Evaporative cooling on home AC

[cbfull]

I came up with this idea many years ago (of course I later discovered that this is not a new idea at all) upon realizing that the outdoor unit (condenser, right?) sits in direct sunlight for many hours each day, causing obvious stress on the compressor and excessive power consumption.

I decided to try aiming a wide spread misting nozzle at the fins to increase the heat transfer, which appeared to have an immediate effect since the water on the fins would evaporate very quickly.

After running it this way for a little more than a week, the AC breaker tripped twice. At first I thought something was being shorted by the water, but I later reasoned that if the rate of heat removal is increased on that end of the cycle, the rate of heat gain should perhaps increase on the other end to restore the overall heat transfer balance between the two extremes.

The first thing I did was completely remove the Filtrete Dust & Pollen 600 filter from the airflow to increase the rate of air flow over the evaporator.

I noticed a fairly big change in the rate at which the house temperture dropped. The temperature inside dropped from 78°F down to 73°F in just a few hours. Our AC has NEVER been able to cool that fast. It used to take almost an entire afternoon just to drop a couple of degrees. Keep in mind this is all while the sun is hitting the outside unit.

I hope to eventually measure the actual power consumption of the system, just to have solid numbers. It also gives me a little peace of mind that the water throughout our house is filtered and softened, so as not to introduce any scaling on the condenser.

What we decided to do about the air flow over the evaporator, was to use the minimum air filtering requirements during the summer when the AC is on, and go back to the allergen reducing filter in the winter.

Anyone else have any experiences (doing the same thing on your own home AC) or comments on my setup or logic? I am a bit partial to effeciency over mechanics when it comes to these things so I need the input.

Thanks everyone,

Craig

[Brian_UK]

It's not uncommon to provide additional water cooling to an air cooled condenser to help through the 'high' temperatures.

Not sure about your breaker tripping but as you say, some solid numbers are needed to help with diagnosis.

[cbfull]

I was thinking that perhaps the imbalance created by very high heat transfer out and fairly low heat transfer into the system, may be causing the system to bog down with too much condensed refrigerant going to the compressor, and not enough leaving causing a back-up or flooding of sorts.

I don't know exactly what can go wrong in these systems, so I thought I better ask some more knowledgeable people. Is my train of logic at least on the right track?

[US Iceman]

I would NOT recommend spraying water onto an air-cooled condenser. Any minerals in the spray water will precipitate onto the condenser coil as the water evaporates. After a short time, your condenser will start to accumulate scale which greatly reduces the heat transfer and dramatically increases the condensing temperature. This is a fast way to shoot yourself in the foot.

Circuit breakers respond to two thing, over amps and over temp. Check your electrical connections to be sure they are tight. Loose connection cause the amperage to increase over the loose connection which cause the temperature to also increase.

If anything, the water spraying onto the condenser should have reduced your amperage well below the normal "trip" point.

I've seen the same problem with the air filters also. The better filters require more static pressure which in turn reduces the total airflow.

PS. I just saw your comments about the use of soft water. Ooops. Are you sure of the compatibility of the soft water and the materials in the condenser?

[Brian_UK]

The condensed refrigerant has already left the compressor, it is now on its way to the evaporator.

[mcjo tech]

This system has been developed before. Rheem had a water cooled assist condenser almost 20 years ago. These systems were prior to SEER ratings so later studies on these systems found them to be in excess of 16 Seer. Some of them are still around today in southwestern USA. The problem with scale buildup and high maintenance costs to remove scale as well as water consumption in a dry environment where water has a high cost lead to their demise although, they did make a model AC-2 for a short period of time. These systems worked best in a dry environment similar to evaporative cooling that is very common in the southwest. Not exactly sure why you would trip a breaker. Don't think it is related. A evaporator freeze condition is much more likely when water is sprayed onto the outdoor coil lowering the suction pressure below freezing. We often use a sprinkler to cool down condenser to provide temporary cooling to a customer when we need to order a new outdoor fan motor. Not the best, but does work in a pinch. Good idea though, We are working on some similar technology that would not cause a scale buildup and could be fitted to exsisting systems and use less water.

[A_Ahed]

Dear Craig
In hot tropical conditions when it is humid the capacity of a typical system increase because :
1-increase in mass flow so a better heat rejection
2-evaporation on fins
But about the problem that u had I reckon as u subcool the liquid ur cooling capacity increases it means u can increase the fans volumetric flow . If u do not increase it u may have a frost formation on evaporator and then decrease in heat transfer and finally lower capacity .
Lemme add if the suction pressure drops due to subcooling (immpossible during summer),it will cause low pressure switch to stop the compressor

[mohamed khamis]

Dear Craig
In hot tropical conditions when it is humid the capacity of a typical system increase because :
1-increase in mass flow so a better heat rejection
2-evaporation on fins


Firstly i agree with u that the air mass flow rate entering air-cooled condenser will be higher in tropical country compared with mild one. However, take into account two things :

1- The ventilation or infiltration latent load in these countries are high and this will burden on the evaporator capacity and lowering the sensible effect. This makes an inefficient performance for the evaporator to carry over the latent load if it is existed as a result of quite short cycling for the thermostat in humid ventilation or infiltration.

2- If the air enters the condenser is moist this tends to a fouling factor for the air to play its role to reduce drastically the condenser performance and periodically maintenance should be occurred. 1 mm of dust drops heat transfer coefficient by 50%.




But about the problem that u had I reckon as u subcool the liquid ur cooling capacity increases it means u can increase the fans volumetric flow . If u do not increase it u may have a frost formation on evaporator and then decrease in heat transfer and finally lower capacity .
Lemme add if the suction pressure drops due to subcooling (immpossible during summer),it will cause low pressure switch to stop the compressor

firstly the enhancement in heat rejection accounts to drop in condensing pressure and
suction pressure as well lowering the refrigerant temperature entering the TXV (this not mean the degree of subcooling is increased). In contrary, the lowering in condensing pressure decreases the degree of subcooling on expense of the increase of the latent heat. the evidence for that u when undercharge the system (low condensing pressure) the degree of subcooling is increased as a result of the area occupied by the phase-change heat transfer.

