Heat Transfered from Air Blower to Water 2

[Ajax001]

Hi Folks,

I would like to estimate the amount of heat transfer I can expect to get when air bubbles are passed through a moving body of water.

For example, we have design applications where an air blower is used to agitate the media inside a biofilter chamber.

The air post air blower can get warm due to the mechanical action of the air blower. This warm air is then passed into the water (typically operating at 16C).

I would like to estimate how much this air could add to the heat gain of the water to consider any impact on temp control equipment.

Does anyone have any experience in that area?

Thanks..

 

[IRstuff]

How big are the bubbles and how long are they in the water?

Assuming they're less than 13 mm in diameter, the heat transfer process within the bubble is purely conductive. You could assume a lumped model of each bubble, determine the heat flow, the length of exposure, and the number of bubbles to determine the heat transfer.

TTFN

FAQ731-376

 

[Ajax001]

Ok - if we assume the bubbles are less then 13mm then wouldn't we use the following formula for conductive heat transfer:

Q = (A / R)x Delta T

Where:

Q = Heat Loss (BTU)
A = Area (sq-ft)
R = Thermal resistance (hr F ft2)/BTU
Delta T = Temp difference b/w air bubble and water

That make sense?

Thanks...

 

[IRstuff]

Depends on how much accuracy you're trying to achieve, which is dubious at best, anyway.

The deltaT is changing over time, since the heat loss reduces the deltaT.

Also, the time of exposure determines the total amount of heat transferred.

Given all of that, I might think that having a new temperature every second might be a reasonable stab.

TTFN

FAQ731-376

 

[Ajax001]

Good points.

I found a paper which has allowed me to get a handle on the average upward velocity of the bubbles in this application. That velocity will be around 20 cm / sec.

The depth in this and most applications is 2m. So, in theory, any one air bubble should spend around 10 seconds in contact with the water column before exiting into atmosphere.

The water temperature is 16C and turnover is very rapid with a retention time of about 30 seconds. As such, the water will carry away any absorbed heat from the air bubbles very quickly.

I am assuming the temp of the air bubbles will be 20C. So the question is I guess - will a 5 mm air bubble reach equilibrium temp with the surrounding water within that 10 seconds?

If we can say that, then maybe the easiest way of going about this would be to simply calculate all of the BTU available in the air within that delta T.

For example, in this application, the air blower is putting out 2,300 SCFM in 2m water. The specific heat of air b/w the range of 16 - 20C is constant at 0.2403 BTU/lbF. By converting the SCFM into lbs of air per min and assuming the air bubble "dumps" all of its heat differential before exiting to the atmosphere, I should be able to calculate heat gain.

Make sense?

 

[Compositepro]

Are you taking evaporation into account?

 

[itsmoked]

You are talking about gargantuan surface area for the thermal mass associated with each bubble. I would expect full equilibrium in that time. I'd expect it in 5 seconds.

Take the motor HP, the blower efficiency and consider all the resulting energy as going into the solution.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[Ajax001]

Hello itsmoked,

Yes, I did some calculations using the following assumptions:

The air flow rate in this case is known to be 2,300 SCFM.
Hole size in the pipe is 1/16" or 1.5m.
I assumed a bubble size of 5mm which is probably bigger then actual given that the rule of thumb is 1-2x hole size I believe.
Once I calculated the volume of one air bubble and divided that into the total SCFM, the number of bubbles was well, gargantuan as you would put it. Total bubble surface area was crazy at over 368,000 sq-ft.

In speaking with the air blower provider, he suggested we use a rule of thumb of 0.5Deg F for every inch of water column to calculate the air temp at outlet of air blower above ambient. Given that water column depth is 80 inches, this amount to 40 Deg F. Assuming the air is entering the air blower at 61 Deg F, it should be exiting the blower (and entering the water column) at 101 Deg F.

So you feel that effectively all of that heat will reach equilibrium with the surrounding water?

Thanks..

 

[itsmoked]

Yes I think it will all go in.

Here's what I think:
Let me try this:

80" / 12 = 6.67ft

psi = 6.67ft x 0.434psi/ft = 2.89psi Call it 3 psi (to get it delivered.)


Blower HP = (Q x p) / (229 x u)

Q = cfm
p = psi
HP = hp
u = efficiency coefficient

In your case:

HP = (2300 x 3.0) / (229 x 0.80) assuming your blower is 80% eff.
HP = 38hp

With 5mm bubbles in 80 inches of water I would expect those bubbles to be leaving the surface at the water temperature regardless of the temperature they entered at. (within reason)

Since 38hp hr = 100,000 Btu, that's what I'd think is being added into your water every hour.

water temp rise degrees F = 100,000 / (8.33 X gallons in tank)

Of course this didn't take into account the temp delta between the blower inlet and the water temp. That would need to be added on top of this.


Your surface area number represents a huge interface between the two media, 8.4 acres for thermal transfer!! And all the mass of interest within 2.5mm of each other.

Yeah I think it all goes in.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[IRstuff]

The thermal time constant is on the order of about 0.09s, so, yes, the bubbles will be equilibrated before they leave the water.

TTFN

FAQ731-376

 

[Ajax001]

Ok, I follow what you are saying but wouldn't it be safe to assume that the majority of the motor heat would be going to atmosphere. The blower in this case is a type HP pressure blower. The motor itself isn't in contact with the air so the majority of the motor heat I should think, would be going to atmosphere.

On the other hand all of the heat from the air will get transfer to the water given your thermal time constant of 0.09s.

That make sense?

So, using the equation Q = (A/R) x Delta T

A = 33,792 sq-m
R = 1.95 m2K/W
Delta T = 22 Deg C

Q = 382,125 Watts or 1,299,226 BTUH

 

[IRstuff]

oops, my bad, the time constant is more like 0.9 seconds.

TTFN

FAQ731-376

 

[itsmoked]

Yes, the blower motor, if it is not in contact with the water AND it's own air dissipated heat is not being sucked into the blower inlet will not be attributed to any water heating.

The pump's horsepower, which is far larger, will show up in the water.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[IRstuff]

You've already stated that the blower exit air temp is 101°F. So that's the heat that the motor has transferred to the air. How, or why, it is, isn't relevant anymore.

To shortcut the calculation, ignore the bubble stuff.

Given 2300 cfm of 101°F air into 16°C water:
flow*dens*C.p*deltaT = 29 kW

where:
deltaT = 22°C <-- 38°C(101°F) - 16°C
dens = 1.2 kg/m^3
flow = 2300 ft^3/min
C.p = 1005 J/kg-K

So assume all that heat goes into the water. Assume some volume of water, say 2m x 1m x 1m exchanges at 30 seconds.

This results in Q.wat*30s/(C.pwat*vol*dens) = 0.1°C

where:
C.pwat = 4.2J/gm-K
vol = 2 m^3
dens = 1 gm/cc

TTFN

FAQ731-376

 

[Ajax001]

I just wanted to say thanks to everyone for your help on this question. With all your input, we have been able to come to a conclusion with this heat transfer question.

Regards...