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Exams > 2020 Q33 — Containment TS Pressure and Temperature Limits

2020 Q33 — Containment TS Pressure and Temperature Limits

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Q33 — Containment TS Pressure and Temperature Limits 007000K3.01 (3.3)
Given:

• Unit 2 is at 100% Power.
• Containment pressure is 0 psig.
• Containment temperature is 99 °F.

Subsequently, the following sequence of events occurs:

• A load rejection results in a reactor trip.
• Following the trip, a Pressurizer Safety Valve opens, and does not completely reseat.
• The PRT rupture disk has relieved to the containment.
• Containment pressure is rising at 0.1 psig every 5 minutes.
• Containment temperature is rising at 1 °F every 5 minutes.

Assuming containment pressure and temperature trends remain constant, which ONE of the following Containment Technical Specification LCO(s), if any, will NOT BE MET ONE HOUR from now?
A. BOTH LCO 3.6.1.4, Containment Internal Pressure, and LCO 3.6.1.5, Containment Air Temperature will be exceeded.
B. ONLY LCO 3.6.1.4, Containment Internal Pressure, will be exceeded.
C. ONLY LCO 3.6.1.5, Containment Air Temperature, will be exceeded.
D. NEITHER LCO 3.6.1.4, Containment Internal Pressure, and LCO 3.6.1.5, Containment Air Temperature will be exceeded.
▶ Show Answer & Explanation
✓ B. Correct. The containment pressure tech spec limit has been exceeded, (0+1.2 = 1.2 psig) greater than 0.3 psig. The containment temperature limit has not been exceeded after one hour, (99 + 12 = 111°F) less than 120°F.
✗ A. Incorrect. Plausible because the containment pressure tech spec limit has been exceeded, (0+1.2 = 1.2 psig) greater than 0.3 psig. Additionally plausible if the candidate believes that the tech spec limit for air temperature is 110°F (99 + 12 = 111°F). Incorrect because the temperature limit is 120°F.
✗ C. Incorrect. Plausible because the candidate may believe that the containment pressure limit is 1.5 psig because the negative pressure limit is -1.5 psig (0 + 1.2 < 1.5). Also plausible if the candidate believes that the tech spec limit for air temperature is 110°F (99 + 12 = 111°F). Incorrect because the temperature limit is 120°F and the pressure limit is 0.3 psig.
✗ D. Incorrect. Plausible because the containment temperature limit has not been exceeded after one hour, (99 + 12 = 111°F) less than 120°F. Candidate may also believe that the containment pressure limit is 1.5 psig because the negative pressure limit is -1.5 psig (0 + 1.2 < 1.5). Incorrect because the pressure limit is 0.3 psig.
Ref: S2.OP-SO.CBV-002(Q), Containment Pressure – Vacuum Relief System Operation and Technical Specifications 3.6.1.4 & 3.6.1.5 | LO: N/A | Source: Bank – Robinson 2016 NRC Exam – Q35 | Cognitive: Comprehension

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