# How much calcite precipitates in your tea kettle- # Kinetic precipitation of calcite # Kinetic loss of CO2(g) from heating water SOLUTION 1; temp 10; EQUILIBRIUM_PHASES; Calcite; CO2(g) -2.0; SAVE solution 2 USER_PRINT; -start; 10 if get(1) = 0 then put(tot("Ca"), 1); -end END RATES CO2(g) # two-film model... # F = 1000A / (z_w / D_w + z_a / (D_a*K_Hx)) * (P_a / (RT * K_Hx) - c_w) # z_w = 0.25 mm. D_w = 1.3e-9m2/s. z_a = 3 mm. D_a = 1e-5m2/s. # K_Hx = (10^1.5 atm.L_w/mol) / (RT = 24 atm.L_a/mol) = 1.3 L_w/L_a # A = air/water interface area, m2. 1000 recalc's from m3 to dm3. -start 1 A = parm(1); 2 P_a = parm(2) 10 RT = 0.08206 * TK 20 K_Hx = 10^(SI("CO2(g)")) / act("CO2") / RT 30 visc = 10^(-(1.37023*(tc - 20) + 8.36e-4*(tc - 20)^2) / (109 + tc)) 40 D_w = 1.3e-9 * TK * 0.891 / (298 * visc) # neglect resistance through air layer = z_a/(D_a*K_H')... 50 F = 1000 * A / (0.25e-3 / D_w) * (P_a / (RT * K_Hx) - act("CO2")) 60 save F * time -end USE solution 2 # Fill the kettle KINETICS 2 # 10 g calcite, s.w. 2.5 g/cm3, 4000 * 1mm cubes, A = 2.4 dm2/dm3 Calcite; -m0 0.1; -par 2.4 0.67; -step 120 in 10 # heating takes 120 seconds CO2(g); -par 0.01 0.316e-3 REACTION_TEMPERATURE; 10 100 in 10 # heat the water USER_GRAPH; -head Temp_C A_cc=2.4 -axis_titles "Temperature / C" "mg Calcite / L" -start; 10 graph_x tc; 20 graph_y (get(1) - tot("Ca"))*100e3; -end END