Figure 1. Condensate flow diagram.

Figure 2. Reduction in load for variable secondary stage airflow and constant primary stage cooling load (100 tons).

The Heat Transfer Process

The process described above is analyzed in two stages. The first stage analysis is the main cooling load and the second stage is the separate heat recovery precooling load. The process is most economical and makes most sense if the first stage latent cooling load is large enough to generate considerable condensate flow, the climate is humid throughout the year (high summer wetbulb temperatures), and the condensate from the first stage coil is drained by gravity to the heat recovery pre-cooling coil (second stage to save on additional pump horsepower).

Table 1A. Basic calculations for enthalpy reduction in stage two.

Psychrometric Cycles

The steady-flow energy equation:
(1) Q_coil = M_air1 (h₁ - h₂)

and mass balance equation:
(2) M_con = M_air1 (w₂-w₁)
Where, Q_coil = The cooling load of the cooling coil in Btu.
M_con = The condensate that results from the cooling of the airstream below its dewpoint (lb).
h = Enthalpy and grains of moisture per pound of air (Btu/lb).
w = Grains of moisture per pound of air.
T = Temperature in °F.
Adding time to both equations yields the rates of heat flow and condensate flow. Let this condensate discharge by gravity to another coil, which is placed in a second incoming airstream M_air2 totally independent of the first airstream cycle. The temperatures of the condensate entering and leaving the second cooling coil are designated as T_con1 and T_con2. Therefore the heat flow rate equation at the second coil is:
(3) Q_con = M_con C_p (T_con1-T_2con)
Where C_p is the specific heat of water. Assuming no external heat gain or loss of the condensate, the heat balance equations between the condensate and the second airstream must balance since the heat lost by the higher enthalpy airstream is gained by the lower enthalpy condensate in the coil. Therefore:
(4) Q_con = Q_air2 and
(5) Q_air2 = Q_con = M_air2 x (h_amb-h_air2)
If we know the quantity of air in the second airstream, we may then solve for h_air2:
(6) h_air2 = h_amb - (Q_con / M_air2)
Applying the above equations to a simple air conditioning process with chilled water as the refrigerant fluid, at standard atmospheric pressure conditions, a psychrometric chart can be used to extrapolate data. Air is cooled to saturation at 55°F. Ambient conditions are chosen for 0.4% ASHRAE values, and the derived equation for flow from equation (3) in gallons per minute (gpm) and Heat loss/Heat gain (Q) from equation (6) in Btuh are assumed to be widely familiar by HVAC engineers in the form:
(7) gpm = Q 490 x (T₂-T₁)
(8) Q = cfm x 4.5 x (h₂-h₁)
Stage 1. Main cooling and dehumidification. An air handler has the capacity to cool a volumetric flow rate of outside air "V" (in cfm) from maximum summer ambient conditions down to 55° saturation. Since mass flow rate is conserved, volumetric flow rate is converted to mass flow rate by dividing by the specific volume and converting from hours to minutes. Using the psychrometric chart, all the entering and leaving conditions including enthalpy and grains of moisture per pound of air are extrapolated. The condensate flow rate, M_con, in pound per minute is obtained from equation (2).

Stage 2. Heat recovery load. The condensate Mcon is piped to another coil totally separate from the first stage. A volumetric flow rate, V_air2, at maximum summer ambient conditions is passed through the coil. Entering conditions for both the condensate and the air are known. If a theoretical maximum heat transfer occurs at the coil, the leaving water temperature will approach the incoming airstream temperature (equal to the maximum summer ambient). The leaving condensate conditions are now known, and equation (7) is used to calculate the cooling capacity of the condensate coil Q_con.

The heat gained by the condensate is equal to the heat lost by the incoming airstream from equation (4). Substituting Qcon for Q_air2 in equation (6), the leaving enthalpy of the outdoor air is calculated. The final enthalpy of the second airstream has been lowered by contacting it with condensate from a previous coil process.

Apart from enthalpy, which was mathematically calculated, the final condition of the second airstream is still unknown. The second airstream did experience a lowering of temperature and moisture content, however, the conditions of the airstream will not reach saturation unless the flow rate of the condensate from stage one is large enough. It is concluded that since M_con, T_con from the first stage are constants, we can achieve maximum pre-cooling of the second airstream if more condensate Mcon is discharged from the first airstream (e.g., a very large primary latent load), and if the mass flow rate of the second air stream is small.

Table 1B. Psychrometrics

Example and Discussion

A standard¹ stage one primary air conditioning cooling tonnage is chosen at 100 tons, 35,000 cfm, and a stage two secondary 100% outdoor air is chosen for all cities at 1,000 cfm (Figure 1). Condensate from stage one enters a pre-cooling coil stage two at 55°. The pre-cooling coil is designed so that the leaving temperature of the condensate (designated as T_con2) approaches the incoming air drybulb (Tambient in this case). Equating the heat gained by the condensate to the heat lost by the ventilation airstream, the leaving air enthalpy h_air2 is calculated from equation (8). Equation variables and the percentage reduction in enthalpy are shown in Table 1.

Table 1 reveals that for any of the locations, there was a reduction in the leaving air enthalpy. The reduction in enthalpy ranges between 26% and 18% for most locations and drops further as wetbulb decreases. The percentage reduction is directly related to wetbulb temperature.

Ultimately, the reader is interested in the maximum load reduction as a result of using the cold condensate to pre-cool the airstream. Table 2 shows the loads for two conditions; the enthalpy of saturation minus the enthalpy of the airstream with and without the pre-cooling coil. Using equation (8), the total load for each condition is calculated.

The percentage reduction in load for cities with wetbulbs as low as 76° ranges between 56% and 45% at peak summer wetbulb conditions. The percentage reduction in the load drops with wetbulb temperature particularly below 76°. Nevertheless, there is a significant amount of heat recovery from the low enthalpy condensate for a wide range of locations. Again, the percentage reduction is directly proportional to wetbulb. One interesting observation is noted. Table 2 shows that certain cities further south, such as Orlando, have equal wetbulb temperatures to Northern cities such as Washington. The percentage reduction for Washington is almost the same as Orlando. This observation shows that the analysis should weigh more on wetbulb temperature rather than geographic location.

What this paper does not touch upon is the duration of the wetbulb temperature throughout the year. Orlando, in this instance, might have a higher wetbulb temperature occurrence throughout the year than Washington. The cost-benefit of installation could be more economically justifiable for projects in Orlando rather than Washington.

Table 1C. Enthalpy reduction.

Sensitivity Analysis

Conclusion

Here we discussed the reduction in load at peak summer conditions. Obviously, it is expected that this system will be more attractive in locations where high wetbulb temperatures occur year-round. Piping the condensate by gravity to the secondary cooling coil adds another benefit of not utilizing motor horsepower for pumping. ES

Footnotes 1. Standard conditions indicate typical design settings at 0.4% ASHRAE summer values, 74° F indoor db thermostat setting, 25% outdoor air ratio, and 55° saturation temperature.