Team 3694 Page 11 of 37 where h is the current thickness of the ice sheet and t is the elapsed time after one timestep. Substituting for Sh with the expression we derived and substituting for the known value of the molecular weight of water yields (18.78-h3sr 6.1121×10-t 0.0003448 S(=h This equation governs the sublimation of the ice To model melting, the second component of ablation, we apply the heat equation. the heat U(x,t=kU(x, 1) (11) where k=0.0104 is the thermal diffusivity of the ice. In order to solve the heat equation for the Neumann conditions, we assume a steady-state Us with the same boundary conditions as U and that is independent of time. The residual temperature V has homogeneous boundary conditions and initial conditions found by U-Us. Thus we can rewrite the heat equation as (x,D)=(x,D)+U,(x,D) The steady-state solution of the heat equation is given by T US=T subject to the constraints 0<x<S(t) and0<t<l month. The following equations follow directly from the heat equation as well (x, t)=kv (x, t)+f, where f is a forcing ter v(O, t)=v(s(),t)=0(necessary conditions for the homogeneous boundary equations
Team # 3694 Page 11 of 37 where h is the current thickness of the ice sheet and t is the elapsed time after one timestep. Substituting for Sh with the expression we derived and substituting for the known value of the molecular weight of water yields ( ) 2 1 257.14 234.5 18.678 2 ( 273.15) 6.1121 10 0.0003448 ( ) + ⋅ × = − + − − T e t S t h T T T ρ ice (10) This equation governs the sublimation of the ice. To model melting, the second component of ablation, we apply the heat equation. The heat equation governs the relationship U (x,t) kU (x,t) t = xx (11) where k=0.0104 is the thermal diffusivity of the ice . In order to solve the heat equation for the Neumann conditions, we assume a steady-state Us with the same boundary conditions as U and that is independent of time. The residual temperature V has homogeneous boundary conditions and initial conditions found by U-Us. Thus we can rewrite the heat equation as: U(x,t) V (x,t) U (x,t) = + s (12) The steady-state solution of the heat equation is given by: x S t T T U T a l s l ( ) − = + (13) subject to the constraints 0 < x < S(t) and 0 < t < 1 month. The following equations follow directly from the heat equation as well: V (x,t) kV (x,t) t = xx + f, where f is a forcing term. (14) V (0,t) = V (S(t),t) = 0 (necessary conditions for the homogeneous boundary equations)
Team 3694 Page 12 of 37 Since no external heat source is present and temperature distribution only depends on heat convection, we take the forcing term f=0. To calculate change in mass balance on a monthly basis, we solve analytically using separation of variables r(xD)=a+∑ae"cos nID (15) T-T a==It+ x=27+7。-7=T+T (16 d nIA Therefore T+7。2(T-7) nJA V(x,1)= (-1)”-1)e 2 (nT) Having found V(x, t) and Us(x, t), we obtain an expression for U(x, t) U(x,)=V(x,)+U,(x,D) (19) Since U is an increasing function of x, and for x> k, U(x, t)>0 for fixed t, the ice will melt for k <x<h. Thus, we seek the solution to U(k, t=0 for k to determine ablation omputationally, we solve this expression using the first 100 terms of the Fourier series expansion and the MATLaB function fero. The solution of this equation for k is the primary computational step for the MATLAB simulation( see Appendix A). The new value of k is used to renew h as the new thickness of the ice sheet, and a consequent time step can begin calculation
Team # 3694 Page 12 of 37 Since no external heat source is present and temperature distribution only depends on heat convection, we take the forcing term f = 0. To calculate change in mass balance on a monthly basis, we solve analytically using separation of variables: ∑ ∞ = − = + 1 0 cos 2 ( , ) 2 2 2 n s n t n s n x a e a V x t π π (15) where x dx s T T T s a s a l ∫ l − = + 0 0 2 = 2Tl + Ta −Tl = Tl + Ta (16) and dx s n x x s T T T s a s a l n l − = + ∫ π cos 2 0 (17) ( cos( ) 1) (( 1) 1) 2 2 − − − = = n n s n n s π π π Therefore, − − − + + = ∞ − = ∑ s n x e n T T T T V x t s n t n n l a a c π π π (( 1) 1) cos ( ) 2( ) 2 ( , ) 2 2 2 1 2 . (18) Having found V(x, t) and Us(x, t), we obtain an expression for U(x, t): U(x,t) V (x,t) U (x,t) = + s (19) Since U is an increasing function of x, and for x > k, U(x, t) > 0 for fixed t, the ice will melt for k < x < h. Thus, we seek the solution to U(k, t)=0 for k to determine ablation. Computationally, we solve this expression using the first 100 terms of the Fourier series expansion and the MATLAB function fzero. The solution of this equation for k is the primary computational step for the MATLAB simulation (see Appendix A). The new value of k is used to renew h as the new thickness of the ice sheet, and a consequent time step can begin calculation
Team 3694 Page 13 of 37 With these two components we can now finalize an expression for ablation and apply it to a computational model. The sum of the infinitesimal changes in ice sheet thickness for each differential volume gives the total change in thickness. To find these changes, we first note that Mass Balance loss due to Sublimation=(h-S)LD Mass Balance loss due to Melting =(S-k)LD (21) where the product LD is the surface area of the ice sheet. Note that in these equations, the mass balance" refers to net volume change. Thus, ablation is given by Mob =(h-S)LD+(s-k)Ld=(h-kLD (22) Mass balance and sea level rise Combining accumulation and ablation into an expression for mass balance, we have M=Mo-Mo=0.025LD-(h-k)LD Relating this to sea level rise, we use the approximation 360 Gt water= Imm sea level rise. Thus SlR=M·pa which quantifies the sea level rise due to mass balance Thermal Expansion A second mode of sea level rise is also considered thermal expansion due to warming According to various literature, thermal expansion of the oceans due to increase in global
Team # 3694 Page 13 of 37 With these two components we can now finalize an expression for ablation and apply it to a computational model. The sum of the infinitesimal changes in ice sheet thickness for each differential volume gives the total change in thickness. To find these changes, we first note that Mass Balance Loss Due to Sublimation = (h-S)LD (20) Mass Balance Loss Due to Melting = (S-k)*LD (21) where the product LD is the surface area of the ice sheet. Note that in these equations, the “mass balance” refers to net volume change. Thus, ablation is given by M ab = (h − S)LD + (S − k)LD = (h − k)LD (22) Mass Balance and Sea Level Rise Combining accumulation and ablation into an expression for mass balance, we have M = M ac − M ab = 0.025LD − (h − k)LD (23) Relating this to sea level rise, we use the approximation 360 Gt water = 1mm sea level rise. Thus, Gt mm SLRmb M ice 360 1 = ⋅ ρ ⋅ (24) which quantifies the sea level rise due to mass balance. Thermal Expansion A second mode of sea level rise is also considered: thermal expansion due to warming. According to various literature , thermal expansion of the oceans due to increase in global