Login | Register

Previous in Forum: Coatings for Underground NG Pipelines   Next in Forum: Design Pressure and Temperature
Close

Comments Format:






Close

Subscribe to Discussion:

CR4 allows you to "subscribe" to a discussion
so that you can be notified of new comments to
the discussion via email.

Close

Rating Vote:







3 comments
Participant

Join Date: Oct 2009
Posts: 2

HVAC Pressurization

10/29/2009 12:56 PM

I want to know how calculate the air flow leakage between two rooms through door to make a diffrential pressure between this rooms equal 15 pa for the following example :-

first room dimension (4.5 m *6 m ) * 3m height and door dimension 1.2 m * 2 m and door clearance is 4 mm

second adjacent room is ( 3 m * 4 m ) * 3 m Height

thanks

Send to a friend Digg this Add to del.icio.us
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.
Commentator

Join Date: Jun 2008
Location: Working in Oman
Posts: 92
Good Answers: 1
#1

Re: HVAC Pressurization

10/30/2009 2:20 AM

If the room is designed as per the clean room specifications proper pressure measuring instruments must have been installed in the rooms. You have to check the number of air changes you have used to arrive the air. Use instruments like u tube ,anemometer ,and pressure gauges to measure various parameters as per the ASHRAE standards and arrive the va

Participant

Join Date: Sep 2009
Posts: 3
#2

Re: HVAC Pressurization

10/31/2009 7:00 AM

100 CFM = 10 Pa

as per this u could know the pressure difference u want 2 maintain

inside u r rooms

Member

Join Date: Jul 2009
Posts: 8
#3

Re: HVAC Pressurization

11/02/2009 4:40 AM

A possible way do create a pressure drop over 2 rooms is to make the difference by means of a different surface of ductwork:

dp = ½ ρ (v2² - v1²).

Suppose v1 = 3 m/s:

dp = 0,5 * 1,3 (v2² - 3²) = 15 , so v2 = 3,75 m/s

suppose that airvolume (Q) to room 1 equals airvolume to room 2, then:

v1 = Q/A1 (A1 = cross section duct 1) and

v2 = Q/A2

so: (Q/A1): (Q/A2) = 3 : 3,75

Then: A1 = 1,25 A2, or A2 = 0,8 A1

So by varying the ductwork cross section you can vary the room pressures.

Of course you can make the pressure difference also by placing another pressure consuming device into the inlet ductwork and then keep A1 equal to A2.

Another possibility to do this in the outlet ductwork from the rooms

3 comments
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.
Copy to Clipboard

Users who posted comments:

mrswamy (1), Rien Dutchman (1), zahir abbas (1)

Previous in Forum: Coatings for Underground NG Pipelines   Next in Forum: Design Pressure and Temperature
You might be interested in: HVAC Ductwork, HEPA Filters and ULPA Filters, Industrial Cooling Systems and Equipment