Ground source heat pumps simulation and specs??

Hi guys,
I am a mech eng student, doing a project on heat pump simulation, since you guys are much more experienced than i am, was wondering if you could supply me with some running specs of a heat pump unit used with a ground loop. preferably with r-22 used

what i need are:
Compressor efficiency,
water temperature at the ground loop outlet,
the working pressure of heat pump system, the high band and the low band
temperature of the refrigerant at the inlet and outlet of the condensor


many thanks in advance

Hi, turbo

I think nobody can give you info.
Because, you did not give any info or data.
Most importantly, you have to do your home work first, then you ask. Too many nice guys in this community to help this kind of students.

tedre

Quote:

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Originally Posted by tedre

Hi, turbo

I think nobody can give you info.
Because, you did not give any info or data.
Most importantly, you have to do your home work first, then you ask. Too many nice guys in this community to help this kind of students.

tedre

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Hi, thanks for your reply.
What info would you guys need?
Its a ground source heat pump with a loop, the aim of the simulation is to find how the physiques of the pipe makes a difference in the system.
hence the heat pump parameters would be constant.
I need some realistic values for the working high and low pressures for an R-22 system...
I am trying my best to do my work but since i have no experience whatsoever on the practical side and there's not many if any website with this sort of technical info i thought maybe i could get some help from the experienced users here

Thanks again

The temperature in the ground will wary dependent on location and season in the year. The return temperature of the fluid from the ground loop can be somewhere between -3*C and up to maybe 8*C in locations where you have some sort of winter. If no real winter, like in Britain, probably not below 0*C. DT ground loop fluid maybe 3-4*C

Evaporation temperature of the refrigerant can be 10*C lower than the actual return temperature dependent on the capasity of the evaporator.

Outlet fluid temperature can be 30 to 35*C when operating on a floor heating system. Condensing temperature roughly 5*C higher than the outlet temperature. DT heating water maybe 4-6*C.

To do this simulation you must understand the basics in refrigeration. Do you understand that basic? Do you know what a Log-ph chart is?

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Originally Posted by SteinarN

The temperature in the ground will wary dependent on location and season in the year. The return temperature of the fluid from the ground loop can be somewhere between -3*C and up to maybe 8*C in locations where you have some sort of winter. If no real winter, like in Britain, probably not below 0*C. DT ground loop fluid maybe 3-4*C

Evaporation temperature of the refrigerant can be 10*C lower than the actual return temperature dependent on the capasity of the evaporator.

Outlet fluid temperature can be 30 to 35*C when operating on a floor heating system. Condensing temperature roughly 5*C higher than the outlet temperature. DT heating water maybe 4-6*C.

To do this simulation you must understand the basics in refrigeration. Do you understand that basic? Do you know what a Log-ph chart is?

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Thank you for your input,
Not heard of log-ph but assuming p is pressure and h is enthalpy??

what I am thinking of doing is to model the ground loop as a big heat exchanger, find the coefficient of heat transfer using an electricital model (using resistors etc), then model the evaporator as another heat exchanger where r-22 meets the ground loop. then using the running pressure low band and the temperature of the super heated vapour to find the enthalpy and entropy before it has entered the compressor, using compressor efficiency equation to find the enthalpy of the r-22 coming out of the compressor and finally using the high band pressure and enthalpy to find the entropy and temperature. then model the condensor and a hot water tank as another heat exchanger, using the high band pressure, the change in enthalpy (Qdot=mdot(h1-h2)to estimate the temperature of r-22 coming out of the condensor and finally since the enthalpy of the fluid going through an expansion valve is the same using that and low band pressure to find the temperature going into the evaporator.

I would appreciate if you could comment on this model if I am going in the right direction or not.

Many thanks

Quote:

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Originally Posted by turbo

Thank you for your input,
Not heard of log-ph but assuming p is pressure and h is enthalpy??
Many thanks

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Yep, you'r right.

It seemed a bit complicated, but you are in the right direction. I will advice you to get a log-ph chart. It is very clarifying to draw the refrigeration prosess on that chart. There are various curves so you se all the refrigerant properties in different plases of the prosess. You can download CoolPack. http://www.et.web.mek.dtu.dk/Coolpac...s/CoolPack.exe It's a very good free program where you can make your own charts and do various calculations.

