Measuring performance

[Larry2]

I've been trying to measure system efficancy without having lots of instrumentation. I have two thermistors installed in the air stream. One is ahead of the coil in the return air. The other is three feet beyond the air handler in the supply duct. I'm measuring resistance of the two thermistors with a Fluke 8050A digital meter. I'm converting resistance readings to temperature rises. The thermistors are Johnson AB99B sensors. They measure within 1 degree F with the fan running and no heat running.

I turned on the heat strips alone and measured voltage and current flow to the heat strip elements and temperature rise. I have a temperature rise of 58* F (32.2 C) at 10.344 KW coil input power. From this I calculated 584 CFM air flow for the 35,325 btu/hr input heat. This air flow seems way off compared with the air handler specs which say it should do around twice this amount.

I repeated the test with the heat strips off and hat pump running. I measured 37.75* (21C) temperature rise. Assuming airflow didn't change from 584 CFM, I calculated 23809 BTU/hr. This is with the outdoor unit power consumption of 4.67KW consumption at the outdoor compressor and fan.

Dividing output BTU gain vs outdoor BTU expenditure, I get 23,809/15947 = 1.49 performance coefficent or an HSPF of 5.1.

This is at outdoor temperature around 32F (0 C).

Do I have an error in my figuring somewhere. The measurements don't match my expectations.

[frank]

Hi Larry

Using your figures with the resistance heater I get 0.263m3/s air volume (584CFM = 0.276m3/s), so your calcs appear to be about right.

With the heat pump running and using the 0.263m3/s I get 6.74kw of usable heat input. When this is divided by the measured input consumption I get a cop of 1.44:1, slightly less than you.

I can't comment on your expectations though :D

[SteinarN]

How have you measured the coil power and the compressor power? Is it an inverter or DOL?

[Larry2]

How have you measured the coil power and the compressor power? Is it an inverter or DOL?

It is not an inverter system. I'm not sure what you mean by "DOL"; I'm not familiar with the abreviation. It is a constant displacement scroll compressor with single-speed fans through out. There is no variable capacity, variable speed, or other gadgetry. It's "bare bones" controlled by turning on or off by thermostat demand. It does have TXV, both indoors and outdoors.

I measured the total power for the compressor plus outdoor fan motor using a true-RMS digital clamp ammeter. I clipped the clamp around each main leg at the power contactor and measured the current. The unit is 240 VAC, 60 Hz, single phase power. Using the meter probes, I measured AC volts between the two legs at the same spot. That gave me the total power consumed in the outdoor section of the heat pump, neglecting the indoor fan motor.

I repeated the same measurement at the electric heat elements by clamping the meter on each leg and them taking voltage between legs. Again, I neglected the indoor fan motor power.

What is surprising to me is that the air flow is so low for a four ton blower-coil unit. The multi-speed blower motor is jumper set to "high" and the air flow specs for a static pressure of 0.6 are more than double what I measure. The table says for 0.6" static pressure, the blower will push 1710 CFM.

I measured an air flow of 577 CFM the other day using the heat strip method. After that, I spent several hours thoroughly cleaning the indoor coil and arrived at 584 CFM. Within the precision I can work with these thermistors, I would say there was no improvement in airflow at all.


I can't comment on your expectations though :D

I'm not sure what expectations should be since I can't find specifications for these units at specific temperatures. On newer gear, I see heating output at 47F and 17F listed but the power required to provide that output is not listed. Only the seasonal average is given. One thing is for sure. Until I get air flow, the unit isn't going to be efficient. That is why I started this exercise.

[nike123]

What is rated (nominal capacity) of that unit, cooling and heating?

You should use 400CFM per 1 ton of refrigeration capacity at indoor coil as guide.

[SteinarN]

I have used that method my self, measuring heater coil power and temperature increase to determine air flow. You should be able to calculate the air flow very accurate with that high temperature increase. I get 0,268m3/s with your numbers.

However, the compressor power is probably less than you calculated. You have not corected for the power factor, which is maybe roughly 0,8 for a equally sized 3-phase motor. I dont know what it may be for a 1-phase motor.

Multiplay 4,67kW by 0,8 gives 3,74kW
Compressor heat capacity is 10,344/32,2x21=6,75kW
6,75/3,74=1,80 COP which isn't that impressive either. However the uncertain number here is the true compressor power, it may be only, lets say, 3,27kW. (power factor 0,7) That will give a COP of 2,06 while somewhat better still is poorly.

What air off temperature do you have in heat pump mode? Is 4 ton roughly 14kW? If it is, then I agree with the air flow beeing very low. You says you have a TXV both at the inndor and outdoor unit?

Edit: Silly me, it's a reversible heat pump. :o

More advanced calculations can be done by measuring the compressor discharge temperature, saturated condenser temperature, suction temperature and saturated evaporator temperature. Assuming thermally insulated compressor it is then possible to calculate the compressor isentropic efficiency and then calculating what the heating COP should be.

More edit:
A generously sized TXV can effectively camuflate a "short of gas" condition. I guess it doesn't have any liquid line sight glas?

[frank]

One thing we seem to agree on is that the airflow is low.
Resistance heat is normally used to compliment the heat pump during low ambients, not replace it.

With the dt you are measuring, it would appear rather high if both the HP and RH were on at the same time, pointing to a low air flow.

Do you have large amounts of ductwork or excessive changes of direction in the ductwork?

[Larry2]

I found some of my missing air flow today. There are two circuits servicing the air handler and two breakers for disconnect mains. When I measured nominally 10 KW of heat on one breaker pair, I assumed the other smaller circuit fed the blower and the control transformer. On closer look today, the second circuit handles an additional 5 KW of heat strips that I overlooked with my meter. Oops :o I need to repeat my measurements. I'm going to buy some LM34 temperature sensors more accurately. Just adding in a 5KW guess on that heat strip and repeating the calculations, I now arrive at 867 CFM (0.409 cubic meters/s. Assuming this air flow, I calculate 35,691 btu/h or 10,451 watts of heating on the compressor for a COP of 2.31. That's a little closer to what I expected. Considering everything that is wrong with the system, that's pretty good.

Still, the air flow is low and as nike123 comments, should be 400 CFM/ton. I have the blower on the highest speed tap and since it's direct-drive, I can't get any more.

Frank, you hit the problem on the nose. The duct work system is excessive and poorly designed. The return air connects to the plenum through three 10" round flexible ducts. They are aproximately 10', 12' and 50' in length. These ducts have a rippled interior surface, so they present a great deal of turbulance that opposes free air flow. The inlet side of a cage blower isn't very tolerant of such restriction.

On the supply side, the trunk splits two directions. Each side has the majority of the registers connected at the far end up to 50 foot away. They are fed from the trunk through 8" flexible ducts.

I'm trying to estimate where my energy losses are in this home. The blower-coil could be located better to shorten the duct runs. It's too late to plan the equipment inside of conditioned space.

SteinarN, I was aware that power factor would effect my calculations of compressor power consumption. I just have no idea what power factor one would expect to find on a compressor.

[ismail]

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