Reservoir Temperature Direct Method Analysis

Evaluate the data in the Direct Methodgraph illustrated in Figure 7-5. These data were measured on a GRI

#2 sample 35-1 that was heated to reservoir temperature. The measured and residual gas contents for this sample were 180.4 and 0 scf/ton, respectively. The time that the core barrel started out of the well was 12/2/90 19:30:00.

The sample was placed into a heated water bath at 12/2/90 22:17:00 at the same time that the canister was sealed.

The first measured point was at 22:37.

The procedure for this problem is as follows. Example 3-5 shows you how to perform the calculations.

1. Prepare a graph of the cumulative desorbed gas content in the right-most column vs. the square root of the desorption time.

This graph has been prepared for you in Figure 7-5.

2. Estimate the temperature recovery time.

This is the elapsed time between the time that the core barrel is lifted off bottom until the canister is placed in a water bath. (22:17)

3. Estimate the end of the temperature recovery time by adding the temperature recovery time to the time that the canister is sealed. The canister was sealed at 12/2/90 22:17:00.

4. Calculate the Direct Method horizontal axis value at the end of the temperature recovery time. The first measured point was at 22:37.

(

i 0

)

dc

DM t t t

t = -

-Where:

t_DM Direct Method horizontal axis value (square root of desorption time), hours^½ t_i measured desorption time i, hours

t_o time zero, hours

t_dc desorption time correction

5. Fit a straight line to the proper portion of the data required for the lost gas estimate.

This line should start near the end of the temperature recovery time for reservoir temperature measurements.

6. Compute the lost gas content from the absolute value of the y-axis intercept.

7. Compute the total gas content by summing the lost gas content, the measured gas content (180.4 scf/ton).

8. Compute the diffusivity from the slope using Equation 3-8.

(3-8) Where:

D/r² = diffusivity, sec^-1

m = slope of the desorbed gas content vs. square root time graph, scf/ton-hour0.5 G_cad = total gas content, scf/ton

9. Compute the sorption time from the diffusivity for a spherical fracture geometry with Equation 3-9.

The value of the shape factor is 15 for a sphere.

(3-9) Where:

τ = sorption time, hours α = shape factor, dimensionless

2

cad 2 203.1G

m r

D ÷÷ø

çç ö è

=æ

r2

3600 D 1

a t =

Temperature Recovery Time

The temperature recovery time is the elapsed time from the time that the sample leaves bottom until the canister is placed in a water bath. The sample started out of the well at 12/2/90 19:30 (19.500 hours). The canister was sealed at 22:17 (22.283 hours). The temperature recovery time is the time difference of 2.783 hours. The end of the temperature recovery time occurred 2.783 hours after sealing or at 22.283+2.783=25.066 hours. This is on the following day (12/30/90) at 1:03:58 A.M. Time zero was at 19:39:08 (19.652 hours). The elapsed time at the end of the temperature recovery time from time zero was 25.066-19.652=5.414 hours. The first measured point was 20 minutes (0.333 hours) after the time of sealing; therefore the desorption time correction is 0.333. The Direct Method horizontal axis value at the end of the temperature recovery time is as follows.

Fit a Straight Line to the Proper Data Interval

The Direct Method graph for sample 35-1 has the appearance of reservoir temperature desorption data since the S-shape characteristic of heated canisters is present. We expect the start of the correct straight line to be near 2.25 hours ^1/2 on this graph. Therefore, select the points between 2.25 and 2.5 that appear on a straight line. The intercept and slope for this line are -208 scf/ton and 115 scf/ton-hour 1/2, respectively.

Figure 7-6. Problem 7-6 Direct Method Graph Solution.

The lost gas content is the absolute value of the intercept, 208.0 scf/ton. The measured gas content is 180.4 scf/ton. Residual gas content was reported to be zero. The total gas content is the sum of the lost, measured, and residual gas contents, i.e., 208.0+180.4+0.0=388.4 scf/ton.

Diffusivity Estimate

The diffusivity is computed from the slope with Equation 3-8.

