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View Full Version : Evaporators - TD, SH & approach - revisited



desA
08-11-2009, 03:04 AM
I would like to investigate evaporator operation in more detail, & would like to understand the relationship between TD, SH & approach temperatures.

We have some known relationships, so far:
1. Magoo rule: SH=(0.6->0.7)*TD
2. Kueba method : SH=0.65*TD
3. Gary method : SH ~ 7.0K

In my research, in trying to reverse-engineer the Magoo rule, have also seen that two possible SH values can exist for each evap operating condition - one low, the other high. (This comes from the mathematics of heat-transfer & is a consequence of the non-linear nature of heat-transfer devices. It is called 'bi-stability').

The high value corresponds closely to the 'Magoo Rule', & the low value corresponds to the 'Gary method'.

There are implications to using either rule, in practice. With the 'Magoo Rule'. As SH is raised, the evap saturated evaporating temp (Te,sat) is lowered. As SH is reduced, the Te,sat value rises. This has implications for both hot & cold air on (Ta,in) operating conditions, with associated Te,sat values.

A request for information:
I would like to conduct a research study into this bi-stability phenomenon & characterise it thoroughly.

For this information, I would like to ask the RE community to assist, if at all possible, with information to be used in this research. I plan to pass the findings of this research back into the public domain via a few academic/industry papers (long term) & back onto RE site.

Required evap information (any units will do):
1. Minimal set:
TD, SH & approach.

2. Full set:
Ta,in = air temp in
Ta,out = air temp out
Te,sat = evap saturated evaporation temp
Te,sup = evap superheated vapour exit temp

3. Other useful info (not essential):
Evap face inlet velocity
RH% at inlet
Location (region, country)
Time (when results taken)

I would really value the support of the RE community. I believe that the results will prove useful to the RHVAC community at large. Thanks everyone for your very kind contributions.

Contributions can be either published directly onto this thread, or sent via pm on the RE site.

desA
08-11-2009, 05:27 AM
There are 3 main cases, that an operational evaporator finds itself in:

1. (APP + SH) > TD
2. (APP + SH) = TD
3. (APP + SH) < TD

desA
08-11-2009, 09:40 AM
Case 1: Evaporator.



Ta,in = 34.8'C
Ta,out = 24.5'C
Te,sat = 14.5'C
Te,sup=24.7'C

Fan at 100%


TD = Ta,in-Te,sat = 34.8 - 14.5 = 20.3K
SH = Te,sup-Te,sat = 24.7 - 14.5 = 10.2K
APP = Ta,out-Te,sat = 24.5 - 14.5 = 10.0K

SH+APP = 10.2 + 10.0 = 20.2K ~ TD
SH/TD = 10.2/20.3 = 0.50

dTlm,e = [(Ta,in-Te,sup)-(Ta,out-Te,sat)]/ln[(Ta,in-Te,sup)/(Ta,out-Te,sat)]
..........= [(34.8-24.7)-(24.5-14.5)]/ln[(34.8-24.7)/(24.5-14.5)]
..........=(10.1-10.0)/ln(10.1/10.0)
..........=0.1/0.00995 (observe the very small numbers - almost 0/0, based on experimental uncertainty)!!!
..........=10.05K (masks the small numbers)

This is close to what I have termed the 'dither point', or apex of the bifurcation curve for this evap.

desA
08-11-2009, 10:08 AM
Case 2: Same evaporator - fan speed reduced to 50%.



