1. The problem asks to derive the field units equation for $p_{wf}$ starting from the dimensionless equation: $$p_{wf} = p_i - \frac{q \mu}{4 \pi k h} \ln \left( \frac{4 k t}{\gamma \phi \mu C_r w^2} + 2 S \right)$$ 2. We start by understanding the variables and units involved: - $p_{wf}$ and $p_i$ are pressures (psi or Pa) - $q$ is flow rate (STB/day or m^3/s) - $\mu$ is viscosity (cp or Pa.s) - $k$ is permeability (md or m^2) - $h$ is thickness (ft or m) - $t$ is time (hours or seconds) - $\phi$ is porosity (dimensionless) - $C_r$ is compressibility (psi^-1 or Pa^-1) - $w$ is wellbore radius (ft or m) - $S$ is skin factor (dimensionless) 3. The goal is to convert the coefficient $\frac{q \mu}{4 \pi k h}$ into field units coefficient $\frac{162.6 q \mu B_o}{k h}$ and rewrite the logarithmic term accordingly. 4. The constant $162.6$ arises from unit conversions and the factor $4 \pi$ in the denominator: $$\frac{1}{4 \pi} \approx 0.0796$$ Multiplying by conversion factors for flow rate, viscosity, and formation volume factor $B_o$ (dimensionless or close to 1) leads to the empirical constant 162.6. 5. The logarithmic argument is split using logarithm properties: $$\ln \left( \frac{4 k t}{\gamma \phi \mu C_r w^2} + 2 S \right) \approx \log t + \log \left( \frac{k}{\phi \mu C_r w^2} \right) - 3.23 + 0.87 S$$ where $\log$ is base 10 logarithm and constants $-3.23$ and $0.87$ come from empirical adjustments and unit conversions. 6. Thus, the field units equation becomes: $$p_{wf} = p_i - 162.6 \frac{q \mu B_o}{k h} \left( \log t + \log \frac{k}{\phi \mu C_r w^2} - 3.23 + 0.87 S \right)$$ 7. This equation is used to analyze transient flow data by plotting $p_{wf}$ versus $\log t$ to find slope $m = 162.6 \frac{q \mu B_o}{k h}$ and estimate permeability $k$ and skin factor $S$. Final answer: $$p_{wf} = p_i - 162.6 \frac{q \mu B_o}{k h} \left( \log t + \log \frac{k}{\phi \mu C_r w^2} - 3.23 + 0.87 S \right)$$