Formation Evaluation Through Casing

Abstract

A number of tools are available for formation evaluation through tubing and casing. They include:

Appendices

Answers to Text Questions

Question #11.1

Uranium

Question #11.2

• (c) Σ increase

• (d) 30 cu

Question #11.3

S _w = 50 %

Question #11.5

• (a) ϕ = 30 %

• (b) S _w = 50 %

• (c) S _w = 55 %

Question #11.6

Σ _w = 110 cu

Question #11.7

• (a) Σ _ma = 13 cu

• (b) Σ _w = 75 cu

Question #11.8

Σ _o = 21.2 cu

Question #11.9

Σ methane = 6.5 cu

Σ _g = 8.0 cu

Question #11.10

Σ _sh = 33 cu

Question #11.11

• (a) ϕ = 25 %

• (b) Σ _wa = 50 cu

Appendix 1: Interpretation of Pulsed Neutron Logs Using the Dual-Water Method

The dual-water method of interpreting pulsed neutron logs is based on the assumption that shales are composed of dry clay, crystalline minerals to whose surface is bound a layer of water. This water is called bound water. A further assumption is that the properties of bound water (e.g., R _w, Σ _w) may be different from those of free water that exists in the effective, interconnected pore space. In particular, the theory of dual-water interpretation proposes that bound water is less saline than free water in most cases. Correct interpretation, therefore, calls for a means to find the amount of (1) dry clay and (2) bound water. The concept of total porosity ϕ _T, that is, the free fluids, ϕ _e, and the bound water, is an important part of the theory. Figure 11.36 illustrates the concepts by mapping bulk volume fractions of a shaly formation.

Fig. 11.36

Dual-water shaly formation

The following relationships pertain:

$$ {\phi}_{\mathrm{e}}={\phi}_{\mathrm{T}}\hbox{--} {V}_{\mathrm{wb}}. $$

$$ {S}_{\mathrm{wT}}=\left({V}_{\mathrm{wf}}+{V}_{\mathrm{wb}}\right)/{\phi}_{\mathrm{T}}. $$

$$ {S}_{\mathrm{we}}={V}_{\mathrm{wf}}/{\phi}_{\mathrm{e}}. $$

$$ {V}_{\mathrm{sh}}={V}_{\mathrm{wb}}+{V}_{\mathrm{dc}}. $$

Essentially, there are five unknown quantities: V _ma, V _dc, V _wb, V _wf, and V _hy. The logs available are Σ, ratio, and GR. The identity:

$$ {V}_{\mathrm{ma}}+{V}_{\mathrm{dc}}+{V}_{\mathrm{wb}}+{V}_{\mathrm{wf}}+{V}_{\mathrm{hy}}=1 $$

adds one more for a total of four measurements. Therefore, one unknown must be eliminated before a solution can be found. The normal way of doing this is to make an assumption about V _wb as a function of V _dc. That is, to assume that a unit volume of dry clay always has associated with it the same amount of bound water. In fact, in “pure shale,” it would be quite common to find a “total porosity” of 30 or 40 % (as reflected by neutron log readings in shales). In this case, the amount of bound water associated with a dry clay can be back calculated.

For example, if a 100 % shale has a total porosity of 35 %, it follows that:

$$ {V}_{\mathrm{wb}}=35\%\;\mathrm{and}\kern0.24em {V}_{\mathrm{dc}}=65\% $$

and hence that:

$$ {V}_{\mathrm{wb}}=\alpha\;.\;{V}_{\mathrm{dc}}, $$

where α is some constant which, in this example, is numerically equal to 35/65 = 0.538. Having reduced the unknowns to four (V _ma, V _dc, V _wf, and V _hy), since V _wb can now be assumed equal to α · V _dc, the solution to the dual-water problem becomes straightforward.

The following steps are required:

• 1. Find all necessary parameters Σ_ma, Σ_dc, Σ_wf, Σ_hy, GR_ma, GR_dc.

• 2. Find ϕ _T and V _dc.

• 3. Solve for ϕ _e and S _we.

Finding Parameters

Crossplot techniques are particularly useful for finding the required parameters. The log data points should be divided into two groups: The 100 % shales and the clean-formation intervals. In clean formations, a plot of Σ vs. ϕ will define Σ _ma and Σ _wf, provided there is sufficient variation in porosity and enough points at 100 % water saturation. Figure 11.37 shows the procedure schematically.

Fig. 11.37

Finding Σ _ma and Σ _wf.