Ta,in = 34.8'C
Ta,out = 20.9'C
Te,sat = 12.75'C
Te,sup=22.3'C

Fan at 50%


TD = Ta,in-Te,sat = 34.8 - 12.75 = 22.05K
SH = Te,sup-Te,sat = 22.3 - 12.75 = 9.55K
APP = Ta,out-Te,sat = 20.9 - 12.75 = 8.15K

SH+APP = 9.55 + 8.15 = 17.7K < TD
SH/TD = 9.55/22.05 = 0.43

dTlm,e = [(Ta,in-Te,sup)-(Ta,out-Te,sat)]/ln[(Ta,in-Te,sup)/(Ta,out-Te,sat)]
..........= [(34.8-22.3)-(20.9-12.75)]/ln[(34.8-22.3)/(20.9-12.75)]
..........=(12.5-8.15)/ln(12.5/8.15)
..........=4.35/0.4277 ('stronger' on numerator & denominator)
..........=10.17K

Has moved away from the bifurcation point (0/0)

desA
08-11-2009, 01:38 PM
Case 3: Same evaporator - fan speed reduced to 27.8%.



Ta,in = 34.3'C
Ta,out = 16.7'C
Te,sat = 11.0'C
Te,sup=19.1'C

Fan at 27.8%


TD = Ta,in-Te,sat = 34.3 - 11.0 = 23.3K
SH = Te,sup-Te,sat = 19.1 - 11.0 = 8.1K
APP = Ta,out-Te,sat = 16.7 - 11.0 = 5.7K

SH+APP = 8.1 + 5.7 = 13.8K << TD
SH/TD = 8.1/23.3 = 0.35

dTlm,e = [(Ta,in-Te,sup)-(Ta,out-Te,sat)]/ln[(Ta,in-Te,sup)/(Ta,out-Te,sat)]
..........= [(34.3-19.1)-(16.7-11.0)]/ln[(34.3-19.1)/(16.7-11.0)]
..........=(15.2-5.7)/ln(15.2/5.7)
..........=9.5/0.9808 (even 'stronger' on numerator & denominator)
..........=9.68K

Has moved further away from the bifurcation point (0/0)

desA
10-11-2009, 12:58 PM
I'd be most grateful if any RE members can contribute evaporator measurements.

I will also continue to add further examples to this thread, exploring what happens to the same evap, with minor system changes.

So far, we have seen the effect of reducing fan speed on the evaporator operating point, in terms of dTlm, TD, SH, & APP.

Chef
11-11-2009, 04:33 AM
desA
Does this also apply to refrigerant systems (ie freezers) or is it just heat pumps? Technically they are almost the same except in one Te falls and in the other Tc rises, that though may be enough to mean the two systems cant be used in your scheme.

You make the call on that one and I will send the data I have.

Chef

desA
11-11-2009, 05:14 AM
desA
Does this also apply to refrigerant systems (ie freezers) or is it just heat pumps? Technically they are almost the same except in one Te falls and in the other Tc rises, that though may be enough to mean the two systems cant be used in your scheme.

You make the call on that one and I will send the data I have.

Chef

Thanks Chef.

To confirm, this will apply, in principle to ANY evaporator. Refrigeration & air-conditioning evaporator data is most welcome.

I suspect that a rising, or falling cycle, will make little difference. For instance, an aircon cycle evap, Te,sat falls as the cycle progresses.

mad fridgie
11-11-2009, 09:14 AM
Hi Des,sorry can not help with steady state conditions.
Again I am not quite sure what you are trying prove (forgive my ignorance) But when you are looking at evap coil performance, surely you need to know your liquid to vapor ratio, which then effects the pressure drop through the evap, this then will affect LMTD (I do not believe using SST as constant is detailed enough to prove your theory. (good enough for most practical installations).
If chef's condensing unit is water cooled and running on a freezer (closed), you should have more stable conditions than ambinet air and heating a tank of water (reduced process variables))
Is Bifurication, a short term effect and what/how if any does influence the process over a time period, in a practical application

mad fridgie
11-11-2009, 09:18 AM
I suspect that a rising, or falling cycle, will make little difference. For instance, an aircon cycle evap, Te,sat falls as the cycle progresses.
Are you predicting that Te, will drop over a time period if the load remains constant, or the Te drops over time as the load reduces due to the previous refrigeration effect.

desA
11-11-2009, 11:30 AM
Hi Des,sorry can not help with steady state conditions.

Of course you can. Use a 'quasi steady-state' situation, where you control the coolant flow in such a way as to slow the heating ramp rate down such that only a small change occurs across the test point. This will suffice.



