69 lines
2.9 KiB
Markdown
69 lines
2.9 KiB
Markdown
---
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tags: tech-tip, HVAC, formulas, psychrometrics, psychrometric-chart
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---
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# Pounds of Water Removed
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This is an article that shows how to calculate the pounds of water removed from an air stream, given the entering conditions (return air
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stream) and the outlet conditions (supply air stream).
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This is useful in the field when you want to calculate the amount of moisture removed from an air-conditioner or a dehumidifier. This
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article assumes that you have knowledge of a psychrometric chart. If you do not have basic knowledge of the psychrometric chart, then here
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are a couple articles to familiarize yourself.
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## Articles
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- [Understand Dew-Point](https://hvacrschool.com/understand-dew-point-absolute-moisture-right-side-psych-chart/)
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- [Impact of Adding or Removing Water from Air](https://hvacrschool.com/the-impact-of-adding-or-removing-water-from-air/)
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## Scenario
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Let's imagine that we have an air-conditioner that has the following measurements taken:
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- Return Air: 75° / 50% RH
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- Supply Air: 55° / 81% RH
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We plot the two values on the psychrometric chart (black line represents the return air conditions and blue line represents the supply air
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conditions).
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We start by finding the corresponding dry-bulb temperature at the bottom of the chart and draw a straight line up to where it intersects the
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relative humidity curve. After that we draw a straight line to the right side of the psychrometric chart to find the grains of moisture per
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pound of air.
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This gives us the following values:
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- Return Air: 66 gr/lb
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- Supply Air: 52 gr/lb
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We can then use the following formula to calculate the pounds of water removed.
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| **Where** | |
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| ------------ | -------------------------------------------------------- |
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| **W** | _Weight of water in pounds per hour_ |
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| **4.5** | _Constant based on density / specific heat of moist air_ |
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| **CFM** | _Airflow in cubic feet per minute_ |
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| **∆G** | _Difference in grains of moisture_ |
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| **7000** | _Constant based on grains of moisture in saturated air_ |
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## Solution
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First, we solve for the difference in grains between the two air streams.
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∆G = 66 - 52 = 14
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Next, we've measured our airflow and have determined to have **797 CFM** of airflow across the evaporator coil, so we can substitute our
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values into the formula.
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So, we are removing about 7 pounds of water per hour at these conditions.
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Another thing to note is that 1 pound of water is approximately 1 pint of water, which can be useful when working with dehumidifiers that
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can often be rated in pints per day.
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I hope you've found this article helpful, thanks for reading!
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