A similar plot for finding Σ _dc and Σ _wb is shown in Fig. 11.38 (all points must come from the shale sections). Note that, on both plots, ϕ _T, derived from the Σ vs. ratio crossplot, is used. This entails an assumption that porosity measured in this way is, in fact, equal to total porosity.

Fig. 11.38

Finding Σ _dc and Σ _sb

Σ _hy can be found by conventional means. Thus, only the gamma ray response to dry clay and response to the matrix remain to be found. It is assumed that neither formation water nor hydrocarbon contribute to the gamma ray response, so it can be written

$$ \mathrm{G}\mathrm{R}=\left(1\hbox{--} {\phi}_{\mathrm{T}}\right)\kern0.24em {\mathrm{GR}}_{\mathrm{ma}}+{V}_{\mathrm{dc}}\;{\mathrm{GR}}_{\mathrm{dc}}. $$

From which it follows that, in shales,

$$ {\mathrm{GR}}_{\mathrm{dc}}=\mathrm{G}\mathrm{R}/\left(1-{\phi}_{\mathrm{T}}\right), $$

and, in clean intervals,

$$ {\mathrm{GR}}_{\mathrm{ma}}=\mathrm{G}\mathrm{R}/\left(1-{\phi}_{\mathrm{T}}\right). $$

For example, in a shale, GR = 110 and ϕ _T = 33 %; but, in a clean section, GR = 25 and ϕ _T = 25 %, so it follows that:

GR_dc = 110/(1 − 0.33) = 149.25, and

GR_ma = 25/(1 − 0.25) = 33.3.

Finding ϕ _T and V _dc

As already stated, ϕ _T is found from the Σ vs. ratio crossplot. V _dc can be found from the GR using:

$$ {V}_{\mathrm{dc}}=\frac{\mathrm{GR}-{\mathrm{GR}}_{\mathrm{ma}}\left(1-{\phi}_{\mathrm{T}}\right)}{{\mathrm{GR}}_{\mathrm{dc}}-{\mathrm{GR}}_{\mathrm{ma}}} $$

Solving for ϕ _e and S _we

Once V _dc and ϕ _T are established, the following relationships hold:

ϕ _e = ϕ _T–V _wb (which also = V _hy + V _wf),

V _ma = 1 − ϕ _T–V _dc, and V _wb = αV _dc,

where α has been established in the shales as ϕ _Tsh /(1 − ϕ _Tsh).

The response of the pulsed neutron log itself can be written as:

$$ \varSigma ={\varSigma}_{\mathrm{ma}}{V}_{\mathrm{ma}}+{\varSigma}_{\mathrm{dc}}{V}_{\mathrm{dc}}+{\varSigma}_{\mathrm{wb}}\alpha {V}_{\mathrm{dc}}+{\varSigma}_{\mathrm{wf}}{V}_{\mathrm{wf}}+{\varSigma}_{\mathrm{hy}}{V}_{\mathrm{hy}}, $$

hence:

$$ {\varSigma}_{\mathrm{hy}}{V}_{\mathrm{hy}}+{\varSigma}_{\mathrm{wf}}{V}_{\mathrm{wf}}=\varSigma -{\varSigma}_{\mathrm{ma}}{V}_{\mathrm{ma}}\hbox{--} {V}_{\mathrm{dc}}\left({\varSigma}_{\mathrm{dc}}+\alpha {\varSigma}_{\mathrm{wb}}\right). $$

The right side of the equation can be evaluated since all the parameters and variables have now been defined. If this quantity is, in fact, Σ*, then

$$ {V}_{\mathrm{wf}}=\frac{\varSigma^{\ast }-{\phi}_{\mathrm{e}}{\varSigma}_{\mathrm{hy}}}{\varSigma_{\mathrm{wf}}-{\varSigma}_{\mathrm{hy}}} $$

By definition,

$$ {S}_{\mathrm{we}}={V}_{\mathrm{wf}}/{\phi}_{\mathrm{e}}. $$

Appendix 2: Radioactive Elements, Minerals, and Rocks

Table 11.5 Natural gamma ray emitters

Table 11.6 Gamma ray lines^a in the spectra of the important naturally occurring radionuclides

Table 11.7 Thorium-bearing minerals

Table 11.8 Uranium minerals

Table 11.9 Potassium, Uranium, and Thorium distribution in rocks and minerals

Table 11.10 Geological significance of natural gamma ratios