Interim reality check

ChrisJ

mattmia2 said:

So basically we are trying to doll up a system that was designed for about a 30 degree differential to work at a 70 degree differential and wondering why it doesn't work very well instead of designing components for a 70 degree differential.

If you run in a vacuum eventually you no longer have air as an insulator and the motor shorts so probably picking a refrigerant that is still around atmospheric pressure at your low temp is important. Or using a 2 stage system with a low temp and a medium temp refrigerant.

There shouldn't be any air in a hermetically sealed refrigeration system.
Several of my Monitor Tops run in a very deep vacuum (28"HG) all of the time by nature. Even the high side is often in a vacuum.

The motor is cooled by oil and refrigerant.

mattmia2

ChrisJ said:

There shouldn't be any air in a hermetically sealed refrigeration system.
Several of my Monitor Tops run in a very deep vacuum (28"HG) all of the time by nature. Even the high side is often in a vacuum.

Or refrigerant is the insulator in this case. If you draw deep enough vacuum there is no longer a gas there to act as an insulator. that is why there are warnings to not operate the compressor while you are evacuating it.

ChrisJ

mattmia2 said:

ChrisJ said:

There shouldn't be any air in a hermetically sealed refrigeration system.
Several of my Monitor Tops run in a very deep vacuum (28"HG) all of the time by nature. Even the high side is often in a vacuum.

Or refrigerant is the insulator in this case. If you draw deep enough vacuum there is no longer a gas there to act as an insulator. that is why there are warnings to not operate the compressor while you are evacuating it.

An insulator to what?
There's a gap between the rotor and stator, and there's insulation around the windings.

There's quite a few reasons not to run a compressor while you're evacuating it.
The lack of cooling from refrigerant for one. Scrolls don't like pulling down into a vacuum but I'm not sure why, but I think it has to do with the unloader and what applies pressure to the scrolls. I don't think it has anything to do with the motor.

A vacuum does cause less refrigerant vapor to be around the motor to help cool it as well.
But I don't think I've ever heard talk about insulation and the windings.

Jamie Hall

On a different aspect... a few posts up there people were debating the results of some calcuations based on some measurements or otherwise fuzzy figures. Any measurement can be put down as a number, but we must remember what the number really means. For instance, if I say that Cedric's fuel consumption is 3.3 gallons per hour, what is meant by that number -- at best -- is "somewhere between 3.25 and 3.35 gallons per hour". Similarly, when I quoted a COP for the heat pump of 2.5, I mean, at best somewhere between 2.45 and 2.55 -- although in this case I'm confident of the number only over a wider range: probably 2.2 to 2.6. This logic applies to any measurement, and there is a convention which is used: we write down only those digits which are not estimates or subject to rounding off. This is the engineer's shorthand way of stating the precision of the measurement ( in some instances engineers use a longhand approach -- they might state Cedric's fuel use as "3.3 gallons per hour, plus or minus 0.2 gallons per hour" for instance). Or his efficiency -- 85%. By cocnvention, that would mean between 84.5 and 8.5. percent; in reality it is likely that the measurement is between 82% and 88% at best.

Now when we do calculations with those measurements, we can only honestly keep the precision of the LEAST precise of the measurements which are involved. In the days of the slide rule this was easy enough to visualize (yeah, I know, what's a slide rule, grampa?). In the days of the calculator and computer, we have to be much more careful about what we are saying, since a calculator will happily provide as many digits as the display has room for -- of which, commonly, only the first one or two have any meaning at all.

Just something to keep in mind.

SlowYourRoll

I just want to clarify something about why heat pump efficiencies drop off in cold weather...

Other people have pointed out other reasons why efficiency drops as temp drops, and they are worth considering, but from what I've read the big reason efficiency drops has to do with available heat. Heat pumps (the kind we're talking about anyway, residential HVAC ones) take heat from ambient air. The colder it is, the less heat there is for the refrigerant to absorb as it boils. And since COP is just heating output / energy input, colder temps result in lower COP. So it doesn't have nearly as much to do with mechanical losses or the temp inside the house and indoor/outdoor temperature differential, it's mainly that the refrigerant can't draw as much heat from 10deg ambient air as it can 45deg ambient air.

Anyway the link i posted earlier has a deeper explanation and links to another study about it. it's pretty cool technology, but that issue of lower efficiency the colder it gets i think makes it more of an auxiliary system to use in conjunction with a main heating system, at least for now.

Hot_water_fan

@Jamie Hall usually costs are expressed as $/MMbtu. Seeing that your number wasn’t expressed like that, I did the math myself. I’m not claiming you know or anyone knows the exact efficiency of either system, but as you saw, people interpreted your calculation as saying the difference was 10x what the same exact calculation shows when you divide by 1000 then round. It’s just to help others understand, so they can draw their own conclusions. 

