Multi-rate test – pressure drop at an interference well: 24-Pet-B5 Reservoir Mechanics, December 2015, Q6

Question 6: Multi-rate test – pressure drop at an interference well (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A three-step rate schedule at the active well (200 STBD, $0\le t<24$ hr; 400 STBD, $24\le t<48$ hr; 600 STBD, $48\le t<72$ hr), an observation well $r=500$ ft away, and the reservoir/fluid properties in the table above. The table's own listed "well oil production rate, $q=300$ STBD" does not match the three-step schedule described in the question text and is not used below; likewise $p_e$ and $r_e$ describe the reservoir's outer boundary condition but are not needed for an interference calculation this far from the boundary over only 3 days (the boundary is not felt on this timescale, confirmed in Step 1).

Find. $\Delta p$ at the observation well, 500 ft away, at the end of the third day ($t=72$ hr).

Approach. Superposition in time: treat the schedule as three rate increments ($\Delta q_1=200$ STBD starting at $t=0$, $\Delta q_2=200$ STBD starting at $t=24$ hr, $\Delta q_3=200$ STBD starting at $t=48$ hr) and sum each increment's own line-source pressure-drop contribution, each evaluated at its own elapsed time since it began.

Step-rate production schedule at the active well (Q6). The observation well sits 500 ft away, silent throughout, recording only the pressure interference from these three rate increments.

Confirm infinite-acting (boundary not felt). At $r_e=10{,}000$ ft the dimensionless time after 72 hr is $t_D=0.0002637kt/(\phi\mu c_or_e^2)\approx3\times10^{-4}\ll100$ – the boundary is nowhere near being felt at the observation-well timescale, so the infinite-acting line-source solution applies to every term.
Dimensionless pressure at $r=500$ ft for each rate increment. Using the exact line-source form $p_D=0.5\,E_1\!\big(948\,\phi\mu c_or^2/(k\,\Delta t)\big)$ (needed since $t_D<100$ at this distance for every term, not the log approximation):

Increment $\Delta q$ (STBD) Elapsed $\Delta t$ (hr) $p_D$

1 (start $t=0$) 200 72 0.687

2 (start $t=24$ hr) 200 48 0.522

3 (start $t=48$ hr) 200 24 0.280
Superpose. $$\Delta p=\frac{141.2\,B_o\mu}{kh}\sum_i\Delta q_i\,p_{D,i}=\frac{141.2(1.32)(0.44)}{25(43)}\Big[200(0.687)+200(0.522)+200(0.280)\Big]$$ $$\boxed{\Delta p\approx 22.7\text{ psi}}$$

Increment	$\Delta q$ (STBD)	Elapsed $\Delta t$ (hr)	$p_D$
1 (start $t=0$)	200	72	0.687
2 (start $t=24$ hr)	200	48	0.522
3 (start $t=48$ hr)	200	24	0.280

Quantity	Result
$p_D$ contributions (Step 1, 2, 3)	0.687, 0.522, 0.280
Pressure drop at observation well, $\Delta p$ (end of day 3)	22.7 psi

Topic. Superposition in time for a multi-rate interference test.

Key relations. $p_D(r,\Delta t)=0.5\,E_1\big(948\,\phi\mu c_tr^2/(k\Delta t)\big)$; superposition $\Delta p=141.2B_o\mu/(kh)\sum\Delta q_i\,p_{D,i}(t-t_i)$.

Why this works. The diffusivity equation is linear, so the pressure response to a step-changing rate history is exactly the sum of the responses to each individual rate increment, each one turned on at its own start time and left running to the observation time – not the response to the final rate alone.

Common pitfall. Using every increment's elapsed time as the same (total) 72 hr, or using the total rate 600 STBD as a single term instead of decomposing into $\Delta q$ increments – both shortcuts silently drop the different exposure times of the earlier, smaller rate steps.

Source. Earlougher, Advances in Well Test Analysis, ch.2 (superposition in time, multi-rate and interference tests); Lee, Well Testing, ch.2.

Question 6 of 7: Multi-rate test – pressure drop at an interference well

Question 6: Multi-rate test – pressure drop at an interference well (20 marks)