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Coolie
13-11-2004, 06:30 PM
In the middle of an assignment for college and need help.

Why use a re-heater on a ahu supplying cold air to a single room.

Brian_UK
13-11-2004, 10:34 PM
Just a small comment to your question Coolie.

Although most of the 'air conditionering' splits sold are so called they are in fact simply 'air coolers'.

To provide full air conditioning (excluding full filtration) requires air flow, cooling, heating and de/re-humidification capabilities.

Good luck with your studies.

Coolie
14-11-2004, 07:35 AM
Thank You guys. The help is always appreciated.

chemi-cool
14-11-2004, 03:20 PM
Hi Brian,

Big central AC or AHU don`t do much more then the small split units.

If you`re lucky, you will find the fresh air louvre open but in most places I find it shot, to prevent from dust entering the air-conditioned area.

Which off the shelf unit also controls humidity?

Chemi :)

rbartlett
14-11-2004, 04:43 PM
To remove moisture from the air, without desiccants, you have to cool much of the air to below dew point. Then, unless you also need to cool the space, you'll have to reheat the air after drying it, otherwise you'll not only dry the air but you'll cool or overcool it too.


this sounds confusing to me..

as far as i remember you cool the air to dew point which will remove moisture -it won't 'dry' the air.

in fact whilst it will have less moisture in weight per m3 ,it will be more humid

in order to 'dry' the air you have to re-heat it to lower the rh. which is either done by re-mixing with room air or re-heating via heaters
(or even the condenser coil upstream as used in domestic and industrial de-humidifiers ..)

I await to be corrected..

cheers

richard

Dan
14-11-2004, 05:09 PM
as far as i remember you cool the air to dew point which will remove moisture -it won't 'dry' the air.

in fact whilst it will have less moisture in weight per m3 ,it will be more humid

Hi Richard. I thought Marc's explanation quite concise. I suppose we are looking at semantics, but if removing moisture cannot be called drying, I am at a loss as how to use the word "Dry." I also think that Relative Humidity is overused as a measure for the moisture content of air.

rbartlett
14-11-2004, 05:26 PM
when you are involved in a college assignment you should endevour to put it on paper rather than dans mind ;-)

cheers

richard

Dan
14-11-2004, 05:52 PM
Too funny. Yep. To go a tad further into the "Relative" aspect of humidity, I think it causes a lot of confusion with technicians new in the field. RH is pretty much only important when we are judging human comfort levels, such as the cooling nature of sweating and the deleterious removal of moisture from the mucosa, etc.

For me, the important criterion is the dew point.

rbartlett
14-11-2004, 06:41 PM
I feel it is important to understand that humidity is relative not only to weight of water present in the air but temperature....

i can remember in college discussing moisture removal and Rh plotting along the dew point etc plus working out on psychometric charts actual amounts of water removed, rh, sh and the effects of reheating/mixing...

obviously greater minds don't think students need to bother to learn this nor understand it ..

as I say what do I know..

cheers

richard

Dan
14-11-2004, 07:35 PM
i can remember in college discussing moisture removal and Rh plotting along the dew point etc plus working out on psychometric charts actual amounts of water removed, rh, sh and the effects of reheating/mixing...

obviously greater minds don't think students need to bother to learn this nor understand it ..


Ouch, Richard! On the contrary. I want students to understand psyhcrometrics - inclusive of the concept of relative humidity, vapor pressure and the wonderful insight and manner of thinking that Willis Carrier brought into being.

I just find relative humidity to be something to be calculated from real values, as opposed to the real values. I can drag the same cubic foot of air throughout the universe and watch its relative humidity go from zero to 100%. If I drag the same cubic foot of air and watch its dewpoint, it will never change.

But aside from human comfort, what is the value of knowing a relative humidity? The hard values are wet bulb, dry bulb and the calculated dew point. It is fun to watch the light go on in a technician's or student's head once you focus on those values.

Coolie
14-11-2004, 08:20 PM
So am I right in my understanding that the reheater is there to warm the air to the desired tempreture if the cooling coil cooled it to much in order to dehumidify it to maintain the moisture content. Well lets see, The following is a copy of what I have written so far. The answer must be no more than 800 words.
The question is to explain how an AHU works.

The air handling unit is an integrated piece of equipment consisting of fans, heating and cooling coils, air-control dampers, filters and silencers. The purpose of this equipment is to collect and mix outdoor air with that returning from the building space. The air mixture is then cooled or heated, depending on the required tempreture, after which it is discharged into the building space through a network of ductwork.

The following is a detailed explanation of how such a system works.


