The introduction of outside air to the air conditioning system is 95°F with a relative humidity of 70%. The air has been conditioned to 70°F with a relative humidity of 80%. Using the psychrometric chart, shown in the illustration, determine the quantity of moisture removed from one pound of the conditioned air. See illustration GS-RA-22.
• Using a psychrometric chart to find moisture content (grains of moisture per pound of dry air) for a given dry‑bulb temperature and relative humidity. • Understanding that moisture removed equals the difference in grains between the entering air state and the conditioned (leaving) air state. • Recognizing that the chart’s right-hand vertical scale is grains of moisture per pound of dry air, not percent humidity.
• On the chart, how do you locate the state of the outside air at 95°F and 70% relative humidity, and what value do you read from the grains-of-moisture scale for that point? • How do you locate the state of the conditioned air at 70°F and 80% relative humidity, and what grains-of-moisture value corresponds to that second point? • Once you have both grain values, what simple calculation gives you the quantity of moisture removed from one pound of the conditioned air, and which answer choice is closest to that difference?
• Be sure you are reading grains of moisture from the right-hand scale, not percent relative humidity from the curved lines. • Verify you are on the correct relative humidity curve (70% for the entering air, 80% for the leaving air) at the correct dry-bulb temperatures (95°F and 70°F). • Confirm that you subtract final grains from initial grains to find the moisture removed per pound of dry air, then match that value to the nearest answer choice.
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