problem 2\na waterflood is under consideratio...
problem 2\na waterflood is under consideration for a narrow \shoestring\ reservoir that is 300 ft wide, 20 ft thick, and 1,000 ft long. the reservoir is horizontal and has a porosity of 0.15 and an initial water saturation of 0.363, which is considered immobile. it is proposed to drill a row of injection wells at one end of the reservoir and flood the reservoir by injecting water at a rate of 338 b/d. viscosities of oil and water are 2.0 and 1.0 cp, respectively.\nestimates of oil displacement rate and cumulative oil displaced are required as functions of injection time.\nthe residual oil saturation is 0.205. base permeability is the absolute permeability to oil at interstitial water saturation, which is assumed to be equal to the absolute permeability. oil and water formation volume factors are 1.0.\n\nproblem 2 (continued)\nrelative permeability data corresponding to to the displacement of oil by water are calculated with the following correlations: (k_{ro}=(1 - s_{wd})^{2.56}) and (k_{rw}=0.75s_{wd}^{3.75}), where (s_{wd}=\frac{s_w - s_{wi}}{1 - s_{or}-s_{wi}})\n(s_{wd}) represents the normalized water saturation.\ncomplete the following:\na) draw the fractional flow curve for the system.\nb) identify the initial water saturation and the flood - front saturation based on drawing the tangent to the fractional flow curve.
Answer
# Explanation:
## Step1: Analyze given reservoir data
Reservoir width $w = 300$ ft, thickness $h=20$ ft, length $L = 1000$ ft, porosity $\phi=0.15$, initial water - saturation $S_{wi}=0.363$, injection rate $Q = 338$ B/D, oil viscosity $\mu_o=2.0$ cp, water viscosity $\mu_w = 1.0$ cp, residual oil saturation $S_{or}=0.205$, and formation - volume factors $B_o=B_w = 1.0$. Also, relative - permeability correlations $k_{ro}=(1 - S_{wD})^{2.56}$ and $k_{rw}=0.75S_{wD}^{3.75}$ where $S_{wD}=\frac{S_w - S_{wi}}{1 - S_{wi}-S_{or}}$.
## Step2: Calculate fractional flow formula
The fractional flow of water $f_w$ is given by $f_w=\frac{1}{1+\frac{\mu_w}{\mu_o}\frac{k_{ro}}{k_{rw}}}$. Substitute $k_{ro}$ and $k_{rw}$ into the formula:
\[
\begin{align*}
f_w&=\frac{1}{1+\frac{\mu_w}{\mu_o}\frac{(1 - S_{wD})^{2.56}}{0.75S_{wD}^{3.75}}}\\
\end{align*}
\]
We need to vary $S_w$ from $S_{wi}$ to $1 - S_{or}$ to calculate $S_{wD}$ and then $f_w$.
## Step3: Draw fractional - flow curve
Plot $f_w$ against $S_w$. For different values of $S_w$ in the range $[S_{wi},1 - S_{or}]$, calculate $S_{wD}$, then $k_{ro}$, $k_{rw}$, and finally $f_w$.
## Step4: Identify initial and flood - front saturation
To find the initial water saturation $S_{wi}$, it is already given as $0.363$. To find the flood - front saturation $S_{wf}$, we draw a tangent to the fractional - flow curve at the origin. The water saturation at the point of tangency is the flood - front saturation. Mathematically, we can also use the method of characteristics for one - dimensional displacement problems. The flood - front saturation $S_{wf}$ satisfies the condition $\frac{df_w}{dS_w}\big|_{S_w = S_{wf}}$ is the maximum slope of the fractional - flow curve.
# Answer:
a) The fractional - flow curve is drawn by calculating $f_w$ for different values of $S_w$ in the range $[0.363,1 - 0.205]=[0.363,0.795]$ using the above - derived formula for $f_w$.
b) Initial water saturation $S_{wi}=0.363$. The flood - front saturation $S_{wf}$ is found by drawing the tangent to the fractional - flow curve at the origin and reading the $S_w$ value at the point of tangency.