Again I am not quite sure what you are trying prove (forgive my ignorance) But when you are looking at evap coil performance, surely you need to know your liquid to vapor ratio, which then effects the pressure drop through the evap, this then will affect LMTD (I do not believe using SST as constant is detailed enough to prove your theory. (good enough for most practical installations).

It seems that a purely macroscopic (outside) view of the evap will be enough. The evap is a single unit with integrated vapourising section & super-heater. Practically, this cannot really be decomposed into two separate series units, due to the interconnected structure of most evaporators.

What I hope to be able to show is that each evaporator has two trajectories that it can follow. Small changes can move the evap from one trajectory to the other, if managed properly.

The two scenarios of SH=0.65*TD and SH=7K present a different scenario. This seems to have implications of the system as a whole.

When the two 'steady' natures (at least) of each evap & condenser (possibly more than 2) are combined, this presents a possibility of at least 4 operating combinations.

Where each evap eventually sits will be a function of operational conditions, as well as refrigerant type & charge. It is beginning to look like charge selection can play an important role in some cases, of switching operating modes.



If chef's condensing unit is water cooled and running on a freezer (closed), you should have more stable conditions than ambinet air and heating a tank of water (reduced process variables))

Quasi-steady conditions are fine.



Is Bifurication, a short term effect and what/how if any does influence the process over a time period, in a practical application

This is what we are setting out to explore. Your data is needed.

It should be noted that fast-ramp unsteady operating conditions will definitely present many more transient possibilities, which will decay to the 'steady' scenarios, with slow-enough ramp rate.

Go & find us some results, MF... :D

desA
11-11-2009, 11:35 AM
Originally Posted by desA
I suspect that a rising, or falling cycle, will make little difference. For instance, an aircon cycle evap, Te,sat falls as the cycle progresses.

madfridgie:
Are you predicting that Te, will drop over a time period if the load remains constant, or the Te drops over time as the load reduces due to the previous refrigeration effect.

As the room temp drops (air onto coil), the Te,sat of the inside evap coil typically follows it downwards, over the cycle.

Chef
11-11-2009, 12:29 PM
So to start the ball rolling with condition number one
Psuction=16PSI
Air temp= -8C
SH=(about)20-25c
Both air and air out are the same as it has no fan.
Its a 134a system.

I have data for Psuction from 25PSI to 15PSI so maybe that will show if moves toward or away from the point of inflexion as it cools down. I will dig it out later.

Chef

desA
11-11-2009, 12:52 PM
So to start the ball rolling with condition number one
Psuction=16PSI
Air temp= -8C
SH=(about)20-25c
Both air and air out are the same as it has no fan.
Its a 134a system.

I have data for Psuction from 25PSI to 15PSI so maybe that will show if moves toward or away from the point of inflexion as it cools down. I will dig it out later.

Chef

Thanks Chef... excellent. I'm assuming that is 16 psig - hope that's correct.


Both air and air out are the same as it has no fan.

Ok, so it's pretty much a natural convection type application. Let's see what pans out.

The additional data would be useful to show a trend & whether something 'interesting' occurs along the trajectory under view.

desA
11-11-2009, 01:04 PM
:D
So to start the ball rolling with condition number one
Psuction=16PSI
Air temp= -8C
SH=(about)20-25c
Both air and air out are the same as it has no fan.
Its a 134a system.

At 16 psig = 1.10316 barg = 2.1164 bar(a) = 0.21164 MPa(a)
R-134a: Te,sat=-8.635'C

Where does the superheater lie physically, relative to the cooled space, in this application?

If it's entirely external (in series after evap space), then it would not influence the evap itself. If that is the case, then we are left with a very simple evap system consisting of Ta,amb=-8'C & Te,sat=-8.635'C.

In this situation:
dT1=0.635K
dT2=0.635K

dTlm = (dT1-dT1)/ln(dT1/dT2) = 0/0... nice :D

desA
11-11-2009, 04:38 PM
A thought...

