs)]
Then:
(Pr – Ps) / Rs = Fs^2 Kr
or
Kr = (Pr - Ps) / (Rs Fs^2)
Consider the OPG contemplated DGR:
(Pr – Ps) = 65 X 10^5 Pa
Rs = 4 m
Assume that to limit the ongoing pumping cost to an acceptable level the maximum acceptable seepage rate Fs must be less than:
Fs = 10^-3 m / hr
Kr = [(Pr – Ps) / (Rs Fs^2)]
= [65 X 10^5 Pa / 4 m] [ 1 / (10^-3 m / hr X 1 hr / 3600 s)^2)]
= (65 / 4) X 10^5 X (3.6)^2 [ 10^6]^2 Pa m^-3 s^2
= 2.106 X 10^19 Pa s^2 / m^3
= 2.106 X 10^19 kg m s^-2 m^-2 s^2 m^-3
= 2.106 X 10^19 kg m^-4
This Kr value is two orders of magnitude larger than is typical for deep mines and dams. Viewed another way, the probability of a randomly chosen DGR site exhibiting this actual Kr value is quite remote.
WELL TESTING:
Recall that:
1 / Fs^2 = Kr Rs / (Pr – Ps)
This equation can be used to compare the seepage rate measured in a well to the projected seepage rate in an unlined cylindrical DGR chamber.
Let Fs for a well be Fsw
Let Fs in a DGR be Fsd
Let Rs for a well be Rw
Let Rs in a DGR be Rd
Then:
1 / Fsw^2 = Kr Rw / (Pr – Ps)
and
1 / Fsd^2 = Kr Rd / (Pr – Ps)
Hence:
(Fsw / Fsd)^2 = Rd / Rw
or
Fsw = (Rd / Rw)^0.5 Fsd
Assume:
Rd = 4 m
and
Rw = 0.1 m
and
Fsd = 1 mm / h
Hence:
Fsw = (4 / 0.1)^0.5 X 1 mm/ h
= 6.32 mm / h
= maximum acceptable seepage rate into a 0.2 m diamater test well
WELL FILL TIME:
Consider a length dH within a well of radius Rw.
The volume of that length is:
dH Pi Rw^2
The rate at which water flows in through the side walls is:
Fsw dH 2 Pi Rw
Hence the time T for the volume to fill up via seepage through the cylindrical side wall is:
T = dH Pi Rw^2 / Fsw dH 2 Pi Rw
= Rw / 2 Fsw
For:
Rw = 0.1 m
and
Fsw = 6.32 X 10^-3 m / h
T = Rw / 2 Fsw
= 0.1 m / (2 X 6.32 X 10^-3 m / h)
= 7.91 hours
Thus after being pumped dry an 8 inch (20 cm) diameter test well uncased lower section must take longer than 8 hours to refill with water in order for the OPG DGR seepage rate to be under 1 mm / hour.
TYPICAL DEEP MINE:
Assume:
Fs = 10 mm/ hr,
Rw = 4 m,
(Pr – Ps) = 65 atmospheres
Then:
Kr = [(Rho / 2) + (2 Kv / Rc)][ 1 / (Np Ac^2)]
= 2.106 X 10^17 Pa s^2 / m^3
Assume that:
(Rho / 2) ~ (2 Kv / Rc)
Recall that:
Rho = 1000 kg / m^3
Hence:
Np Ac^2 ~ (1000 kg / m^3) / (2.106 X 10^17 Pa s^2 m^-3)
= (10^-14 / 2.106) kg Pa^-1 s^-2
Assume:
Ac = .785 X 10^-10 m^2
corresponding to 10 micron diameter circular cross section capillary channels.
Then:
Np = (10^-14 / 2.106) kg Pa^-1 s^-2 / (Ac^2)
= [(10^-14 / 2.106) kg Pa^-1 s^-2] [1 /(.785)^2 X 10^-20 m^4)]
= 0.770 X 10^6 kg s^-2 S^2 m^2/ m^4 kg m
= 0.770 X 10^6 m^-3
This connected pore density is consistent with 10 micron diameter channels connecting pores which occur about every ~ 1.1 cm over the exposed surface.
HYDRO ELECTRIC DAM:
Consider high hydroelectric a dam such as the WAC Bennett Dam operating at a unit gradient:
dP / dX = (Fs)^2 Kr
Also from gravitational energy theory, for a unit hydraulic pressure gradient:
dP / dX = Rho g
Where:
g = gravitational acceleration
Hence:
Rho g = Fs^2 Kr
or
Kr = (Rho g) / Fs^2
For Fs = 1 mm / hr:
Kr = (1000 kg / m^3 X 9.8 m / s^2) / (10^-3 m / hr X 1hr / 3600 s)^2
= 3.6^2 X 9.8 X 10^15 kg m^-2 s^-2 m^-2 s^2
= 1.27 X 10^17 kg m^-4
This Kr value is two orders of magnitude less than what OPG needs for its DGR in order to have an acceptable pumping rate.
