Air at a dry-bulb temperature of 90 degrees Fahrenheit and
60 percent relative humidity is cooled to 45 degrees F and is completely
saturated. Flow of air is maintained at
5,000 cubic feet per minute. Find the
amount of refrigeration heat that needs to be supplied to bring this change to
the condition of the air.
Solution:
Let us use the Psychrometric Chart suitable for Normal
Temperatures ranges.
Locate the Point 1 representing 90 degrees dry-bulb and 60
percent saturation.
Draw a horizontal line to read a humidity value (W1)
of 0.0183 pounds of water per pound of dry air.
Move along the wet bulb line and extend it further to read
the enthalpy of saturation, and incorporate the enthalpy deviation. Humid enthalpy is found to be 41.82 BTUs per
pound.
Please note that humid volume value is plotted as curves on
the Psychrometric Chart at angles with negative slope. These values are 13, 13.5, 14, and 14.5
cubic feet per pound of dry air. For
case in hand, its value is 14.26 cubic feet per pound of dry air.
The humid air is cooled to 45 degrees F and is completely
saturated. Point 2 lies on the
saturation curve. Humidity value (W2)
is found to be as 0.0063 pounds per pound of dry air.
Note that water condensation occurs when humid air is cooled
to a temperature below its initial dew point.
It is assumed that the humid air is uniformly processed. Although water condenses at various
temperatures below its initial dew point, all condensed water is cooled to the
final air temperature and is drained from the system.
Please note that the enthalpy of exit water is 13.09 BTUs
per lb.
Heat given by humid air is absorbed by water and the
refrigeration system. The enthalpy
received by drained water is 13.09 times 0.12 equals 1.5708 BTU per pound of
dry air. The enthalpy given by humid
air is 41.82 minus 17.65 equals 24.17 BTUs per pound of dry air. Refrigeration requirement is found to be
24.17 minus 1.5708 or 22.5992 BTUs per pound of dry air.
Mass flow rate of air can be calculated from the volumetric
flow rate and specific volume information at entering stage. This value is found to be 5,000 divided by
14.26 equal 350.63 pound per minute.
The refrigeration requirement is 22.5992 times 350.63 equals
8,474 BTUs per minute. Since one ton of
refrigeration equals heat removal rate of 200 BTUs per minute, the
refrigeration demand equals 8,474 divided by 200 or 42.37 tons.