Convert a solar DHW tank (or fuel oil) into a thermodynamic water heater with a small swimming pool heat pump

[Christophe]

Yes at the end of the compressor ON cycle (OFF at 39 ° C) and at high circulator speed 22W ...

It's stupid not to have the real flow rates ... that would make things easier ... (my mano is not precise enough, at 7W the needle does not move, and probably not well placed to be able to find them with the curves...)

[sicetaitsimple]

Yes at the end of the compressor ON cycle (OFF at 39 ° C) and at high circulator speed 22W ...

It's stupid not to have the real flow rates ... that would make things easier ...

That's for sure! If, for example, between the minimum position (7W) and the maximum position (22W) the flow rate is multiplied by 2, your provisional conclusions are called into question.

[Christophe]

For a known delta P, we have the flow rate ratios at the bottom of the image of the circulator curves here: water-pumping-filtration / transforming-a-solar-DHW-tank-into-thermodynamic-water-heater-with-a-small-swimming-pool-t17035-40.html

The pressure drop of the heat pump is given at 0.8 m (given by Poolex) ... let's say that the rest of the circuit has an identical loss (??) ... i.e. 1.6 m in all ...

Going from 7W to 22W we therefore have respective flow rates of the order of 2.1 m3 / h to 3.4 m3 / h ... and as 2 m3 / h this is the minimum flow requested by the heat pump, we know why it does not does not fault ... even at minimum speed.

Suddenly we can estimate the Power of the I / O heat pump since I have the precise temperatures and the flow ...

If not good bin is that the pressure drops are different ... Suddenly if we assume the power of the heat pump at 3 kW, bin we can have the real pressure drop!

I'll let you check and calculate that

ps: I do not see how this calls into question the fact that the heat exchange must be done correctly in the heat exchanger of the balloon, too high a speed would limit the heat exchanges at this level compared to a lower speed!

[sicetaitsimple]

Going from 7W to 22W we therefore have respective flow rates of the order of 2.1 m3 / h to 3.4 m3 / h ... and as 2 m3 / h this is the minimum flow requested by the heat pump, we know why it does not does not fault ... even at minimum speed.

Well no .... According to your first readings at minimum speed on page 12, the water deltaT is 7,3 °. As your heat pump has a maximum power of 4kW, and that P = q * 1,16 * deltaT, this means that your flow at minimum speed is less than or equal to 0,5m3 / h.

[Christophe]

It's BAH NO it's that the pressure drops are higher than that!

For the rest, I haven't done the math yet ...

[Christophe]

Considering all of this, I wonder if the best performance / lifespan setting isn't:

a) low speed circulator
b) Heat pump at 40 ° C (the COP is reduced a little but the tank heats up more and therefore there are fewer heat pump cycles therefore less consumption and therefore less wear)
c) Delta greater than 3 ° C to be determined (currently 5)

Why wait to test? So that will be the current setting and I zap the end of the high speed test which was done at 50% ...

This is not because for 37.5 ° C max in the 2nd floor shower I got this as data today:

Duration: 122 hour
Consumption: 6,38 kWh
Average power: 52.3 W

I am going to revert the high setpoint of the heat pump to 39 ° C since it does not change anything in "comfort" ... and that it increases consumption ...

[Christophe]

Going from 7W to 22W we therefore have respective flow rates of the order of 2.1 m3 / h to 3.4 m3 / h ... and as 2 m3 / h this is the minimum flow requested by the heat pump, we know why it does not does not fault ... even at minimum speed.

Well no .... According to your first readings at minimum speed on page 12, the water deltaT is 7,3 °. As your heat pump has a maximum power of 4kW, and that P = q * 1,16 * deltaT, this means that your flow at minimum speed is less than or equal to 0,5m3 / h.

Weird this matter of flow ... if I had so much pressure drop that I would see a little circulator on / off on the mano ...

In addition it is 3 kW max ... Frankly there I dry ...

P (W) = Q (L / s) * 4,18 (kJ / L. ° C) * DeltaT (° C)

Q = 3 / (4,18 * 7.3) = 0.1 L / s or 360 L / h ... logically the heat pump should not start since it requires 2 m3 / h ...

According to the Biral curves, I would be at 10m / 1 bar of pressure drop in this mini circuit? I can't believe it for a second!

So this old differential thermometer is probably out of order (I told you it was over 10 years old ... and not used for roughly the same length of time) ... but I had recalibrated it before taking these measurements (but only on 1 point) ...

In short, this does not change much: technically and energetically it works!

But I would still like to understand!

[izentrop]

With 2 m of coil and not even 1 m of height difference, the pressure drops must be minimal.

[Christophe]

That's it izy!

Well it's my differential thermometer which is in PLS ... I have inverted the probes and I have different data and the delta T ° is divided by 2 !!

I'll buy a new one ...

[sicetaitsimple]

Well it's my differential thermometer which is in PLS ... I have inverted the probes and I have different data and the delta T ° is divided by 2 !!

Everything is explained!