# Pitched ceilings? Calculating house cross

```Air speed in tunnel houses is determined by
the following equation:

Speed
= Cfm / Cross-sectional area of the house
The greater the air speed
Factors affecting air speed in tunnelventilated houses
As a result…
for a given
number
of tunnel fans
The
smaller the
cross-sectional
area…
The larger
the cross-sectional
area…
The slower the air speed
Michael Czarick
The University of Georgia
How do we calculate house cross-sectional
area?

Cross-sectional area
Pitched ceilings?
= width X ceiling height
ceiling height
width
Calculating house cross-sectional area


Cross-sectional area
= width X average ceiling height
= width X (peak + side wall)/2
When trying to determine how much air
speed/cooling in a tunnel house will have…

We start out with our minimum summertime air
exchange rate to assure a uniform house temperature:



peak
side wall
10 cfm/ft2 (dropped ceiling) – rule of thumb
11 cfm/ft2 (open ceiling) – rule of thumb
Or, ideally the total required cfm would be calculated using
+5 F
width
1
Example:

40’ X 500’ dropped ceiling house
40’ X 500’
(dropped ceiling house)

House width = 40’
Side wall = 8’
Peak ceiling height = 11’
Fan capacity = 200,000 cfm

Speed



40’ X 500’ X 10 cfm/ft2
11’
8’
40’




WInd Chill (F)
Wind-chill effect at 85oF
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
50
100 150 200 250 300 350 400 450 500 550 600
Air Velocity (ft/min)
Wind-chill effect at 85oF
Wind Chill (F)
If we want more air speed/cooling we could

We take the original formula:

Speed

0
Cfm = desired speed X cross sectional area

For instance if we wanted 600 ft/min…
Cfm
= 600 ft/min X 380 ft2
= 228,000 cfm


Another option would be to decrease side
wall height…

0
50
= cfm / cross-sectional area
Rearrange to allow us to determine how much fan capacity we
need to have to obtain our desired air speed:


16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
= Cfm / (width X (peak + side wall)/2
= 200,000 / (40’ X (11’ + 8’)/2)
= 200,000 / (40’ X 9.5’)
= 200,000 / 380 ft2
= 526 ft/min
Lower ceiling…smaller cross-sectional area…higher air
speed
For instance, what if we lowered the ceiling one foot?
100 150 200 250 300 350 400 450 500 550 600
Air Velocity (ft/min)
2



House width = 40’
Side wall = 7’
Peak = 10’
Wind-chill effect at 85oF
10’
7’
40’

Speed




= Cfm / (width X (peak + side wall)/2
= 200,000 / (40’ X (10’ + 7’)/2)
= 200,000 / (40’ X 8.5’)
= 200,000 / 340 ft2 (Before 380 ft2)
= 588 ft/min
Of course if we increased side wall height air
speed would decrease

1’ higher side wall and peak
Wind Chill (F)
40’ X 500’ with shorter side wall
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
0
50
100 150 200 250 300 350 400 450 500 550 600
Air Velocity (ft/min)
40’ X 500’
(tall side wall)



House width = 40’
Side wall = 9’
Peak = 12’
12’
9’
40’





Wind Chill (F)
Wind-chill effect at 85oF
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Speed
= Cfm / (width X (peak + side wall)/2
= 200,000 / (40’ X (12’ + 9’)/2)
= 200,000 / (40’ X 10.5’)
= 200,000 / 420 ft2 (Before 380 ft2)
= 476 ft/min
Keep in mind…
short

Air speed is determined by fan capacity and cross
sectional area…not house length
tall
0
50
100 150 200 250 300 350 400 450 500 550 600
Air Velocity (ft/min)
3
Air speed in tunnel houses


Air speed in open ceiling houses
Though shorter houses require fewer fans to obtain the
desired air exchange rate…
The fact is that if you want the same amount of bird
cooling (heat removal) in your 300’, 400’ and 500’
houses and they all have the same cross-sectional area
they all require the same tunnel fan capacity
40’ X 500’ Open ceiling house




House width = 40’
Side wall = 8’
Peak = 16’
Fan capacity = ?
