Obstacles in a cubic array

The essential result of the theory of this paper is more easily demonstrated with a simplified version of the theory. Assume that only one-size obstacle exists in the model bed of Figure 1. Assume the ice–rock interface between the obstacles can support no shear stress. Assume no water is produced at the bed by geothermal heat and no water reaches the bed from the upper ice surface. Assume the obstacles are of hemispherical shape and are arrayed in the simple cubic pattern shown in Figure 2.

Fig. 2.

Looking down on a single obstacle-size glacier bed where the obstacles are arranged in a simple cubic pattern. Water cavities and connecting channels are also shown.

All the mechanical work of sliding occurs around the obstacles. This work is converted into heat that produces melting of ice near the obstacles. The water produced flows in a linked cavity–channel system also shown in Figure 2.

In Figure 3 is shown a cross-section of a water-flow path. The path consists of a cavity of length λ_ca, and approximate radius r_ca = L/2, an exit channel of length λ_ch and radius r, and a constricted channel through high-pressure ice of length λ_hp and radius r_hp.

Fig. 3.

Cross-section of water-flow path through water cavities and connecting channels of Figure 2.

Let P _i be the average overburden ice pressure. (P_i = ρgh, where ρ is the density of ice, g is the acceleration of gravity, and h is the thickness of the glacier.) Let P _u be the average ice pressure on the up-stream side of an obstacle and P _d be the pressure on the down-stream side. The pressure P_u exists over a distance of the order of L in front of an obstacle and P_d over a similar distance behind the obstacle. If the sliding velocity is V, the pressure P _u is given by the equation

where the dimensionless constant c _v is of order of 0.1 to 1 and B is the constant of Glen’s power-law creep equation for ice:

In Equation 2, σ is a uniaxial tensile or compressive stress. (The exponent n is approximately equal to n =≃. In Equation (1) and in the equations below, the exponent n is set equal to n = 3. The constant B is equal to (Reference PatersonPaterson, 1981; Reference WeertmanWeertman, 1983).)

The length λ_ca of the water-filled cavity behind the obstacle of Figures 2 and 3 is of the order of

where the dimensionless constant c_ca is approximately equal to c_ca = 6c_v and P _w = P _d is the pressure of the water within the cavity. When P_w = P _i – P _u , by symmetry of the stress field about an obstacle, there must be complete closure of ice on the down-stream side of an obstacle. Hence the constant c_ca is of the order of but less than 1 to insure that λ_ca < L when P _w = P _i–P _u (This is the case if the effective stress σ that causes creep around the up-stream side of the obstacle is or (1/3) (P _u – P _i).

Equation (3) is obtained by noting that the closure velocity dr_ca/dt of the cavity is given by (Nye, 1953; see also equation Equation (11) in Reference WeertmanWeertman, 1972)

Since ice moves past the obstacle at the velocity given by Equation (1), integration of Equation (4) leads to

Here x is distance from the obstacle measured in the down-stream direction. At a distance of the order x = λ_ca, where λ_ca, is given by Equation (3), the cavity is almost closed.

Because water is continuously produced at the bed, it must, in any approximate steady-state situation, continuously flow out from beneath the glacier. In this situation, water cannot accumulate in the lee-side cavities. Hence, as shown in Figures 2 and 3, an exit channel must exist at the rear of the cavity. (Although exit channels are R-channels, they are called exit channels rather than R-channels in this paper to emphasize that they are a continuation of a cavity.)

Let W _cc be the water flux passing through a linked cavity-channel water-flow path. The radius r_ch of the exit channel required to transport the water flux out of a cavity is

where P’ _ch is the pressure gradient within the exit channel and η is the viscosity of water (η = 0.0018 Pa s). The water flow is assumed to be laminar. (Equation (6) is the usual equation of flow of a fluid through a pipe.) The water pressure within the exit channel is virtually the same as within the cavity because the drop in water pressure from one water cavity to the next cavity down-stream from it is equal to P’ _av L, where P’ = –dPi/dx and x is distance down the glacier. Here P _av is the average value of the water-pressure gradient down the glacier. Since P’ _av = ρgα, where α is the surface slope and P _w is of the order of magnitude of P _{i =} ρgα the pressure drop between two cavities is very small compared with the water pressure itself.