Yes the cooling capacity of TXV is increased responding to the lowering temperature of subcooling with provision of the same pressure difference between the condenser and evaporator and therefore the evaporator capacity is also increased. This produces a more drop in discharge air temperature without frosting. It is very scarce for AC application for frosting to occur even the suction pressure is lowered as this condition. the normaly Tev is 7.5 to 10°C and to the frosting occurs this means the air filter is blockaged

Cheers

[Abby Normal]

Firstly i agree with u that the air mass flow rate entering air-cooled condenser will be higher in tropical country compared with mild one. However, take into account two things :

1- The ventilation or infiltration latent load in these countries are high and this will burden on the evaporator capacity and lowering the sensible effect. This makes an inefficient performance for the evaporator to carry over the latent load if it is existed as a result of quite short cycling for the thermostat in humid ventilation or infiltration.

2- If the air enters the condenser is moist this tends to a fouling factor for the air to play its role to reduce drastically the condenser performance and periodically maintenance should be occurred. 1 mm of dust drops heat transfer coefficient by 50%.



firstly the enhancement in heat rejection accounts to drop in condensing pressure and
suction pressure as well lowering the refrigerant temperature entering the TXV (this not mean the degree of subcooling is increased). In contrary, the lowering in condensing pressure decreases the degree of subcooling on expense of the increase of the latent heat. the evidence for that u when undercharge the system (low condensing pressure) the degree of subcooling is increased as a result of the area occupied by the phase-change heat transfer.

Yes the cooling capacity of TXV is increased responding to the lowering temperature of subcooling with provision of the same pressure difference between the condenser and evaporator and therefore the evaporator capacity is also increased. This produces a more drop in discharge air temperature without frosting. It is very scarce for AC application for frosting to occur even the suction pressure is lowered as this condition. the normaly Tev is 7.5 to 10°C and to the frosting occurs this means the air filter is blockaged

Cheers

Warm humid tropical air is less dense. A water molecule is considerably lighter than an oxygen molecule or a nitrogen molecule. A cold wind is a higher mass flow rate than a hot wind, given equal velocity.

When the load is very latent, you want the sensible capacity to be greatly reduced. There is plenty of heat to grab from humid air. If you have too much sensible and not enough latent you are either oversized or in an application that needs reheat.

Surely the load of outside air in Malaysia must be about 80% latent.

Lowered sensible effect means it keeps running trying to remove moisture. Maybe something was lost in the translation.

[mohamed khamis]

Warm humid tropical air is less dense. A water molecule is considerably lighter than an oxygen molecule or a nitrogen molecule. A cold wind is a higher mass flow rate than a hot wind, given equal velocity.

Yes it is correct, it was established in my thought the humid air is more dense than the dry as a result of increasing in the vapor pressure. howvere, when i checked it out on the pschrometery chart i found u r right. An i am still trying to find out explanation for this, oh may be the increase in number of moelculs is overbalanced the increase in pressure so the density drops.

density = total pressure / (N * M*R*T)

although the more humid air the more increase in total air pressure (as a vapor content increases) but it is trivial increase. On the other hand, the number of molecules N increases also, this leads to a drop in density.

However as u mentioned "A water molecule is considerably lighter than an oxygen molecule or a nitrogen molecule" ok it is correct but the molecule are summed according to Daleton's LAW for thermal equilbruim mixture.

In Gas turbine power plant, they sometimes inject water droplets in air path before entering compressor and the purpose of that as they claimed the air flow is increased in the cycle due to the air density is increased from air humidifications so my thought came from this. However, now i can say the air density yes increases but not from increasing the humidity level but from the adiabatic cooling from air washing, the air temperature drops and so the density increase against a slight increase in pressure.



When the load is very latent, you want the sensible capacity to be greatly reduced. There is plenty of heat to grab from humid air. If you have too much sensible and not enough latent you are either oversized or in an application that needs reheat.

No .... if u have very latent load u need oversized coil with reheater. when the coil is oversized that means u increase the sensible effect of the coil to get more dehumidification. ok i will give example to understand what i mean:

Suppose u have two coils one has discharge air temperature of 15°C and the other has 12°C and both coils are installed in the same conditions (high latent with realtively small sensible cooling load). And both system thermostat is set on 24°C. Which system will have ability to carry over much latent heat?

Obviously the second one (12C) because the thermostat
time cycling on is bigger plus the moisture removal is higher and this system has also to carry over much sensible heat. However, to avoid overcooling u have to equip heater to resume the air at 15°C and after u surely dehumidified better the moist air and when the set-point of thermostat reaches the indoor humidity is located in comfort zone of 50 to 60% rH.

Cheers

[Abby Normal]

Yes it is correct, it was established in my thought the humid air is more dense than the dry as a result of increasing in the vapor pressure. howvere, when i checked it out on the pschrometery chart i found u r right. An i am still trying to find out explanation for this, oh may be the increase in number of moelculs is overbalanced the increase in pressure so the density drops.

density = total pressure / (N * M*R*T)

although the more humid air the more increase in total air pressure (as a vapor content increases) but it is trivial increase. On the other hand, the number of molecules N increases also, this leads to a drop in density.

However as u mentioned "A water molecule is considerably lighter than an oxygen molecule or a nitrogen molecule" ok it is correct but the molecule are summed according to Daleton's LAW for thermal equilbruim mixture.

In Gas turbine power plant, they sometimes inject water droplets in air path before entering compressor and the purpose of that as they claimed the air flow is increased in the cycle due to the air density is increased from air humidifications so my thought came from this. However, now i can say the air density yes increases but not from increasing the humidity level but from the adiabatic cooling from air washing, the air temperature drops and so the density increase against a slight increase in pressure.



No .... if u have very latent load u need oversized coil with reheater. when the coil is oversized that means u increase the sensible effect of the coil to get more dehumidification. ok i will give example to understand what i mean:

Suppose u have two coils one has discharge air temperature of 15°C and the other has 12°C and both coils are installed in the same conditions (high latent with realtively small sensible cooling load). And both system thermostat is set on 24°C. Which system will have ability to carry over much latent heat?

Obviously the second one (12C) because the thermostat
time cycling on is bigger plus the moisture removal is higher and this system has also to carry over much sensible heat. However, to avoid overcooling u have to equip heater to resume the air at 15°C and after u surely dehumidified better the moist air and when the set-point of thermostat reaches the indoor humidity is located in comfort zone of 50 to 60% rH.

