air / water heat transfer 1

[platypus83]

i have a question about delta t for heating coils in fan coil units
it seems like a stupid question but anyway

if a heating system is designed as 82deg c flow and 71 deg c return therefore an 11 deg c delta t. If all the coils in the system are also designed for 11 deg delta t at design flow rate of water. Assuming that heating water flows and airflows across coils are at design, what would the effect of running the boiler at 70 deg c be?
would the return water temp be 11 deg c lower? (i.e. 69 deg c)
would the air still achieve its design heat gain across the coil? assuming max air temp design is 24 deg c

basically what im wondering is how important is the flow temp of the water?
i know in design Q = mc delta t is used for boiler size, coil sizes, airflow and water flowrates but that only uses a delta t and not an actual temp.

 

[IRstuff]

Why would you assume that everything would stay the same? If someone dropped your body core temperature by 12°C, would everything work the same?

Your system appears to be designed to supply a certain amount of heat from 82°C water to 24°C air, i.e., a 58°C temperature delta. Reducing that delta to 46°C means less heat supplied, so the return cannot have dropped 11°C.

TTFN

FAQ731-376

 

[ChasBean1]

If you set the m*c*dT air equal to m*c*dT water and drop your inlet water temperature, all else being constant, you'll find you don't achieve your peak (design) air temp.

Which is often fine for most conditions. When it becomes not fine is at design OA conditions, when your reduced air temperature causes control valves to be fully open and eventually not enough flow to meet demand. If you think of it on the demand side, the building will just require one Q for the given OA condition regardless of what temperature the boiler provides. But if you want to reduce temp (fine for warmer weather) your flow (m) has to go up to keep the same Q...

Hopefully helpful? CB

 

[platypus83]

ok still not sure

if air design conditions are as follows (example only)
air flowrate = 0.3 m3/s
outside air = -2, design supply air = 22 (air DT = 24 deg c)
Heating requirement to do this 8.82kW

Therefore fan coil unit heating coil size is 8.82kW, heating system has 11 deg c DT (82 flow, 71 return). (which are the standard f & r temps for LTHW systems)
so water flowrate to do this = 0.191 l/s

assuming we are achieving these air and water flowrates and they are constant, and assuming it is -2deg c outside.
so at 82 deg c water flow we will have 22 deg c air.

what will happen when water flow temp from boiler is reduced to say 70 deg c?
what will air temp now be?
how to calculate this?
how to calculate new DT?
would it not be DT= q/mc....in that case 8.82/0.191*4.2= 11

this is what confuses me. has Q changed?
does Q stay constant because that is what the coil physically is?
Or is a coil actually specified/supplied for kW, and a DT only at design water flow temp (i.e. 82)
in that case Q of the coil changes with flow temp, and then DT will also change so flow will need to increase to keep DT at the desired 11.
could somebody clarify please!

 

[MintJulep]

Q is not constant. For a given physical heat exchanger at fixed flow rates of both fluids it is a direct function of the temperature difference between air temperature and water temperature.

Imagine if the entering water temperature were the same as the entering air temperature.

 

[IRstuff]

OK, you need to dig out your HT textbooks.

Q is proportional to dT. Reduce dT, you reduce Q.

TTFN

FAQ731-376

 

[platypus83]

ok that makes sense thank you.

so if water flow temp drops from 82 to 70 and both flowrates across coil remain the same and outside temp remains same
we know that Q and DT have both changed
How do i work out what they have changed to? as i cannot use Q=mcDT

is coil output directly proportional to flow temp of water?
i.e. 82deg c = 8.82 kW = 100%
so 70 deg c = 7.5kW = 85%

if it was then the new DT could be worked out with
DT = Q/MC
7.5/0.191*4.2 = 9.35

is this correct??

 

[IRstuff]

How can you say Q=mcdT in one sentence, and not use that equation in the next sentence?

70-24 / 82-24 = 79.3%

TTFN

FAQ731-376

 

[platypus83]

IRstuff - i am just looking for clarity on this as my knowledge is not extensive. sorry to drag this on

in the equation you used above you got the ratio between a 70 deg water flow temp to heat air by 24 deg - to - an 80 deg water flow temp to heat air by 24 deg
this was 79.3%

so if the 100% output of the coil at 82 deg c was 8.82kW then with 70 deg c flow it is now 6.99kW

so new water DT = 6.99/0.191*4.2 = 8.72 deg c (was 11)

is this correct, and is this the only way to work out the new DT?

the reason why I am confused is because Q = mcDT does not allow me to work out new DT because all I know is M and C.
so is this ratio method the only way?

 

[MintJulep]

Read this:

http://www.me.gatech.edu/energy/laura/node5.html

 

[sailcat753]

You are dealing with two different variations of q=mcdeltaT
The water side can be written q=500gpmdeltaT
The air side can be written q(sensible heat)=1.08cfmdeltaT
Set both sides of equation equal to q or
500gpmdeltaT(Water)=1.08cfmdeltaT(Air)
Assuming gpm and cfm are constant,
Changing the deltaT(Water) will effect the deltaT(Air)

 

[platypus83]

mintjulep - im not sure if log mean temp difference is what i need? (what can this be used for anyway) and the fan coil has a counterflow heat exchanger

sailcat 753 - yes i know what you have said above but my query is how do i work out the new DT of the water if the flow temp has changed from the design temp, if water flowrate remains constant? see above example @ 18:31

 

[walkes]

Phoenixpark,

If you need the same temperature for the air side, to get the capacity out of the water, you will likely require a different coil. Possibly more rows (or fins per inch), or a different construction.

So give a coil supplier your desired conditions, and they will give you the deltaT and other useful info.

 

[MintJulep]

LMTD can be used to determine the performance of heat exchangers - which appears to be exactly what you need.

 

[lowedogg]

heat transfer effectiveness of the coil will decrease as the delta between the air and water decreases. read the heat exchanger section of your heat transfer textbook. like MintJulep says, LMTD determines heat transfer for a heat exchanger, it is not a simple delta T equation due to the effectiveness of the heat exchange surface.

 

[IRstuff]

The whole point of LMTD is to come up with a single temperature difference to use in the heat exchange equation. Otherwise, you'd be doing a distributed analysis.


TTFN

FAQ731-376

 

[LordP]

Hmm, I have a similar question but this doesnt seem to have been answered definitively.
Heres my version of the problem:
Nearly all manufacturers data on chilled water coils is for 6/12 degrees C water.
I am looking at the impact of using 9/15 chilled water in a datcentre to save energy (More hours free cooling and is above dew point so no re-humidification load) and need to know how much the recirc units will be de-rated when using the higher temperature.
Yes I know I should ask the manufacturer but thats no fun...
Given that the airflow over the coil, coil surface area, air temp, and max water flow rate in the coil are all identical, is there no way to calculate the reduction in heat transfer efficiency?

 

[IRstuff]

Your deltaT is 3°C higher

TTFN

FAQ731-376

 

[LordP]

The equations in the LMTD link give sensible answers, Thanks.
Although they asume that surface heat transfer cooeficients are constant, so I wouldnt use them where the air flow rate also changes.