Abby Normal wrote:

Infiltration will only be a factor when the systems are cycled off,
under less than a design load.

That's what controls are for :-) Internal schemes can work better than
outdoor coolers, maintaining the same indoor air conditions with less
water and fan energy...

But that Mastercool MMB10 cooler has a much higher cooling capacity
with outdoor vs mixed indoor and outdoor air, which cuts the net output
way down from 69K to 3K Btu/h, so we need a lot more wet surface, eg
a damp slab or basement or crawlspace or watered indoor wetland/planter.

With the arrangement below, the cooler moves 2K cfm of air with
wi = (155x0.00545+(2K-155)0.012)/2K = 0.0115 and Pai = 0.5428 "Hg,
so it might evaporate 4.78 lb/h. If 4.5C(wi-wo) = 4.78, C = 162 cfm,
which makes wi = 0.001147, which makes Pai = 0.5147 "Hg, which makes
P = 4.79, and so on...

For more cooling capacity with dry outdoor air, we might put one near
a window inside a house with a $55 Lasko 2155A 16" 90 W 2470 cfm intake
fan in the window and use the fan thermostat to turn on the cooler when
the room temp rises to 80 F and a humidistat to turn on the fan when
the indoor RH rises to 56%, with 1-way plastic film dampers in a box
between the cooler and the window to force outdoor fan air to flow
through the cooler pad when the window fan is running and make indoor
air flow through the cooler when the window fan is not running, like
this, viewed in a fixed font like Courier:

| |
| |
---------
| |llld| |
|c| d| | outdoors
|o| d|f|
2K cfm ==|o| d|a| == 170 cfm With the window fan off, indoor air
|l| d|n| would flow in through left and right
|e| d| | dampers lll and rrr. With the fan on,
|r| d| | ddd would open and the fan air would
| |rrrd| | force lll and rrr closed.
---------
| |

The box above isn't worth building, if it only increases the cooling
capacity from 3K to 3.2K Btu/h, and if the MMB10 uses 5 lb/h of water
and 3.5x120 = 420 watts of electrical power, we are likely better off
with a $69 5K Btu/h 10 EER 500 watt window AC.

20 PAO=.1*EXP(17.863-9621/(460+110))'outdoor vapor pressure ("Hg)
30 PWO=EXP(17.863-9621/(460+78))'cooler vapor pressure ("Hg)
40 WO=.62198/(29.921/PAO-1)'outdoor humidity ratio
50 WI=.012'indoor humidity ratio
60 PWC=EXP(17.863-9621/(460+80))'cooler water vapor pressure ("Hg)
70 FOR N=1 TO 5
80 WC=(WO*C+WI*(2000-C))/2000'cooler humidity ratio
90 PAC=29.921/(1+.62198/WC)'cooler air vapor pressure ("Hg)
100 P=6.9*(PWC-PAC)/(PWO-PAO)'water evaporation rate (lb/h)
110 C=P/(4.5*(WI-WO))'outdoor airflow (cfm)
120 Q=1000*P-(90-80)*C'sensible cooling with 90 F outdoor air (Btu/h)
130 PRINT N;C,P,Q
140 NEXT N

1 161.9926 4.657094 3037.168
2 170.067 4.889222 3188.552
3 170.4698 4.900802 3196.104
4 170.4899 4.901381 3196.481
5 170.4909 4.901409 3196.5

Nick