Difference between revisions of "Notes/vanderVeen Aug5"

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(Exercise in deriving force balance in the horizontal)
m (Exercise in deriving force balance in the horizontal)
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<math> \frac{\partial}{\partial x} \int_b^h \sigma_{xx} dz - \sigma_{xx}(s)\frac{\partial s}{\partial x} + \sigma_{xx}(b)\frac{\partial b}{\partial x} + \sigma_{xz}(s) - \sigma_{xz}(b) = 0 </math>.
 
<math> \frac{\partial}{\partial x} \int_b^h \sigma_{xx} dz - \sigma_{xx}(s)\frac{\partial s}{\partial x} + \sigma_{xx}(b)\frac{\partial b}{\partial x} + \sigma_{xz}(s) - \sigma_{xz}(b) = 0 </math>.
 +
  
 
We can work on the first term, by writing out what <math> \sigma_{xx} </math> is.
 
We can work on the first term, by writing out what <math> \sigma_{xx} </math> is.
  
 
<math> \frac{\partial}{\partial x} \int_b^h \sigma_{xx} dz = \frac{\partial}{\partial x} \int_b^h [\R_{xx} - \rho g (h-z)] dz </math>
 
<math> \frac{\partial}{\partial x} \int_b^h \sigma_{xx} dz = \frac{\partial}{\partial x} \int_b^h [\R_{xx} - \rho g (h-z)] dz </math>
 +
  
 
We don't know anything about how <math> R_{xx} </math> varies in z, so we won't touch that.  But we can integrate the second term:
 
We don't know anything about how <math> R_{xx} </math> varies in z, so we won't touch that.  But we can integrate the second term:
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<math> \frac{\partial}{\partial x} \int_b^h R_{xx} dz - \frac{1}{2} \frac{\partial}{\partial x}[\rho g H^2] - R_{xx}(s)\frac{\partial s}{\partial x} + R_{xx}(b)\frac{\partial b}{\partial x} - \rho g H\frac{\partial b}{\partial x}  + R_{xs}(s) - R_{xz}(b) = 0 </math>
 
<math> \frac{\partial}{\partial x} \int_b^h R_{xx} dz - \frac{1}{2} \frac{\partial}{\partial x}[\rho g H^2] - R_{xx}(s)\frac{\partial s}{\partial x} + R_{xx}(b)\frac{\partial b}{\partial x} - \rho g H\frac{\partial b}{\partial x}  + R_{xs}(s) - R_{xz}(b) = 0 </math>
 +
  
 
We can simplify the terms involving <math> \rho g H </math> now.  Those are:
 
We can simplify the terms involving <math> \rho g H </math> now.  Those are:
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<math> - \frac{1}{2} \frac{\partial}{\partial x}[\rho g H^2] </math> and <math> - \rho g H\frac{\partial b}{\partial x} </math>.  
 
<math> - \frac{1}{2} \frac{\partial}{\partial x}[\rho g H^2] </math> and <math> - \rho g H\frac{\partial b}{\partial x} </math>.  
  
If we differentiate the first term, we get <math> - \frac{1}{2} \frac{\partial H}{\partial x}[\rho g H] </math>.  We should note that since <math> H = s - b </math>, it is also true that <math> \frac{\partial H}{\partial x} = \frac{\partial s}{\partial x} - \frac{\partial b}{\partial x} </math>.   
+
If we differentiate the first term, we get <math> - \frac{1}{2} \frac{\partial H}{\partial x}[\rho g H] </math>.   
 +
 
 +
We should note that since <math> H = s - b </math>, it is also true that <math> \frac{\partial H}{\partial x} = \frac{\partial s}{\partial x} - \frac{\partial b}{\partial x} </math>.   
  
 
This lets us compare the first and second <math> \rho g H </math> terms: their sum is just <math> - \rho g H \frac{\partial s}{\partial x} </math>.  This is the definition of the driving stress, <math> \tau_d </math>.
 
This lets us compare the first and second <math> \rho g H </math> terms: their sum is just <math> - \rho g H \frac{\partial s}{\partial x} </math>.  This is the definition of the driving stress, <math> \tau_d </math>.
  
