Posts: 668
Joined: Sat Jan 23, 2016 3:21 pm
Delivery Date: 23 Jul 2014
Leaf Number: 015896
Location: Estonia, Europe

Heat pump efficiency - invented a way to calculate

Thu Feb 09, 2017 6:24 pm

Guy's, I have awesome way to calculate Heat Pump efficiency. I was thinking about that a lot and I found
a solution that will work in the less efficient temperature zone (at temperatures between 0C and -15C, 32F and 5F)
Without any modifications to Leaf! I'll just write down the basics right now. Later will do measurements (others are welcome).

Set recirculation to fresh manually, set temperature to max, set air direction to defrost, set fan speed manually to appropriate fixed value, close side vents manually. All this is extremely important to get valid data. I would also open one window slightly so cabin would never warm completely to set value (otherwise climate draw might get too small).

Install thermometer with external sensor into the defrost vent. Rely on outside temperature reading or install another thermometer outside (more precision recommended, I think I'll find another sensor for outside as outside temperature sensor is not near air intake and is rounded value, that is also slow to update)
Launch LeafSpy and have AC load and PTC load screen active.
Wait until consumption gets stable, wait until temperature reading is stable. Write data down.

And now the way to measure HP efficiency.
Premise: switching on AC button will actually switch off heat pump functionality and will not generate anything at all (no heat nor cold) if outside temperature is below 1C and fresh air mode is selected (not partial). Already tried that and it is true. Can be verified with LeafSpy.
This means one can switch on/off heat pump without changing other parameters, SV/SL Leaf will act like S-trim Leaf that has only air PTC heater.

Now, as the air quantity with and without heat pump is constant we don't need that in our formula.
What changes is output air temperature.

So, in short, how this thing will work.
HEAT mode enabled
Air intake -10C, air output +40C, PTC load 0W, HP load 1500W
AC+HEAT mode enabled
Air intake -10C, air output +30C, PTC load 2500W, HP load 0W.
As the air quantity is fixed we know delta temperature is 50C and 40C. Heat generated with HP in this scenario is 25% more than with PTC in the second scenario. But we also know that in second scenario heat generation rate was exactly 2500W. Therefore with HP we generated 2500*1,25=3125W of heat. COP in this scenario is exactly 3125/1500=2,1 (210% efficiency).

This all can be done at heavy load as well.
Full fan speed, outside -10C, air output 35C, PTC load 1250W, HP load 1800W, delta temperature 45C
Full fan speed, outside -10C, air output 30C, PTC load 3750W, HP load 0W delta temperature 40C
Heat generated in Hybrid mode: 45/40*100=112,5%, 12,5% more
Heat generation rate in Hybrid mode: 3750W*1,125=4220W, subtract PTC 1250W, HP heat production rate: 4220-1250=2970W
Compressor power draw was 1800W, heat generation rate was 2970W, 2970/1800=1.65, 165% efficiency.

Due to the fact result is expressed in ratio (percentage) it is OK to use Fahrenheit.
Heat pump's efficiency depends on dozens of variables which are all excluded here. atmospheric pressure, air humidity, return gas temperature, compressor load, vehicle speed, etc.
We can only extract few parameters: outside temperature, fan speed, compressor load, vehicle speed.
Higher fan speed and vehicle speed will result in higher COP, lower outside temperature and higher compressor load will result in lower COP.
To keep things simple I will generate some groups of Low Medium and High:
Fan speed: L 1-2, M 3-4, H 5-7, Compressor load: L 0-750W, M 750-1500W, H 1500-2250W, vehicle speed L 0-30kmh, M 30-60kmh, H 60-90kmh.
I think vehicle speed is the least important as at L and M fans are running which should keep airflow optimal (like in H). I will keep it in mind but exclude when I gather data. Outside temperature is critical data and will be raw and not rounded up.

Just facts on which all of this is based on:
1 calorie is the amount of energy that is required to heat 1 gram of water for 1 degree (not Fahrenheit).

1 calorie = 4.186 joules, 1 joule per second is exactly 1 watt

We just have another substance (not water). Due to humidity fluctuations results should only fluctuate 3%, which is very little and should be ignored.
Short range EVs <30kWh -- Medium range: 30-60kWh -- Long range: >60kWh
Charging: Trickle <3kW -- Normal 3-22kW -- Fast 50-100kW -- Supercharging >100kW

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