When calculating air speed in tunnel houses

16’
8’


40’
(open ceiling)

House width = 40’
Side wall = 8’
Peak = 16’
Fan capacity = 220,000 cfm

Speed







40’ X 500’ X 11 cfm/ft2
16’
40’
= Cfm / (width X (peak + side wall)/2
= 220,000 / (40’ X (16’ + 8’)/2)
= 220,000 / (40’ X 12’)
= 220,000 / 480 ft2
= 458 ft/min
8’
10 cfm/ft2 (dropped ceiling) – rule of thumb
11 cfm/ft2 (open ceiling) – rule of thumb
Wind-chill effect at 85oF
WInd Chill (F)
40’ X 500’
We start out with cfm required to insure our minimum
summertime air exchange rate to assure uniform house
temperatures:
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
dropped
open
0
50
100 150 200 250 300 350 400 450 500 550 600
Air Velocity (ft/min)
4
To increase air speed we need to add more
fan capacity




Cfm = desired speed X cross sectional area
Cfm = 528 ft/min X 480 ft2
= 253,440 cfm

Deflectors/Baffle curtains
Same sized dropped ceiling house (8’ side wall, 11’ peak)
an air speed of 528 can be obtained with only 220,000
cfm (20 percent less fan capacity).
What happens to air speed if we install a 10’
tall deflector curtain?
Air deflectors

Is there another option?
Air deflectors can be used to temporarily decrease cross
sectional area and therefore temporarily increase air
speed.
10’
Determining wind speed in a open ceiling
house with air deflectors

House width = 40’
Side wall = 10’
Peak = 10’
Fan capacity = 220,000 cfm

Speed






How do you determine what size deflector
you need?

Deflector height = cfm / (desired velocity x width)
10’
= Cfm / (width X (peak + side wall)/2)
= 220,000 / (40’ X (10’ + 10’)/2)
= 220,000 / 400 ft2 (Before 480 ft2)
= 550 ft/min
5
What if we wanted 600 ft/min in our open
ceiling house?




House width = 40’
Side wall = 8’
Peak = 16’
Fan capacity = 220,000 cfm
Air deflectors

?
8’
Though air deflectors can be very effective in increasing
bird cooling they can cause more problems than they
solve if not installed properly!
40’




Deflector height = Cfm / (600 X width)
= 220,000 / (600 X 40’)
= 220,000 / 24,000
= 9.2’

It is important to keep in mind that air speed is only
increased in the immediate vicinity of the deflector
General guidelines:


Deflectors should be no closer to the floor than 9’.
No more than 40’ on center…ideally 25’ on center.

9’
40’

These guidelines will help to insure that bird cooling is
maximized throughout the house
As the deflector is lowered air speed in the
vicinity of the deflector increases
Slightly uneven bird cooling

25’
9’ deflector

7’
But, air speed will only increase
in the immediate vicinity of the deflector
500 ft/min
500 ft/min
400 ft/min
500 ft/min
400 ft/min
40’
25’
6
Lowered further…air below deflector will
again increase but only a few birds are

6’
Significant differences in cooling

40’

300 ft/min
600 ft/min
300 ft/min
Large deflectors 60’ on center
40’ some differences in cooling
Large differences in cooling

600 ft/min
600 ft/min
25’
How does deflector spacing affect air
speed/bird cooling uniformity
6’ deflector
60’
To minimize cooling differences…

“Smaller” deflectors should ideally be installed 25’ on
center
7
Smaller deflectors 25’ on center
Smaller deflectors 25’ on center

If you are going to use deflectors..



An example of this is mini-deflectors in a
dropped ceiling houses
Keep in mind that less is more… and it is best to install a
greater number of smaller deflectors than fewer larger
deflectors
What about deflectors in dropped ceiling houses?
Mini-deflectors

More uniform wind speeds/cooling
Air speed profile in 50’ X 560’ totally
enclosed house
Many houses have 500 ft/min but may be growing a large
bird and want to try 600 ft/min
Furthermore, air speed is higher at the ceiling than the
floor because the ceiling is smooth…less friction than the
rough bird covered floor
684
789
764
845
705
745
810
805
812
711
705
748
741
692
619
8
(air speed 18” above floor)
550
500
450
400
350
s1
Distance from floor (ft)
Air velocity tends to be higher at the ceiling
than floor…with birds it gets worse
10
9
8
7
6
5
4
3
2
1
0
650
700
750
800
Air speed (ft/min)
850


30
20
10

320
340
360
380
400
420
440
460
12:00 PM
8:00 AM
10:00 AM
6:00 AM
4:00 AM
2:00 AM
air speed
fan
How do we discourage air flow along the ceiling?