 
So now the balance equation is  
 
So now the balance equation is  
 +
 
<math> \frac{\partial}{\partial x} \int_b^h R_{xx} dz - R_{xx}(s)\frac{\partial s}{\partial x} + R_{xx}(b)\frac{\partial b}{\partial x}  + R_{xs}(s) - R_{xz}(b) - \tau_d = 0 </math>
 
<math> \frac{\partial}{\partial x} \int_b^h R_{xx} dz - R_{xx}(s)\frac{\partial s}{\partial x} + R_{xx}(b)\frac{\partial b}{\partial x}  + R_{xs}(s) - R_{xz}(b) - \tau_d = 0 </math>
 +
  
 
We will apply the zero-stress surface boundary condition, which says that
 
We will apply the zero-stress surface boundary condition, which says that
 +
 
<math> R_{xz}(s) - R_{xx}(s) \frac{\partial s}{\partial x} = 0 </math>.  
 
<math> R_{xz}(s) - R_{xx}(s) \frac{\partial s}{\partial x} = 0 </math>.  
  
 
So we can lose those terms, and the balance equation becomes
 
So we can lose those terms, and the balance equation becomes
 +
 
<math> \frac{\partial}{\partial x} \int_b^h R_{xx} dz + R_{xx}(b)\frac{\partial b}{\partial x} - R_{xz}(b) + \tau_d = 0 </math>
 
<math> \frac{\partial}{\partial x} \int_b^h R_{xx} dz + R_{xx}(b)\frac{\partial b}{\partial x} - R_{xz}(b) + \tau_d = 0 </math>
 +
  
 
And, finally, we'll substitute the basal drag, <math> \tau_b </math>, for its definition:
 
And, finally, we'll substitute the basal drag, <math> \tau_b </math>, for its definition:
 +
 
<math> \tau_b = R_{xz}(b) - R_{xx}(b)\frac{\partial b}{\partial x} </math>
 
<math> \tau_b = R_{xz}(b) - R_{xx}(b)\frac{\partial b}{\partial x} </math>
  
 
to arrive at the final answer,
 
to arrive at the final answer,
 +
 
<math> \frac{\partial}{\partial x} \int_b^h R_{xx} dz - \tau_b + \tau_d = 0 </math>
 
<math> \frac{\partial}{\partial x} \int_b^h R_{xx} dz - \tau_b + \tau_d = 0 </math>

Revision as of 11:56, 5 August 2009

Crash Course in Glacier Dynamics

Kees van der Veen, University of Kansas August 5, 2009 Portland Summer Modeling School


What's the objective of an ice sheet model?

  • Understand evolution of ice sheet given some forcing (global warming, etc.)


Fundamental equations: conservation of xxx

  • Mass
  • Energy
  • Momentum


Conservation of Mass: Continuity Equation

  • What comes in (flux, basal freezing if, accumulation if) to some control volume must go out (flux, basal melting if, ablation if).
  • Assumption: ice is incompressible, so density is constant. Mass conservation ~ volume conservation
  • Assumption: ignore firn layer (100-150m in Antarctica, less in Greenland)

 M \Delta x + H(x)U(x) - H(x + \Delta x) U(x + \Delta x) = \frac{\Delta H}{\Delta t} \Delta x

 \frac{\Delta H}{\Delta t} = -\frac{H(x + \Delta x)U(x+\Delta x) - H(x)U(x)}{\Delta x} + M

Shrink timestep & spatial step to infinitessimal to write as differential equation

 \frac{\partial H}{\partial t} = -\frac{\partial}{\partial x} HU + M


Conservation of Momentum: Newton's second law

  •  F = ma , with zero acceleration
  • so the sum of all forces must be zero.
  • stresses are easier to work with than forces: stress is force per unit area
  • Nine stress components:  \sigma_{ij}
  • i: plane perpendicular to axis (x)
  • j: direction of stress
  • Stress tensor is symmetric, so  \sigma_{ij} = \sigma_{ji} and there are really only six distinct stress components
  • 3 equations with 6 unknowns


Force balance in z

 F_z = 0

 \sigma_{zz}(z + \Delta z) \Delta x \Delta y + \sigma_{xz}(z+\Delta z) \Delta z \Delta y + \sigma_{yz}(z + \Delta z) \Delta x \Delta z = \rho g \Delta x \Delta y \Delta z

 \frac{\partial \sigma_{xz}}{\partial x} + \frac{\partial \sigma_{yz}}{\partial y} + \frac{\partial \sigma_{zz}}{\partial z} = -\rho g

Force balance in x

 \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{xy}}{\partial y} + \frac{\partial \sigma_{xz}}{\partial z} = 0

Force balance in y

(we won't concern ourselves with the transverse y direction, which also means we can eliminate the y-derivative terms in the z and x equations above)

Newton's First Law: action / reaction

  • What drives glacier flow? Gravity is the "action".
  • What is the response? Resistance to flow is the "reaction".