I wonder if the dTlm='0/0' situation, in your simple example, is telling us something about a forced-convection evap? I have a suspicion that it is.

I've seen the situation at times, where dT1~dT2 - for large parts of a longish test run - even at slightly different operating conditions. It's like the evap gradually maneuvers itself into that position & then parks there.

More thought processes to follow...

Chef
12-11-2009, 02:16 AM
I wonder if the dTlm='0/0' situation, in your simple example, is telling us something about a forced-convection evap? I have a suspicion that it is....

So all of the other data points will also result in a dTlm=0/0 so cannot be used.

It has long been taboo in mathematics to divide by zero, even if the case never crops up during the calcs being done at the time it does mean the equation will tend to infinity when the denominator gets infinitesimally small. for small postive denominators it would approach +infinity and for negative denomintors it would approach -infinity. ie its an unstable equation and if this is part of your bifurcation calculations it seems you might want to relook at that.

What is the equation you are using to get your bifurcation plot?

Chef

desA
12-11-2009, 02:55 AM
So all of the other data points will also result in a dTlm=0/0 so cannot be used.

It has long been taboo in mathematics to divide by zero, even if the case never crops up during the calcs being done at the time it does mean the equation will tend to infinity when the denominator gets infinitesimally small. for small postive denominators it would approach +infinity and for negative denomintors it would approach -infinity. ie its an unstable equation and if this is part of your bifurcation calculations it seems you might want to relook at that.

What is the equation you are using to get your bifurcation plot?

Chef

The dTlm=0/0 case is a strict case in the limit as:

dTlm = lim(x,y ->0+)[x/y], or lim(x,y ->0-)[x/y]

What happens in ALL heat-transfer theory is that we !choose! to remain on either side of the dTlm=0/0 divide. There are two sides to the theory, if you go back into the heat-transfer notes. For the F-LMTD method, nothing special is mentioned, for the NTU-eps method, the matter is 'managed' by the Cmin value. It masks the singularity, which, in practice, seems to turn out to be a reality, rather than a pure mathemagical trick. NTU-eps theory can be simply derived from F-LMTD theory.

What will be seen, from the working evap temp measurements, is that a dTlm value will exist sometimes with negative numerator & denominator & at other times with positive values. The form will depend on which terminal dT is selected as reference. In my case, I elect to refer to dT1=(Ta,in-Te,sup) & dT2=(Ta,out-Te,sat)=APP

Practically, in your example - in terms of APP, SH, TD:
SH=0K
APP=0.635K
TD=0.635K
APP+SH = 0.635+0=0.635K =! TD

Magoo
12-11-2009, 04:29 AM
Des A,
My and Kueba rule is based on a balanced system, a factor that most selection people forget.
The evaparator duty and expansion valve capacity is balanced at designed evaporator duty. Despite all theories the TX vav will deliver a given flow ae at a given PD across the TX vav. Constant conditions are hard to control as a stright line condition. MOP vavs make things slightly less harsh on compressors.
But component selection is critical.
the Kueba rule works incredibly well under pull down and design conditions, that has been my experince over the last 40 years in the business.
But I will stand to be corrected.

desA
12-11-2009, 05:01 AM
Des A,
My and Kueba rule is based on a balanced system, a factor that most selection people forget.
The evaparator duty and expansion valve capacity is balanced at designed evaporator duty. Despite all theories the TX vav will deliver a given flow ae at a given PD across the TX vav. Constant conditions are hard to control as a stright line condition. MOP vavs make things slightly less harsh on compressors.
But component selection is critical.
the Kueba rule works incredibly well under pull down and design conditions, that has been my experince over the last 40 years in the business.
But I will stand to be corrected.

Thanks very much Magoo, for weighing in. Excellent review.

From what I've been able to observe from your & the Kueba rule, is that this seems to place the evaporator on a stable operating curve. This curve is fairly flat. To me, this is the point I'm aiming for.