Cheers

Don't try so hard

Consider a container holding a finite number of molecules, oxygen with a mass of 32 and nitrogen with a mass of 28.

Now start dropping in molecules of water vapour with a mass of 18. Drop in one water molecule, displace either a heavier oxygen or nitrogen molecule. Air gets lighter as you humidify it.

There are many strategies for dealing with a latent load but it goes back to air bypassing or making contact.

You can have a deeper row coil to increase contact, or slow the air speed down so less bypasses.

Smaller coils will run colder, air making contact will release more mositure.

Oversizing the coil does a few things. With DX it raises the suction pressure and therefore the temperature, so the air that makes contact gives up less moisture. It also gives air more surface area to contact and therefore reduces the amount of air that bypasses the coil.

Over sizing the coil and not slowing down the airflow can end up reducing latent capacity.

You need reheat when you have an 'impossible process', meaning that a coil alone cannot supply air at both the correct temperature and moisture content. In situations where there is no ADP possible for what you need. It leaves you no choice but to over cool the air to get the mositure out, then reheat the air back up to the dry bulb temperature needed.

[Abby Normal]

If you pay attention to the latent load, and you control how air moves in and out of a building, I think you will find that the confort zone can be extended down to 40%, even in a humid tropical location.

[mohamed khamis]

Well ... u have a big confusion about the mixture calculations and coil design. This is my reply


Don't try so hard

Consider a container holding a finite number of molecules, oxygen with a mass of 32 and nitrogen with a mass of 28.

Now start dropping in molecules of water vapour with a mass of 18. Drop in one water molecule, displace either a heavier oxygen or nitrogen molecule. Air gets lighter as you humidify it.


firstly u know how to deal the mixture or Delatons' Law...I think if u know that u did not say all the previous
When u have mixture of gases as u mentioned with molecular Weight(M) not mass as u mentioned.The total mass of the mixture is

m = No2 *Mo2 + Nn2*Mn2 + Nh2o * Mh2o

As u said Mo2 = 32

Mn2 = 28

Mh2o = 18

ok now if u drop in water vapor in ur mixture the number of molecules of water is surely increased without any action in those for oxygen or nitrogen neither increase nor decrease. As u might know the percentage for dry gas (oxygen, nitrogen, co2, argon,...ect) in the atmosphere is fixed but the water vapor content is different from zone to zone. AS U claimed the additional amount of water vapor will displace the heavier component which is the oxygen (32). This means here we can find oxygen to survive and perhaps we have a lot of deceased due to lack of oxygen :D.

these things are called basic fundamentals for thermodynamics.



There are many strategies for dealing with a latent load but it goes back to air bypassing or making contact.

You can have a deeper row coil to increase contact, or slow the air speed down so less bypasses.


Loll..can u tell me bypass factor function in what parameters.

bypass factor = 1/EXP(NTU)

NTU = UA/Cmin

In principle;

U have a fixed value of surface area (A) corresponding to the required cooling capacity. So a deeper coil row means a smaller frontal area and u need a certain air mass flow rate.

A = Afrontal * Nrows (approx.)

As Nrows increases the Afrontal decreases to get the same Area, A.

air mass flow rate = density * Afrontal* air velocity

As Afrontal decrease the air velocity increases to fix mass flow rate .

Back to ur claim,

1-u r saying "a deeper row coil to increase contact" ok it is accepted if u fix the frontal area.

2- u r saying "slow the air speed down so less bypasses" it is not accepted because u will deteriorate the air coefficient heat transfer and so the overall heat transfer U and increase the bypass factor as the previous equation.



Smaller coils will run colder, air making contact will release more mositure.

On what basis u define smaller coil if u mean smaller frontal area this drives u to get smaller CFM and it is defined from cooling load calculation unless u will offest it by incraesing frontal air velocity.



Oversizing the coil does a few things. With DX it raises the suction pressure and therefore the temperature, so the air that makes contact gives up less moisture. It also gives air more surface area to contact and therefore reduces the amount of air that bypasses the coil.

NO...If u want to select an evaporator with excess of cooling capacity u select the TD is bigger than required this in turn to a smaller suction temperature and pressure for the same evaporator inlet air temperature. The excess of area enhances the contact factor and enhance the moisture removal as a smaller Tev.



Over sizing the coil and not slowing down the airflow can end up reducing latent capacity.


Again No...the latent heat is removed Coincidentally with the sensible heat if increase the sensible effect u get more humidification to limit not causing overcooling. So the reheater is added to negate the overcooling process.

Eventually u need to read two these articles:

http://www.trane.com/commercial/library/vol33_2/ADMAPN011EN_0704.pdf

http://www.cdhenergy.com/presentations/ASHRAE%20KC%202003%20Part-Load%20Dehumidification,

Cheers:)

[Abby Normal]

Well ... u have a big confusion about the mixture calculations and coil design. This is my reply




firstly u know how to deal the mixture or Delatons' Law...I think if u know that u did not say all the previous
When u have mixture of gases as u mentioned with molecular Weight(M) not mass as u mentioned.The total mass of the mixture is

m = No2 *Mo2 + Nn2*Mn2 + Nh2o * Mh2o

As u said Mo2 = 32

Mn2 = 28

Mh2o = 18

ok now if u drop in water vapor in ur mixture the number of molecules of water is surely increased without any action in those for oxygen or nitrogen neither increase nor decrease. As u might know the percentage for dry gas (oxygen, nitrogen, co2, argon,...ect) in the atmosphere is fixed but the water vapor content is different from zone to zone. AS U claimed the additional amount of water vapor will displace the heavier component which is the oxygen (32). This means here we can find oxygen to survive and perhaps we have a lot of deceased due to lack of oxygen :D.

these things are called basic fundamentals for thermodynamics.



Loll..can u tell me bypass factor function in what parameters.

bypass factor = 1/EXP(NTU)

NTU = UA/Cmin

In principle;

U have a fixed value of surface area (A) corresponding to the required cooling capacity. So a deeper coil row means a smaller frontal area and u need a certain air mass flow rate.

A = Afrontal * Nrows (approx.)

As Nrows increases the Afrontal decreases to get the same Area, A.

air mass flow rate = density * Afrontal* air velocity

As Afrontal decrease the air velocity increases to fix mass flow rate .

Back to ur claim,

1-u r saying "a deeper row coil to increase contact" ok it is accepted if u fix the frontal area.

