Chapter

5.1. Calculate the difference between the surface velocity and the bed velocity in a glacier 300 m thick with a surface slope of 0.046. Use n = 3

• the infinitely wide approximation,

• an approximation based on Raymond's estimate of the appropriate shape factor for Athabasca Glacier (0.58), and

• the semicircular approximation.

Which result comes closest to the values measured by Raymond on Athabasca Glacier, and why?

5.2. At the equilibrium line on the Barnes Ice Cap Trilateration Net, the surface velocity is 6.7 m a-1, the ice is 185 m thick, and the surface slope is 0.07. Using B = 0.317 MPa a1 /n (appropriate for ice at about —5 °C) and n = 3, calculate and plot a velocity profile through the ice cap. What is the basal velocity. Is your result consistent with the above ice temperature?

5.3. An infinitely wide glacier has a velocity of 1 m a—1 at the surface and

0.7 m a—1 at a depth of 16 m. Determine the thickness of the glacier. Assume ub = 0 and n = 3.

5.4. (a) An ice sheet has a surface profile given by h = .Jcx, where h is the height in meters and x is the distance from the margin, also in meters. Differentiate this to obtain an expression for the surface slope, S. By inserting this in the expression for the basal drag, T = pghS, show that T is independent of x. Obtain a numerical value for T if c = 16 m.

(b) Let the ablation rate be bn m a—1. By equating the discharge through any cross section to the volume of ice lost by melting downglacier from that cross section, show that the average horizontal velocity in the ablation zone is: u = Q^fx where ^ = bn/^/c. (Note that ^ is negative because bn is negative. Thus, u is negative, consistent with the fact that it is in the

(c) At x = 1500 m the glacier flows over a bump in the bed, 0.5 m high, and quarries a cobble from the lee slope of the bump. The ice closes under the cobble, so at the start of its journey to the margin it is 0.5 m above the bed. Determine the x and z coordinates (z vertical) of the point where the cobble will melt out, and its time en route. Plot its path. Assume plug flow and incompressibility. Use bn = 0.6 m a—1 and c = 16 m, and assume the ablation zone is 2 km wide. Hint: use the incompressibility condition, du/dx = —dw/dz, and the result from Problem 5.4b to get w(x).

Then use the definition of velocity, u = dx/dt, and the initial condition, x(t = 0) = xo, to integrate the expression for u to get x(t) = (xo1/2 + 1/2^t)2. Then use w = dz/dt, and the initial condition, z(0) = zo to obtain z(t) = zo(xo/x)1/2.

(d) The accumulation zone in the above problem is 10 km wide (Figure P1). At the end of the Pleistocene a mammoth dies 500 m from the divide. Determine the x and z coordinates of the point where he melts out, and his

^Divide Equilibrium line —^ Margin-^

Accumulation zone

Ablation zone

2 km

Figure P1.

time en route. Assume the glacier has had a balanced budget for the last 10 000 a, and that the ablation rate is 0.6 m a-1 over the 2 km wide ablation zone, as before. Plot the path. Hint: do as in Problems 5.4b and 5.4c, remembering that the horizontal velocity is now u = -bn(L — x)/«Jcx, where L is the distance from the margin to the divide. You will encounter an integral /«Jx/(L — x) dx which may be transformed using r = L — x and then evaluated using the tabulated integral:

(e) If the mammoth was 3 m long and, when he died, he was lying down with his tail 3 m closer to the divide than his head, determine the time required for his body to pass completely beneath the equilibrium line, and his approximate length when he is at this point in his journey.

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