ChrisJ

SlowYourRoll said:

I just want to clarify something about why heat pump efficiencies drop off in cold weather...

Other people have pointed out other reasons why efficiency drops as temp drops, and they are worth considering, but from what I've read the big reason efficiency drops has to do with available heat. Heat pumps (the kind we're talking about anyway, residential HVAC ones) take heat from ambient air. The colder it is, the less heat there is for the refrigerant to absorb as it boils. And since COP is just heating output / energy input, colder temps result in lower COP. So it doesn't have nearly as much to do with mechanical losses or the temp inside the house and indoor/outdoor temperature differential, it's mainly that the refrigerant can't draw as much heat from 10deg ambient air as it can 45deg ambient air.

Anyway the link i posted earlier has a deeper explanation and links to another study about it. it's pretty cool technology, but that issue of lower efficiency the colder it gets i think makes it more of an auxiliary system to use in conjunction with a main heating system, at least for now.

I don't know.
To me, that's not the reason, I think Jamie was more on track honestly. But, it's far from my area of expertise for sure.

For example, methane has absolutely no problem boiling violently at -200F at atmospheric pressure. There's plenty of heat there for it.

0F is only cold to animals etc, it's not nearly as cold as absolute zero.

mattmia2

ChrisJ said:

SlowYourRoll said:

I just want to clarify something about why heat pump efficiencies drop off in cold weather...

Other people have pointed out other reasons why efficiency drops as temp drops, and they are worth considering, but from what I've read the big reason efficiency drops has to do with available heat. Heat pumps (the kind we're talking about anyway, residential HVAC ones) take heat from ambient air. The colder it is, the less heat there is for the refrigerant to absorb as it boils. And since COP is just heating output / energy input, colder temps result in lower COP. So it doesn't have nearly as much to do with mechanical losses or the temp inside the house and indoor/outdoor temperature differential, it's mainly that the refrigerant can't draw as much heat from 10deg ambient air as it can 45deg ambient air.

Anyway the link i posted earlier has a deeper explanation and links to another study about it. it's pretty cool technology, but that issue of lower efficiency the colder it gets i think makes it more of an auxiliary system to use in conjunction with a main heating system, at least for now.

I don't know.
To me, that's not the reason, I think Jamie was more on track honestly. But, it's far from my area of expertise for sure.

For example, methane has absolutely no problem boiling violently at -200F at atmospheric pressure. There's plenty of heat there for it.

0F is only cold to animals etc, it's not nearly as cold as absolute zero.

you can increase surface area and airflow to increase the heat you extract. some of the efficiency loss is inherent to the greater delta t but not all of it.

As far as insulation of electrical components in a vacuum, the air is the insulator in 2 conductors otherwise uninsulated and spaced apart. If there was no air or other insulator electrons would be free to move between the 2 if there was enough potential. That is how a vacuum tube works. the heater helps but the vacuum is the reason electrons flow from the cathode to the anode.

SlowYourRoll

ChrisJ said:

I don't know.
To me, that's not the reason, I think Jamie was more on track honestly. But, it's far from my area of expertise for sure.

For example, methane has absolutely no problem boiling violently at -200F at atmospheric pressure. There's plenty of heat there for it.

0F is only cold to animals etc, it's not nearly as cold as absolute zero.

The equation for thermal conduction is...
[heat flux density in Watts/m^2] = [material's conductivity] * [temperature gradient]

So the radiant power (Watts) passing through a material scales with temperature gradient. Doubling the temperature gradient doubles the radiant power.

Freon boils at -22°F. That seems to be the refrigerant currently being used for heat pumps. That's pretty close to 0°F. And that's all the energy the refrigerant can absorb, is whatever energy comes from raising it from boiling point to whatever temp it hits the compressor at. If it were -21°F the refrigerant would've absorbed almost no thermal energy going into compression stage.


[side note: I used the thermal conduction equation because it shows how things scale. You'd use a different equation if you were really trying to calculate actual values, but that equation would have the same ultimate result...the energy transferred scales with temperature gradient/differential. I can update the post if you want to see it with that equation, but the convection/conduction equations are all very similar, they're all rooted in atoms banging into each other. Radiation is a different story...]

mattmia2

And surface area.

SlowYourRoll

mattmia2 said:

you can increase surface area and airflow to increase the heat you extract. some of the efficiency loss is inherent to the greater delta t but not all of it.

Increasing surface area and airflow only help if the current setup isn't adequately extracting all the heat there is to extract. Freon boils at -22F. If outside air is 40F and the Freon vapor is reaching the compressor at only 20F then yes, you can increase surface area or increase airflow to get that Freon vapor closer to 40F. If outside air is 5F and your Freon vapor is reaching the compressor at 5F you could blast that pipe with gale force winds and the vapor would still be 5F when it hits the compressor.

