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Supplementary Methods 2
Calculation of Basal Hydraulics.
In this section we detail the calculations whose results are given in the main text. Our purpose is to
show that the potential energy released by the subglacial lake discharge is sufficient to melt a tunnel
whose cross-section can maintain the peak discharge rate.
A simple two-lake hydraulic model of the situation is shown in Supplementary Figure 4. Assuming
that the motion is steady, and that any forces associated with motion in the lakes (but not the tunnel)
are negligible, it is simple to show that a balance of forces requires
TL TU

 p
AL AU
 
(1)
where  is the change in the ‘basic’ hydraulic potential
(2)
  g i h  w z 
between the tunnel entrance at Lake L and channel exit at Lake U, T L and TU are the total vertical
forces on the ice due to deformation, AL and AU are the areas of the lakes, p is the additional
pressure drop in the channel that arises when flow occurs, h is the thickness of ice, z is the
elevation of the bed, g is the acceleration due to gravity, and i and w are the densities of ice
and water respectively. Eqn. (2) assumes the water volumes above the tunnel entrances are very
small in comparison with the ice volumes.
Eqn. (1) shows that not all of the change in hydraulic potential is available to drive the flow. Some
of it is required to deform the ice roof as the discharge proceeds. However, if the pressures TL / AL
and TU / AU associated with the roof deformation are negligible, eqn. (2) reduces to
(3)
  p
In the Rothlisberger theory of tunnel drainage (see e.g. reference [15] of the main text), eqn. (3) is
supposed to hold between any two points along the tunnel. A semi-circular cross-section S is
assumed and is related to the discharge Q by the empirical relation [15, eqn. 4.7.17)]
Q  2  / 2 S 4/3m1   / w g
1/3
1/2
(4)
where   is the gradient of hydraulic potential with respect to channel length, and m is the
Manning coefficient. When using this formula, we suppose that S , Q and   are uniform along the
channel, and that   is well approximated by  / l , where l is the distance between the lakes. We
take Q = 50 m3 s-1,  = 1.47 x 106 Pa, l = 290 km, m = 0.08 m-1/3 s, w =1000 kg m-3 and g = 9.81
m s-2 in eqn. (4) to determine S = 26 m2.
We now return to the question of whether the pressures associated with the deformation may be
ignored, which we do by considering the fate of the potential energy released in the discharge. This
energy is used to heat water, melt ice, and deform the roof. The total energy released is
Et  V 
(5)
3
15
where V is the volume of the discharge. With V =1.8 km , this provides 2.7 x 10 J. To obtain the
energy used to heat water, we assume the water is maintained close to the pressure melting point,
and in this case
Ew  cvV
 1
T
1
i h   
L
 w i 
(6)
where cv = 4.2 kJ kg-1 is the specific heat capacity of water, T  273 K is the melting point at the L
site, h = 449 m is the change in ice thickness between the sites, and L = 3.3 x 105 J kg-1 is the
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latent heat of water. (Eqn. (6) uses the Clausius-Clapeyron equation). These values provide Ew =
2.4 x 1011 J, that is, a small quantity in comparison with Et .
To obtain the energy of deformation at Lake L, we suppose the motion is uniform throughout the
discharge and that the deformation is accommodated by simple shear in the vertical around the
circumference of the lake, extending horizontally over an ice thickness. The total energy of
deformation is
z h
z h
1/ n
 ε 
(7)
Ed  Cd  dzτ ( z )  Cd  dz 

