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Psychrometrics

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grnblr

Electrical
May 1, 2002
14
I have a room having a window air conditioner and a separate axial fan throwing in 100% fresh air When I draw the plot in a psychrometric chart, in order to arrive at delta H from the plot, should I consider the enthalpy values (H1 and H@)as those between OA conditions and coil ADP or enthalpy figures between ADP and room conditions

Please clarify
Thanks in advance
 
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lol, from the room air drawn into the unit and the supply air leaving the unit. Sounds like you are using a room as a mixing box, hope it is a dry climate.

Take the "V" out of HVAC and you are left with a HAC(k) job.
 
What happens to the 100% fresh air? Do you have an exhaust fan? Give us the scheme of your setup and you will get better replies from the members.

 
If you are supplying fresh air straight into the room then the fresh air becomes a space load. The oil entering conditions will be the same as the space conditions and not a 'mixed' condition as per your standard psychometric chart lines.
 
Take the return air temp to the coil and the supply air temp off the coil. Wet bulb and dry bulb.
 
when the room has become a mixing box, i think the parameter that will change will be the return air to the coils?
 
Thanks all and sorry for the delay

I was away for almost a week. The situation is simple. The application is a class room of 35 students. The room is fitted with with a window AC which is having no provision to feed in fresh air. I want to provide 100% fresh air into the room through suitable capacity axial fresh air fan(not drawn in through the coil)directly thrown into the room. There is no exhaust fan - only bleeding through door leaks.

After calculating the heat load I was therefore trying to locate the ESHF line on the psychro chart and also calculate the difference in enthalpys so as to use the formula Q = 4.5 x cfm x delta Grain. Like Marcoh said I think I will have to correct the ventilation load which I assumed in my Heat load calculation as the dehum air quantity which I got from the heat load calculation.

I am still not clear about the two enthalpy points H1 and H2 in the chart and how to arrive at the return air temperature

Thanks

 
Unless you are in a dry climate, this sounds disasterous. You may want to at least try an ERV for the fresh air.

Try this approach maybe for your situation. Hopefully you have a full set of ratings for that window unit, so you can see the total/sensible capacities at several entering db & wb conditions.

DO a mositure balance. Pick an RH that you hope to maintain

Work out how much moisture needs to be in the room air.

Add up how much mositure is added to the room, namely generated by those 35 kids, plus being brought in with the fresh air.

Subtract how much moisture is being displaced through the exfiltration, theoretically this displaced air is at the design condition.

The remianing amount of moisture is what this window unit must remove. If you are anywhere humid, it is going to be over whelmed with moisture.

Try iterating this a few times, each time start off by raising the humidity that the room will hold at. If you are anywhere humid, I think it is going to end up being 80% RH in that class room.

Take the "V" out of HVAC and you are left with a HAC(k) job.
 
add dehumidifiers

Take the "V" out of HVAC and you are left with a HAC(k) job.
 
I see few problems in the original scheme.

1. If I consider 15cfm fresh air per student then the fresh air is around 525cfm. You may require about a square feet leak area to push this for a room pressurisation of 5Pa. Pushing through doors may not be a possibility because you may have to have 60sq.ft to 100sq.ft doors depending upon the door construction.

2. As you are mixing the air in the room, i.e after the recirculating air being treated, the mixing may happen anywhere from the coil off temperature to the room condition. If the recirculated air is at a temperature less than that of dew point of ambient air, you will have condensation. So, moisture removal from the ambient air using the window A/C should be avoided and I deviate here from AbbyNormal's suggestion.

3. Homogenous mixing of fresh air with treated air is a problem.

If this an inevitable situation, I would use either of the two methods.

First one is to find out the grain difference by required humidity control(only room latent load) and then dropping both fresh air and recirculating air temperatures to a common ADP and then using combined cfm to plot ESHF line.

Second one is to draw ESHF line based upon only recirculating air flowrate and then cooling fresh air only to room condition.

Both these methods require treating fresh air separately.

AbbyNormal said:
Subtract how much moisture is being displaced through the exfiltration, theoretically this displaced air is at the design condition.