The water pressure P _w within the exit channel must satisfy the equation

The constant E is the one that converts mechanical work to the amount of ice melted (E = 3.2 – 10^-9 m³ J^-1 ≡ 3.2 – 10^-9Pa^-1). (Equation (7) is the same as Equation (17) of Reference WeertmanWeertman (1972). Note that there are typographical errors in Equation (17) which can be found and then corrected by combining equations (3) and (12) of Reference WeertmanWeertman (1972).) Equation (7), which was originally derived by Reference RöthlisbergerRöthlisberger (1972), is found by insuring that creep closure of water channels is exactly balanced by tunnel enlargement from melting of ice of the channel walls by the heat dissipated in flowing water.

If Equations (6) nd (7) are combined:

At a distance of about L up-stream from an obstacle the exit channel enters a region of high-pressure ice. The channel radius in this region constricts from the value of r_ch to the value r _hp at a velocity

Integration of this equation gives

where x is measured from the start of the high-pressure ice region. The length λ_tr of the transition region in which the channel radius changes from the value of r_ch to the value of λ_hp is, from Equation (10), equal to

Since c_ca is of the order of 0.6 to 6 and P _w < P _i, the transition length λ_tr is smaller than L. Hence, except under conditions when P_w = P_i, the length of channel λ_hp in which the channel radius is constricted to a radius appreciably smaller than r_ch is of the order of .

The radius r_hp of the channel within the high ice-pressure region is equal to

where P’ _hp is the pressure gradient in the constricted channel.

The water pressure P _w within the constricted channel must satisfy the equation

Combining Equations (12) and (13) gives

The pressure gradients P’ _ch , P’ _hp, and P’ _ca, where P _ca is the gradient within the cavity, obviously must obey the equation

This equation, since the pressure gradient in the cavity , reduces to

On insertion of Equations (3), (8), and (14) into Equation (15b) gives

Re-arranging Equation (16a):

where

Equation (16b) gives the water pressure P _w when the sliding velocity V, the average pressure gradient P’ _av, and the water flux W_cc are specified. (Note that if V is specified then, through Equation (1), (P _u – P _i ) is specified.)

Let λ_ca = γL. From Equation (3), when γ ≫ 1 with c _ca = 1,

Combining Equations (16b) and (17) gives, when s 1,

When γ is small compared with s, Equation (18) reduces to

The pressure difference (P _i – P _w) is equal to (P_i – P_w) = (P_u – P _i ) when γ ≃ 1. This is the condition that no cavity forms behind an obstacle because the ice closes in behind an obstacle as fast as it spreads apart ahead of the obstacle. (Note that the ice pressure P _d behind an obstacle is equal to P _w.)

When γ is almost equal to s, Equation (18) reduces to (with c_ca ≃ 1)

Once the sliding velocity is fixed, the pressure P _u is also fixed. With P _u known, Equation (18) then determines, in a steady-state situation, the cavity length λ_ca = γL as well as the water pressure P _w It should be noted in Equation (18) that the steady-state cavity length actually decreases when the water supply W_cc is increased. The steady-state water pressure P_w decreases as the cavity length decreases and thus decreases when the water supply is increased.

The physical reason for this steady-state behavior is simple. The longer is the cavity the greater is the pressure gradient in the exit channel. The greater is the pressure gradient, for the same water flow, the smaller must be the radius of the exit channel. But the smaller is the radius the greater must be the pressure difference (P _i – P _w ) in order to increase the inward-wall velocity of the channel that contracts the outward-wall velocity produced by melting. The radius of the exit channel is greatest when the cavity length is smallest. The water transport through the exit channel increases when either or both the channel radius r_ch is increased or the pressure gradient P_ch is increased. Since, in the equations above, a decrease in P _ch requires an increase in r_ch, or vice versa, an increase in water transport actually comes about because the increase in radius r _ch is more effective in increasing the water flow than the decrease in pressure gradient P’ _ch is in decreasing the flow.

The largest water flux that can transport in a steady-state situation of Figures 1, 2, and 3 is given by Equation (18b) with λ set equal to approximately 1 to 3. Thus

At this maximum water flux the value of the water pressure has its smallest value.