 AT z 
z
z
where C = 9 x 104 m is the lake circumference, d =3 m the total displacement,  the shear stress, ε
the strain rate, A is the ice flow-law parameter and n is the flow-law exponent, that we take to
equal 3. Eqn. (7) acknowledges that while the motion is uniform with depth (a consequence of mass
conservation), the stress will depend on depth through the temperature dependence of the flow-law
parameter. We use eqn. 2.4.1 of [15] to describe the temperature dependence of the flow-law
parameter. Because the lakes are located near a divide, and accumulation is small, we suppose the
temperature varies linearly with depth from -50˚ C at the surface to the melting point at the bed. We
take ε  Q /( AL h) , with AL = 600 km2 and h = 3249 m. With these assignments, eqn. (7) provides
E d = 7.8 x 1013 J.
This energy is small in comparison with the total energy released by the discharge. Since we do not
have an estimate of the area or circumference of Lake (or lakes) U, this energy of deformation is
less well-constrained. Nonetheless, it seems reasonable to suppose that it is of the same order. In
this case, all the energy sinks other than latent heat are small. Ignoring them, the total volume of
melted ice is ET / L i  and the area of a uniform tunnel with this volume is ET / Ll i , which
provides an area of 28 m2. This is the area towards the end of the discharge, when the discharge is at
its maximum, and its value compares well with the area estimated from the peak discharge rate via
eqn. (4).
These calculations show that the phenomenon we observe may be consistently described by the
theory of jökulhlaups associated with temperature ice bodies. This said, there is considerable
uncertainty in the parameters we use, principally due to the uncertain geometry. In particular, the
roof deformation may be more important than the calculations above imply. We illustrate this by
considering the uncertainty in some parameters.
To determine the hydraulic potential via eqn. (2) requires the ice elevation and depth. The elevation
is accurately known from satellite radar altimetry, but the depth is only poorly known from sparse
radio echo soundings. An examination of these data in the vicinity of Lake L shows that the basal
depth we have used (-285 m) might be too high by as much as 750m. Similarly, our data allow a
smaller lake area, and a smaller average displacement, than we estimate, perhaps together by a
factor of 2. If one re-calculates on this basis the total energy released, one obtains 8 x 1014 J.
This figure remains large in comparison with the energy of deformation (7.8 x 1013 J) we calculated
above, but this too may be questioned in a similar fashion. It may be, for example, that we have
underestimated the role of longitudinal stresses in the deformation of the roof. If we supposed this
energy might perhaps be as large as 4 x 1014 J, there then remains only 4 x 1014 J to melt ice. Thus it
is possible these energies are of the same order, and that the dynamics of the deformation is
important to the evolution of the discharge. If so, this situation is distinct from that of temperate
jökulhlaups.
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Were only 4 x 1014 J to melt ice which would generate a tunnel only 7 m2. This is now small in
comparison with that deduced from eqn. (4), but here too there is considerable uncertainty. We used
a value of Manning coefficient corresponding to a rough-walled tunnel. A smoother tunnel may be
more appropriate, and using a value of 0.01 m-1/3 s, one obtains from eqn. (4) an area of 5 m2..
Again, the 4/3rds power in eqn. (4) arises from the assumption of a semi-circular cross-section.
Other sections may be postulated, and it is also possible that the ‘tunnel’ (or tunnels) is a channel
(or channels) cut into the bed-rock (see e.g. [5]) whose ice-roof provides the meltable area. In short,
the data allow for a wide range of detailed possibilities.
Turning now to the termination of the discharge, Nye's theory of jökulhlaups (see e.g. [15]) gives
the change in tunnel area (assumed independent of position) as
dS mr A
n
(8)

 n S i gh  p 
dt ice n
where mr is the melt rate per unit length of tunnel and p is the water pressure. The quantity in
brackets is the effective pressure and the second term in eqn. (8) describes the closure rate of the
tunnel due to the flow of ice. (Eqn. (8) admits to the fact that eqn. (3) only holds approximately
within the tunnel.) Using the first set of parameters above, estimating the melt rate with
mr  Q / Ll  provides 8.3 x 10-7 m2 s-1. With A = 2.5 x 10-24 Pa-3 s-1 (the value at the melting
point) and S = 26 m2, the effective pressure required to balance growth through melting with
closure through ice flow is 590 kPa. While the effective pressure need not equal zero at the
initiation of the discharge [4], this is a very large value in comparison with the change in pressure at
Lake L ( i gd ) of 26 kPa. This calculation is not greatly altered if the second set of parameters is
used, and thus it appears to be robust. For this reason, we think it unlikely that the flow was stopped
due to the closure of the tunnel.