2- u r saying "slow the air speed down so less bypasses" it is not accepted because u will deteriorate the air coefficient heat transfer and so the overall heat transfer U and increase the bypass factor as the previous equation.



On what basis u define smaller coil if u mean smaller frontal area this drives u to get smaller CFM and it is defined from cooling load calculation unless u will offest it by incraesing frontal air velocity.





NO...If u want to select an evaporator with excess of cooling capacity u select the TD is bigger than required this in turn to a smaller suction temperature and pressure for the same evaporator inlet air temperature. The excess of area enhances the contact factor and enhance the moisture removal as a smaller Tev.



Again No...the latent heat is removed Coincidentally with the sensible heat if increase the sensible effect u get more humidification to limit not causing overcooling. So the reheater is added to negate the overcooling process.

Eventually u need to read two these articles:

http://www.trane.com/commercial/library/vol33_2/ADMAPN011EN_0704.pdf

http://www.cdhenergy.com/presentations/ASHRAE%20KC%202003%20Part-Load%20Dehumidification,

Cheers:)

Keep the number of moles constant as you are not pressurizing the container. Only so many molecules fit into the container. If you add molecules with a mass of 18, you displace molecules with a mass of 28 and 32. Moist air is less dense.

The pyshcrometric chart tells you the volume of moist air that holds a certain mass of dry air. That volume also contains water vapour.You need to understand this before yuou quote me Dalton's Law of partial pressure. Make an effort to understand the concepts you quote before you quote them.

I skimmed over the Trane link on the woes of Part Load Dehumidification, and they appear to recycle a 1998 ASHRAE IAQ paper by Hugh I. Henderson. They comment on how constant fan on an evaporator is an evpaorative cooler when the compressor cycles off. Water on the coil is re-evaporated.

Perhaps in your Trane link read the part about "Avoid using SAT reset..." with DX, if you over size the coil you are inadvertantly raising the SAT.

Samller coil will remove more moisture with DX as it has to run colder. Slight reduction in total capacity, significant increase in latent capacity.

Example of a smaller coil. Evaporator coil sized for a nominal 2 tons of air conditioning would have a face area of about 1.6 square feet. Connect this coil to a condenser with a nominal 2.5 tons capacity. Suction pressure will balance out low, most likely an SST of 40F or even a little cooler, perhaps the total capacity would be a little shy of 2.5 tons, most likely SHR be 0.70 or a little less.

With DX, the evaporator capacity drops as the suction temperature rises. The compressor is the opposite, its capacity increases as suction pressure rises.

A larger coil then permits heat transfer at higher temperatures. Higher temperature means less moisture removal from air that contacts the coil surface.

This is half the secret of the high SEER air conditioning. Make the evaporator larger to get the heat transfer at a higher temperature and suction pressure. Higher suction pressure means you can use a smaller compressor. Increase condensing surface to lower condensing pressure. Smaller compressor, less differential between evaporating and condensing pressures, less work the compressor has to do. An unfortunate side effect of all this is now, the system wants an SHR of upwards to 0.80 to make its high SEER rating. Latent removal tends to suffer.

Write some of these down and then later observe how you are not stuck with 60% RH anymore. Properly pay attention to the latent load and conytrol how air moves in an out of the building and enjoy 40% RH in Malaysia

[Abby Normal]

your second link was invalid

[mohamed khamis]

Keep the number of moles constant as you are not pressurizing the container. Only so many molecules fit into the container. If you add molecules with a mass of 18, you displace molecules with a mass of 28 and 32. Moist air is less dense.

The pyshcrometric chart tells you the volume of moist air that holds a certain mass of dry air. That volume also contains water vapour.You need to understand this before yuou quote me Dalton's Law of partial pressure. Make an effort to understand the concepts you quote before you quote them.

Well,,, Unfortunately u can not differentiate between closed system and open system. The example which u gave me is for closed system, which the process of humidification undergoes Constant Volume and if add more moisture (without increase pressure )the density will inevitably of the mixture is increased due to increase molecules number. And the pyshcrometric chart that u r taking about was made for constant pressure air for open system. If u add steam with saturation temperature equals to air dry bulb temperature the density of dry air will be decreased and i told u before I agree with u the moist air is less dense. But i disagree with ur explanations , my explanation here is the volume of the dry is increased as increase in added moisture but the mass of dry air is fixed constant so the net product is the less density for the dry air. Regarding to Delton's law tell me what is the wrong in what i said. can u tell me its formula.? u might know if add o2 with N2 in the same container

Vtotal = Vo2 = Vn2

Ttotal = To2 = Tn2

Ptotal = Po2 + Pn2

mtotal = mo2 + mn2 = no2*Mo2 + nn2*Mn2

dens total = densio2 + densn2

This is Delton law and if u have another law pls quote it here and thanks in advance

Regrading to the latent load and its effect, tomorow i will compelete with u because now here so late.

Cheers:)

[Abby Normal]

Well,,, Unfortunately u can not differentiate between closed system and open system. The example which u gave me is for closed system, which the process of humidification undergoes Constant Volume and if add more moisture (without increase pressure )the density will inevitably of the mixture is increased due to increase molecules number. And the pyshcrometric chart that u r taking about was made for constant pressure air for open system. If u add steam with saturation temperature equals to air dry bulb temperature the density of dry air will be decreased and i told u before I agree with u the moist air is less dense. But i disagree with ur explanations , my explanation here is the volume of the dry is increased as increase in added moisture but the mass of dry air is fixed constant so the net product is the less density for the dry air. Regarding to Delton's law tell me what is the wrong in what i said. can u tell me its formula.? u might know if add o2 with N2 in the same container

Vtotal = Vo2 = Vn2

Ttotal = To2 = Tn2

Ptotal = Po2 + Pn2

mtotal = mo2 + mn2 = no2*Mo2 + nn2*Mn2

dens total = densio2 + densn2

This is Delton law and if u have another law pls quote it here and thanks in advance

Regrading to the latent load and its effect, tomorow i will compelete with u because now here so late.

Cheers:)The air we breathe is an open system it moves in and out of buildings. The number of molecules inside remains constant unless of course there is pressurization or temperature changes.

If you add moisture to the air in a building it displaces O2 and N2 out of the building. Use daltons law, ideal gas law, and come up with the maximum number of water molecules you can have in that air, it is related to the vapour pressure of the water.

