Thermodynamics of ice Ian Hewitt, University of Oxford - PowerPoint PPT Presentation

Thermodynamics of ice Ian Hewitt, University of Oxford hewitt@maths.ox.ac.uk

Example temperature profiles Energy equation - Derivation - Boundary conditions Simple solutions - Surface seasonal wave - Robin’s solution - Horizontal advection Thermo-mechanical coupling - Surges - Ice streams

Why is temperature important? ✓ ◆ − Q Flow law coefficient ε ij = A τ n � 1 τ ij ˙ A = A 0 exp RT Coefficient in Glen’s flow law varies by around 3 orders of magnitude over range of glacial temperatures. Knowing the temperature is crucial to predicting how fast the ice deforms. Basal conditions Thermal conditions at the bed exert primary control on basal sliding. Many areas of ice sheets have basal temperature at or very close to the melting point.

Example temperature profiles Cuffey & Paterson 2010

Energy equation The energy equation describes how the temperature evolves in space and time ✓ ◆ = k r 2 T + τ 2 ✓ ∂ T ◆ ∂ t + u · r T ρ c η viscous advection conduction r dissipation density ρ ≈ 910 kg m − 3 ≈ specific heat capacity c ≈ 2 × 10 3 J kg − 1 K − 1 ≈ × k ≈ 2 . 1 W m − 1 K − 1 thermal conductivity In components ✓ ∂ 2 T ∂ x 2 + ∂ 2 T ∂ y 2 + ∂ 2 T + τ 2 ✓ ∂ T ◆ ◆ ∂ t + u ∂ T ∂ x + v ∂ T ∂ y + w ∂ T ◆ = k ρ c ∂ z 2 ∂ z η ✓ ◆ ∂ x, ∂ ∂ ∂ y ⌧ ∂ shallow aspect ratio ∂ z

Derivation of energy equation (one dimension) Consider the change in energy over time of a section of ice between and ◆ x + ∆ x ∆ t x q = ρ cuT − k ∂ T heat flux ∂ x ◆ + S − q ( x + ∆ x ) + = q ( x ) − Fourier’s law ε ij = τ 2 x + ∆ x heat source x S = τ ij ˙ η Change in energy = flux in - flux out + heat source rate of work done [ ρ c ∆ T ] ∆ x = [ q ( x ) − q ( x + ∆ x ) + S ∆ x ] ∆ t Divide by and let ∆ t, ∆ x → 0 ∆ t ∆ x → ρ c ∂ T ∂ t = − ∂ q ∂ x + S from mass = k ∂ 2 T ∂ x 2 + τ 2 ✓ ∂ T ◆ ∂ t + u ∂ T ∂ u ρ c ∂ x = 0 conservation ∂ x η ✓ ◆

Boundary conditions ice

Surface boundary condition ✓ ◆ Condition at surface expresses surface energy balance. It is related to the surface mass balance (kinematic) condition. Q (1 − a ) − εσ T 4 + h T ( T a − T ) = k ∂ T T < T m ∂ z conduction into ice net shortwave radiation (account for albedo) (Fourier’s law) longwave radiation sensible heat transfer (Stefan-Boltzmann law) (turbulent heat transfer coefficient - depends on wind speed) ) = k ∂ T ∂ z Q (1 − a ) − εσ T 4 + h T ( T a − T ) = When melting occurs, we must also account for latent heat fluxes. Conduction term is relatively small - energy balance effectively determines surface temperature − T = T s ( t ) r

Basal boundary condition A number of different thermal conditions are possible at the bed ice T < T m frozen bed T = T m temperate ice above bed (requires viscous heating) temperate bed - freezing temperate bed - melting

Basal boundary condition Condition at bed expresses basal energy balance. G = � k ∂ T T < T m ∂ z geothermal heat flux conduction into ice G = � k ∂ T ∂ z � � ∂ z If bed is at melting point G + τ b u b � mL = � k ∂ T T = T m ∂ z frictional heating basal melt rate (freeze-on if negative) L ⇡ 3 . 4 ⇥ 10 5 J kg − 1 latent heat

Polythermal ice Temperate ice (ice at the pressure melting point) can result from - heating caused by viscous dissipation. - heating caused by refreezing of infiltrating surface melt water. Many mountain glaciers are entirely temperate - referred to as temperate glaciers . Some areas of ice sheets have temperate ice near the bed - they are referred to as polythermal. The energy equation in temperate ice becomes an equation for water content φ τ 2 1 ∂φ T = T m ∂ t + u · r φ = �r · q + ρ w L η viscous dissipation now causes internal melting + additional assumptions for how water moves (eg Aschwanden et al 2012, Schoof & Hewitt 2015).