The smallest water flux that can transport in a steady-state situation is given by Equation (18c). Thus

The water pressure P _w has its greatest value when the water flux is given by Equation (19b).

What happens if the water flux W _cc is smaller than the value given by Equation (19b) ? Obviously, no steady state involving water flow only through a cavity channel system is possible. (This is also true if the water flux is larger than that given by Equation (19a). However, from the calculations given below, it is very unlikely that the water flux is ever so great that W _cc > (W _cc)_max.) But a steady-state solution is possible in which the water leaks out of the cavities. Consider Figure 4. Cavities extend from the rear of one obstacle to the front of the next down-stream obstacle. The high-pressure ice region closes off a cavity. The excess water must leak out of the cavity into a water film (see Fig. 4). When water leaks out of the cavities, the water pressure in a cavity is equal to the ice-overburden pressure. (Note that in Equation (18c) P _w ≃ P _i when s ≫ 1. In the next section it is shown that the term s actually should be replaced with a term S where S ≫ s ≫ 1.)

Fig. 4.

Same figure as Figure 2 when cavities extend to the next obstacle and the water flux is too great to be carried by a cavity–channel system.

The conclusion to be reached from the equations of this section is that, if the water flux is smaller than a critical one, the water pressure rises to a value equal to the ice-overburden pressure and water leaks from the cavities into a water film between the cavities.

Obstacles in a random array

A somewhat more realistic bed than one consisting of an array of obstacles in a simple cubic pattern is one in which the obstacles are placed in a random array (see Fig. 5). Let the average spacing between obstacles of size L again be equal to sL. On simple probability grounds, the length an exit channel must have for it to join a down-stream cavity is now SL, where S = s ² /Γ. The constant Γ takes into account how close a channel must approach a down-stream obstacle to be “captured” by that obstacle’s cavity. A reasonable estimate for Γ is Γ 2–3. The channel formed behind one obstacle that is “captured” by a down-stream obstacle, as in the previous section, passes through a region of ice which is subjected to much higher than the nominal ice pressure P _i It must pass through a region in which the ice pressure is of the order of P _u The separation distance D between exit channels is D ΓL. (For the bed of the previous section D = sL.)

Fig. 5.

Looking down on a glacier bed with randomly spaced obstacles. Large obstacles are not shown. Each obstacle has a lee-side water cavity with an exit channel that connects with a down-stream water cavity.

Let the analysis of the previous section be repeated for the random obstacle bed. When it is, it is found that the only formal change needed in the equations is to replace the term s with the term S. (Of course, since there are now more water paths, since the separation distance between paths is decreased from D = sL to D = ΓL, the water flux W _cc in each path is decreased by the amount Γ/s if the total water flow down the glacier bed is unchanged.) Equation (19a) remains unchanged. Equation (19b) becomes

Sliding velocity, shear stress, and friction stress

Equation (1) relates the up-stream ice pressure P _u with the sliding velocity V. The sliding is a result of a shear stress τ (τ = ρgα) that acts across the bed. By the usual balance-of-force argument, the shear stress satisfies the equation

if the rock–ice interface is frictionless and the bed is that of Figures 1 through 5 with only one size of obstacles. (If very large obstacles are present, as shown in Figure 1 Equation (20) remains valid because large obstacles offer so little resistance to the sliding motion.) The value of P _u estimated from Equation (20) is too large because in a real bed obstacles that are both smaller than and larger than the controlling obstacles help support the shear stress across the bed.

The effect of obstacles smaller and somewhat larger than the controlling obstacle size can be mimicked in a single-obstacle-size bed model by allowing the ice-rock interface to support a finite shear stress τ_f, where τ_f = ξ’τ, where the term ξ’ lies in the range 0 < ξ^’ < 1. The influence of a water film of thickness w can also be mimicked by considering that the “friction stress” is given by the equation

where ξ = ξ’ (1 – (w/2L) if w < 2L and ξ = 0 if ξ > 2L. (In this approximation, if the water film has drowned the controlling obstacles and the obstacles are twice the con-trolling obstacle size, the “friction” stress is reduced to a value τ_f = 0.)

