1. ERT 460 CONTROLLED ENVIRONMENT ENGINEERING
CONTROLLED ENVIRONMENT: Moisture
Lecture 4 (Week 4)
Dr Mohammud Che Husain

2. Controlled Environment: Moisture
Mass transfer of moisture based on plant and animal, latent heat of production and latent heat of vaporization, heat and mass transfer of aquaculture ponds and tanks.

3. 1.0: Introduction Environment parameters to be consider: Temperature
Light intensity & Solar radiation RH
Wind
CO2
NH3

4. Comfortable ambient tempt. & RH
For animals <25oC , 50 – 80%
For plants <30 oC , 60 – 70%
Optimum RH for GH crops:
RH; 60 – 70% for better plant growth
To high (above 80%): occurrence of fungal diseases
To low (less 20%): cause wilting due to high evaporation rate

6. Livestock need good in-house environment (temp, RH, light, O2)

7. b. Dairy & beef cattle housing

8. Conducive environment for fish & worker (hatcheries & growout)
Water quality: temp (20 – 30 oC), DO content (5-10 mg/l), pH ( ). Light

9. Evaporation (conversion of sensible heat to latent heat)
Animal house/barn:
from wash water & animals wastes
from animals (respiratory system & body surface)
GH:
from the wet floors & benches
Surface of the potting medium/soil
Plants transpiration (major component) in GH
Aquaculture house:
from tanks and wet floor.

10. In animal & plant house, moisture production can be estimate.
In animal house, moisture production is proportional to
the number
type
size of animals
Moisture production within GH is a function of
Plant population,
light intensity &
water management practices
To remove moisture can be done by provides ventilation

11. Moisture production of various animals (in Appendix 5-1)

13. Moisture Production (MP)
Example 1:
Problem: Determine the rate at which moisture is generated and be removed from a swine barn housing 100, 40kg; 100, 60kg; and 100, 80kg pigs. Air temperature is expected to be 20°C.
Solution:
Data from Appendix 5-1 may be used. At 20°C, moisture production (MP) is
which is for 100 of each size pig,
Moisture = 100 ( )(10-6kg/mg) = kg/s
Size, kg
Moisture Production (MP)
mg/kg-s
mg/pig-s 40
0.61 24 60
0.47 28 80
0.39 31

14. Example 2:
Problem: Determine the rate at which moisture is generated and must be removed from a dairy barn housing 100 cows averaging 570kg, when the indoor air temperature is 10°C.
Solution: By the data in Appendix 5-1, at 10°C MP of a 500kg dairy cow is 0.28mg/kg-s. MP rate from a 570kg dairy cow can be estimated as
Moisture = 0.28(570/500)0.734 = 0.31mg/kg-s
Note: This formulation for moisture production is assumed to mirror the form which has been found to represent sensible heat production from mammals.
MP from the herd will be
Mp = 0.31mg/kg-s (570kg)(100cows)(10-6kg/mg) = 0.018kg/s

16. 2.0: Mass balance of moisture
The rate of water vapor is carried into building by ventilation air
The rate of water vapor produced by animals/plant/water
The rate of water vapor is carried out from building by ventilation air
The steady state mass balance for Fig 5-2 is
individual terms of the energy and mass balances will now be examined in detail.
mp + mvi = mv0

17. MC = HR (W) X mass of air
mass of air (mair) = air density (ρ) X vol. flow rate (v)

18. This calculation provide min. ventilation rate.
If moisture production rate (mp) is know, the ventilation rate required can be calculated using Eq 5-2
This calculation provide min. ventilation rate.
If higher ventilation rate is required, a humidity ratio lower than design condition….will result,
This presents no problem, for design humidity conditions are usually chosen to be the max. desired. If an exact humidity level is desired, irrespective of the ventilation needed for temperature control, moisture must be added to or removed from the air by mechanical means (a humidifier or a dehumidifier)
mp + mvi = mv0
mair = mwater / (Wi – Wo)
Eq 5-16