[Abby Normal]

Use the p-chart

Consider a space with a volume of 500 cubic meters filled with air at 25C db and 20% RH. Work out the mass of dry air and water vapour held in this volume of 500 cubic meters.

Now consider air at 25C db and 100% RH inside that volume of 500 cubic meters. Again work out the mass of dry air and water vapour inside that volume. Is there less dry air now than compared to 20% RH?

[mohamed khamis]

Hi Abby




Perhaps in your Trane link read the part about "Avoid using SAT reset..." with DX, if you over size the coil you are inadvertently raising the SAT.

Ok u know what does it mean by this sentence, u asked me to exert an effort to understand the concepts then quite it and it is very of great to say to u the same phrase.

The meaning of “Avoid using SAT reset”, In cooling coil capacity control there are many types of control parameters, one of them is the supply air temperature SAT is fixed constant or reset at a certain value whatever is happened to the cooling load by modulation system capacity. So the article author recommend don not use this type of control when u have a big latent heat during low sensible load. Because if u have SAT at a design value the system will cycle off quickly (because the supply air is considered wormer in this case) and prevents the latent load to be removed efficiency.

Again, plz read carefully the article, the oversizing here it means the CFM is bigger than the design CFM therefore from thermodynamic principles higher mass flow rate lead to small DT

Q = m* C*DT and cause the supply air temperature SAT to be wormer and aggravate the problem of removing moisture



Samller coil will remove more moisture with DX as it has to run colder. Slight reduction in total capacity, significant increase in latent capacity.


Yes it is correct but it is not open-ended it has limitation :

1- For the same cooling capacity, the smaller coil size corresponds to the larger TD i.e. evaporating temperature is smaller but to the limit above the zero to avoid frosting, am I right or wrong . If u say yes correct the TEv is bove the zero by 3 or 4 K, okay what about partial-load condition, the TEv must be dropped and it will drop ineviatably under zero to handle the drop in the cooling load imposed, the frosting will be popped in ur coil. If u say no the TEv should above that and this is the normal design it is of 7.4 for AC purpose.
2- As u mentioned in the last paragraph, the decrease in TEv will make the compressor has a large size and consumes a lot of power and this is not favourbly design in terms of SEER
3- The decrease in TEv accounts to decrease in refrigerating effect and in turn to a higher refrigerant velocity for the same design conditions and therefore an excessive pressure drop in evaporator and burden on the compressor power




A larger coil then permits heat transfer at higher temperatures. Higher temperature means less moisture removal from air that contacts the coil surface.

Yes it is correct and I agree with u but there is a comprise should be taken between the compressor power and removal moisture as I abovementioned.


Cheers

[Abby Normal]

Hi Abby




Ok u know what does it mean by this sentence, u asked me to exert an effort to understand the concepts then quite it and it is very of great to say to u the same phrase.

The meaning of “Avoid using SAT reset”, In cooling coil capacity control there are many types of control parameters, one of them is the supply air temperature SAT is fixed constant or reset at a certain value whatever is happened to the cooling load by modulation system capacity. So the article author recommend don not use this type of control when u have a big latent heat during low sensible load. Because if u have SAT at a design value the system will cycle off quickly (because the supply air is considered wormer in this case) and prevents the latent load to be removed efficiency.

Again, plz read carefully the article, the oversizing here it means the CFM is bigger than the design CFM therefore from thermodynamic principles higher mass flow rate lead to small DT

Q = m* C*DT and cause the supply air temperature SAT to be wormer and aggravate the problem of removing moisture


Moisture removal comes down to humid air making contact with a coil surface that is at or below the dewpoint of that humid air mixture. The colder that surface is, the more moisture condenses out.

The air does not care what the NTU is, nor do I care how many formulae you quote. m x CP x DT describes sensible heat trransfer, we are talking about dehumidifying.

Raising the SAT, raises the coil temperature and reduces how much moisture condenses out. Over sizing a DX coil will make it run warmer and will reduce how much mositure condenses out of the air contacting it.

Over sizing the coil for a larger face area, gives you more area for the air to contact (but it is still warmer) so with more area to make contact with, it means less air bypasses the coil surface.

But the oversized coil is warmer, and is a "BASIC THERMODYNAMIC" move in the opposite direction as to what causes moisture to condense out of air.




Yes it is correct but it is not open-ended it has limitation :

1- For the same cooling capacity, the smaller coil size corresponds to the larger TD i.e. evaporating temperature is smaller but to the limit above the zero to avoid frosting, am I right or wrong .


When you are dehumidifying mechanically you will always have to stop at the barrier of freezing the condensate.




If u say yes correct the TEv is bove the zero by 3 or 4 K, okay what about partial-load condition, the TEv must be dropped and it will drop ineviatably under zero to handle the drop in the cooling load imposed, the frosting will be popped in ur coil. If u say no the TEv should above that and this is the normal design it is of 7.4 for AC purpose.

Is TEv a thermal expansion valve?

Part load condition, if you do not have an inverter the compressor cycles off does it not?

If you have it set up properly, you should not have the coil freezing up unless no one changes the filters. You should have a pretty steady load in Malaysia, especially with some ventilation air mixed in, year round.



2- As u mentioned in the last paragraph, the decrease in TEv will make the compressor has a large size and consumes a lot of power and this is not favourbly design in terms of SEER

This is where properly controlling how air moves in and out of the structure is important.

If you neglect occupancies with a high density of people like an Assembly hall, or worse yet a night club where the occupants are very active, if you did not have to worry about the humidity in the outside air that infiltrated into a building, it would be almost impossible to oversize an air conditioner to the point of causing high RH.

Building the structure tight, pressurizing the sturcture with dry air, will greatly reduce the amount of mositure removal needed.

High seer residential equipment, will respond to a call to dehumidify by acting like low SEER equipment. Air flow is slowed down so that the coil runs colder, the space is overcooled by perhaps 3F. Adding reheat other than hot gas reheat to the high SEER equipment blows the energy budget also.



3- The decrease in TEv accounts to decrease in refrigerating effect and in turn to a higher refrigerant velocity for the same design conditions and therefore an excessive pressure drop in evaporator and burden on the compressor power

If TEv is the SST, saturated suction pressure and not the Thermal expansion valve I do not really see what your point is here.