Quote:

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Originally Posted by SteinarN

Yep, you'r right.

It seemed a bit complicated, but you are in the right direction. I will advice you to get a log-ph chart. It is very clarifying to draw the refrigeration prosess on that chart. There are various curves so you se all the refrigerant properties in different plases of the prosess. You can download CoolPack.
It's a very good free program where you can make your own charts and do various calculations.

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Thanks alot for the link, seems to be a great program. just wondering if there are any tutorials for it on the web? because the help doesnt explain much...i got the log-ph graph..when it comes to drawing the cycle, what are the sub cooled liquid temperatures in K? (it seems to only allow between 0-100 which is very cold is it supposed to be that way?)

Quote:

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Originally Posted by turbo

Thanks alot for the link, seems to be a great program. just wondering if there are any tutorials for it on the web? because the help doesnt explain much...i got the log-ph graph..when it comes to drawing the cycle, what are the sub cooled liquid temperatures in K? (it seems to only allow between 0-100 which is very cold is it supposed to be that way?)

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I dont think there is any tutorial on the web. All data in the program is metric only.

You can chose between °C and °K when specify subcool and superheat. If you chose °C, then it is like this:

You first specify a condensing temperature or saturated pressure in bar, lets say 35°C. You next specify the actual liquid temperature, lets say 33°C. The program will draw the cycle with a liquid temperature of 33°C. In this case you have a subcool of 2°K. Each "°K" is worth the same as each "°C", but the zero point for "°K" is minus 273,15°C.

We use "°K" when specify a temperature difference without bothering what the actual temperature might be.
An example in "°K":

You still specify the condensing temperature or pressure, lets say it's still 35°C. You next specify 2°K in subcool. The program will draw the cycle with a liquid temperature of 35 minus 2 is 33 °C. If you change the condensing temperature to, lets say 30°C and leave the subcool, then the program will draw the cycle with a new liqiud temperature of 30 minus 2 is 28 °C.

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Originally Posted by turbo

what I am thinking of doing is to model the ground loop as a big heat exchanger,...

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What type of ground loop are you considering? A borehole with pipes inserted in it, a buried direct expansion coil, or other...

If you decide on what you want to use, try a Google search for it. You should be able to find some Masters or PHD thesis on the modeling used, which might be helpful to you.

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Originally Posted by turbo

I would appreciate if you could comment on this model if I am going in the right direction or not.

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Seems like you are taking a proper approach.

hi guys,
I have made a model in matlab,
my model calculates if the ground temp is 10*C, after steady state with
100meter of pipe in the ground,
1.2 soil conductivity
0.039 pipe con
0.58 water con
radius of pipe 0.017m
mdot=0.05 kg/s

it gives me around 7.39*C outlet water temperature after endless loops while mixing with R-22 in a perfect heat exchanger

is that right? what sort of differences should i expect by halfing the ground loop?

also for the refrigeration loop I have assumed condensor puts out 100% liquid...is this the case? shud I make a different assumption to make it realistic?

my second loop simulates the R-22 in a system with high band 1.5MPa, low band 0.5 MPa with 0.15 saturation T and once its gone through the evaporator it will be 6.07*C and through a compressor with 0.65 isentropic efficiency it shows a temperature of ~82*C and once its passed through the condensor with 100% liquid out it gives out about 10KW with an mdot of 0.05 kg/s

could you tell me if the model is ok in terms of the assumptions i have made?

and also in order to run the ground loop at 0.05kg/s what sort of an electrical input does the system need?

also whats the electrical consumption of a compressor used in similar systems and what sort of mass flow rate does it provide? (in the refrigeration loop)

I'm guessing you extract roughly 7kW from the ground loop. That is 70W/m. It is a high figure.

I understand the mass flow in the ground loop is 0,05kg/s water. 7kW at that flow requires a DT of 33°K. It is a totally impossible number. Adjust the flow rate to a DT of roughly 3-4°K.

Water outlet (return from ground loop) seems reasonably.

Condenser puts out 100% liquid, again reasonable.

At 7,4°C water inlet to the evaporator, the refrigerant outlet should be roughly 2-3°K lower, that is roughly 5°C instead of your 6°C. Rest of your second loop seems reasonable.

