Transient Radial Flow — Bottom-Hole Pressure After 2 Days: 24-Pet-A3 Fundamental Reservoir Engineering, December 2015, Q3

Question 3: Transient Radial Flow — Bottom-Hole Pressure After 2 Days (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. $r_e=3000$ ft, $c_t=6\times10^{-6}$ psi$^{-1}$, $\mu=2$ cP, $B_o=1.25$ bbl/STB, $k=100$ mD, $h=100$ ft, $p_i=3000$ psia, $\phi=0.20$, $r_w=0.33$ ft, $q=250$ STBD, $t=2$ days.

Find. The bottom-hole flowing pressure $p_{wf}$ after 2 days of production.

Check: no skin factor is stated for this well, so $s=0$ (undamaged well) is assumed.

Approach. Since $r_e$ is large relative to the flow radius reached in 2 days, treat the well as an infinite-acting line source: compute the dimensionless time $t_D$, confirm the semi-log ($\ln$) approximation is valid ($t_D>100$), obtain $p_D$, then apply the transient inflow equation.

Dimensionless time. $t_D=\dfrac{6.33\,k(\text{D})\,t(\text{day})}{\phi\mu c_t r_w^2}=\dfrac{6.33(0.100)(2)}{(0.20)(2)(6\times10^{-6})(0.33^2)}=\boxed{4.84\times10^{6}}$, far above the $t_D>100$ threshold for the log approximation.
Dimensionless pressure. $p_D=\tfrac12(\ln t_D+0.809)=\tfrac12(\ln(4.84\times10^6)+0.809)=\tfrac12(15.393+0.809)=\boxed{8.101}$.
Pressure drop and $p_{wf}$. $\Delta p=\dfrac{141.2\,q\mu B_o}{kh}(p_D+s)=\dfrac{141.2(250)(2)(1.25)}{(100)(100)}(8.101+0)=8.825\times8.101=71.5$ psi. $\boxed{p_{wf}=p_i-\Delta p=3000-71.5=2928.5\ \text{psia}}$.

Quantity	Value
Dimensionless time, $t_D$	$4.84\times10^{6}$
Dimensionless pressure, $p_D$	8.101
Pressure drop, $\Delta p$	71.5 psi
Bottom-hole flowing pressure, $p_{wf}$ (2 days)	2928.5 psia

Concept & Theory

Topic: Transient (infinite-acting) radial flow, line-source solution.

Key relations: $t_D=\dfrac{0.0002637\,k\,t}{\phi\mu c_t r_w^2}$ (field units, $k$ in md, $t$ in hours), and for $t_D>100$ the exact Ei-function solution collapses to $p_D=\tfrac12(\ln t_D+0.80907)$; $\Delta p=\dfrac{141.2\,q\mu B}{kh}(p_D+s)$.

Why this works: $t_D\approx4.8\times10^6$ means the pressure transient has been diffusing for far longer than the exponential-integral solution needs to settle into its logarithmic asymptote — using the full $-Ei(-1/4t_D)$ form here would change $p_D$ by a negligible amount.

Common pitfall: Applying the log approximation when $t_D<100$ (early-time transient), where the exact line-source Ei-function form is required instead — always check $t_D$ before choosing which form to use.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 6 (fundamentals of transient well testing, line-source solution).

Question 3 of 7: Transient Radial Flow — Bottom-Hole Pressure After 2 Days

Question 3: Transient Radial Flow — Bottom-Hole Pressure After 2 Days (20 marks)