Sorption Time Estimate

Sample Number

Ash Content

Moisture Content

Inorganic Content

Total Air-Dry Gas Content

Total Dry, Ash-Free

Gas Content

Lost Gas Content

Measured Gas Content

Residual Gas Content

Diffusivity Sorption Time

fraction fraction fraction scf/ton scf/ton scf/ton scf/ton scf/ton 1/sec hours 34-1 0.4845 0.0696 0.5541 228.8 513.1 76.0 152.8 0.0 1.26E-06 14.7 34-2 0.3948 0.1308 0.5256 267.3 563.4 45.5 221.7 0.0 2.81E-07 65.8 34-3 0.1098 0.0958 0.2056 529.3 666.2 92.9 436.4 0.0 2.70E-07 68.5 34-4 0.2846 0.1190 0.4036 339.4 569.2 75.9 263.6 0.0 5.79E-07 32.0 34-5 0.2047 0.0823 0.2870 443.2 621.6 72.5 370.7 0.0 3.13E-07 59.1 Average 0.2957 0.0995 0.3952 361.6 586.7 72.5 289.1 0.0 5.41E-07 48.0

Problem 7-7. Multiple Sample Analysis

Evaluate the data in Table 7-5 to determine the organic fraction gas content estimate for the Valencia Canyon 32-1 well.

Table 7-5. Valencia Canyon 32-1 Gas Desorption Summary.

The procedure for this problem is as follows. Figure 4-1 of this book illustrates a typical analysis graph.

1. Construct a graph of the total air-dry gas content vs. the inorganic content for each sample.

This graph has been prepared for you in Figure 7-7.

2. Fit a line to the data and extrapolate the line to a zero inorganic content value.

The value of the intercept is the organic fraction gas content.

0 100 200 300 400 500 600 700 800 900 1,000

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Ash + Moisture Content, fraction

Problem 7-7. Solution

Figure 7-8 illustrates the analysis graph for the five Valencia Canyon 32-1 desorption samples.

0 100 200 300 400 500 600 700 800 900 1,000

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Ash + Moisture Content, fraction

Regression Line

Upper 95% Confidence Interval Lower 95% Confidence Interval Individual Sample Data

Figure 7-8. Problem 7-8 Gas vs. Inorganic Content Solution Graph.

The intercept and slope of the solid line are 687.5 scf/ton and -824/9 scf/ton, respectively. The organic fraction gas content is equal to the intercept value.

Compute the density corresponding to the inorganic contents listed in the following table. Assume that the equilibrium moisture content is equal to 1.74%. Use ash and organic fraction densities of 2.497 and 1.295 g/cm³, respectively.

Ash Content Density fraction g/cm³

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Table 7-6. Problem 7-7 Ash Content Values.

Solve this problem by computing the density with Equation 4-2.

(4-2) Where:

ρ ⁼ density, g/cm³ ρ_a = ash density, g/cm³

ρ_o ⁼ organic fraction density, g/cm³ ρ_w ⁼ sorbed water density, g/cm³ w_a = ash content, weight fraction w_w = moisture content, weight fraction

( )

¹

w w o

w a a

a 1 w w w

w

-úû ê ù

ë

é + - + +

= r r r

r

Problem 7-8. Solution

At an ash content of 0.5, the solution is as follows.

( )

¹

( )

¹

w w o

w a a

a

1 0174 . 0 295

. 1

0174 . 0 5 . 0 1 497 . 2

5 . w 0

w w 1

w ^-

-úûù

êëé + - + +

ú = û ê ù

ë

é + - + +

= r r r

r

[

^{0 .2002}+^0.3727+^0.0174

]

¹ =^{1. 694g/cm3}

=

-r

The results for all of the data are listed in Table 7-7.

Ash Content Density fraction g/cm³

0.0 1.288 0.1 1.353 0.2 1.425 0.3 1.505 0.4 1.594 0.5 1.694 0.6 1.808 0.7 1.938 0.8 2.089 0.9 2.264 1.0 2.473

Table 7-7. Problem 7-7 Solution.

Note that the relationship between density and ash content is a hyperbola as illustrated by Figure 7-9.

1.0 1.5 2.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Inorganic Content, fraction

Density, g/cc

In document GRI - CBM Gas in Placasdfe Analysis (Page 99-110)