Question 15 OSE01 - Chief Engineer - OSV

The Correct Answer is B **Explanation for why option B ("2 psig") is correct:** The maximum allowable pressure drop ($\Delta P$) across the fuel primary strainer is given as 2 psig. The formula for pressure drop is: $$\Delta P = P_{inlet} - P_{outlet}$$ We are given: * Maximum allowable pressure drop ($\Delta P_{max}$) = 2 psig * Strainer inlet pressure ($P_{inlet}$) = 4 psig To find the minimum allowable outlet pressure ($P_{outlet, min}$), we use the maximum allowable pressure drop: $$P_{outlet, min} = P_{inlet} - \Delta P_{max}$$ $$P_{outlet, min} = 4 \text{ psig} - 2 \text{ psig}$$ $$P_{outlet, min} = 2 \text{ psig}$$ Therefore, the minimum allowable outlet pressure before recommended servicing is 2 psig. **Explanation for why the other options are incorrect:** * **A) 2" Hg:** This value uses incorrect units (inches of Mercury, " Hg, which measures vacuum or differential pressure in low-pressure systems) and does not represent the minimum absolute pressure required. Furthermore, 2" Hg is a very small pressure (approximately 0.1 psig) and is not the result of the calculation. * **C) 6 psig:** This value would be the result if the pressure drop were *added* to the inlet pressure (4 psig + 2 psig = 6 psig). Since a strainer causes a pressure *drop*, the outlet pressure must be lower than the inlet pressure. * **D) 6" Hg:** This value uses incorrect units and is not the result of the calculation. 6" Hg is a low pressure value (approximately 0.3 psig) and does not represent the minimum positive pressure required on the outlet side.