1. The constant 162.6 appears in the field units form of the transient flow equation for pressure drop in a reservoir: $$p_{wf} = p_i - 162.6 \frac{q \mu B_o}{kh} \left( \log t + \log \left( \frac{k}{\phi \mu C_r^2 w} \right) - 3.23 + 0.87 S \right)$$ 2. This constant arises from converting the original equation from SI units to field units (oilfield units) and from the natural logarithm to the base-10 logarithm. 3. Starting from the original transient flow equation in SI units: $$p_{wf} = p_i - \frac{q \mu}{4 \pi k h} \ln \left( \frac{4 k t}{\phi \mu C_r^2 w} + 2 S \right)$$ 4. To convert the natural logarithm $\ln$ to the common logarithm $\log_{10}$, use the relation: $$\ln x = 2.303 \log_{10} x$$ 5. Substituting this into the equation: $$p_{wf} = p_i - \frac{q \mu}{4 \pi k h} \times 2.303 \log_{10} \left( \frac{4 k t}{\phi \mu C_r^2 w} + 2 S \right)$$ 6. The factor $\frac{2.303}{4 \pi}$ simplifies numerically: $$\frac{2.303}{4 \pi} = \frac{2.303}{12.566} \approx 0.1833$$ 7. When converting to field units, the flow rate $q$ is in STB/day, permeability $k$ in millidarcies, viscosity $\mu$ in centipoise, thickness $h$ in feet, and pressure in psi. The unit conversion factors combine to give a multiplier of approximately 888.6. 8. Multiplying $0.1833$ by 888.6 gives: $$0.1833 \times 888.6 \approx 162.6$$ 9. Therefore, the constant 162.6 is the product of the conversion from natural to common logarithm and the unit conversion factors to field units. 10. This constant allows the transient flow equation to be expressed conveniently in field units with base-10 logarithms, making it easier to analyze well test data. Final answer: The constant 162.6 comes from the factor $\frac{2.303}{4 \pi}$ multiplied by unit conversion factors to field units, approximately $0.1833 \times 888.6 = 162.6$.