Homework #3               Glaciology, GEOL 4888,  Spring 2019                        Humphrey

1. On the lee side of Medicine Bow Peak, the wind in the last storm has deposited a 1 meter deep snow drift with a temperature of -15C. In this following week the sun will melt about 2cm of water equivalent from the surface. Is there enough energy in the melt water to ripen the snow pack to 0C? Assume the snow has a density of 300kg/m3. (hint, use the cold content Index we discussed in class)

2.   (To do this question you will need to plot the results on a computer, or if you want to do it by hand, it must be done on graph paper with correctly scaled axes.)

Let’s look at the deformation velocity expected in the accumulation region of Greenland.  Assume the ice is approximately a ‘slab glacier’, uniform in all directions.  With a coordinate system x-downglacier, and z-down from the surface, the shear stress at any depth is due to the weight of the overlying ice and the surface slope. In other words, the shear stress is given by r_ice g sin(a)Z, where alpha is the surface angle from horizontal and Z is the depth below the surface. If the ice is 1800m thick, with a surface angle of about 1/3 of a degree:

a. plot the value of the shear stress versus depth? Label the axes, and plot height above the bed as the Y-axis value, and stress on the x-axis [hint so you are not totally wrong, your shear stress at depth should be on the order of 10⁵ Pa]

Now to make the discussion specific, let’s look at a region where the ice temperature is about -20C.  See table below (notice how much A varies with temperature!). The equation we developed in class for velocity versus depth (the signs have been adjusted so that Z is positive downward):

V_surface - V_(depth=z) = A/2 * (r_ice g sin(a))³ * z⁴

Calculate the total expected deformation velocity over the full depth. For consistency, use 900kg/m3 for the ice density. The deformation velocity is not known at the surface, but if the ice is not sliding, the actual velocity is equal to 0 at the bed. In the accumulation region, the ice is mostly frozen to the bed and the sliding velocity is zero. This allows you to calculate the surface velocity as the total difference between the bed (0) and surface velocity.

b.    Plot, using graph paper or a computer, the velocity profile. Plot depth on the vertical, with the bed at 0 on the y-axis, and velocity on the x-axis. [hint your answer surface velocity should be of order 1 or 10 m/year]

Table of Values of A(T), s^-1Pa^-3	Temperature (Celsius)
2.4 x 10^-24	0
9.3 x 10^-25	-5
3.5 x 10^-25	-10
1.2 x 10^-25	-20