Compressor capcity increases with suction pressure and has nothing to due with friction of refrigerant flowing through an evaporator coil.

You are talking about it like it is going to be like a restriction. The pressure drop of an evaporator coil is minute with respect to the pressure differential needed to pass the flow of refrigerant through a TX valve.






Yes it is correct and I agree with u but there is a comprise should be taken between the compressor power and removal moisture as I abovementioned.

Cheers

I am talking from the point of view that I deal with a climate similar to yours and that maintaing humidity under 50% is possible.

If you are viewing the comfort zone as being 50 to 60%, it sounds to me like you are settling for the best you can do with your oversized coils.

[Abby Normal]

I never seem to be able to edit typos lately or use the quick replies on this forum but in the above post I made a slight error

I wrote

If TEv is the SST, saturated suction pressure and not the Thermal expansion valve I do not really see what your point is here.


What I meant to say was

If TEv is the SST, saturated suction temperature and not the Thermal expansion valve I do not really see what your point is here.

[mohamed khamis]

Ok it sounds this thread will be over-lengthened



Moisture removal comes down to humid Air making contact with a coil surface that is at or below the dewpoint of that humid Air mixture. The colder that surface is, the more moisture condenses out.

It is not incomplete claim; the coil surface must be less than the dew point temperature for entering air with sufficient cooling capacity to dehumidification to occur. There are adherent parameters to achieve dehumidification, try this link to understand more;

http://www.refrigeration-engineer.com/forums/showthread.php?t=8447&highlight=point



The Air does not care what the NTU is, nor do I care how many formulae you quote. m x CP x DT describes sensible heat trransfer, we are talking about dehumidifying.


Well ….. care or don not care it does not matter and make any difference with me but I have to use formula as u r talking about scientific demonstration. As u mentioned now the air does not care about the NTU and I can not find any answer to reply u except using formula which bothers u, however:

Suppose u have two coils (A &B) with different NTU of 3 and 4 and both of coils subject to the same operating conditions dbti = 27 and 60% Rh and Cmin = 2 with surface area of 10 m2 and evaporating temperature is of 5C, tell me which of these coil will discharge or supply colder air A or B?, if u can answer this question u can reply urself otherwise I have to use the bothered formula.

Regarding to “m x CP x DT” yes it is good to note it but I gave it to show to u the higher mass flow leads to small DT and it does not matter with dehumidification process where the equation will be “m x Dh” and the Dh is the difference enthalpy which functions in the difference temperature and thanks for great note.






Raising the SAT, raises the coil temperature and reduces how much moisture condenses out. Over sizing a DX coil will make it run warmer and will reduce how much mositure condenses out of the Air contacting it.

Also I don not find any answer to this claim, I do believe u don not know what is the meaning of SAT which I have written it in the previous post.

The SAT is the supply air temperature and tell me how u will raise it to raise the coil temperature and I think u want to say rasing the coil temperature will raise the SAT and at this is moment I agree.

Regading to the oversizing which I meant is the system has excess area over it design area at the same TD i.e.

Q = U * A *TD

If we have for example: at Tai (design air temperature) = 27 and RH = 60 % and Tev (evaporating air temperature) = 7 this TD = 20 and U = 2.00 kW/m2.K and Q = 40 kw so the A = 1 m2 and assume air flow rate (ma) of 1.0 kg/s

So we exceed 10 to 20 % over this area to have more capacity of the coil i.e. the cooling capacity will be 48 kW and the A = 1.2 for the same TD so when u select the compressor u select with cooling capacity of 48 and give the same design evaporating temperature. Note that the increase in A will be accomplished by increase in number of rows not frontal area to avoid increase in CFM.

No at partial load conditions the air temperature is dropped to let us say 20C and the same Rh of 60% what is the corresponding Tev and what is the mount of removal humidity in both cases

Case A ; the overdesign of 20% = A = 1.2

Case B : the perfect design = A = 1.0

Assume the Q and U will drop by the same amount of reduction for the Tai (29.3% from 27C to 20 C)

Case A

Q = 33.936 kW and U = 1.414

Therefore, TD = 33.936 / (1.414*1.2) = 20

Therefore Tev = 0 C

Assume the SAT (supply or discharge air temperature) equals to the Tev = 0 therefore,

Q = ma *1.005* (20 – 0) + ma * 2501 * (Wi – Wo)

From psychometric chart at condition of 20 C and 60 % the Wi = 8.7 g/kg therefore,

33.936 = 1.0 *1.005*20 + 1.0 * 2501* (8.7 – Wo)/1000

Thus Wo = 3.167 g/kg


Case B

Q = 28.28 kW and U = 1.414

Therefore, TD = 28.28 / (1.414 *1.0) = 20

Therefore Tev = 0 C (Note that it is the same like overdesign)

However

Q = ma *1.005* (20 – 0) + ma * 2501 * (Wi – Wo)

From psychometric chart at condition of 20 C and 60 % the Wi = 8.7 g/kg therefore,

28.28 = 1.0 *1.005 * 20 + 1.0*2501*(8.7 – Wo)/1000

Wo = 5.429 g/kg >> 3.167 g/kg (that means the overdesign will make the air to be more dehumidified than the perfect design)

I think after this example no argument about this issue. If u will say that u mean smaller design to increase TD and increase the humidity as u repeated it each post u have limitation with TD u can not overstep it. I wish it is clear

Cheers

[mohamed khamis]

When you are dehumidifying mechanically you will always have to stop at the barrier of freezing the condensate.


How do u do that and u want to design the evaporating temperature close to zero otherwise u have a relatively constant cooling load at ur design conditions and this another issue.




Is TEv a thermal expansion valve?

Part load condition, if you do not have an inverter the Compressor cycles off does it not?

If you have it set up properly, you should not have the coil freezing up unless no one changes the filters. You should have a pretty steady load in Malaysia, especially with some ventilation Air mixed in, year round.

I think from intuitive perception TXV is the terminology is the thermostatic expansion valve while Tev is the temperature and it is evaporating temperature.

I don not what u mean by inverter here but if the frosting is occurred the compressor low pressure cut out will shut down the compressor and we say adieu for system to work in partial load conditions

Cheers

[mohamed khamis]

I never seem to be able to edit typos lately or use the quick replies on this forum but in the above post I made a slight error

I wrote

If TEv is the SST, saturated suction pressure and not the Thermal expansion valve I do not really see what your point is here.


