problem 2\na waterflood is under consideratio...
problem 2\na waterflood is under consideration for a narrow \shoestring\ reservoir that is 100 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:\nkro=(1 - swd)^2.0 and krw = 0.78swd^3.75, where\nswd=\nsw - swi\n1 - swi\nswd 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: Identify given parameters
Reservoir width \(w = 300\) ft, thickness \(h=20\) ft, length \(L = 1000\) ft, porosity \(\phi=0.15\), initial - water saturation \(S_{wi}=0.363\), residual - oil saturation \(S_{or}=0.205\), injection rate \(Q = 338\) bbl/D, oil viscosity \(\mu_o=2.0\) cp, water viscosity \(\mu_w = 1.0\) cp, formation - volume factors \(B_o=B_w = 1.0\), base permeability \(k\) (absolute permeability to oil at interstitial - water saturation).
## Step2: Calculate pore - volume of the reservoir
The pore - volume \(V_p\) of the reservoir is given by \(V_p=\phi\times w\times h\times L\).
\[V_p = 0.15\times300\times20\times1000=900000\] ft³. Since \(1\) bbl \( = 5.615\) ft³, \(V_p=\frac{900000}{5.615}\approx160285\) bbl.
## Step3: Calculate relative - permeability equations
The relative - permeability of water \(k_{rw}=(1 - S_{wn})^{2.0}\) and relative - permeability of oil \(k_{ro}=0.78S_{wn}^{3.75}\), where \(S_{wn}=\frac{S_w - S_{wi}}{1 - S_{wi}-S_{or}}\).
## Step4: Calculate fractional flow of water \(f_w\)
The fractional flow of water is given by \(f_w=\frac{1}{1+\frac{k_{ro}\mu_w}{k_{rw}\mu_o}}\). Substitute \(k_{rw}\) and \(k_{ro}\) into the \(f_w\) formula and calculate \(f_w\) for different values of \(S_w\) to draw the fractional - flow curve.
## Step5: Determine initial and flood - front saturation from the fractional - flow curve
To find the initial water saturation \(S_{wi}\) and flood - front saturation \(S_{wf}\), draw a tangent to the fractional - flow curve. The initial water saturation is the given \(S_{wi}=0.363\). The flood - front saturation \(S_{wf}\) is the water saturation at the point of tangency of the tangent line to the fractional - flow curve.
## Step6: Calculate oil displacement rate and cumulative oil displaced
The oil displacement rate \(q_o\) and cumulative oil displaced \(N_p\) can be calculated using the following relationships based on the fractional - flow theory and injection rate \(Q\). The oil displacement rate \(q_o=(1 - f_w)Q\). The cumulative oil displaced \(N_p=\int_{0}^{t}q_o dt\).
# Answer:
The fractional - flow curve needs to be plotted by calculating \(f_w\) for different \(S_w\) values using the relative - permeability equations. The initial water saturation is \(S_{wi}=0.363\), and the flood - front saturation is determined from the tangent to the fractional - flow curve. The oil displacement rate \(q_o=(1 - f_w)Q\) and cumulative oil displaced \(N_p=\int_{0}^{t}q_o dt\) are functions of injection time \(t\) and can be calculated once \(f_w\) is known as a function of \(S_w\).