Jamie Hall

Oh dear. Well, the thermodynamics of mechanical refrigeration is absolutely fascinating -- but not, I think, what we need to get into too deeply. But there is a fundamental -- and inevitable -- reason for the change in COP for a given set of refrigerant and low and high temperatures. The basic principle is simple enough. A liquid refrigerant is exposed to a heat source -- usually air but by no means always -- at a temperature which is low, but higher than the boiling point of the refrigerant at whatever the pressure is at the time (depending on the refrigerant, this could be as low as say -250 Celsius, if need be!). This absorbs heat from the source in the form of latent heat in going from a liquid to a gas. Now we take that gas, which is at a low pressure we compress it (we'll come back to that). As the pressure rises, the boiling point of the refrigerant also rises, but it is still a gas as we don't let any heat come out of it here. Then we take than high pressure gas and expand it in a condenser back to the original low pressure. In the condenser, heat is released from the gas as it condenses and that heat is removed -- again, usually by an air flow, but not always. That is exactly the same heat as the gas absorbed when it boiled in the evaporator, back in the beginning, just coming at you at a higher temperature, plus some additional heat added by the compressor itself. That additional heat comes from the work the compressor does, which is exactly related to the difference in pressure from the low side to the high side. The greater the difference in pressure, the more work the compressor has to do. And looking back, the difference in pressure is directly related to the difference in temperature from the low side to the high side -- the greater that difference, the greater the pressure difference required and the greater the work the compressor has to do.

Now the COP of any refrigeration system is defined simply (using consistent units) as the heat delivered per unit of time divided by the power absorbed by the compressor.

If you followed the above somewhat complicated main paragraph, it becomes obvious that the power absorbed by the compressor is, barring other losses, always directly related to the pressure difference and thus temperature difference between the low side and the high side. It's the difference which counts. Not the absolute values of either one. Further, the choice of refrigerant has very little to do with the COP -- all usable refrigerants behave closely enough to ideal gasses that that isn't a factor.

The only thing the choice of refrigerant affects is the pressure required on the low side to allow evaporation (boiling) to occur at the desired temperature, and the pressure required on the high side to allow condensation to occur.

That's the basics. There are losses, of course, so the actual COP of a system can never be as high as the simple math would suggest. But the fundamental relationship is there.

mattmia2

SlowYourRoll said:

mattmia2 said:

you can increase surface area and airflow to increase the heat you extract. some of the efficiency loss is inherent to the greater delta t but not all of it.

Increasing surface area and airflow only help if the current setup isn't adequately extracting all the heat there is to extract. Freon boils at -22F. If outside air is 40F and the Freon vapor is reaching the compressor at only 20F then yes, you can increase surface area or increase airflow to get that Freon vapor closer to 40F. If outside air is 5F and your Freon vapor is reaching the compressor at 5F you could blast that pipe with gale force winds and the vapor would still be 5F when it hits the compressor.

so you use a greater volume of refrigerant.

or a 2 stage system

or a low temp refrigerant.

pecmsg

Look at any Performance Curve for any compressor and there’s your answer. 

The lower the temperature the lower the pressure. The lower the pressure the lower the compression ratio!

Jamie Hall

Just to muddy the waters a bit more -- the exact same thermodynamics apply in the case of the evaporation taking place at a high pressure, and then allowing the resulting gas to expand (rather than compress) in a machine and condensing it at a low temperature. It it happens that the refrigerant in that case is water, we call the resulting machine a "steam engine"...

mattmia2

pecmsg said:

Look at any Performance Curve for any compressor and there’s your answer. 

The lower the temperature the lower the pressure. The lower the pressure the lower the compression ratio!

So you need a larger bore and stroke to the compressor (or the equivalent in other designs)

SlowYourRoll

mattmia2 said:

so you use a greater volume of refrigerant.

or a 2 stage system

or a low temp refrigerant.

yes, that's some of what people are working on now to get better COP at low temps.

I was just trying to explain to people how COP drops at lower temps mainly because of the temperature difference between refrigerant boiling point and ambient temp. other people were suggesting lower COP had to do with temperature effects on the compressor mechanism itself (which probably has some effect) or other factors, and i was just trying to clarify what is the predominant factor in heat pump COP.

Jamie Hall

SlowYourRoll said:

mattmia2 said:

so you use a greater volume of refrigerant.

or a 2 stage system

or a low temp refrigerant.

yes, that's some of what people are working on now to get better COP at low temps.

