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content/articles/2023-09-08-pounds-of-water-removed.md

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+---
+tags: tech-tip, HVAC, formulas, psychrometrics, psychrometric-chart
+---
+
+# Pounds of Water Removed
+
+This is an article that shows how to calculate the pounds of water removed from an air stream, given the entering conditions (return air
+stream) and the outlet conditions (supply air stream).
+
+This is useful in the field when you want to calculate the amount of moisture removed from an air-conditioner or a dehumidifier. This
+article assumes that you have knowledge of a psychrometric chart. If you do not have basic knowledge of the psychrometric chart, then here
+are a couple articles to familiarize yourself.
+
+## Articles
+
+- [Understand Dew-Point](https://hvacrschool.com/understand-dew-point-absolute-moisture-right-side-psych-chart/)
+- [Impact of Adding or Removing Water from Air](https://hvacrschool.com/the-impact-of-adding-or-removing-water-from-air/)
+
+## Scenario
+
+Let's imagine that we have an air-conditioner that has the following measurements taken:
+
+- Return Air: 75° / 50% RH
+- Supply Air: 55° / 81% RH
+
+We plot the two values on the psychrometric chart (black line represents the return air conditions and blue line represents the supply air
+conditions).
+
+![chart](/articles/images/2023-09-08-pounds-of-water-removed.png)
+
+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
+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
+pound of air.
+
+This gives us the following values:
+
+- Return Air: 66 gr/lb
+- Supply Air: 52 gr/lb
+
+We can then use the following formula to calculate the pounds of water removed.
+
+![formula](/articles/images/2023-09-08-formula.png)
+
+| **Where**    |                                                          |
+| ------------ | -------------------------------------------------------- |
+| **W**        | _Weight of water in pounds per hour_                     |
+| **4.5**      | _Constant based on density / specific heat of moist air_ |
+| **CFM**      | _Airflow in cubic feet per minute_                       |
+| **∆G** | _Difference in grains of moisture_                       |
+| **7000**     | _Constant based on grains of moisture in saturated air_  |
+
+## Solution
+
+First, we solve for the difference in grains between the two air streams.
+
+∆G = 66 - 52 = 14
+
+Next, we've measured our airflow and have determined to have **797 CFM** of airflow across the evaporator coil, so we can substitute our
+values into the formula.
+
+![solution](/articles/images/2023-09-08-solution.png)
+
+So, we are removing about 7 pounds of water per hour at these conditions.
+
+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
+can often be rated in pints per day.
+
+I hope you've found this article helpful, thanks for reading!