Question 16 DDE01 - Designated Duty Engineer - Unlimited HP

The Correct Answer is A. **Explanation for Option A (2 psig):** The problem provides the following information: 1. Maximum allowable pressure drop ($\Delta P$) across the strainer: 2 psig. 2. Strainer inlet pressure ($P_{in}$): 4 psig. The pressure drop is defined as the difference between the inlet pressure and the outlet pressure: $$\Delta P = P_{in} - P_{out}$$ We are looking for the minimum allowable outlet pressure ($P_{out, min}$) which corresponds to the maximum allowable pressure drop ($\Delta P_{max}$). $$P_{out, min} = P_{in} - \Delta P_{max}$$ Substituting the given values: $$P_{out, min} = 4 \text{ psig} - 2 \text{ psig}$$ $$P_{out, min} = 2 \text{ psig}$$ Therefore, the minimum allowable outlet pressure before recommended servicing is 2 psig. **Explanation of Incorrect Options:** * **B) 2" Hg:** This answer uses the incorrect units ("Hg" instead of "psig") and represents a pressure value significantly lower than 2 psig (1 psig $\approx$ 2.04 "Hg), making it dimensionally and numerically incorrect in the context of the required result. * **C) 6 psig:** This value represents the result of adding the pressure drop to the inlet pressure ($4 \text{ psig} + 2 \text{ psig}$), which would imply a pressure increase across the strainer (an impossibility), or it would be the inlet pressure if the outlet pressure were 2 psig (which is the correct answer). * **D) 6" Hg:** This answer uses incorrect units ("Hg" instead of "psig") and the incorrect calculation (adding the values), making it both dimensionally and numerically incorrect.