Removing heat from wastewater flow

[cfree3]

I need to calculate how much energy it would use to reduce the temperature of a wastewater steam that is running approximately 80 gpm from 90 F to 75 F.
I am not an engineer so I don't know how to go about the calculation.

Once the energy cost is figured out, what is the best type of equipment to do this job?

Thanks for any help out there

 

[DRWeig]

What's the stream composition? If it's all water, it'll take about 80*(90-75)*500 = 600,000 BTU/hr net after losses.

If significant portion is not water, let us know and wait to hear from others in this forum...

Good on ya,

Goober Dave

 

[cfree3]

It is mostly water with a high degree of fine suspended solids. Very little big solids which are screened out prior to where we would need to take the heat out.

We can either take heat out as the water flow through a pipe or we could insert a cooling coil into a collection sump.

 

[zdas04]

cfree3,
I had to do a similar calc on a project once and the math is amazingly complex. You need to determine a "thermal entry length" which is the length of pipe at entry conditions required to acheive 80% of the required dT at a constant temperature sink. If you can determine that (which is more trial and error than anything else) then you can figure out how long the pipe would have to be to get to your require outlet temperature with a real sink (as the fluid temp approaches the sink temp, the rate of heat transfer decreases rapidly). I spent several weeks developing this term on a flow stream that was several orders of magnitude easier than what you are trying to do. Anything you do with a heat transfer coeeficient (look in the HVAC references) and a constant dT will end up being pretty wrong on a real flow.

DRWeig,
It always amazes me when people remember equations like you used. I have no problem remembering that Q=mcdT, but to know that m*c for water is gpm*500 BTU/gal-F-hr is just beyond me. I used to have hundreds of those simplifying equations in my head and they've all evaporated. MathCad does it too easy.

David

 

[DRWeig]

David,

Thanks for the comment -- actually, I have to admit that as I get older, I find the need to go back and re-derive the constants in some equations because I forgot them...

I happen to have the 500 constant pinned to my wall, though, since my company sells a lot of BTU meters and such.

I agree with you that designing the actual heat exchanger for the OP's problem would be a challenge!

Good on ya,

Goober Dave

 

[cfree3]

Maybe it would be easier to use a glycol chiller to run to a coil that is imersed in the sump.

However I would need to figure out what the electricity demand would be an what tonnage the chiller would have to be rated.

Would I need to have the sump volume and retention time to run the calculations?

 

[sterl]

I believe you are trying to upgrade the available heat or is the real objective to cool the drain water??? If its the evaporator side of a heat pump, the target temperature of the heated fluid determines what your compression arrangement will look like; lots of reasonably inexpensive pieces will operate pretty steadily over a long time if you are looking at 125 Deg. F. as a condensing temperature...You get past 140 and longevity gets a whole lot more challenging. And takes a bigger machine....

I've used a Baudelot cooler (a used piece from a scrapped out dairy) to pull heat out of wastewater and avoid fouling, plugging and so forth of the heat exchanger. Fine solids don't bother it a bit, though you do have to rework the water distribution pans and etc it won't ake a hit from a rock bigger than a golf ball. ....That for comfort heating with the other end a finned coil in a duct, like a domestic AC coil. I have also boosted both boiler feed and hydronic glycol,

 

[chicopee]

Take the difference in enthalpy of the fluid(liquid or steam)at 90F and 75F and you'll have a theoretical value.

 

[KiwiMace]

Alfa Laval has wide gap heat exchangers. I've used them before for removing heat from wastewater with sticky entrained solids. The percentage of solids in the water is usually so low that the water calc will be fine. A few percent of solids is enough to make water go non-newtonian and ruin your heat exchange calcs.

 

[cjw81]

Goober Dave, can you tell me where the constant 500 is derived from?
Thanks

 

[KiwiMace]

I'm not Goober Dave, so sorry for speaking out of turn, but:

60 seconds in a minute x 8.33 lb/US gallon of water = 499.8

 

[KiwiMace]

Sorry, that 60 is min/hour, to be pedantic.

 

[DRWeig]

Good on ya, Kiwi.

Goober Dave