As the leaking air is at room condition, I think, moisture load on the A/C will not be reduced. You can only reduce moisture load on the A/C if the leaking air contains more moisture than the room condition.

 
What goes in must go out quark, it is a simple mass balance.

Exfiltrating air takes moisture with it. To keep a constant moisture level, the AC has to remove the difference between what comes in from the ventilation, and that generated internal, less the amount of moisture that leaves the space in the exfiltration.

No other way of looking at it.



Take the "V" out of HVAC and you are left with a HAC(k) job.
 
Quark for discussion comsider this.

Evaporative cooling. They pump in humid cool air into an Arizona home and open a couple windows to let the air pressure and of course the humidity to leave.

You are pumping gallons of water per hour into that space, yet humidty only builds so high, because the air leaving the space takes mositure with it.

Take the "V" out of HVAC and you are left with a HAC(k) job.
 
100 % outside air unit can suipply dry air into a space. Keeps RH low by displacing humidity out of the building.

Take the "V" out of HVAC and you are left with a HAC(k) job.
 
Its so humid here quark, it is almost cheaper for me on jobs with an dense active occupancy to do it with 100% outside air. The air flow for ventilation is high as it is, cool this down to a 50 dewpoint reheat it to about 60F and still have enough sensible cooling capacity,

Take the "V" out of HVAC and you are left with a HAC(k) job.
 
Contractors following manual j to size air conditioning often are given 'outdoor grains of moisture' that is actually the difference between the amount of moisture held in room air to the amibient moisture.

Example, conditioned air could hold 64 grains. A humid ambient could have 150 grains. The moisture in the outside air is carried as 150-64=86 grains. Looking at it this way, the moisture in exfiltration or exhaust is autmatically accounted for when the space holds at 50%.

Following this procedure in dry places, often the outdoor grains is 'negative'

Take the "V" out of HVAC and you are left with a HAC(k) job.
 
Abby,

I considered cooling and dehumidification and that is why I said not to consider the moisture of exhausted air. We will precondition the air to a level, below the set condition by number grains generated in the controlled space. So, the air carries the moisture that is being generated in the controlled space and goes out. As we have to follow this step through out, the grains carried out by air shouldn't be considered. I hope I made my point clear now.

 
The OP has an open system, pressurizing a room with untreated outside air and is trying to describe the entire process with a single line on the chart. Not going to happen.

If you want to approximate what the room condition is going to be, you will have to do a moisture balance. If the climate is humid, it is going to balance out with a high RH. If the climate is dry, it will be a low RH.

If you only consider how much more moisture is in the outside air, compared to the room air, the mositure in the exhaust is already accounted for. If you do not follow this methodology and never consider the mositure leaving in exhaust, you have an automatic latent factor of safety.

Take the "V" out of HVAC and you are left with a HAC(k) job.
 
This is not a good design dumping outside air into the room and making the classroom a mixing box but for the sake of answering the question here were my thoughts.

Hmmm I thought this was a simple draw a line from the outdoor air point to the cclt point on the psych chart but what I'd do for this is figure out first if your TWU CFM is greater than your outside air CFM.
If you supply 100 cfm of outside air per 3'x7' door (anything more and you may start hearing whistling, not good for classroom environment) and let's say you have 2 doors.
So that's 200 cfm of outside air.
What is your air conditioner rated for? basically your air conditioner is now handling 200 cfm of outside air plus the load from the room.
I believe the nominal tonnage rating on those units are based on 75db 50%rh with a cclt of 55F saturated but you will need to verify that.
Using the nominal tonnage you can figure out what your TWU can handle in delta h terms.
For new england 200 cfm of outside air works out to roughly 14000 btu/h of total cooling at worst case ashrae design data. So that would be 14,000 btu/h less available to TWU. It would be several iterations on the psych chart to figure out the new supply temp. but it's definitely not a straight forward line between 2 points since your room condition is also your entering coil condition.
That's my take on it I'd like to know if there is an easy solution to the problem.
 
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