Ground Water Heat Pump - reality check/napkin calculations.

izhadano

Hello,
I became interested in heat pump heating and did my homework to understand the nuts and bolts of the most efficient one, - Ground Source Heat Pump (GSHP). Looks like reality check doesn’t support GSHP optimistic prognosis.

Back in 2013 article concerns were made about GSHP high installation cost, maintenance, equipment cost rising, inflated efficiency claims, lack of performance field data, flawed rating methods, etc. Some of these concerns are resolved today, but the pumping power question was not raised and disputed yet.
Simple math reveals strong controversy with GSHP efficiency claims.

Please, correct inaccuracies in my calculations if any and share the real life experience you may have.

The basics of heat pumps was presented on a Wall in 2014, clear explanation of heat pump CoP is below
The key to the efficiency of a GSHP is the Coefficient of Performance: the “CoP”. In spite of the first law of thermodynamics, which tells us that energy can neither be created nor destroyed, a GSHP in a good installation can yield up to four units of heat for each unit of electricity consumed. The heat pump is not creating this energy, but merely separating a medium temperature from the ground into warmth (which can be used for heating) and cold (which can be returned to the ground).
The CoP will vary with each installation, but the lower the output temperature to the heat distribution system the higher the CoP will be. If an output temperature of 60°C (140F) is needed to heat radiators the CoP is likely to fall to level of only 2.5. If the heat distribution is to a well-designed underfloor heating system that works well at an output temperature of 40°C (104F) then the CoP can rise to a level of 4.
The input temperature is also critical to the CoP of the heat pump. The higher the input temperature from the ground, the lower the amount of work needed from the heat pump, the higher the CoP will be. In fact, the critical factor is the “uplift” between the source temperature and the output temperature.

GSHP benefits are mostly advertised in comparison to the most expensive type of heating - electrical heating. Similarly, I’ve started napkin calculations to compare electrical heater and heat pump operating in the favorable conditions:
GSHP with COP=4 is to bring 3 tons (36000 btu/hr = 10.55 kWh) heat into a house.
Heat pump power usage would be 10.55/4 = 2.64 kWh
The rest of the heat (7.91 kWh) is free delivered by groundwater, - energy savings of 300% !!!! – sounds impressive …

What about pumping power? Somehow this additional electricity usage usually either silenced completely or mentioned as negligent.

A typical GSHP manufacturer states that with a groundwater temperature of 55°F, the 3-ton (36,000 BTU) unit requires a flow rate of 2.5 to 3 GPM/ton or 7.5-9.0 GPM total. With a 45°F (7C) source water, the same unit now requires a flow rate of 10GPM/ton (= 30 GPM total).
In design standards, the pumping power is defined as a power density: 19 W/gpm for hot water.

Assuming groundwater temperature of 55°F, required pumping power is 7.5 GPM or
7.5x19W/GPM = 142.5 W @ 7.5 GPM = 8610 Wh @ 7.5 GPM= 8.61 kWh @ 7.5 GPM
The total for GSHP and pumping 11.25 (2.64 + 8.61) kWh exceeds electric heating requirements 10.55kWh.

Should we consider colder weather and less favorable operating condition with output temperature of 60° C (140°F) and groundwater temperature of 45°F, COP would drop from 4 to 2.5, and groundwater flow rate will increase 4-folds from 7.5 to 30 GPM total:
• Heat pump power usage would be 10.55/2.5 = 4.22 kWh
• Pumping power would be 30x19W/GPM = 570 W @ 7.5 GPM = 34.2 kWh @ 30 GPM
The total for GSHP and pumping 38.44 kWh (4.22 + 34.2) exceeds 3.5 fold electric heating requirements 10.55kWh. That explains why the GSHP should be backed up by a furnace/boiler for colder weather.

Questions:

Is estimated pumping power of 19W/GPM too high? Anybody has real life results on GSHP pumping power sub metering? Maybe the groundwater pump is not ON all the time? - That will partially solve the puzzle. For how long per hour?

Pumping power inside the house was not included in these calculations. How much water is circulated on the inner loop? How much power is required additionally?

I’ve read a lot of ads on heat pumps efficiency, energy savings and comfort. Many customers are very happy with GSHP. I personally enjoyed Mini-split Air Source Heat Pumps comfort in the hot summer. Still, for heating, existing fuel based heating system have to be kept and maintained as backup for cold weather. Investing $30-40K specifically for a mild weather system which is actually not more economical than electric heating sounds unreasonable. Room electric heaters will do the trick at a fraction of cost and higher reliability…

Jamie Hall

It's Sunday morning, so I don't have time to go into detail -- but in several places up there you have managed to confuse watts or kilowatts (power) or BTUh (power) with kwh or BTU (both energy). Easy to do, but mixing the two in calculations can have unfortunate results -- like making an error of a factor of 60 in your pumping energy requirements.

george_42

I am only interested in the results. I have 3500 sq ft home with geothermal water to air and I have an amp meter on this unit only and at 10 cents per kw I have been able to heat or cool this house for less than $100 per month. Home is in central Pa and is insulated very well . Six inch studded walls with fiberglass and ceiling with open foam.

EBEBRATT-Ed

@izhadano

Indoor pumping is roughly 2 gpm/ton, outdoor pumping is roughly 3 gpm/ton. The pumps only need to run when the equipment is on.

