×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Gas spray cooling temperature control
2

Gas spray cooling temperature control

Gas spray cooling temperature control

(OP)
I am cooling gas in a bypass around a CO Boiler from 1200 deg F to about 600 deg F using water sprays.  I am looking for a way to calculate the evaporation time in order to locate the downstream temperature sensor for the temperature control which will adjust the water rate.  Any rules of thumb would be helpful too.

RE: Gas spray cooling temperature control

kpar:

Biggest variable here is the type of nozzle you and use, the pressure drop across it and the nozzle arrangement (i.e., whether or not you have a hose stuck into the side of the duct, or a bank of fog nozzles.  However at those temperatures, it won't take long to flash off the water, assuming you have even a halfway decent spray setup.  I would suggest you do a quick heat transfer calculation for a representative droplet in your gas stream.  Bet you find total flash time is about 1.5 - 2.5 seconds.  

RE: Gas spray cooling temperature control

2
kpar: Fizzhead is correct.

There is one formula I found for the approximate time, t, a droplet takes to evaporate in stagnant surroundings:

t=(λ)(ρL)(D2)/[8 (kg)(LMTD)]

λ: latent heat of evaporation
ρL: liquid density
D: original drop diameter
kg: gas thermal conductivity
LMTD: logarithmic temperature difference


Let's work out an example with rough figures:

λ: 2,400 kJ/kg
ρL: 1000 kg/m3
D: 300 microns = 0.0003 m
kg: 0.05 W/(m.K)
LMTD: 700 K

t=(2,400,000)(1000)(0.00032)/[8(0.05)(700)]~0.8 s

Estimates as this, may enable one to see whether at the prevailing gas velocities there is enough residence time to accomplish full evaporation and the desired gas quenching.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources