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[Hardware] Seeing CDCS Reliable Test Cram - No Worry About EXIN EPI Certified Data Centre S

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【Hardware】 Seeing CDCS Reliable Test Cram - No Worry About EXIN EPI Certified Data Centre S

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EXIN EPI Certified Data Centre Specialist Sample Questions (Q105-Q110):NEW QUESTION # 105
EMF shielding material needs to be installed as EMF levels from the transformer room into the computer room are measured at 100 mG. The transformer room is ~10 meters away, separated by a corridor. Where should shielding be installed?
  • A. Shielding is not required as 100 mG is within acceptable levels
  • B. As close as possible to the computer room
  • C. As close as possible to the transformer room
  • D. It does not matter; either close to the transformer room or computer room is okay
Answer: C
Explanation:
The most effective EMF mitigation is to install shielding as close as possible to the source of radiation. By blocking or redirecting magnetic flux at the origin (the transformer room walls), the overall field propagation into adjacent areas is minimized. If shielding were placed at the computer room, the field would already have spread over the intervening space, requiring more material and higher cost.
Standards such as IEEE Std 299 (EMC Shielding Effectiveness) and IEC 61000 emphasize source-based mitigation. Additionally, ANSI/TIA-942 requires EMF shielding where magnetic flux exceeds recommended ICT thresholds (generally <5 mG for sensitive tape/disk storage).
Although 100 mG is often tolerated by modern equipment, legacy magnetic storage can be affected, so shielding is still prudent. Hence, the correct location is at the transformer room wall.
References: IEEE Std 299 (EMI Shielding), ANSI/TIA-942-B §6.6.4 (EMF Requirements), IEC 61000 EMC standards.

NEW QUESTION # 106
A computer room with a raised floor has been designed with racks in a hot/cold aisle setup.
What should you recommend for the placement of down-flow air conditioners?
  • A. Air conditioners perpendicular to (at the end of the row of) the Cold-Aisle
  • B. Air conditioners perpendicular to (at the end of the row of) the Hot-Aisle
  • C. Air conditioner placement has no influence on cooling effectiveness and efficiency. Hence, they can be placed at any convenient location.
  • D. Air conditioners should always be placed at both sides of each row of racks
Answer: A
Explanation:
In a hot/cold aisle configuration, placing down-flow air conditioners perpendicular to the cold aisle ensures that cool air is directed efficiently into the cold aisles where server intakes are located. This layout allows for optimal cooling performance by aligning the airflow directly with the equipment intakes, minimizing hot spots and enhancing cooling efficiency.
Detailed Explanation:
With a raised floor design, cold air from the air conditioners is supplied into the cold aisle, where server intakes are located. Positioning the air conditioning units perpendicular to the cold aisles ensures that cool air is delivered directly into these aisles, preventing air mixing and optimizing cooling. This setup takes full advantage of the airflow management strategy inherent to the hot/cold aisle configuration.
EPI Data Center Specialist References:
EPI guidelines on cooling emphasize that down-flow air conditioners should be positioned to maximize the effectiveness of cold aisle delivery, which improves cooling efficiency and helps maintain consistent temperatures across server racks.

NEW QUESTION # 107
You are allowed to use a calculator for this question.
A computer room has a net volume of approximately 2,500 m³ / 88,287 ft³.
The temperature is 20 °C / 68 °F.
The required design concentration is 7%.
The S-Factor is 0.1359 (metric) / 1.885 (imperial).
Calculate the amount of gas required for this computer room based on FM200. What is the correct weight?
  • A. Approximately 820 kg / 1,800 lbs
  • B. Approximately 1,390 kg / 3,000 lbs
  • C. Approximately 1,640 kg / 3,600 lbs
  • D. Approximately 410 kg / 900 lbs
Answer: A
Explanation:
The amount of FM200 gas required can be calculated using the formula:
Weight of Gas=Net Volume×Design Concentration×S-Factor        ext{Weight of Gas} =         ext{Net Volume}         imes         ext{Design Concentration}         imes         ext{S-Factor}Weight of Gas=Net Volume×Design Concentration×S-Factor Using metric units:
Net Volume: 2,500 m³
Design Concentration: 7% (or 0.07)
S-Factor: 0.1359
Calculation:
2,500 m3×0.07×0.1359=821.325 kg2,500 ,         ext{m}
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