Go to the "Water Furnace" web site they have some calculators that may help

izhadano

Jamie Hall Member Posts: 16,136
7:14AM
It's Sunday morning, so I don't have time to go into detail -- but in several places up there you have managed to confuse watts or kilowatts (power) or BTUh (power) with kwh or BTU (both energy). Easy to do, but mixing the two in calculations can have unfortunate results -- like making an error of a factor of 60 in your pumping energy requirements.

Jamie,
thanks for pointing to potential problems.

To clarify this:
A kilowatt-hour, expressed as kWh or kW·h, is a measure of energy that is equivalent to 1,000 watts of power for a 1-hour time period. Thus, to convert watts to kilowatt-hours, multiply the power in watts by the number of hours, then divide by 1,000.
So, if 19 W power is required to pump 1 gallon per minute, it's 1140 W per hour to pump 60 gallons per minute and 7.5 more power (8610 W per hour) for 7.5 GPM. Any error by factor of 60 in this calculus ?
I found though that variable speed pump can reduce power usage significantly - 22 gpm while only taking 116 watts - that's roughly 5W/GPM. Still, tarnish the claim of 300% energy savings and not enough for compete with electric heaters in cold weather.

Regarding confusing BTUh (power) with kwh or BTU (both energy), - I followed the example 6-4 from "T H E R M O D Y N A M I C S, AN ENGINEERING APPROACH, EIGHTH EDITION". - can be downloaded for free. IMHO, converting BTU/hr into kWh is as legitimate as converting kJh in kWh in example below.
EXAMPLE 6–4 Heating a House by a Heat Pump
A heat pump is used to meet the heating requirements of a house and maintain it at 208C. On a day when the outdoor air temperature drops to 228C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine (a) the power consumed by the heat pump and (b) the rate at which heat is absorbed from the cold outdoor air.
SOLUTION The COP of a heat pump is given. The power consumption and the rate of heat absorption are to be determined.
Assumptions Steady operating conditions exist.
Analysis (a) The power consumed by this heat pump, shown in Fig. 6–24, is determined from the definition of the coefficient of performance to be
Wnet,in = Qh/COP_HP = 80,000 kJ/h/2.5 =32,000 kJ/h (or 8.9 kW)
(b) The house is losing heat at a rate of 80,000 kJ/h. If the house is to be maintained at a constant temperature of 20C, the heat pump must deliver heat to the house at the same rate, that is, at a rate of 80,000 kJ/h. Then the rate of heat transfer from the outdoor becomes
Q = Qh - Wnet.in = (80,000 - 32,000) kJ/h = 48,000 kJ/h
Discussion. Note that 48,000 of the 80,000 kJ/h heat delivered to the house is actually extracted from the cold outdoor air. Therefore, we are paying only for the 32,000-kJ/h energy that is supplied as electrical work to the heat pump. If we were to use an electric resistance heater instead, we would have to supply the entire 80,000 kJ/h to the resistance heater as electric energy. This would mean a heating bill that is 2.5 times higher.
This explains the popularity of heat pumps as heating systems and why they are preferred to simple electric resistance heaters despite their considerably higher initial cost.

Thanks,
Igor

Hot_water_fan

@izhadano Hi Igor! You've typed a lot today.

Your conclusion is correct that when you're considering geothermal systems, you need to account for pumping energy as the loops can require a lot. This often brings their efficiency down close enough to an Air-Source heat pump that the additional $$$ isn't worth it. This is frequently discussed on GreenBuildingAdvisor.com. Each installation will have varying pumping requirements. Geothermal systems have a lot more variation than an air-source installation.

So, if 19 W power is required to pump 1 gallon per minute, it's 1140 W per hour to pump 60 gallons per minute and 7.5 more power (8610 W per hour) for 7.5 GPM. Any error by factor of 60 in this calculus ?

Yes you're off by a factor of 60, as Jamie explained. It's an easy mistake to make!
Think about it this way. How many kWhs does a 100w light bulb use over the course of an hour?

Jamie Hall

The problem, I think, is that BTU/hr is a measure of power, as is a kilowatt. BTU/h is 0.0003 kilowatts, to a first approximation.

So to start off, take our 3 ton unit -- 36,000 BTU/h (the use of the word "ton" is one of those archaic units which should be taken out and drowned...). To get that much power with resistance heat would take roughly 11 kilowatts. If we have a COP of 2.5, then of that 11 kw only about 4.3 kw will be being drawn from the grid. In the meantime, to push that water around, we are using about 140 watts for the pump using the 19 watts/gpm and 7.5 gpm. If we run the show for an hour, we will have used about 4.4 kwh of energy. A lot better than the resistance energy usage!

Hope this helps -- I think (it's still Sunday, and I shouldn't be working at all!) the problem was in figuring the power vs. energy of the pump...

Sort of off-topic, but regarding the power used in the pumping as "lost" isn't really quite correct; the energy used shows up as a temperature rise in the piping. Not much, to be sure, but that's where it goes.

izhadano

Jamie Hall and Hot_water_fan:
Many thanks for the clarification. I've overestimated the pumping power a by a factor of 60, - made the GSHP approach looks unfeasible. My mistake.
Warm regards,
Igor