Unformatted text preview: orner straight line
corresponding to ∆t = 1 hr. Form this, the following equation drops out: In the above equation t represents the shut-in time, which is the effective time of 36.48 hrs.
pressure at the end of the flow period corresponding to ∆t = 0 hrs. This is as 2970 psi.
represents a point on the Horner straight line corresponding to ∆t = 1 hr. This is found as follows: This point is then found on the straight Horner line: 14 Horner Plot - Case Study 6
3550 3500 pw 3450 3400 3350 3300 3 250
1 10 100 1000 (t+∆t)/∆t Therefore, from the above,
for skin: . Inputting these values into the equation and solving A close look at the data on the Horner plot clearly reveals the presence of a nearly by fault. We can see
the pressure values on a single slope (the shifted Horner line) during early times. At later times, the
pressures transition to falling on a new straight line with double the slope of the previous line. Doubling
slope is a clear indication of a sealing fault close to a well. Knowing this, we can use the equations
derived in part 1 of this case study to determine the distance to this sealing fault. 15 Using the equations as determined in the initial derivations and explained in step 7 we can now
calculate the distance to the fault. Specifically, we I will use the Shifted Horner straight line for early
times since the formula contains R, the distance to the fault. The formula requires us to pick a point on
the shifted Horner line. Since previously we have used a point corresponding to ∆t = 1 hr, I will use this
point again in this analysis. The corresponding Horner ratio is: Which corresponds to
following equation allows us to solve for R: , as shown in the previous plot. Inputting this into the To solve this I will first try to use the
see if the approximation is valid: approximation, solve for R, and then check to For the assumption to be valid X must be less than 0.01. Checking this assumption: Therefore the assumption is invalid and another method must be utilized. Thankfully, the function
“expint” in matlab evaluates the exponential int...
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- Spring '12