DHW Cylinder

[Clunsy]

Hi All.

The following query may seem fairly basic for you mechanical guys, but please forgive me - im electrical!

Im currently assessing the need for a DHW Cylinder in an industrial plant. The cylinder is 2m x 1m, so i've calculated the volume to be 1570 litres. As the diverting valve is manually open it is curently being heated 24/7 from a gas boiler. Could someone please advise me how i would go about calculating its daily energy consumption in kWh.

By my calculations, only around 700 litres of DHW is being used per day, so while sorting the controls would be a good first step, im wondering would the best idea be to do away with the cylinder completely and either use point of use electric water heaters of install a small combie boiler.

Any help would be much appreciated.

 

[desertfox]

hi clunsy

Not enough information in your post we need the temperatures of the water that your heating ie start temp say 20 degrees Centigrade and final temp 100 degrees Centigrade for example and the power of the current heat source.
having got that information a typical calculation would go something like this:-

Q = m*Cp*(T1-T2) where Q= heat input

Cp = specific heat capacity
4190 joules/(kg*K)

T1-T2 = temp change in
Kelvin


So your 1570 litres = 1567.65kg of water

therefore to heat that amount of water from 20C to 100C

then Q = 1567.65*4190* (373-293)Kelvin

Q = 525.5 Mjoules

Thats your heat input required and because I have plucked figures out of the air its a very large amount.

Now to finish lets say it takes 24 hours to heat up.

Watt = joule/second and 24 hours is 24*60*60= 86400secs

power watts = 525.5 * 10^6/(86400) = 6.083KW

so thats the power required to heat the water over the day.

This isn't an answer to your specific question but if you get the figures I mentioned at the start you should be able to work from my example.

desertfox

 

[Clunsy]

Thanks very much for your responce Desertfox.

The heat source is a 560kW gas boiler, via a primary pump. As well as the DHW cylinder, it is also serving the building's radiator circuits, AHU's and a large heat exchanger for the plant's process.

From an efficency point of view, i believe the plant is currently very poorly run. Due to lack of controls knowledge the maintenance staff switch everything on on a monday morning and off again on a friday night.

I estimate that around 720 litres of hot water is used from the cylinder per day:
- 60 staff (over 3 shifts), each washing their hands 3 times a day, using 30 seconds hot water per wash, with a flow rate of 8 litres/minute.

However the primary pump runs on hand 24 hours a day 5 days a week.

Hence, your query regarding starting temp (say 10 degrees C) and regulated temp (60 degrees C) is relevant for a monday, however the rest of the week i am trying to calculate the energy wasted keeping it up to temperature.

*In fact, because the diverting valve is currently manually open the cylinder temp is not the required 60 degrees C, but rather the same temp as the boiler.

Again all advice/guidance will be very gratefully received.

 

[Clunsy]

2 values im not sure of are the size of the primary pipework and the size of the primary pump (possibly around 2kW?).

For the purpose of the exercise im happy to use estimates for these.

Again, thanks.

 

[desertfox]

hi again

I guessed that you might have a cut off temp and then the heater cuts in again, well at what temp does the heater cut in again? you can use the same calculation above just change the temps to suit your situation so monday its the same calcultion except your figures for temp are 10 and 60, the rest of the week whatever temp the water drops too before the heater cuts in again so if its say 30 then temps between 30 and 60

desertfox

 

[Clunsy]

If controled right, thats how it should work - cut in (say 55 degC), cut out (say 65 degC). But as the primary pump is currently left on manual and the cylinder valve is manually open, the cylinder is never cooling below the boiler flow temperature (around 80 degC).

Effectively im trying to prove the energy wastage associated with the this unstatisfactory set up, so i need to prove the hourly kW consumption associated with keeping the cylinder at 80 degC.

 

[desertfox]

hi again

Well won't it be the 56KW onstantly on, they will be no heat input into the water unless it drops through heat losses and stuff like that, what sort of insulation you got round the tank.
I would start with the insulation round the tank and work out the heat loss through that.
so if your 56KW is on for 1 hour won't that be 56KWh

desertfox

 

[Clunsy]

The boiler is 560kW. The cylinder is only a very small part of its load.

With standard (not great) insulation would it be fair, roughly speaking, to assume losses of around 5 degC/hour?

If i understand your worked example the total heat required to heat the cylinder from 20 to 100 degC is 145kW, regardless of whether its 145kWh for 1hr or 6kWh for 24 hrs?

If ive got that right (and please correct me if im wrong), then i think the following would be a fair calculation of the energy consumption of the this DHW cylinder in 1 week:

Calculate the kWhs required to bring it up to temp on a monday. Then calculate the kWh value required to top up the 5degC differential each hour. Multiply this figure by the number of hours eg.
Monday 3hrs for initial rise
21hrs top ups
Tuesday - Friday 24hrs top ups.

Total top ups = 117hrs

Total Energy:
(117 x 9.1kW)+ 145kW = 1208.7kW/week

I cant thank you enough for you guidance, you help's been excellent!

 

[desertfox]

hi Clunsy

The heat input required is in Joules in my example using my original figures the heat input would be 525500000 joules of heat for your case your heat input would probably be round about half of that.
i think your confusing pwer and heat input lets use your figures:-

monday 10C to 60C

therefore 1567.65kg * 4190 * (333K-283K)= Q = 328422675J

now I don't know how long the tank takes to heat but lets say which is unlikely that all your power source is used to do this therefore:-

328422675J/(56000)watts = time

do this you get an answer in seconds divide by 3600 secs to get hours and the answer would be 1.629 hours.


so without allowing for any heat losses, then the full power source ie 56KW would take just over one and a half hours to heat the full tank.
Hope that helps.
What we need if the power supply is supplying other stuff is the heat loss from the tank approx to the atmosphere, you really need to measure the temp on the outside of the tank after monday and knowing the temp inside is 60 we can calculate roughly whats lost through the walls for starters, we then can workout what heat is going back in the keep temp constant. So insulation type and thickness would be good as well as outside temp on insulation.

desertfox

 

[Clunsy]

I totally agree that those factors would be required for an exact answer, but for the time being would you agree that 5degC per hour would be a fairly standard loss from a tank that size, with average quality insulation?

 

[desertfox]

Hi Clunsy

Well if you wish we can assume that I don't have feel for that, ok so it would go like this:-

Q = 1567.65*4190*(-5K)=-32842267.5

so power loss in 1 hour = 32842267.5/3600 = 9122.85 watts

or 9.122KW in 1 hour

so a day = 9.122Kw * 24 = 218.948 KW

seems a lot to me but thats the answer

desertfox

 

[Clunsy]

Perhaps its not the most scentific way to approach it, but ive got the value of 5degC loss/hour from the assumption that it would take a similar cylinder around 8 hours to cool from 60 to 20degC.

On reflection, perhaps it would be closer to 10 or 12 hours, but i dont think we're far away.

That aside, would you agree with the rest of my methodology in the total weekly energy consumption calculation below?
Total Energy:
(117hrs x 9.1kW)+ 145kW = 1208.7kW/week

You've been a fantastic help today - i'll know where to come again if ive mechanical queries!

 

[desertfox]

Hi Clunsy

I agree except for the 145KW for heating the tank, going from 10 to 60 degrees Centigrade would only take half hour or just over to heat up but if thats what it is then thats it, you know the plant better than I.

desertfox

 

[Clunsy]

Well spotted. The 145kW was from you original calc.

I feel like ive learned a fair bit from this conversation, so thanks again.

 

[desertfox]

hi Clunsy

your welcome

desertfox