I was just trying to explain to people how COP drops at lower temps mainly because of the temperature difference between refrigerant boiling point and ambient temp. other people were suggesting lower COP had to do with temperature effects on the compressor mechanism itself (which probably has some effect) or other factors, and i was just trying to clarify what is the predominant factor in heat pump COP.

Indeed. A good deal of work (the mental sort, not the thermodynamic sort!) is being done to improve the internal efficiency of the compressors -- and therefore the COP (more of the power input showing up in the compressed gas, and less in losses in the compressor). However, the laws of thermodynamics limit this. That said, recent compressor designs are much better than older ones in this regard. The resulting improvements in real world COP sometimes mislead the cheerleaders into thinking that there is no limit to that, but there is -- and some of the newer designs are getting dismayingly close to that limit.

Hot_water_fan

Do extremely low temp COPs matter much? I’d argue no - they’d be nice but the existing hyper heat models are in the good enough territory. 

ChrisJ

@mattmia2

Hmm..
You bring up a good point about vacuum tubes.

But maybe the distance between connections is big enough and the voltage is low enough.  Also there's no heated cathode to boil off electrons.

SlowYourRoll

Hot_water_fan said:

Do extremely low temp COPs matter much? I’d argue no - they’d be nice but the existing hyper heat models are in the good enough territory. 

It basically means that you have to pair a heat pump with a more conventional heat source like steam or forced air, but as an add-on technology alongside your existing heating I don't think those low COPs matter.

I'm sort of thinking of it like turbocharging an engine. It undeniably improves performance and efficiency in some engine ranges, it has a few quirks that makes it not useful at other ranges, overall its pretty awesome to have, but I'm definitely not removing my engine and replacing with just a turbo.

Hot_water_fan

@SlowYourRoll exactly - thinking about most Americans here, they’ll replace an AC with a heat pump. If they want, they can keep the furnace or use electric resistance. Efficiency gains are nice but won’t make or break this- the hours spent in the negative teens are low.

ChrisJ

@mattmia2

I'm being told by a very good friend that flashover is an issue especially on 480v compressors when you get into a really deep vacuum, like 50 microns.   So the good news is no system is ever going to do that under normal and useful operation.

EdTheHeaterMan

Just the Facts?

@Jamie Hall Ask Alexa "play Radio Dragnet" and you will get the pre TV radio show from the 40s and 50s

With Cigarette adds an all!

EdTheHeaterMan

Hot_water_fan said:

@Jamie Hall usually costs are expressed as $/MMbtu. Seeing that your number wasn’t expressed like that, I did the math myself. I’m not claiming you know or anyone knows the exact efficiency of either system, but as you saw, people interpreted your calculation as saying the difference was 10x what the same exact calculation shows when you divide by 1000 then round. It’s just to help others understand, so they can draw their own conclusions. 

I used to like mathematics until this discussion popped up. I'm wondering if the reciprocal of the square root of the COP with the hypotenuse of the bilateral equation might in fact be the reason for the quantum equation that Sheldon was never able to verify could be all in Cedric’s combustion chamber.

Jamie Hall

Something like that, @EdTheHeaterMan !

hot_rod

I think one could, and has parsed or manipulated numbers until the cows come home?

Carnot started the COP ball rolling back in the 19th century and it could be argued his COP, Carnot COP is still a target to shoot for.

@ChrisJ is probably the only one alive that owns a Carnot refrigerator

Over the years the RPA funded multiple studies trying to get a radiant to forced air comparison, labs, universities, etc. And the numbers are and should always be challenged.

Accurate, stable, agreed upon, peer reviewed data probably doesn't exist much anymore, especially in the times we live in.

I'd challenge that good old Cedric runs 85% efficient every time he, or she fires and steams?

Jamie Hall

"I'd challenge that good old Cedric runs 85% efficient every time he, or she fires and steams? "

I dare say the only time he runs at 85% is when @Charlie from wmass is here, giving him the evil eye.

ChrisJ

@hot_rod

I need some time to figure out if that was a shot or not.

Might take weeks.  

Hot_water_fan

@hot_rod the Carnot Sink temperature is the leaving water/air temp? Or the water/air temp entering the indoor part? Or something else entirely? 

hot_rod

Hot_water_fan said:

@hot_rod the Carnot Sink temperature is the leaving water/air temp? Or the water/air temp entering the indoor part? Or something else entirely? 

Source is the air, water or ground, depending on the type of HP.
Sink is where the energy is being delivered, the space in the building, tank of water, etc.

Hot_water_fan

Thanks - so which temp column counts as the sink temp here? 

Jamie Hall

Carnot efficiency -- which only sort of applies -- would use the cold side as Tc, and the hot side as Th, and give you the theoretical maximum efficiency of a heat engine operating between those two points. In the case of refrigeration, keep in mind that we are interested in the power in, not the power out, of the process.