Mini deflectors

300
s6
900
Air velocity profile of a 40’ X 500’ dropped
ceiling house (post side walls)
280
s3
Mini-deflectors

600
10:00 PM
300
12:00 AM
900
Air speed (ft/min)
600
8:00 PM
850
650
6:00 PM
700
750
800
Air speed (ft/min)
700
4:00 PM
650
750
12:00 PM
600
800
86
84
82
80
78
76
74
72
70
68
66
64
62
60
2:00 PM
10
9
8
7
6
5
4
3
2
1
0
50’ X 560’ House - All fans operating
Temperature (F)
Distance from floor (ft)
Air velocity tends to be higher at the ceiling
than floor
480

15’ in length
Installed 5’ off the side
wall
12” double hemmed
curtain (18”)
PVC pipe in top and
bottom
Suspended just below feed
line/water line cables
9
12” mini-deflectors 25’ on center
18” mini-deflectors 25’ on center
30
30
20
20
10
10
280
300
320
340
360
380
400
420
440
460
480
30
30
20
20
10
280
300
320
340
360
380
400
420
440
460
480
280
300
320
340
360
380
400
420
440
460
480
10
280
300
320
340
360
380
400
420
440
460
480
Last mini-deflector
Last mini-deflector
18” mini-deflectors 50’ on center
Fan placement and air speed
30
20
10
280
300
320
340
360
380
400
420
440
460
480
280
300
320
340
360
380
400
420
440
460
480
30
20
10
Last mini-deflector

40’ X 500’ (200,000 cfm)
40’ X 500’ Dropped ceiling house

Fan capacity = 200,000 cfm
House width = 40’
side wall = 8’
peak ceiling height = 11’

Speed



11’
8’
40’




= Cfm / (width X (peak + side wall)/2
= 200,000 / (40’ X (11’ + 8’)/2)
= 200,000 / (40’ X 9.5’)
= 200,000 / 380 ft2
= 526 ft/min
10
Bi-directional tunnel house


40’ X 250’ Dropped ceiling house
Same fan capacity, but put &frac12; of the fan capacity in each
end
What will be the air speed now?

Fan capacity = 100,000 cfm
House width = 40’
side wall = 8’
peak ceiling height = 11’

Speed



11’
8’
40’




WInd Chill (F)
Wind-chill effect at 85oF
Bi-directional tunnel house

16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0

0
50
= Cfm / (width X (peak + side wall)/2
= 100,000 / (40’ X (11’ + 8’)/2)
= 100,000 / (40’ X 9.5’)
= 100,000 / 380 ft2
= 263 ft/min
Same fan capacity, but put &frac12; of the fan capacity in each
end
What will be the temperature difference between the
100 150 200 250 300 350 400 450 500 550 600
Air Velocity (ft/min)
Simmons &amp; Lott 1981
What about a single direction tunnel house
with extra pad &frac12; down the house?
Five pound bird 85oF
60
Total heat loss
55
50
45
40
35
30
100
200
300
400
500
600
Air speed (ft/min)
11
But what effect would this have on air
speed…and therefore bird cooling?

75% - 25 % split pad
150,000 cfm
395 ft/min
Tunnel fan end (550 ft/min)
220,000 cfm
550 ft/min
Tunnel inlet end (395 ft/min)
100.0&deg;F
100
100.0&deg;F
100
95
95
90
90
85.0&deg;F
What will this do to air speed?
85.0&deg;F
Fan placement and air speed

40’ X 500’ (200,000 cfm)
526 ft/min
?
&frac14; the fans
&frac14; the speed
130 ft/min
12
What effect do the birds have on the average
air speed?
40’ X 500’
(Dropped ceiling house)




Fan capacity = 200,000 cfm
House width = 40’
side wall = 8’
peak ceiling height = 11’
11’
8’
40’





40’ X 500’
(Dropped ceiling house)




Fan capacity = 200,000 cfm
House width = 40’
side wall = 8’
peak ceiling height = 11’
Speed
= Cfm / (width X (peak + side wall)/2
= 200,000 / (40’ X (11’ + 8’)/2)
= 200,000 / (40’ X 9.5’)
= 200,000 / 380 ft2
= 526 ft/min
What effect do the birds have on air speed?
10.5’
7.5’
40’

Speed




= Cfm / (width X (peak + side wall)/2
= 200,000 / (40’ X (10.5’ + 7.5’)/2)
= 200,000 / (40’ X 9’)
= 200,000 / 360 ft2
= 555 ft/min
40’ X 500’
(Dropped ceiling house)




Fan capacity = 200,000 cfm
House width = 40’
side wall = 8’
peak ceiling height = 11’
9’
6’
40’





Speed
= Cfm / (width X (peak + side wall)/2
= 200,000 / (40’ X (9’ + 6’)/2)
= 200,000 / (40’ X 7.5’)
= 200,000 / 300 ft2
= 667 ft/min
13
```