Surface seasonal wave Near the surface, suppose we may ignore advection (ok if accumulation not too large) ∂ t ∂ z ≈ ∂ t = κ∂ 2 T ∂ T κ = k κ ≈ 10 − 6 m 2 s − 1 thermal diffusivity ∂ z 2 ρ c Impose temperature oscillation at surface T = T 0 − ∆ T cos ω t z = 0 ∂ T far field ∂ z ! 0 z ! 1 � � r ω T = T 0 � ∆ T exp( � α z ) cos( ω t � α z ) α = 2 κ r r r r κ P Depth of temperature variation z ∗ = 2 κ π P = 2 π for period of oscillation ω r ≈ Eg. P = 1 d z ∗ ≈ 0 . 17 m ∗ ≈ z ∗ ≈ 3 . 2 m P = 1 y Cuffey & Paterson 2010

Conductive profile Away from the surface, consider steady state temperature profiles. T = T s z = H The simplest case is if conduction dominates ice k ∂ T 0 = k ∂ 2 T ∂ z = G z = 0 ∂ z 2 − T = T s + G ( H − z ) k ⇣ Depth For thicker ice, the bed is at the melting point k 1 − z ⇣ ⌘ T = T s + ( T m − T s ) = T s H z = H Depth = T s z = H z = 0 G Temperature T < T m ≈ − ≈ Melting Eg G ≈ 60 mW m − 2 T m − T s ≈ 50 K z = 0 G Temperature H c = k ( T m − T s ) T < T m ≈ 1750 m G

Example temperature profiles Cuffey & Paterson 2010

Ice divides T = T s z = H Near an ice divide, advection is mostly vertical - a simple assumption is linear vertical velocity (Robin 1955) surface accumulation w = � az H Energy equation balances advection and conduction ∂ z = κ∂ 2 T w ∂ T ∂ z 2 k ∂ T ∂ z = G z = 0 r ✓r ✓r  ◆ ◆� T = T s + G πκ H a a Depth erf − erf 2 κ H H 2 κ H z = T s 2 a k z = H ✓r  Pe = aH Peclet number measures importance of advection Pe κ z = 0 G Temperature T < T m

Horizontal advection Colder interior temperatures are a result of horizontal advection ∂ z = κ∂ 2 T u ∂ T ∂ x + w ∂ T ∂ u ∂ x + ∂ w ∂ z = 0 ∂ z 2 Generally requires a numerical solution… But it is easy to see schematically why this can produce colder interior ice Cold Warmer

Numerical calculations Simulations of ice flow require a good estimate of ‘initial’ temperature. These estimates are challenging due to uncertainty in forcing parameters (surface temperature history, geothermal heat flux, etc). Improvements in data assimilation are ongoing. Basal temperature T − T m ( p ) [K] Pattyn 2010

Surging Thermo-mechanical coupling may be responsible for interesting dynamical phenomena. 2 ✓ ∂ T ◆ = k r 2 T + 2 A ( T ) τ n +1 ∂ t + u · r T ρ c One possibility is thermal runaway (Clarke et al 1977) - an increase in temperature causes increase in viscous dissipation that increases temperature further - positive feedback. Mechanism for surging? Probably not, in the absence of sliding (Fowler et al 2010). MacAyeal: Binge/Purge Cycles of the Laurentide Ice Sheet 779 But basal sliding and frictional heating can lead to surging (eg Payne 1995). os• &o This is essentially the basis for the ‘binge-purge’ Osl A 0 binge purge model of Heinrich events (MacAyeal 1993). y sediment Osl / A 0 thawed frictional heating MacAyeal 1993 Osl A 0 Legend: y •glaciological equilibrium profile ature-depth profile ice column adiabatic lapse rate Osl / trigger point (melting point) Fig. 2. A conceptual view of the temperature-depth profile O(y) in an ice column during the binge/purge cycle of the Laurentide ice sheet. Vertical elevation from the base of the ice column is denoted by y and 0 represents temperature. The annual average sea level atmospheric temperature is denoted by Osl. The melting temperature of ice is represented by the black triangles. The four graphs surrounding the central circle display the sequence of states through which the ice column evolves during a complete cycle. Time passage is represented by counterclockwise progression through the sequence of graphs.

Ice streams Field observations show ice-streams are associated with a temperate bed (Engelhardt et al 1990). Themo-mechanical instability may provide a mechanism for forming ice streams on an otherwise uniform bed. Hindmarsh 2009

Summary Temperature is important for determining ice flow and basal conditions. Temperatures generally increase with depth as a result of geothermal and frictional heating, and viscous dissipation. Simple analytical solutions of the energy equation help explain qualitative features of observations. Thermo-mechanical coupling has important dynamical consequences for ice flow.