Problem 3.5.5

Saturated steam, at 230 degrees Fahrenheit, is injected into a humid air flowing at a temperature of 70 degrees Fahrenheit dry-bulb and 45 degrees F wet-bulb. Mass flow rate of dry air is reported as 200 pounds per minute. Determine the steam flow rate, in pounds per hour, required to produce air having a dew point of 55 degrees F. What is the final state of humid air?

Solution:

Let us use the Psychrometric Chart suitable for Normal Temperatures ranges.

Locate the Point 1 representing outdoor air at 70 degrees dry-bulb and 45 degrees wet-bulb.

Mass flow rate of dry air is 200 times 60 equal 12,000 pounds per hour. Humidity value can be read as 0.0007 pound of water per pound of dry air.

Final condition of the air is represented by dew point of 55 degrees F, thus providing a humidity value of 0.0092 pound of water per pound of dry air.

Moisture increase in the air is 0.0092 minus 0.0007 equals 0.0085 pounds per pound of dry air. Steam addition is 12,000 times 0.0085 equals 102 pound per hour.

Now refer to Figure 1, Abridgement of Chart. Use the protractor to join the origin with enthalpy over humidity ratio of h_g equals 1157 to obtain a reference line. Draw a condition line from Point 1 parallel to the reference line. Draw a horizontal line having dew point of 55 degrees F. Intersection of this horizontal line and the condition line provides the final Point 2. Its dry- and wet-bulb temperatures can be read.