You find the electric consumption in CoolPack --> Cycle Analysis--> One Stage Cycles--> DX evaporator. Run a simulation where you put in the refrigerant, cooling power, evaporating and condensing temperatures, any subcooling, superheat and compressor isentropic efficiency. The program then calculates the mass flow and compressor power consumption.

SteinarN, Thank you very much for your input.

the following are some more simulations
tcout = at outlet of ground loop *C (after steady state)

7.3944


tcin = at outlet of evaporator *C (after steady state)

6.3500


Qout = maximum heat output at condensor KW

10.2979


COP = of the refrigeration loop

4.5754


Wdotc = work dot of the compressor (KW)

2.2507


so in order to calculate the COP of the whole system i need a work dot for the pump in the ground loop, i could not find it in the coolpack program.

the above is found using a 50 meter loop in the ground, mdot of 0.05 for both loops.

the thing is the output mainly depends on the mdot of the refrigeration loop since
Qdot = mdot*(h2-h1[condensor's ends])
I have made an assumption the evaporator is a perfect heat exchanger so the temperature of the water and refrigerant at its outlets are equal.

What is your mass flow in the ground loop? You says 0.05, is it kg/s, and is it water?

As the mass flow in the evaporator loop increases, the temperature in the ground loop and the evaporating temperature/pressure will decrease rapidly therebye make it difficult to sustain that increased mass flow.

Your pressures, are they absolute or relative pressures?

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Originally Posted by SteinarN

What is your mass flow in the ground loop? You says 0.05, is it kg/s, and is it water?

As the mass flow in the evaporator loop increases, the temperature in the ground loop and the evaporating temperature/pressure will decrease rapidly therebye make it difficult to sustain that increased mass flow.

Your pressures, are they absolute or relative pressures?

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yes, mass flow rate in the ground loop is 0.05 kg/s and its water. pressures are absolute pressures. modelling a perfect heat exchanger means the amount of heat trasnfered is from water is equal to the heat given to the refrigerant hence
Qdot = mdot (ground loop) *Cp(water)* DT (ground loop at condensor ends)
Qdot also = mdot (refrigeration loop) * latent heat of evap + mdot*Cp (R-22)*DT (of refrigeration loop

using this method since the Cp of water is around 4 times that of R-22 and latent heat is minor means with a change in temperature of water at 1 degrees the refrigerant will have a DT of around 4 *C if the mass flow rates are alike. hence why my system seems to work on such low DT, obviously it will have to be alot higher if the component dont work in a perfect way
is this right?

I see your point now, my bad i had used the wrong figure for latent heat

tcout =

4.9994


tcin =

0.1103


Qout =

0.9820


COP =

4.9385


Wdotc =

0.1989

results now are more like this,
a TD of almost 5 degrees. and now much output since if the mdot of the refrigeration loop is increased the liquid will not change to vapour completely.

I'm having some trouble to picture your system. I'm not that familiar with your abbreviations. However if I understand your numbers correct, it seems reasonably.

The only thing is your water outlet from the evaporator is the same as the evaporating temperature of the refrigerant. This is not possible in real life. However for a demonstration purpose to show how changes in the system works, it is ok.

Edit:
You didn't state your refrigerant mass flow in your latest post. If it is roughly 0.0062kg/s, then all seems to be correct.

More edit:
I'm assuming 5 bar evaporating and 15 bar condensing as you stated in an earlier post. Also zero subcool and isentropic efficiency of 0,65. At those conditions your COP is slightly to high (should be 4,77).

Quote:

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Originally Posted by SteinarN

I'm having some trouble to picture your system. I'm not that familiar with your abbreviations. However if I understand your numbers correct, it seems reasonably.

The only thing is your water outlet from the evaporator is the same as the evaporating temperature of the refrigerant. This is not possible in real life. However for a demonstration purpose to show how changes in the system works, it is ok.

Edit:
You didn't state your refrigerant mass flow in your latest post. If it is 0.0062kg/s, then all seems to be correct.

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its actually 0.005 kg/s which is actually close to ur figure. ill just need to make a small change since the tcin = 0.11 where it cant be less than 0.15*C but it seems its working

thank you very much for your help, much appreciated.

Don't take circulation pump operation cost as a freebie. It's easy to write that cost off as unimportant, but as you push efficiancy up, smaller loads can add up. And I'm not saying you ignored that; I've not followed closely. I do read of excess spent on pumping spoiling the desired results.