For the sake of this example lets say that the conditioned space needs to be maintained at 22c db and 50% saturation all year round. In order to understand the process that is undergone to maintain this condition we need to start at one point, i.e. the room.
Air from the room is removed by means of the return air fan situated in the ductwork. A certain amount of this air will be re-circulated back into the room by means of air-control dampers. This figure will vary depending on the design. For this example well use a ratio of 70:30. In other words, 70% of the air extracted from the room will be re-circulated back into the room while the remaining 30% will be discharged into the atmosphere.
Therefore 30% of the air supplied into the room will be fresh.


The fresh air will pass through a damper then a filter media to remove excess dirt that is contains.
The re-circulated air will be mixed with the fresh air in a mixing chamber that is situated after an anti frost heater which heats the fresh air to above 0c. This anti frost heater will therefore only work during the winter months when the outside ambient falls bellow freezing.

Once the fresh air and the re-circulated air have mixed together in the mixing chamber, it will pass through a pre-heater coil.
The pre-heater coil is only used in the winter and only on an A.H.U utilising an air-washer to humidify the air. When this coil is utilised the function of it is to heat the air up to the required tempreture before it passes through the air washer. This air washer works by spraying water droplets into the duct. The droplets evaporate into the air and hence the air becomes humidified. The water that is not evaporated is collected in a sump underneath the washer and is then, by means of a pump, re-circulated to the nozzles of the air washer. The water level in the sump is maintained by using a float valve.
The air-washer reduces the dry bulb tempreture. It is for this reason that a re-heater is required to warm the air up to a few degrees above the required supply tempreture. When the air passes through the ductwork it will inevitably lose heat to the atmosphere, hence it is heated to above the required supply tempreture.
The air is now suitable to be supplied by means of the supply fan.

In the case of steam humidification the pre-heater is not used. The air passes directly from the mixing chamber through the heater coil where it is heated to the desired tempreture. The steam humidifier is located after the heater battery and injects steam into the airflow. The isothermic humidification does not lower the dry bulb tempreture so a re-heater is not required. The supply fan now supplies the air to the room.






During the summer months a few of the parts described in the above paragraphs are not utilised. These include the frost heater, the pre-heater (in the case of the air-washer system) and the humidifier.

I will start describing the cooling process from the point where the air leaves the mixing chamber.
The cooling process is far simpler than the heating process because when the mixed air leaves the mixing chamber it still passes through the pre-heater and the humidifier but these are turned off. Once the air reaches the cooling coil it passes through the coil and exits at the required tempreture. The re-heater is not generally used in cooling mode, but will come on when the moisture content is too high and the cooling coil is kept on to dehumidify the air. The re-heater will then warm the over cooled air to the desired tempreture.

The air is cooled to below the required supply tempreture to allow for heat gain through the ductwork. The air is now fed into the room by means of the supply fan situated after the re-heater.

The system will be controlled by a series of thermistors and sensors located at strategic points throughout the whole system giving readings of the condition of the air. The necessary adjustments will be made according to these readings.

Brian_UK
15-11-2004, 01:25 PM
Which off the shelf unit also controls humidity?Chemi, haven't you seen those super efficient dry functions on some of the splits :cool: :D

Coolie
15-11-2004, 06:07 PM
So I see that I've started something and in the process have been totally ignored. I'm not complaining as all of the posts make for interesting reading and have helped a lot. It's just that I wanted to know what some of you thought about my answer to the question.

Gary
15-11-2004, 08:19 PM
So am I right in my understanding that the reheater is there to warm the air to the desired tempreture if the cooling coil cooled it to much in order to dehumidify it to maintain the moisture content.

That is correct.



For the sake of this example lets say that the conditioned space needs to be maintained at 22c db and 50% saturation all year round.

As Marc has pointed out, you may want to think in terms of hard values as opposed to calculated values. In this case, the hard values would be 22C/72F DB temperature and 11C/52F dewpoint temperature. If you trace the DB and dewpoint lines on a psychrometric chart, they intersect at the calculated value of 50% RH.

The hard values more accurately define your twin objectives of maintaining 22C/72F DB and 11C/52F dewpoint.

In order to raise or lower the DB in the room, the DB of the supply air must be higher or lower (respectively) than the air in the room.

In order to raise or lower the dewpoint in the room, the dewpoint of the supply air must be higher or lower (respectively) than the dewpoint of the room air.

We raise the dewpoint by adding moisture to the air. We lower the dewpoint by removing moisture from the air.

In order to remove moisture from air, we lower the temperature of that air below its current dewpoint, causing moisture to drop out of it. We continue to lower its dewpoint by further lowering its temperature until the dewpoint is at or below our target dewpoint (11C/52F).

IOW, if our target dewpoint is 11C/52F, our cooling coil must be cold enough to lower the temperature of the air below 11C/52F. I have seen many instances where humidity levels have not been achieved because the air was not sufficiently cooled as it passed through the coil. It is the temperature of the coil, not the reheat, which dehumidifies the air.