What I meant to say was

If TEv is the SST, saturated suction temperature and not the Thermal expansion valve I do not really see what your point is here.

Never mind i got it earlier and in fact there is a differnce between the Tev (evaporating temperature) which in the TD and SST which it is supposed to use to when u select the compressor for the system. the difference between them the superheat degree and if there is suction-line heat exchanger the difference will be larger.

Cheers:)

[Abby Normal]

SAT is the supply air temperature. If you raise the SAT you are raisng the coil temperature

i answered a lot of your concerns in detail, please review.

[mohamed khamis]

SAT is the supply air temperature. If you raise the SAT you are raisng the coil temperature

i answered a lot of your concerns in detail, please review.

Ok I don not have any comments for this phrase and i wish u or any body explain it to me and i am so grateful in advance. And I am so thankful for ur time and good discussion.

Best regards:)

[Abby Normal]

Trane's link on part load dehumidification strongly recommended not resetting (aka raising) the SAT.

If you keep air flow and entering air condition constant,then to raise the supply air temperature raise the supply air temperature means the coil is warmer. Moisture removal is reduced.

Oversizing a DX coil raises the suction pressure and results in a warmer coil.

Did you ever go over the example of the the 500 cubic meter volume and realize that when the air is humid there is less dry air 'inside the container'?

You are digging up yet another basic heat transfer equation to try and prove your point. You are now using "Q = U x A x DT".

If I wanted to figure out how much heat was conducting through a wall, I would use that equation. I would not use that equation for a coil with sensible and latent heat transfer.

For a sensible coil or a chiller barrel with next to no superheat, you could use Q= U x A x LMTD, and the LMTD term would be calculated differently depending on parallel flow, counter flow,or if it was a chiller barrel.

When it is a cooling coil with moisture condensing out and therefore mass not conserved, it is not useful equation other than to explain what happens when you pipe the coil backwards.

Basic fundamental of mechanical dehumidifying is to have air contact a surface at or below its dewpoint, the air making contact leaves saturated at that coil surface temperature. The colder the surface the air contacts, the more mositure is condensed out of the air making contact.

Oversizing DX coils will result in higher suction pressures and it is a step backwards when trying to dehumidify.

In your climate you probably think you are doing good if you can holder under 60% RH, and when I first relocated to the tropics 9 years ago, I thought that was the case. The truth is, you can hold under 50% also, and as long as it is not a dense active occupancy, you can hold under 50% without desicants or reheat.

[mohamed khamis]

Trane's link on part load dehumidification strongly recommended not resetting (aka raising) the SAT.

I have replied u about what is the meaning of avoiding SAT resetting and raising in part load to get better humidification and I again I have to repeat here.

The meaning of “Avoid using SAT reset”, In cooling coil capacity control there are many types of control parameters, one of them is the supply air temperature SAT is fixed constant or reset at a certain value whatever is happened to the cooling load by modulation system capacity. So the article author recommend don not use this type of control when u have a big latent heat during low sensible load. Because if u have SAT at a design value the system will cycle off quickly (because the supply air is considered wormer in this case) and prevents the latent load to be removed efficiency.



If you keep Air flow and entering Air condition constant,then to raise the supply Air temperature raise the supply Air temperature means the coil is warmer. Moisture removal is reduced.

How to raise the supply temperature? Bear in mind The supply temperature SAT is dependent and the coil temperature is the independent i.e. the increase in coil surface temperature accounts for an increase SAT and vice versa. U could not say I shall increase the SAT and this leads to increase in coil temperature the SAT has no authority on coil surface. Even if u have bypass flow in chilled water coil or Dx coil the SAT increases with any change in coil surface temperature.



Oversizing a DX coil raises the suction pressure and results in a warmer coil.

Yes it is completely correct for the same cooling capacity, there is misunderstanding between us i.e.

If I have two design for cooling capacity of 40 kW (the example I gave u early) and U = 2 kW/m2.C and I have two areas of A = 1.0 m2 and the other is 2.0 m2, and I have rather big latent load, blindly I will select the smaller one (A = 1.0 m2) which corresponds to the bigger TD of 20 K to get lower supply air temperature SAT and more dehumidification and both of us until now agree. However, after I select the area I will increase the number of rows by additional one-row to increase the area to get more Cooling capacity on the same conditions i.e. the same TD. When I go to select the compressor I will select it at the SST corresponds to the TD but the cooling capcity is more than 40 kW as a result of the increase in rows number. I wish u understand what I mean now by oversizing the coil capacity.





Did you ever go over the example of the the 500 cubic meter Volume and realize that when the Air is humid there is less dry Air 'inside the container'?

Really I could not swallow your explanation for why the dry air becomes less dense when it is humidified at constant pressure. However, I agree with u about this concept after I realized by my self on psychometric chart. However, ur explanation of the water vapor is lighter than the dry gas and this is correct and the lighter replaces the heavier I could have abilty to understand it, as I know the lighter floats the heavier and this explains why the clouds floats in the dry gases and the ice over the water too.

My explanation is this chart was established at specific condition at sea level (atmospheric pressure of 1 bar) and any excess of humidity the partial vapor pressure increases and to sustain the same value of the atmospheric pressure the partial dry gases must decrease and so it density. However the mixture density and mass will increase with no doubt.

Dry gas density = (atmospheric pressure – vapor pressure )/ Ngas *R*T



You are digging up yet another basic heat transfer equation to try and prove your point. You are now using "Q = U x A x DT".

If I wanted to figure out how much heat was conducting through a wall, I would use that equation. I would not use that equation for a coil with sensible and Latent Heat transfer.

For a sensible coil or a Chiller barrel with next to no superheat, you could use Q= U x A x LMTD, and the LMTD term would be calculated differently depending on parallel flow, counter flow,or if it was a Chiller barrel.

When it is a cooling coil with moisture condensing out and therefore Mass not conserved, it is not useful equation other than to explain what happens when you pipe the coil backwards.

Without digging up or down, Ur object the using of "Q = U x A x DT" and u r saying it is only valid for sensible heat transfer only and this is my answer:

I disagree with u it is valid and can be used in the simultaneous heat and mass transfer with some modification which is

U = 1/hiAi + SHF/ (ho Ao fin effeci) + conduction thermal resistance.