Exercise in deriving force balance in the horizontal

Integrate the force balance equation in x over the depth of the ice column to eventually derive an expression relating driving stress, basal drag, and longitudinal stress gradients.

Mathways, start with:

 \int_b^h \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{xy}}{\partial y} + \frac{\partial \sigma_{xz}}{\partial z} dz = \int_b^h 0 dz

The integral in  \sigma_{xz} is easy because it is just the difference of the stress at the surface and the bed -- the two dz's cancel. But we have to use the Leibniz rule to work with the  \sigma_{xx} term, because the limits of integration h and b are really h(z) and b(z) - they depend on z, which we're trying to integrate over. Shucks.

Using the Leibniz Rule,  \int_b^h \frac{\partial \sigma_{xx}}{\partial x} dz = \frac{\partial}{\partial x} \int_b^h \sigma_{xx} dz - \sigma_{xx}(s)\frac{\partial s}{\partial x} + \sigma_{xx}(b)\frac{\partial b}{\partial x} .

The entire balance equation is thus

 \frac{\partial}{\partial x} \int_b^h \sigma_{xx} dz - \sigma_{xx}(s)\frac{\partial s}{\partial x} + \sigma_{xx}(b)\frac{\partial b}{\partial x} + \sigma_{xz}(s) - \sigma_{xz}(b) = 0 .


We can work on the first term, by writing out what  \sigma_{xx} is.

 \frac{\partial}{\partial x} \int_b^h \sigma_{xx} dz = \frac{\partial}{\partial x} \int_b^h [\R_{xx} - \rho g (h-z)] dz


We don't know anything about how  R_{xx} varies in z, so we won't touch that. But we can integrate the second term:

  \int_b^h \rho g (h-z) dz = \frac{1}{2} \rho g H^2

So the entire force balance equation is now

 \frac{\partial}{\partial x} \int_b^h R_{xx} dz - \frac{1}{2} \frac{\partial}{\partial x}[\rho g H^2] - R_{xx}(s)\frac{\partial s}{\partial x} + R_{xx}(b)\frac{\partial b}{\partial x} - \rho g H\frac{\partial b}{\partial x}  + R_{xs}(s) - R_{xz}(b) = 0


We can simplify the terms involving  \rho g H now. Those are:

 - \frac{1}{2} \frac{\partial}{\partial x}[\rho g H^2] and  - \rho g H\frac{\partial b}{\partial x} .

If we differentiate the first term, we get  - \frac{1}{2} \frac{\partial H}{\partial x}[\rho g H] .

We should note that since  H = s - b , it is also true that  \frac{\partial H}{\partial x} = \frac{\partial s}{\partial x} - \frac{\partial b}{\partial x} .

This lets us compare the first and second  \rho g H terms: their sum is just  - \rho g H \frac{\partial s}{\partial x} . This is the definition of the driving stress,  \tau_d .

So now the balance equation is

 \frac{\partial}{\partial x} \int_b^h R_{xx} dz - R_{xx}(s)\frac{\partial s}{\partial x} + R_{xx}(b)\frac{\partial b}{\partial x}  + R_{xs}(s) - R_{xz}(b) - \tau_d = 0


We will apply the zero-stress surface boundary condition, which says that

 R_{xz}(s) - R_{xx}(s) \frac{\partial s}{\partial x} = 0 .

So we can lose those terms, and the balance equation becomes

 \frac{\partial}{\partial x} \int_b^h R_{xx} dz + R_{xx}(b)\frac{\partial b}{\partial x} - R_{xz}(b) + \tau_d = 0


And, finally, we'll substitute the basal drag,  \tau_b , for its definition:

 \tau_b = R_{xz}(b) - R_{xx}(b)\frac{\partial b}{\partial x}

to arrive at the final answer,

 \frac{\partial}{\partial x} \int_b^h R_{xx} dz - \tau_b + \tau_d = 0