Equation (20) is altered when the friction stress is given by Equation (21). Equation (20) is changed to

The sliding velocity V is found from Equations (1) and 22 and is equal to

where τ’ = s²(τ – τ_f).

The amount of water w_f produced per unit time between two of the exit channels (see Fig. 5) is equal to

In Equation (24) M _gt is the geothermal heat (expressed in mechanical units) and ʌ is the distance to the head of the glacier. (The approximation is made in Equation (24) that the sliding velocity V does not vary down the glacier.)

The critical water flux (W_cc)_min given by Equation (19b) required that Pi= P _w Thus, from Equation (23),

and the sliding velocity is equal to

The following equation is obtained for the critical distance ʌ from the head of a glacier at which a water film can no longer form at a glacier bed (when W_cc given by Equation (24) is set equal to the critical value. (W _cc ) given by Equation (19c) and Equation (25) and (26) are substituted into the resultant equation).

In the approximation that M_gt 0, this equation reduces to

Since τ_f =ξτ, where 0 < ξ <1, Equation (27b) is reduced further to

The critical water flux (W_cc)_max given by Equation (19a) requires that (P_u – P_i) = (P_i – P_w). When (P _u – P_i )= (P _i = P_w), according to Equation (23),

and the sliding velocity is equal to

The following equation, which is analogous to Equation (27c), is obtained for the critical distance ʌ

The value of s required for a given sliding velocity can be found after noting that τ^’ = (1 – ξ)s ² τ. Thus

when V is given by Equation (26) and

when V is given by Equation (29).

The values of the critical distance ʌ given by Equations (27c) and (30) are listed in Table I for different values of ξ and s required to have V = 50 m year^-1 = 1.6 – 10^-6ms^-1.

Table I.

Critical Distance Λ

The constants used in the calculations: n = 0.0018 Pa s ; τ = 100 kPa; E = 3.2 – 10^-9m³J^-1 ≡ 3.2 – 10^-9 Pa^-1; r = 3; L = 10 mm; c _ca = 1; c_ca = 1; c _v = c _ca /6; S = s ² Γ; p’_av = 333 Pa m^-1; and B = 9.6 – 10^-25 Pa^-3s-1.

In Table II are given values of ʌ from Equation (27c) for two fixed values of s and different values of ξ. The different sliding velocities for these values of s and f, are also listed in this table.

Table II.

Critical Distance Λ

See Table I for values of the constants.

The values of the term ξ listed in Tables I and II (0.95 ≤ ξ ≤ 0.6) require that obstacles other than the con-trolling obstacles support 40–95% of the basal shear stress. For a bed with a full spectrum of obstacle sizes, we estimated (Weertman, 1964) the controlling obstacles supported only about 30–50% of the basal shear stress.

The following result in Tables 1 and II should be noted: (a) The value of ʌ, for the critical water flux (W_cc)_min predicted by Equation (27c) is extremely large for a glacier bed in which controlling size obstacles support an appreciable fraction of the basal shear stress. Consequently, it is unrealistic to expect that this critical length is ever exceeded for glaciers of unexceptional length. Thus, in a glacier there is insufficient water generated at the base of a glacier to keep open a continuous linked cavity-channel system. (If the high-pressure ice regions did not exist, an open linked channel-cavity flow-path system would exist regardless of the magnitude of the water flux.) (b) The distance ʌ, for the critical water flux (W- _cc ) predicted by Equation (30) is orders of magnitude larger than that given by Equation (27c). This length should never be exceeded.

The implications of these results are: the water pressure P _w in water cavities is equal to the ice-overburden pressure P _i The water cavities have the maximum length possible. They start behind one obstacle and end in front of the next down-stream cavity. The water cavities close up within the high-pressure ice regions. Water generated at an obstacle must flow out of a water cavity into the water film that exists in the regions between cavities. (A water film itself is unstable (Reference WalderWalder, 1982). Periodically, the film should collapse into R-channels. These R-channels themselves are unstable and they should collapse and the water film re-form (Weertman and Birchfield, 1983[b]). The cycle of water film into R-channel into water film should repeat itself.)