The sensible heat factor SHF modulates the correlation to be used during mass transfer and gives some what accurate results but not as it is used by therlkeld 1972 and try this link to verify by urslef:

And I use it with u to shorten the writing and explain in simple terms and on the same time u also object the using the simplified formula and what is happened if I use to u two lines formula as in my Msc thesis and before u say u must use LMTD instead of TD.

By the way if the system has no superheat as u mentioned in the chiller it does not matter the kind of flow is parallel or counter or even counter- parallel flow .




Basic fundamental of mechanical dehumidifying is to have Air contact a surface at or below its dewpoint, the Air making contact leaves saturated at that coil surface temperature. The colder the surface the Air contacts, the more mositure is condensed out of the Air making contact.

From Basic fundamental of mechanical dehumidifying, to have dehumidification the surface temperature must be under the entering air dew point temperature. It is impossible to the air to exit form the coil in saturation form except this coil has 100% effectiveness or contact factor of 1.0. It can reach saturation line in case of dehundifcation by direct contact by spraying chilled water temperature and with that the exit Rh is 98%.


[QUOTE=Abby Normal;74721
In your climate you probably think you are doing good if you can holder under 60% RH, and when I first relocated to the tropics 9 years ago, I thought that was the case. The truth is, you can hold under 50% also, and as long as it is not a dense active occupancy, you can hold under 50% without desicants or reheat. [/QUOTE]

In effect, I don not understand what u mean. Clarify it please

Cheers

[mohamed khamis]

sorry this is the website link

http://www.coolit.co.za/coilsim/coildesign.htm

Sorry for the invalid previous link

cheers

[Abby Normal]

You are missing the big picture. I am well aware of the formulae, but you are sounding like a salesman, who resorts to more and more technobabble, when he does not understand what he is talking about.

[mohamed khamis]

You are missing the big picture. I am well aware of the formulae, but you are sounding like a salesman, who resorts to more and more technobabble, when he does not understand what he is talking about.

Unfortunately it sounds u forgot that me who show u this formula when u was asking about how to treat with the humidification process:cool:, try this link to remind u if u forgot that;):

http://www.refrigeration-engineer.com/forums/showthread.php?t=8447&highlight=contact+factor&page=3

thank u so much for great ethics for calling me a salesman who does not understand what he sells, and for ur information I have two articles in reputable international Journals which are talking about the optimal design of evaporator and condenser design . If u want i send copies of them and u'r welcome.


If u r salesman who knows well what he sells, can u answer this equation? If u make comparison between two designs by using inaccurate formula to show which design is more efficient in terms dehumidification capability, is it a big deal? or u did not find any word to reply me except that.

As i see u r very expert in designing evaporator and condenser, stat the full governing equations for the design in either sizing or rating design without the talent of mouse clicking for the ready-made software...;)? if u can do that, u can make judgment on the others

Have a nice day

[mohamed khamis]

I forgot to tell also i have one which is published on well known International Heat Transfer Conference 13 which is set each 3 years and also is talking about evaporator design optimization using thermoeconomic approach.
http://www.edata-center.com/proceedings/IHTC13,69cdef06449c53d9,455ef27822d28638.html

[mohamed khamis]

By the way before i forget ..can u explain to me why in the refrigerant-side in the evaporator we don not care about the mass transfer coefficient and we care only with it on the air-side although both sides (air and refrigerant)there is a mass transfer phenomenon, boiling in refrigerant-side and condensation in air-side?. If u know this plz tell me because i want to increase the volume of my sales.:D

Cheers

[Abby Normal]

Unfortunately it sounds u forgot that me who show u this formula when u was asking about how to treat with the humidification process:cool:, try this link to remind u if u forgot that;):

http://www.refrigeration-engineer.com/forums/showthread.php?t=8447&highlight=contact+factor&page=3


If you look at that thread, I was the one who mentioned the concept of the bypass factor in the first place. You can calculate it based on temperatures, or using line segments on the chart as I mentioned later on.

On these forums, it is a red flag everytime someone mentions NTU or LMTD when talking about cooling with dehumidification.

You have a familiar posting pattern, you regurgitate a technical link, make incorrect comments on it, then you have to back track.

Do not try so hard to impress people. If you cannot explain a concept in simple laymans terms you do not understand the concept to begin with.

Salesman are notorious for this. When challenged on claims they make they try to sound more and more technical. Almost like trying to shame the layman into feeling stupid if he keeps asking the question that the salesman cannot really answer.

[mohamed khamis]

If you look at that thread, I was the one who mentioned the concept of the bypass factor in the first place. You can calculate it based on temperatures, or using line segments on the chart as I mentioned later on..

That's to tell u unfortuntly u r rush to reply without deep thinking, when i mentioned to u this thread to tell u that the concept of using LMTD method ot NTU method can be done with dehumdifcation with some specific treatemnt using of SHF and i don not mention contct factor to u. Think little bit then answer pls to save atime. See with urself and read and rwalize what u wrote firstly and then post it.



You have a familiar posting pattern, you regurgitate a technical link, make incorrect comments on it, then you have to back track.

quote to me one mistake i mentioned in my all posts and or at least reply me on my question if u see urself r quilifed and u can evaluate the others.



Do not try so hard to impress people. If you cannot explain a concept in simple laymans terms you do not understand the concept to begin with.

i did not try to impress anybody and i expalin to u by all means but u want to convince me that the lighter gas can replace the heavier and i don not know where the heavier will be existed after that, if it is the layman terms that u mean sorry i can not do it



Salesman are notorious for this. When challenged on claims they make they try to sound more and more technical. Almost like trying to shame the layman into feeling stupid if he keeps asking the question that the salesman cannot really answer.

All my repect to the laymen and if in effect they are stupid what the saleman can do woth them if he tries to expalin by all means and no response can be gotten from them and he repeats the answer of the repeated questions and gave figurtive speeking. after all that what can u expect from the salesman, personally i will say to him don not try hard the hard stone can not discharge water even u hammer strongly. Once again I do appreciate ur worth explantion for the scentific isues and i wish i can get ur level.

Cheers

[Abby Normal]

you are a student, that explains it

[mohamed khamis]

Really I am honoured to hear that from you and many thanks for kindness with me.

Best regards:)

[farview]

i learn a lot here thank you all :)