Quote:
Originally Posted by MEMPHIS91
...I saw in another post how you said to calculate the COP. What would be best way to measure water temperature accurately?...
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[NOTE: After I wrote all this, I realized that you only wanted to know about water temperature. I answered a lot more, but I'll let it stand for other readers. As to your question, I would think that if you removed the sacrificial anode rod and made measurements through that, it would work well. You are likely to have temperature stratified water, so I suppose a measure at the top and at the bottom, and averaging them would be a fair approximation.]
Calculating COP with a water heater is fairly easy.
The general idea is: COP = (energy out) / (energy in)
One BTU is the amount of energy to raise one pound of water one degree F.
You will need to know what the weight of water is in your tank. If you know what the volumetric capacity of the tank is, you can convert volume of water to the weight of water.
The density of water is about 62.4 pounds/cubic foot.
The density of water is also 8.343 pounds/gallon.
So, I will guess that your tank is 50 gallons, so 50 x 8.343 = 417.15 pounds of water.
Next, you will need to know the initial temperature is in your tank. For this, you could just stick a thermometer in, maybe through the hole in the top for the sacrificial anode rod, and take a reading. I will just guess that the starting temperature is 50 degrees F. You will want to know how many degrees the water is heated to. So you will subtract your beginning temp (50 F) from your final temp.
Next you will need to be able to measure the amount of power used. A Kill-a-Watt meter is an accurate and cheap way to do this. It can tell you how much power (kW-h) is used, and it will also keep track of time, if you plug it in at the beginning off the test.
So now, you have the initial conditions.
I'm not sure what method you are going to use to stop the heat pump when the water reaches the desired temperature. Even if you don't have this arranged yet, if you run the system for an hour and a half or so, you'll get useful information.
Now, having written down the initial temp, you start the test...
Plug it in and marvel at your creation... but don't get too carried away, because you are conducting a scientific test.
Make sure that the Kill-a-Watt doesn't get interrupted during the test, or it will 'forget' the data.
After the run time (maybe 1.5 hours), record the power power that the Kill-a-Watt indicates (kW-h). Also record the run time from the Kill-a-watt. As I recall, the Kill-a-Watt measures watt hours, but when the count goes over 999 watts-h, it goes to kW-h. So, when you do your calcs you will want to make sure that you are in watts... so 1.234 kW-h will be the same as 1234 W-h.
Just guessing here... let's just say that at the end of the test, your water temp is 85 degrees. This may be a bit tricky because you will have stratification with hotter water on top and colder water on the bottom. So do an average of top and bottom temps.
POWER OUT (power produced by your heat pump)
Your power-out will be (1042.9 pounds) * (85-50 degree F)
... so power-out = 1042.9 * 35 = 14600.25 BTU
Converting this to the electrical energy equivalent you divide by 3.412
elec power = 14600.25 / 3.421 = 4279.1 watts.
POWER IN (power consumed by your heat pump)
Read the kW-h from the Kill-a-Watt.
let's just say that you read: .9 kW-h
This would be the same thing as 900 W-h
then you multiply by time (1.5 hour) to get watts
Watts = (900 W-h) * (1.5 h) = 1350 watts
CALCULATE COP
COP = (power-out) / (power-in)
COP = (4279.1 watts) / (1350 watts)
cop = 3.2
I think that your efficiency will be better.
Best,
-AC