This is in contrast to the typical a/c operation, where the evaporator follows an unstable operating curve. This curve rolls off fairly sharply.

Could I ask you to perhaps comment on typical values you've observed for SH, TD, APP in the applications you work in? This would be very, very useful.

What I'm beginning to suspect is that the a/c folks make use of the unstable curve to allow the evap to track a rapidly-changing room environment at start-up. This presents a problem though, when an evap is placed into an uncontrolled environment - as for an AWHP. This can force the Te,sat to rise outside the compressor operating envelope.

Chef
12-11-2009, 08:47 AM
The dTlm=0/0 case is a strict case in the limit as:

dTlm = lim(x,y ->0+)[x/y], or lim(x,y ->0-)[x/y]

What happens in ALL heat-transfer theory is that we !choose! to remain on either side of the dTlm=0/0 divide.

Your really going to have to explain the equation and what it means so that we can all understand it, if you have access to LaTeX then you could write it in the form of integrals with limit functions.

And on the basis I cant yet understand or define dTlm=0/0 I am not sure how to 'choose' which side of it my systems lies - so I need to understand this.

What happens if you include the pressure drop across the evaporator plate - there will be 2 pressure loss factors, friction and acceleration as the quality changes through the evap plate. Lets assume for now that this is 3PSI for the 2 components combined.
What does that do to the calcs for my plate?

Chef

mad fridgie
12-11-2009, 09:59 AM
Thanks Des for the explantion, to me what you are looking at is typical (in a practical application) I believe the issues you are relating to is not directly related to the evap, but more related to the refrigerant control, what you are decscibing is very typical of a capilary system and a TXV internally equalised. The SH is not related to Te (SST) but is directly proportinal to TXV outlet pressure.
The practical application is that at high evap pressure drop you have an actual higher SH,you also have the effect of mass flow through through the TXV at different Pressure drops, also allowing that a mechanical TXV will not shut 100%.
I am always willing to change my mind if the facts change!

Magoo
12-11-2009, 08:59 PM
Hi Des A,
The AWHP application is probably the most difficult, due to trying to maintain max heat recovery, and varying heat source conditions. Not too dissimilar to a heat pump.
The prime consideration should the compressor design conditions, to maintain optimum efficiency and longevity.
You will have to maintain heat source and heat dicipation conditions by variable speed control. Just like the average heat pump.
Variable compressor speed , variable or stepped condenser fan speed, variable evap fan speed. By doing this you can reasonably maintain system balance. You could possibly use a CPU board out of a heat pump and apply control functions to a AWHP.

desA
13-11-2009, 12:02 AM
Thanks Magoo.


Hi Des A,
The AWHP application is probably the most difficult, due to trying to maintain max heat recovery, and varying heat source conditions. Not too dissimilar to a heat pump.

Agreed. The AWHP, in Asia, really has to deal with some fairly tough conditions & high TD's. The ever-changing (high) humidity also adds to the complexity. Trying to get everything into a compact footprint also has its fair share of challenges.



The prime consideration should the compressor design conditions, to maintain optimum efficiency and longevity.
You will have to maintain heat source and heat dicipation conditions by variable speed control. Just like the average heat pump.

I think that about sums it up fairly well. I've focused on system simplicity, so far & have tried to keep as far away from electronics as possible. Hot, humid air & electronics are not overly friendly. Looks like I'll have to look further into the integration aspects, though.



Variable compressor speed , variable or stepped condenser fan speed, variable evap fan speed. By doing this you can reasonably maintain system balance. You could possibly use a CPU board out of a heat pump and apply control functions to a AWHP.

This would be a good place to start - good idea.

I agree that system balance at each point in the heating cycle is crucial. The system goes through stages along its heating path where it clearly has to re-balance itself. Minimising, or smoothing this change, will surely add to improved COP's.

