Question 43

The Correct Answer is A. ### Explanation for Option A (2 psig) This problem is a simple calculation involving pressure drop. 1. **Given Maximum Pressure Drop ($\Delta P_{max}$):** The maximum allowable pressure drop across the strainer is $2 \text{ psig}$. 2. **Given Inlet Pressure ($P_{in}$):** The pressure at the strainer inlet is $4 \text{ psig}$. 3. **Relationship:** Pressure drop ($\Delta P$) is defined as the difference between the inlet pressure and the outlet pressure ($P_{out}$): $$\Delta P = P_{in} - P_{out}$$ 4. **Finding the Minimum Allowable Outlet Pressure ($P_{out, min}$):** To find the *minimum* allowable outlet pressure, we must use the *maximum* allowable pressure drop ($2 \text{ psig}$), as servicing is required once this drop is reached. $$P_{out, min} = P_{in} - \Delta P_{max}$$ $$P_{out, min} = 4 \text{ psig} - 2 \text{ psig}$$ $$P_{out, min} = 2 \text{ psig}$$ Therefore, the minimum allowable outlet pressure before recommended servicing is $2 \text{ psig}$. ### Explanation of Why Other Options Are Incorrect * **B) 2" Hg:** This option uses the wrong unit. Pressure drop and system pressure are given in psig (pounds per square inch gauge), which is standard for measuring moderate fluid system pressures. Inches of mercury (" Hg) is typically used for measuring vacuum or very low pressures, but it is not directly comparable to the given psig values without conversion, and it does not represent the correct numerical answer. * **C) 6 psig:** This option incorrectly *adds* the pressure drop to the inlet pressure ($4 \text{ psig} + 2 \text{ psig} = 6 \text{ psig}$). The outlet pressure must always be less than the inlet pressure in a flow system due to friction and restriction (pressure drop). * **D) 6" Hg:** This option is incorrect because it uses the wrong unit (" Hg) and incorrectly adds the numerical values.