Abe
16-11-2004, 12:12 AM
I dont often interject during "discussions" as these, though I do read them religiously and attempt to understand the principles.....

But on this occasion I have a situation at home whch may be of relevance.

I have a room in the house, ( my home built early 1960's) double brick with no cavity or insulation between bricks. This room receives very little sunshine, There are 2 external wall which are cold, ( ie: facing outdoors) a central heater radiator in the room which provides a modicum of warmth.

After this summer rains........I noticed mould growth on the walls..........I put this down to very wet saturated conditions, the
mild warm from central heater radiator and the cold wall producing a climate which is condusive for mould growth.

The solution I am pursuing is this, I am fitting a ventilation fan with a humidity sensor to extract the water laden air from the room. ( I am attempting to make the air in room drier)

I am installing a combination boiler ( Vaillant 100 000 btu) which will run hotter and provide a hotter radiator then the existing pumped type central heating system) The idea behind this is, I dont feel the existing radiator provides sufficient heating in room to combat effects of cold coming in through the walls.

The other solution I may have is fit a small air conditioner, and use LG';s humdity function operation to extract the water from the room

Am I on the right course?? or thinking???

eggs
16-11-2004, 12:44 AM
i have a similar mould problem, but i think the a/c solution is curing the symptom and not the cause.
My house is an old 1930's jobbie, currently looking more like a building site than a real building site.
Within the last 6 mths, i have had 6 of the rooms replastered, so as you can imagine the air in here is very moist.
The A/C in the bedrooms has been running virtually non stop in dehum mode, the fans in the kitchen and bathrooms same as,but still i get mould on the walls.
For now i am going to ignore it, and see what happens when i complete the work and the house gets a chance to dry out properly.

cheers

eggs

Gary
16-11-2004, 02:26 AM
A few things to keep in mind:

1. If you are exhausting air from the house, a similar amount of outdoor air is finding its way into the house to replace it. Nature hates a vacuum.

2. Reducing the temperature of air below its dewpoint causes moisture to drop out of the air, thereby lowering its dewpoint.

3. Dewpoint is one side of the relative humidity (RH) coin. The other side is drybulb temperature. As you raise the DB, without changing the dewpoint, you lower the RH. As you lower the DB, again without changing the dewpoint, you raise the RH.

Assuming you want to lower the humidity in your house:

In the summer, when the dewpoint of the outdoor air is high, exhausting air from the house is going to cause high dewpoint outdoor air to enter the house, adding to the indoor humidity. Better to run an A/C, dehumidifying the indoor air, and keep the outdoor air outdoors.

In the winter, the outdoor air temperature has been lowered by mother nature, and a lot of moisture has dropped out in the form of rain and/or snow. The dewpoint of the cold outdoor air is low. If you want to dry out the house, now is the time to turn on those exhaust fans and bring the outdoor air indoors. The downside is that it will cost you to heat that air.

So, assuming you want the RH to be low:

In the winter, bring in low dewpoint outdoor air and/or raise the room temperature. (However, in very cold weather, you may find it more economical to dehumidy the indoor air, than to heat the incoming outdoor air)

In the summer, dehumidify the air and/or raise the room temperature. Do not bring in high dewpoint outdoor air.

frank
21-11-2004, 10:45 PM
If you want to impress the tutor with the temperature drop in the ductwork you can use the following formula:
t2 = (t1(2000vpc-UA) +2UA tai)/UA +(2000 vpc)

where

t2 = air leaving temperature
t1 = air entering temperature
v = volume of air in duct
p = density of air(assume 1.205 kg/m3 @ 20 deg C)
c = sensible heat capacity of the air (1.02 kj/kg)
U = U value of the duct (assume 2W/M2/deg C)
A = Area of duct
tai = Ambient temperature around ductwork

Note that the density of the air needs to be calculated at the duct air entering temperature by the formula p1,t1 = p2,t2

so for p2 = (p1t1)/t2

as an example we can calculate the temperature drop in a 30m length of circular ductwork 0.5m in diameter in which we have 0.49m3/s flowing and the duct is running through a ceiling void where the ambient is 10C. Assuming that the air entering the duct is 50C we can calculate the temperature drop as (Air entering - Air leaving).

P2 is determined thus: (1.205 x (273+20))/(273+50) = 1.09

now to incorporate that into the formula for t2 as follows

t2 = 50(2000x0.49x1.09x1.02) - (2xPix0.5x30) + ((2x2x(Pix0.5x30) x 10))/2x(Pix0.5x30) + (2000x0.49x1.09x1.02)
= 44.52C

therefore the temperature drop is 50 - 44.52 = 5.48K

Now,

This can be reduced with the introduction of ductwork insulation and to calculate this you need to subtract the U value of the insulation value from the ductwork U value to give a new overall U value

:)

Hope your HNC maths are up to it :D