Effects on the glaciers
Description 
Variables
Calculations
Calculations
Equations
(ID 7434)
(ID 8249)
Examples
(ID 95)

effect015
(ID 7410)

effect010
(ID 7405)

effect012
(ID 7407)

effect014
(ID 7409)

effect011
(ID 7406)

effect013
(ID 7408)

effect037
(ID 7430)

effect009
(ID 7404)

effect032
(ID 7425)

effect016
(ID 7411)
To calculate the ablation rate (melting speed), we\'ll assume that the glacier has a height h and is at a temperature $\Delta T$ below the melting point. The energy captured by a layer of height $\Delta x$ is partly conducted into the glacier, contributing to the melting of the layer and its warming. If l is the latent heat and $\rho_e$ the ice density, a volume element with surface $S$ and height $\Delta x$ will require the energy
$\Delta Ql = S\Delta x l \rho_e$
to melt.
To heat it up to the melting temperature $\Delta T_m$, it will require
$\Delta Q_c = S\Delta x\rho_ec\Delta T_m$
where c is the specific heat. Lastly, thermal conduction will remove heat
$\Delta Q_{\lambda}=\displaystyle\frac{\lambda S\Delta T_b}{h}\Delta t$
where $\lambda$ is the thermal conductivity, $\Delta T_b$ is the base-surface temperature difference, and $\Delta t$ is the elapsed time.
Therefore, the total heat will be
$\Delta Q_l + \Delta Q_c + \Delta Q_{\lambda} = (1 - a_{ev})(1 - \gamma_v)S I_s\Delta t$
which, after replacing with the expressions, becomes
$S\Delta xl\rho_e + S\Delta x\rho_ec\Delta T_m + (\lambda/h)S \Delta T_b \Delta t = (1 - a_{ev})(1 - \gamma_v)S I_s\Delta t$
Solving for
| $ v_a =\displaystyle\frac{(1 - a_{ev} )(1 - \gamma_v ) I_s - ( \lambda / h ) \Delta T_b }{ \rho_e (l + c \Delta T_m )}$ |
Hence, an increase in temperature leads to an increase in the ablation rate.
(ID 7432)
The accumulation rate, denoted as
| $ v_c =\displaystyle\frac{ \Delta x }{ \Delta t }$ |
(ID 7612)
Solar radiation is partly reflected and partly absorbed by the surface. If $I_s$ is the radiation flux, $a_{ev}$ is the Earth\'s visible albedo, and $\gamma_v$ is the coverage factor, the absorbed fraction is
$(1 - a_{ev})(1 -\gamma_v)I_s$
The heat supplied is partly conducted into the glacier\'s interior and partly contributes to melting a layer of thickness $\Delta x$ in a time $\Delta t$.
In this way, the glacier\'s surface would decrease at an ablation rate (melting speed)
$v_a =\displaystyle\frac{\Delta x}{\Delta t}$
due to the melting effect, while it would grow at an accumulation rate $v_c$ (snow deposition speed) due to the effect of snow being deposited on its surface. Therefore, melting would occur if the total velocity
| $ v_b = v_c - v_a$ |
turns out to be negative.
(ID 7434)
La taza de balance de masa que se calcula de la taza de acumulaci n y la taza de ablaci n
| $ v_b = v_c - v_a$ |
permite estimar la variaci n en la altura especifica del glaciar (en un lugar en particular)
| $\Delta h=v_b\Delta t$ |
(ID 8249)
ID:(582, 0)
