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Tayters
18-10-2012, 11:28 PM
Customer wants to use an air to water to heat water in a tank. Time to heat up isn't too much of an issue as the water is used in a hostel so has all day or night to heat.

I don't want to get to bogged down in the whys/ do's and don'ts and specifics of the system, just need pointing in the right direction for the theory behind the sizing then I can work stuff out as required.

The tank holds 600lit.
Air to water capacity at ambient encountered 7.1kW
Nominal flow rate 28.7 lit/min.

I want to work out how long it will take to heat the water but can't quite fathom it,

Lets say I want to raise the water from 5*C to 55*C.
Q = mass x shc x dt
Q= 600 x 4.18 x 50
Q = 125,400kJ

Now how do I work out the time the heat pump will take to heat the water.

If the duty is 7.1kW then this is 7.1kJ/s, divide this into 125400 gives 17662 sec or 4.9hrs.

Or...

Lets assume I lose 5K between water into tank and water out.
Flow rate is 28.7lit/min = 0.48lit/sec = .48kg/sec

Q = 0.48 x 4.18 x 5
Q = 10kJ/s added to the tank.

Dividing the same way gives around 3.5 hrs to heat the water.

Second way seem wrong as total duty is 10kJ/s but unit is rated at 7.1kW (therefore 7.1kJ/s)but this way I get to use flow rate which must have some bearing. Sure, td will change, hey maybe 55*C is too ambitious but just looking for the method if anyone can help then check specifics of the air to water system where required.

Thanks for reading,
Andy.

Tesla
19-10-2012, 12:22 AM
Hi Tayters
If you use a water to water heat exchanger taking the remaining heat from the waste shower water to preheat the cold water you can reduce the kW required by half. The water comes out of rose at say 42degC and goes down the drain at say 38degC - The fresh water should be near ground temp say 10degC, so the fresh water could be preheated to over 20degC. Search GFX heat exchanger - you could make one easily from 5/8" Cu tube. If you calc on instantanious instead of a tank get the total max flow rate in L/s. I did an exel spread sheet on show heatload calcs - I will try to find it and post.

Tesla
19-10-2012, 04:34 AM
Bumber
My harddrive is dead. I will have to try to get the data retrieved from it. I had an awesome spread sheet of where I could change any parameters and the calcs were all sorted with formulas. Sorry I will still try to post the spread sheet.

icecube51
19-10-2012, 08:04 AM
Hi Tayters,

If you want to do it, this is how i do the maths.

600Lt X 4.19KJ/Kg.K X 50°C = 104.75 kWh
time 3600

so if you have a heatpump of say 10kW the sum is then 104.75 kWh / 10kW = 10H 47min 50 secs to heat up 600Ltr tank
To calculate the losses of say 2°C a day you do the same as above and top it up by the previus.
you become than 104,75 kWh + 1,39 kWh = 106,14 kWh

That is if you use the 600Ltr water a day.
Otherwise it is the amount of Kwh needed to heat the tank for the first time and then calculate the amount of water/ day is used. You can come to now this bi putting a metering device on the cold water line.

Hope this was usefull?

Ice

mad fridgie
19-10-2012, 10:36 AM
Do not mix flow temps and total water mass.

So heating 600L of water is around 35Kwhr, so your run time is about right.

The heat pump flow rate will be multi-pass (the water only rises in temp by a very little), on would hope your are not loosing 5C when flowing!

Tayters
20-10-2012, 12:32 AM
Thanks for the replies all.

Tesla: good idea about the waste water, wouldn't have considered that one. Yes if you manage to get the file that would be great.

Ice: Yes, useful thanks. It confirms my working although is that equation right as I get a different answer with those figures of 34.83kWh? When I divide that by 7.1 though I end up with 4.9hrs so i suppose we have worked out the same thing different ways.

MF: Ha, 5K was an (un)educated guess of loss through the hot water tank. Never played with a system but I take it then water though the outdoor HE will start off low and gradually build up temp? Are you saying then that the first way is correct and the second with the flow rate is the wrong method? Thinking about it you would mass flow if working out the heat emmited from a radiator for example?