These predicted results require that water transported through Nye channels is a minor fraction to the total water flow. The analysis that led to these predictions contained the assumption that surface melt water which descends to the glacier bed does not dominate the basal water hydraulics. If surface water increases the water flux by several orders of magnitude in the equations above, the value of ʌ is reduced several orders of magnitude, Open-cavity channel-flow paths are then quite likely to exist. A more abundant water flow would lead in this situation to a decrease in the water pressure.

Large obstacles

Large obstacles have been ignored so far. Suppose a set of large obstacles of size L^* ≫ L exist in addition to the controlling obstacles (see Fig. 1). What are the conditions for cavity formation behind them and a linked cavity channel system?

For a given sliding velocity, Equation (1) is replaced with

where P_u ^* is the up-stream ice pressure at the obstacle. Equation (3) becomes

where λ_ca ^* is the length of the cavity behind the large cavity and P_w ^* is the water pressure within the cavity.

Other equations derived for the obstacles of size L are changed in similar ways. Equation (27c) for the minimum water flux is altered to

Equation (30) for the maximum water flux is changed to

The value of ξ^* in Equation (34) and (35) is equal to 0 if w ≥ 2L. It is equal to

if w ≪ L and long cavities are behind both small and large cavities and

and long cavities exist only behind the small obstacles. The value of ξ ^* given by Equation (36a) is estimated by noting that the effective stress s²(1 – ξ) τ, causing flow past obstacles of size L is, from Equation (1), proportional to (V/L)^1/3. For the large obstacles, the effective stress s ² (1 – ξ ^*) is proportional to (V/L^*) ^1/3 according to Equation (34). Since the sliding velocity is the same for the large and small obstacles, has the value given above. The value of ξ^* given by Equation (36b) is found in a similar way. If no cavity forms behind a large obstacle, the effective stress, which is proportional to (V/L ^*)¹/³, is (1/2)s²(1 – ξ^*.)τ.

Table III, which is the analogue of Table I, gives values of ʌ^* predicted (with the constants used in Table I) by Equation (34), (35), and (36b) when L^* = 1 m and w ≪ L = 10 mm. Note that ξ^* 1. It can be seen from Table III that the critical lengths ʌ^* are orders of magnitude smaller than those given in Table I.

Table III.

Critical Distance Λ^*

The constants used in the calculations are the same as in Table I except for: L* = 1 m; (1 – ξ*) = 2(1 – ξ)(L/L*)^1/3. The value of γ* = λ_ca*/L* is at Λ = 5 km.

The length λ_ca = γ ^*L^* of a water cavity behind a large obstacle is found from the equation (which is the analogue of Equation (18a))

Using Equation (32) and (37), the ratio γ^* = λ^*/L ^*, at a distance ʌ from the head of a glacier, is given by the equation

In Equation (38) the term (ʌ)_min is the value of ʌ when the water flux W_cc = (W _cc )_min. When , Equation (38) becomes

When , Equation (38) becomes

Table IV presents values of ʌ^* and V for the case of ξ ^* = 0 using some of the values of s given in Table III.

Table IV.

Critical Distance Λ^*

See Table I for values of other constants except for L^* = 1 m. The value of γ ^* = λ_ca ^*/L^* is at ʌ = 5 km.

(The Table IV results apply when the only obstacles holding up the glacier are the large obstacles.)

The values of γ^* in Tables III and IV are calculated using Equation (39). It is seen in these two tables that in-creasing the amount of the basal shear stress that is supported by the larger obstacles increases the length of the cavities behind the larger obstacles.

Effect of melt water from the upper surface

Suppose surface melt water that descends to the glacier bed is able to escape from R-channets and mingle with the water generated at the bed. For mathematical simplicity let the surface melt-water generation rate be constant and equal to (ϛ – 1) times the melt generated at the bed. Equations (17c) and (30) become

and

In Tables I and II the critical distances ʌ are reduced by a factor of ϛ. From these tables it is seen that the presence at the bed of abundant surface melt water can re-duce the critical distance ʌ to values sufficiently small that water can flow through a linked cavity channel system. (Before it can be concluded that this can happen, it is still necessary to show that the surface melt water can escape from R-channels and mingle with the water produced at the bed.)