6.Oil reservoirs below ground frequently are in contact with underground water and, inconnection with an oil drilling operation, you are asked to compute the solubility of waterin a heavy oil at the underground conditions. These conditions are estimated to be 140°Cand 410 bar. Experiments at 140°C and 1 bar indicate that the solubility of steam in theoil is x = 35 x 10-4 (x, is the mole fraction of steam). Assume Henry's law in the formf = H(T)x1, where H(T) is a constant, dependent only on the temperature, and f is the fu-gacity of H,O. Also assume that the vapor pressure of the oil is negligible at 140°C. Datafor H,0 are given in the steam tables.

Solution The solubility of water in oil is described by f1 = H(T)4 Henry's constant can be evaluated…

Answered: Oil reservoirs below ground frequently…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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6.
Oil reservoirs below ground frequently are in contact with underground water and, in
connection with an oil drilling operation, you are asked to compute the solubility of water
in a heavy oil at the underground conditions. These conditions are estimated to be 140°C
and 410 bar. Experiments at 140°C and 1 bar indicate that the solubility of steam in the
oil is x = 35 x 10-4 (x, is the mole fraction of steam). Assume Henry's law in the form
f = H(T)x1, where H(T) is a constant, dependent only on the temperature, and f is the fu-
gacity of H,O. Also assume that the vapor pressure of the oil is negligible at 140°C. Data
for H,0 are given in the steam tables.

Expert Solution

Step 1

Solution

The solubility of water in oil is described by

f₁ = H(T)4

Henry's constant can be evaluated at 1 bar where,

* f₁ = 1 bar then,

H(T) = f₁ / x₁ (x₁=35×10^-4)

= 1/35×10^-4 (t=140° C)

= 286 bar

* To obtain f₁ at 410 bar and 140° C use the steam table alternatively get f at saturation (3.6 th bar) and use the pointing factor to correct to 410 bar.

at 140° C

RT In f₁ = Δg₁ → 410 bar = Δh₁ → 410 bar => TΔS₁ → 410 bar

From steam tables book

RT In f₁ = 282 J/g

R=8.314 J/mole.k

=> 8.314 J/18 grams.k and

T = 413k since 1 mole of steam = 18 grams

In f₁ = 1.48

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