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How long to heat DHW with ASHP


AdyHoggs

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Hi, First post on here, I wondered if anyone could help me with my question. I’m having a Valliant 5kw ASHP fitted soon and will also have a 200 Litre Valliant cylinder installed for hot water. I was wondering how long it would take to heat the hot water from cold up to 48 c, as I have Economy 7, and was hoping to take advantage of my cheap night rate, thanks for any help, Ady

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Welcome.

There is an easy answer 1 and 3/4 hours.

 

The better answer is to understand what it all means.

 

I shall start with the ASHP power, the 5 kW (that is lower case k for 1000, and upper case W for watt, watt is lower case, unless at the start of the sentence or referring to James Watt).

Now a watt [W] is a joule per second.  A joule, J, is the SI unit for energy and is named after James Prescott Joule.  Any SI unit that is in capitals like W or J, is a derived unit.

A joule is derived from the kilogram (kg), the metre (m) and time (s).  The kilogram is the odd one out in the SI system as it is the base unit, even though it is made from 1000 grams.  So 1 J = 1 kg.m2.s-2

So all that is really saying is that you are moving energy with respect to time. 5 kW = 5000 J.s-1.

 

This may seem a bit pointless, until you get to the bit about specific heat capacity of materials.  Which is coming now.

 

All materials have the capacity to store energy, so if you heat up a stone in the sun, it has absorbed some solar energy and increased in temperature.  Always remember that temperature not energy.

You can look up what the different amounts of energy that are needed to raise a material by 1 K or 1°C (note that it is an upper case K as it is named after a person William Thomson, better known as Lord Kelvin).  It is generally better to use the kelvin scale, even though the scales match, once the offset is taken into account, 0°C is 273 K, well close enough).

 

Now liquid water is a strange material in that it has a high density compared to its solid and gaseous states, 1000 kg.m-3 (at 277 K, 4°C).  This works out nicely, and close enough for all intents and purposes at 1 kg per litre.

 

The energy needed to increase water 1 kg, by 1 K is 4.18 kJ.  This is usually written as 4.18 kJ.kg-1.K-1 or 4.18 J/kg.K.

 

Now you have 200 litres of water at mains temperature (we shall call that 283 K or 10°C) and you want to raise it up 38 K, to 321 K

So now it is just a matter of doing the arithmetic.

 

Energy (kJ) needed = 4.18 [kJ.kg-1.K-1] x 200 [kg] x (321 [K] - 283 [K)

Energy (kJ) needed = 4.18 [kJ.kg-1.K-1] x 200 [kg] x38 [K]

Energy (kJ) needed = 31,768 

 

All those letters, except the kJ cancel out to leave just the energy figure.

 

So that is the energy required, assuming perfect energy transfer and no losses.

 

Now remember that the power of your ASHP is 5 kW, which is 5 kJ.s-1.

If you divide the energy needed, 31,768 kJ by the energy input, 5 kJ.s-1, you get left with the number of seconds.

So

 

Times (s) = 31,768 [kJ] / 5 [kJ.s-1]

Time (s) = 6,353.6

 

Now we know that there are 60 seconds in a minute, and 60 minute in an hour.

 

Time (minute) = 6,353.6 / 60

Time (minute) = 105.89

 

Now as that is below 120 minutes, or two hours, if we take away 60 minutes, the remainder is the minute part of the second hour.

 

105.89 - 60 = 45.89

 

Call it 46 minutes.

Now add on the 1 hour.

1 hour 46 minutes.

 

You will find that people often talk about their water cylinder storing some number of kWh (be careful with this one, it is as typed, not Kw/h, KW per Hour, or kill wot our).

All a kWh is, is a constant amount of power, the kW part, multiplied by the time it is delivered, or consumed.  That is why it is kWh, 1000 [k] x power [kW] x time [hour].

Now there are 3,600 seconds in an hour (60 minutes x 60 seconds).

If we divide the kilo joules needed by 3,600 seconds, we get the kWh needed.

 

kWh = 31,768 [kJ] / 3,600

kWh = 8.82

 

If you divide 8.82 [kWh] by 5 [kW] you get 1.76 hours.

Which is 105.89 minutes.

 

In reality, there are losses, and the closer the sink temperature (the water in the cylinder) and the source temperature (the water from the ASHP) get to each other, the less energy is transferred, so it will actually take longer, but that is another lecture in thermodynamics.

 

(as usual, I may have made an error somewhere, and I am sure others will pull me up on it)

Edited by SteamyTea
That (expletive deleted)ing strikethrough happened again. And after a second edit to put right, it has come back. Sort this out Mods.
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2 hours ago, SteamyTea said:

(as usual, I may have made an error somewhere, and I am sure others will pull me up on it)

I did

2 hours ago, SteamyTea said:

This is usually written as 4.18 kJ.kg-1.K-1 or 4.18 J/kg.K.

Should be 4.18 kJ/kg.K or 4.18 J/g.K

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Just now, joe90 said:

I wish I understood it, tho i did understand the 1 and 3/4 hrs bit ?

Read it untill you don't understand it.

Then start reading from the start again.

How I do it. Can sometimes take me hours to read a technical essay or report.

Worth listening to those links I posted up about water.

I am not going to tell people what the really interesting part was. That really requires a bit of effort to make it more valid, and useful.

 

https://forum.buildhub.org.uk/topic/23721-water/

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The one caveat here is that the ASHP power also depends on i) outside temperature ii) what mode the heat-pump is in, and iii) the flow temperature (which increases as the tank heats up).  The 5kW specified is just nomimal.