WAC Bennett Dam is 186 m X 2068 m = 384,648 m^2
At a seepage rate of 1 mm / hr the downstream seepage flow rate would be:
384,648 m^2 X 10^-3 m / hr = 384.648 m^3 / hr
= 384.648 m^3 / hr X 1 hr / 3600 s
= .106 m^3 / s
In real life the actual down stream seepage water volumetric flow rate prior to regrouting is thought to have been of the order of 1.0 m / s suggesting that the average seepage rate through the dam was about 10 mm / hr and the actual minimum Kr value was four orders of magnitude less than the Kr value required by the OPG proposed Bruce DGR. Since the WAC Bennett Dam is a highly engineered structure the quality of the rock at the OPG proposed Bruce DGR location must be extraordinarily high to meet the Kr requirement of the Bruce DGR.
INCREASE IN SEEPAGE RATE WITH TIME:
Major hydroelectric dams, such as the WAC Bennett Dam, are characterized by a gradual decrease in Kr with time which causes a gradual increase in the seepage rate. This seepage rate increase is due to erosion by seepage water which gradually washes fine sealing particles out of the dam structure. The practical experience with the WAC Bennett Dam was that it had to be regrouted after 30 years of operation. Some smaller dams that work at lower pressures have operated for up to 90 years before regrouting.
The issue for the OPG proposed Bruce DGR is that due to the high pressure gradient near the DGR while the DGR is open the seepage rate will gradually increase over time. This increase in the seepage rate will constrain the DGR open period, which in turn will strongly affect the economics of the Bruce DGR. Due to the location and geometry of the Bruce DGR it is almost impossible to significantly reduce the future seepage rate of the Bruce DGR rate via grouting.
The useful life of the Bruce DGR could be substantially extended by addition of a rigid water tight liner, which would almost stop seepage. However, the cost of that liner and its related required additional excavation is believed to be prohibitive.
SEEPAGE PUMPING:
The seepage pumping costs associated with the OPG proposed Bruce DGR are substantial. As a minimum the seepage pumping system requires 14 pumps, each the equivalent of the high pressure cold water booster pump used in a 30 storey apartment/condominium building. The pumps must be connected in pairs such that if one pump fails its companion will automatically take over. The pumps must all be protected against loss of suction pressure and the pumps andrelated piping and valves must all be protected against both gravitational pressure and water hammer. The pumps will require mechanical closets at approximately 100 m intervals in the access shaft. The pump system will require fully redundant piping, cables and emergency power backup systems, because any sustained loss of power to the pumps or sustained flow obstruction would result in flooding of the Bruce DGR.
The Bruce DGR is specified to have a nominal storage volume of 200,000 m^3 and is composed of nearly round tunnels each with a radius of about:
Rd = 4 m
Let L = the total tunnel length.
Then the volume of the DGR is:
Pi (4 m)^2 L = 200,000 m^3
or
L = 200,000 m^3 / [Pi (4 m)^2]
Thus the surface area of the DGR is about:
Pi 2 (4 m) L = Pi 2 (4 m) 200,000 m^3 / [Pi (4 m)^2]
= 200,000 m^3 / 2 m
= 100,000 m^2
Assume a seepage rate of 10^-3 m / hr.
Then the seepage flow rate Fv is given by:
100,000 m^2 X 10^-3 m / hr = 100 m^3 / hr
If the pumping system had no energy losses the required pumping power would be:
(100 m^3 / hr) X (1 hr / 3600 s) X (1000 kg / m^3) X (9.8 m / S^2) X 680 m = 185.111 kW
However, there are energy losses in the electrical distribution, pump motors, pumps and piping which cumulatively are about 50%. Hence the electricity required to drive the pumps is about:
2 X 185.111 kW = 370.222 kW
Thus the minimum annual electricity requirement for pumping at a seepage rate of 1 mm / hr is about:
370.222 kW X 8766 hr / year = 3,245,368 kWh.
The value of this electricity at $0.10 / kWh is:
$0.10 / kWh X 3,245,368 kWh / yr = $324,536.80 / yr
If the actual seepage rate is 7.42 mm / hr and if the actual DGR volume is doubled to 400,000 m^3 the projected electricity cost for pumping increases to:
7.42 X 2 X $324,536.80 / yr = $4,816,126 / yr + 13% HST = $5,442,222.50 / yr
The pump system depreciation and maintenance cost is additional to this ongoing electricity cost.
All of these seepage pumping related costs can be avoided by choosing a DGR location that is high above the local water table, dry and accessible so that the DGR gravity drains and there is no need for mechanical pumping.
This web page last updated October 30, 2013
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