19. Example 3
A poultry house is designed for 30,000 leghorn laying hens (1.8 kg, average). Determine the ventilation rate required to maintain indoor air at 23oC and 70% RH when outside air is -20oC and 55% RH. The humidity ratio, Wo and Wi are and kg/kg, respectively. The air density is 1.03 kg/m3.
Solution:
From Appendix 5-1:
At 18oC, MP = 0.97 mg/kg-s
At 28oC, MP = 1.19 mg/kg-s
So, at 23oC, MP is estimated by interpolation=
MP or mwater = (5/10)( ) = 1.08 mg/kg-s
The total MP for 30,000
= (30,000)(1.8)(1.08)
= 58,320 mg/s
= kg/s

20. Moisture production of various animals (in Appendix 5-1)

21. Using Eq 5-16, the air mass flow rate = mair = mwater/(Wi-Wo)
=( kg/s)/( )kg/kg
= 4.3 kg/s (dry air).
Each kg of dry air hold kg of water, thus the exhaust ventilation rate contain
( ) = kg/s.
The vol. ventilation rate,
vair = (4.314 kg/s)/(1.03 kg/m3)= 4.44m3/s

22. ERT 460 CONTROLLED ENVIRONMENT ENGINEERING
CONTROLLED ENVIRONMENT: Air Quality
Lecture 5 (Week 5)
Samera Samsuddin Sah

23. Controlled Environment: Air Quality
Sources and production of gaseous and particulate matter especially in plants and animal structures.

24. 1.0: Introduction Environment parameters to be consider: Temperature
Light intensity & Solar radiation RH
Wind
CO2
NH3

25. Comfortable ambient CO2 level
For animals < 2500 ppm
For plants < 1500 ppm
CO2 produced by crops:
CO2; ppm for optimal photosynthesis
affect the plant canopies
CO2 produced by animals:
Is by-product of metabolism as is heat & moisture
Additional CO2 produced from decomposition of waste and digestive processes of ruminants
1L of CO2 : 24.6 kJ total heat
345 – 2500 ppm for animals growth

26. Crops need good environment in greenhouse (temp, RH, light, CO2)

27. Mass Balance of CO2
The ambient CO2 level is approximately 345 ppm MCO2 = Mair(CCO2)mass Where M is mass and (CCO2)mass is the mass concentration, kg/kg of CO2 Vp + Vvo = Vvi Where vp (unit L/s) = qtotal / 24.6kJ/L Vvo = Vair (CCO2) volume out Vvi = Vair (CCO2) volume in

28. Moisture production of various animals (in Appendix 5-1)

29. Example 1:
Problem: A poultry house id designed for 30,000 leghorn laying hens (1.8kg, average). Calculate carbon dioxide level when ventilation is 4.3m3/s
Solution:
CO2 production depends on total heat production which at 23°C is
qkg of bird = (5/10)( ) = 6.7 W/kg
30,000 birds average 1.8kg, thus total heat production is
qtotal = (30,000)(1.8)(6.7) = 361,800 W

30. Vvi = Vair (CCO2) volume = (4.3 m3/s) (CCO2) volume
The conversion factor from total heat to carbon dioxide production leads to a CO2 production rate of
Vp = (361.8kW)/(24.6kJ/L)= 14.7L/s = m3/s
Vvo = Vair (CCO2) volume = (4.3 m3/s) ( )
= m3/s
Vvi = Vair (CCO2) volume = (4.3 m3/s) (CCO2) volume
The balance for CO2 is
Vp + Vvo = Vvi
= 4.3 (CCO2) volume
(CCO2) volume = = 3764 ≈ 3800ppm

31. Vvo = Vair (CCO2) volume = (4.3 m3/s) (0.000345)
Vvi = Vair (CCO2) volume = (4.3 m3/s) (CCO2) volume
The balance for CO2 is
Vp + Vvo = Vvi
= 4.3 (CCO2) volume
(CCO2) volume = = 3764 ≈ 3800ppm

32. REFERENCESS
Albright (1990). Environment control for animals and plants. ASAE Textbook.