 

We are also having a Vaillant ASHP installed (7kW aroTHERM Plus) and I had exactly the same query so I have the datasheets and found the following:

- The Vaillant aroTherm Plus's comes, from factory, configured to run in "eco" mode for DHW which means 50% compressor speed.  This can be disabled.

- The power output of the 5kW model (eco mode) at 2C external temp is 3.6kW.

- The power output of the 5kW model (max mode) at 2C external temp is 7.3kW.

 

The same datasheets quote a time for heating a Vaillant 188L tank from 10-40C as 57 minutes.  That would imply approx 1hr 12min for 48C.  (they don't specify external temperature or DHW mode, but it seems fairly obvious that this is not eco mode.)

 

image.thumb.png.5d879f047f13e9f767060f54513a4de0.png

 

Edited by Dan F
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9 minutes ago, AdyHoggs said:

So the kWh needed to heat the hot water is just under 9, so does that mean it will use 9 units of electric, thanks, Ady

No, as COP (coefficient of performance) is 3->4.  See the table on the left).

 

I think you're safe to assume around 3kW in the winter and closer to 2.25kW in the summer.  (Assuming arotherm plus monobloc)

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It is worth remembering that it is very rare that a hot water cylinder will drop to the incoming mains temperature.  Mine rarely goes below 30°C and I only heat my water to around 45°C, so about 3.5 kWh.

This is double checked with my energy meter and is about right (I am all electric so easy to monitor).

When I get back home, after a week away, my cylinder will be cold, I shall check the temperature if I remember as I have some probes on it, so easy to do.  Then I can see what the overnight electricity usage is.

My system is a vented system, so the 100 litres in the loft header tank is often above 20°C.  This is, in effect, passive solar heating.  That changes during the winter though.  Not had a full year logging the loft tank temperatures yet, but the data is being collected.  I have posted up my water temps somewhere on here.

 

Here it is.  I cannot remember what time period this chart comes from, I should have put it into the title.

 

image.png.a202e32d89f5a9f2d49b7ef0a79302ab.png

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26 minutes ago, OldSpot said:

Is there a way to work out how long the tank will take to cool when in use?

Yes.

It is identical to working out the heat loss through a wall.

You need to calculate the total surface area, work out the U-Value from the materials used. Usually just the insulation type and thickness is close enough. Then work out the temperature differences between the water (take a mean temp) and the surroundings.

Then multiply the lot out.

That will give you the power loss. Then calculate how long it will sit idle for i.e no input or usage.

When I get some free time I may write this out with a worked example.

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3 hours ago, OldSpot said:

I meant is there a simplistic way to work out how long it takes for the UFH to reduce the temperature of the water in the tank by drawing off (hot) water from the tank

Yes.

Just calculate the energy in the tank, calculate the energy needed to raise the slab temperature, then work out at what rate you want to deliver the power.

 

So if you start with 10 kWh in the tank, and it takes 1 kWh to raise the slab 1°C, and you transfer at 3 kW, then it will 20 minutes.  The tank will therefore last 200 minutes, or 3 hours 20 minutes.

 

The thing that is important is the starting temperature of both the tank and the slab, plus (well minus really) the losses they both experience.

 

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  • 3 weeks later...
On 21/10/2021 at 23:01, Dan F said:

We are also having a Vaillant ASHP installed (7kW aroTHERM Plus) and I had exactly the same query so I have the datasheets and found the following:

- The Vaillant aroTherm Plus's comes, from factory, configured to run in "eco" mode for DHW which means 50% compressor speed.  This can be disabled.

- The power output of the 5kW model (eco mode) at 2C external temp is 3.6kW.

- The power output of the 5kW model (max mode) at 2C external temp is 7.3kW.

 

There are three modes - Eco (45% on the 12kW but that may vary with the actual unit fitted) - achieves about 3.8 at 48°C; Normal which simply runs 100% until satisfied (achieves about 2.5 but offers much shorter recovery times); and balance which runs 100% until the flow temp is satisfied (cylinder set point plus 5K) then backs down the compressor to reduce overshoot. The latter extends heating time of a 250 litre cylinder by about 20 mins in my experience and achieves a COP of maybe 3 or so.

 

The system is able to recover surprisingly quickly with enough coil area in the tank. For example when the water is cold the 12kW ASHP will deliver about 17kW into the tank, which reduces as the cylinder temp and hence flow temps increase.

 

HTH

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Just curious..  can you run ASHP into a directly heated tank (eg one that doesn't have a coil on the input). I can't think of a reason why not.

 

Edit: Does the loop between ASHP and a TS have to be pressurised? 

Edited by Temp
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3 minutes ago, Temp said:

Just curious..  can you run ASHP into a directly heated tank (eg one that doesn't have a coil on the input). I can't think of a reason why not.

 

Edit: Does the loop between ASHP and a TS have to be pressurised? 

 

Yes as it will lock-out if the primary circuit pressure is below a threshold. A buffer tank is a tank without a coil - the only downside will be cost to fill it with all that glycol.

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In simplistic terms, can plumb the existing to a plate heat exchanger and the AHSP on the other side of it. That way the two are separated meaning there is less glycol needed (only between the ASHP and the heat exchanger) and everything else can stay as it is. When work is needed on the house side, it can also be drained and filled as normal. The downside is that the heat exchanger creates a temperature step so costs a little bit of efficiency.

Edited by J1mbo
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