Thanks again... :)

Magoo
13-11-2009, 01:38 AM
Hi Again des A.
Have you heard or come across the CO2 eco-cute AWHP, currently developed in Japan. Aparently it has been market tested for approx., ten years and is due for global release. As in buy at local super market and owner installed. Cash and carry theory.
I understand that in Australia there is a time date that every household will have a AWHP, for domestic hot water.
All part of the global warming senario. Mainly to cut down coal/and gas power generation CO2 overload as current.
Could be a good time to chime in with similar system.

desA
13-11-2009, 03:06 AM
Could be a good time to chime in with similar system.

I agree. The cash-&-carry concept is something I like very much.

The CO2 systems have been heavily-government subsidised. They will, predictably, be reasonably expensive, & be pretty difficult to maintain. With ultra-high pressures will come fatigue failures, especially with aluminium components.

Good point. Affordable, CO2 systems are my next compass grid point.

mad fridgie
13-11-2009, 04:25 AM
Ecocute units are very expensive, if much of the technology was used in for example a R410a or R134a systems, then I would predicate better savings.
Two Stage compression,
High efficiency motors (pumps and fans)
And high spec control scenrios
Only at the very cold ambient temps and cold inlet water temps do they come into there own.
Cash and Carry, I think will end up a reality, the manufactures will need to sort out in advanced service regiems.
Many in the HVACR trade are refusing to under take warranty to work that has not been installed by themselves (Every man and his dog are now installing heat pumps/AC with out being able to support warranty and is be passed back to the manufactures and agents)

desA
13-11-2009, 10:54 AM
Case 4: Kueba evaporator - market plus SPA 022D. (Source : Kuba published performance figures)



Ta,in = 15.0'C
Ta,out = 7.8'C
Te,sat = 5.0'C
Te,sup=11.5'C

TD = Ta,in-Te,sat = 15.0 - 5.0 = 10.0K
SH = Te,sup-Te,sat = 11.5 - 5.0 = 6.5K
APP = Ta,out-Te,sat = 7.8 - 5.0 = 2.8K

SH+APP = 6.5 + 2.8 = 9.3K < TD=10.0K
SH/TD = 6.5/10.0 = 0.65

dTlm,e = [(Ta,in-Te,sup)-(Ta,out-Te,sat)]/ln[(Ta,in-Te,sup)/(Ta,out-Te,sat)]
..........= [(15.0-11.5)-(7.8-5.0)]/ln[(15.0-11.5)/(7.8-5.0)]
..........=(3.5-2.8)/ln(3.5/2.8)
..........=0.7/0.2231
..........=3.14K

Fairly close to 'dither point'.

desA
14-11-2009, 02:19 AM
Case 4A: Kueba evaporator - market plus SPA 022D. (Source : Kuba published performance figures)



Ta,in = 15.0'C
Ta,out = 4.9'C
Te,sat = 1.0'C
Te,sup=10.1'C

TD = Ta,in-Te,sat = 15.0 - 1.0 = 14.0K
SH = Te,sup-Te,sat = 10.1 - 1.0 = 9.1K
APP = Ta,out-Te,sat = 4.9 - 1.0 = 3.9K

SH+APP = 9.1 + 3.9 = 13.0K < TD=14.0K
SH/TD = 9.1/14.0 = 0.65

dTlm,e = [(Ta,in-Te,sup)-(Ta,out-Te,sat)]/ln[(Ta,in-Te,sup)/(Ta,out-Te,sat)]
..........= [(15.0-10.1)-(4.9-1.0)]/ln[(15.0-10.1)/(4.9-1.0)]
..........=(4.9-3.9)/ln(4.9/3.9)
..........=1.0/0.2283
..........=4.38K

Moved marginally further away from 'dither point'.

desA
14-11-2009, 02:21 AM
Define:


dT1 = (Ta,in-Te,sup)
dT2 = (Ta,out-Te,sat) = APP

Questions:
1. Why is dT1 always greater than (or equal to) dT2 (APP)?
2. Where is the control point on the evap?
3. Where is the sensing point on the evap?
4. What effect does fan speed play?
5. What is the relationship of SH to Te,sat?
6. The 'dither point' represents dT1=dT2. What gives?