I assume you are not studying a burried DX copper system. A compressor burn out could wreck the larger share of the system investment and a leak in burried coils into soil could be expensive to mitigate.

Turbo
if you drop me a PM if you are able with an email address I have an electronic copy of some heat pump training materials I could send you. Quite a big file but has everything you could possibly want to know about heat pumps.

Ian

Quote:

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Originally Posted by Larry2

Don't take circulation pump operation cost as a freebie. It's easy to write that cost off as unimportant, but as you push efficiancy up, smaller loads can add up. And I'm not saying you ignored that; I've not followed closely. I do read of excess spent on pumping spoiling the desired results.

I assume you are not studying a burried DX copper system. A compressor burn out could wreck the larger share of the system investment and a leak in burried coils into soil could be expensive to mitigate.

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Well, to be honest with you I have sort of assumed it as a freebie but only because i have not a clue about how much energy does a pump need to operate such a system, im sure many factors of the ground loop have effects on the operation costs such as the length the pipe diameter, the pressure drop and etc so I would be grateful if you could give me an idea about the pumps energy consumption.

I am working on a model which estimates the amount of heat that can be extracted from a given system. there are many assumptions in my model (after all its an undergraduate research) such as the heat sink being able to take the maximum amount of heat out, the vapour being trasformed into liquid by the time it reaches the end of the condensor, no pressure drop in the ground loop system and a perfect heat exchanger between the ground loop and the refrigeration loop.

Thanks for your input, look forward to hearing from you.

Quote:

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Originally Posted by Pooh

Turbo
if you drop me a PM if you are able with an email address I have an electronic copy of some heat pump training materials I could send you. Quite a big file but has everything you could possibly want to know about heat pumps.

Ian

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Pls check your pm,

much appreciated.

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Originally Posted by turbo

Pls check your pm,

much appreciated.

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Dont seem to be able to send a private message,
pls send it to turbomr[at]msn[dot]com

Quote:

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Originally Posted by turbo

Well, to be honest with you I have sort of assumed it as a freebie but only because i have not a clue about how much energy does a pump need to operate such a system, im sure many factors of the ground loop have effects on the operation costs such as the length the pipe diameter, the pressure drop and etc so I would be grateful if you could give me an idea about the pumps energy consumption.

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I have only spent a few afternoons looking into this topic and never delved into the math side. But what I have learned through reading on the internet is that many residential ground source systems are installed operate nicely but at less than optimum efficiancy because excess energy is spent pumping fluid through the heat exchange field faster than was necessary or optimum.

The more gallons per minute pumped through the field, the higher the pumping cost is. I wanted to draw to your attention that there is an important energy cost to pumping a fluid through the heat exchanger field.

As you push towards higher efficiency numbers, a 100 or 200 watts excess spent will harm your performance numbers. Four hundred watts taken as a freebie to simplify calculations could throw your calculations off base.

I wish I could help quantify the pumping cost for your project, but it's beyond me. Pumping cost is not a freebie though. Fluid doesn't flow through pipework without energy cost. Pipe caliber, pipe surface roughness, length, division through multiple pipes through manifolds, fluid viscosity etc will all be inputs to this calculation which ufortunately I don't know how to calculate.

There are some ground source energy resources on the web that might help you out. If you contact these people, especially at the educational email addresses or government addresses provided, my experience is these professionals or professors sometimes answer questions if you don't ask too many at once. I recommend emailing or phoning manufactures of ground source systems too. Educational sources will be more free with the information you need, since what they have studied is not proprietary.

You have an interesting project. Keep us updated with your progress.

Hi,
Been reading with allot of interest, I have just started looking at Ground source heat pumps and downloaded the cool pack program. I have a large selection of recorded data that I need to review and try and make sense of. When I do if you would like to share information I would be most happy to.
the system I am currently studying has 12 bore holes and 3 heat pump systems. Hope to speak to you again soon. Cheers James.

if you are still looking for real world measurements, try: our cool house dot com (i'm new, and can't post the link)
they are a couple of eco-geeks, very proud (rightfully) of their custom home with geothermal heat.
there are live metrics on the site, with history. this will give you some real world numbers to compare your calculations with.

edit by Brian_UK

Interesting site, worth a look, link is...
http://www.ourcoolhouse.com/monitor/monitor.htm