LECTURES ON GEOCHEMICAL !N1 ERPRETATION OF ...

- 29 -As a second example. consider a situation in which a wellhead steamwaterseparator operates at 1.99 bars (28.9 psia) vapor pressu~ of water(and, therefore, at a temperature of 120·C from steam tables) . The separatedLiquid is then cooled quickly in an ice bath so that no additionalvapar separates, and the silica concentration in the resulting water isfound to be 400 mg/kg. The enthalpy of liquid water at 120·C is 503.7 JIg(point C in Fig. 4) and that of the coexisting steam is 2706 JIg (point Gc)'The straight line drawn between C and Gc intersects the quartz solubilitycu~e at point 0, which characterizes the initial reservoir water.Data for constructing the quartz solubility curve on an enthalpy- silicagraph are given in Table 2. Alternately, an equation that expresses thesolubility of quaC'tz in liquid wateC' at the vapor pressure of the solutionas a func t ion of the enthaply of the liquid water solvent isD H2 3S ::s 01 + D2H + + °4 H34• DSH ,(8)wheC'e S is silica concentC'ation in mg/kg, H is enthalpy of liquid water inJIg, and 01 through 06 are constants given in Table 1. Equation (8)should not be used with enthalpies larger than 1670 Jig.Equations expressingthe enthalpy of liquid water and steam at the vapoC' pressuC'e of thesolutions in the temperature range 50· to 340·C areL = Al A t 2 A t 3 4 5 - 1 A t - 2 A 10g t (9)• A2t • • + Ast + A6 t + A7 t3 4 • 8 • 9andG = B1 + B2t + B t 23+ B4t 3 + Bst 4• B t 56 • B 7 t-1 • B t - 28+ B 910g t, (10)where t is tempeC'ature in degrees Celsius, L and G are enthalpies of liquidwateC' and steam, respectively, and 't through A9 and B1 through B9 areconstants given in Table 1.enthalpies when t < SO·C.Do not use equations (9) and (10) to calculateIn the temperature range O· to SO·C.L 4.1868 t. (11)