×
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

simple air flow problem

simple air flow problem

simple air flow problem

(OP)
Administration kills engineering. Design calculation ability is a muscle, if not trained it atrophies...

I am curently designing a mobile generator unit in a closed area.I need to determine the amount of air flow requiered to maintain an ambiant temp of no more than 40ºc.

Known:
desired ambiant temp:  40ºc
outside temp (cooling air) :could reach up to 30ºC
kw regected by components inside habitacle :205Kw

unknown:

air flow volume needed to maintain ambiant temp at no more than 40ºc in order to select exhaust fan and inlet damper/loovers?

air speed could reach up to 1000ft/min

I need reminding of practical equations/formulas and basic notions?

RE: simple air flow problem

As the specific heat of air is close to 1000 J/kg°C you simply get
Q=205000/1000/10=20.5 kg/s or some 16 m3/s.
This seems quite a huge amount (unless you are in a wind tunnel): an area of some 3 m2 is required to stay within your speed limit.

prex

http://www.xcalcs.com
Online tools for structural design

RE: simple air flow problem

The heat rejection is quite impressive. At the present cost of electricity, this means the generator would be throwing away more than USD 15 an hour. Although the general geometric dimensions of the generator and its hot parts as well as the overall energy output were not given, one can appreciate from the lost heat it seems quite a big unit.

It appears it would be impossible to keep the temperature of the room at 40oC only by convection with the imposed limitations on a continuous basis.

Even with radiation effects included the temperature of the hot surfaces become very large or, alternatively, the temperature difference between them and the surroundings are excessive, or both.

For example, htc for convection (20) at the given "wind" speeds, and for radiation (10), together would result in a combined htc of about 30 W/(m2*deg C).

Assuming an air heat up of 10oC, to estimate its flow rate we can use the balancing equation:

205,000 W(=J/s) =  (m)*(Cp)*(10) = (m)*1010 J/(kg*oC)*(10oC) = (30)*(A)*(Tgen-Tsurroundings)

Then, m = 205,000/10100 = 20.3 kg/s, or some 36,000 cfm, which would need a generator cross-section of 36 ft2 to get a "front" air velocity of 1,000 fpm as a given maximum.

On the other hand, the product (A)*(Tgen-Tsurroundings) = 205000/30 = 6833, means either a very large area for the hot surface, or a very high temperature for this surface, or both.

A finned-tube chiller could be considered to remove this heat to the required level on a continuous basis.
I've stressed the continuity of the operation since nothing was said about it in ddace's presentation.






RE: simple air flow problem

(OP)
I am now worried about this continuous basis information.

But to answer the questions yes it is a large unit: 2.5 Mw

the 205kw regected includes every thing dissipated in the enclosure.

Please explain why convection is insufficient on a continous
basis ?

RE: simple air flow problem

I guess that the mobile generator may consume 205 kW fuel. often they are watercooled and the exhaust gases are of course blown out, only a fraction will be left in the room as losses. Ask the supplier of the generator abou this figure. My guess is 40kW. If the outdoor air is 30 C. Then the air flow will be 40 kW/(40-30)= 4 kg/s

RE: simple air flow problem

The continuity of the operation refers to around-the-clock operation, since the night outside air may be much cooler than 30 deg C, meaning a smaller more amenable cooling air flow rate.

A heat release of 205 kW, means 8.2% of the generated energy. Kindly refer to lilling's comments for clarity.

 

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