1. ERT 460 CONTROLLED ENVIRONMENT DESIGN II
CONTROLLED ENVIRONMENT: Temperature
Lecture 3 (Week 3)
Dr Mohammud Che Husain

2. Controlled Environment: Temperature
The temperature in plant, animal and aquaculture structures from sensible heat balance for steady state heat transfer.

3. 1.0: Temperature in plant/animal/ aquaculture structures
Environment parameters to be consider:
Temperature
Light intensity & Solar radiation RH
Wind speed
Gases (CO2 & NH3)

4. Temperature: major factor negatively influence all livestock/aquaculture/crops production. This will affect:
Food/nutrient intake
Metabolism (conversion of nutrients into products) such as:
Meat, Milk & Egg
Heat dissipation (alter the partition of energy)
Heat that causes a change in temp of an object is called SENSIBLE HEAT.

5. Greenhouse for crop

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

7. Aquaculture Building

8. 2.0: Sensible heat
heat exchanged by a body that changes the temp. & some macroscopic variables of the body, but leaves unchanged certain other macroscopic variables ( volume/pressure).
When an object is heated, its temp. rises as heat is added. The increase in heat is called sensible heat.
Similarly, when heat is removed from an object & its temp. falls, the heat removed is also called sensible heat.

9. 2.1: Sensible energy balance (SEB)
example of SEB is shown below:
The energy flow term are:
qs = SH gain from animals
qm = SH gain from mechanical sources (motor/lighting)
qso = SH gain from sun (thro window/glass panel)
qh = SH gain from heating system
qvi = SH gain thro ventilation air coming into house
qw = SH loss thro wall/ceiling/window/door
qf = SH loss thro floor
qe = SH loss to latent heat (water evaporation from floor/leaves)
qvo = SH loss thro in ventilation air leaving the house

10. 2.2: Steady state SEB Gain = Loss
Gains – Losses = change of storage (if conditions are steady-state, there is no change in storage)
 Steady-state SEB is rearranged in the form:
Gain = Loss
qs + qm + qso + qh + qvi = qw + qf + qe + qvo

11. Energy balance can be used for the determination of:
heating & cooling
ventilation rate
indoor air temperature
insulation requirement
Hence, the component of energy balance must be quantified.

12. 2.2.1: Sensible heat produced by animal, qs
Animal must live within “Comfort zone” & must prevent from ‘Thermal stress zone’
Animals lose heat to surrounding by radiative ,convection transfer and also conductive heat transfer (body to floor).
Sensible heat lost primarily from the outer surface of animals
Latent heat lost from respiratory tracts.

13. Albright, L. D. (1990). Environment control for animals and plants
Albright, L. D. (1990). Environment control for animals and plants. ASAE Textbook

14. Example 1:
Determine the sensible heat production from 100 dairy cow each at 590kg wt housed in a barn at 12 oC.
Solution: refer to Appendix 5-1:
At 10 oC 500 kg cow lost SHP = 1.5 W/kg
At 15 oC 500 kg cow lost SHP = 1.2 W/kg
Interpolating linearly for 12oC,
(qs )500 kg = (2/5)( ) = 1.32 W/kg
590kg cow lose heat >500 kg cow by the ratio: (590/500)0.734
(qs )590 kg = 1.32 W/kg (500) (590/500) = 745 W/cow
If 100 cow, total sensible heat : 745 W/cow X 100 cow = 74.5 kW

15. 2.2.2: Sensible heat from mechanical source, qm
Lighting is the largest source heat
Source from Motor are assume negligible unless if it operate continuously

16. Example 2:
A poultry building is lighted by fluorescent lights generated 40 w/m2 of floor area. The barn is 12m wide by 40m long. Estimate the sensible heat generated from this source.
Solution:
Floor area = 12m X 40m = 480 m2
Total lighting capacity = 480 m2 X 40 w/m2 = 19,200 w = 19.2 kw
Assume additional 20% heat generated by ballast use: 1.2 X 19.2 Kw = 23 kW

17. 2.3: Sensible Heat Balance (SHB)
qs + qm + qso + qh + qvi = qw + qf + qe + qvo
Can be rewritten into :
qs + qm + qso + qh = ∑UA (ti-to) + FP (ti-to) +1006ρV(ti-to) + qe
Where:
∑UA = sum of heat loss conduction (from wall/ceiling/window)
ti = Air temp. inside
to = Air temp. outside
FP = perimeter heat loss conductance, W/k
ρ = air density, kg/m3
V = ventilation rate
Animal House:
i) qs & qe are combined into 1 = qs
ii) transmitted solar input to barn is usually neglected
iii) heat from mechanical sources is assumed negligible
Supplemental heat in barn is not frequently used.
When this simplifying assumption are accepted, the energy balance reduces to:
qs = (∑UA + FP +1006ρV)(ti-to)
Ventilation rate,
V = {qs - (∑UA + FP)(ti-to)}/((1006ρ)(ti-to))

18. Example 3
Determine the ventilation rate (m3/s) required to maintain a dairy barn at 15oc, given following condition:
No of cow = 60 , Av. wt = 580 kg
Wall area = 274m2, R-value = 2.05 m2K/W
Ceiling area = 520m2, R-value = 1.97m2K/W
Window area = 12m2, R-value = 0.30m2K/W
Door area = 15m2, R-value = 0.49m2K/W
Perimeter length = 110m, heat loss factor = 1.5 W/m K
to = -5oC
Assume no heat from light & motor, little solar heating

19. Heat loss conductance from wall/ceiling/window/door:
Solution:
Data from Appendix 5-1 will be use to calculate total sensible animal heat production:
(qs )580 kg = 1.2 W/kg (500) (580/500) X (60 cows) = 40,000 W
Air density , ρ = 1.14 kg/m3
FP = (110m) x (1.5 W/m K) = 165 W/k
Heat loss conductance from wall/ceiling/window/door:
∑ UA = ∑ (Ai/Ri) = 468 W/k
 V = {qs - (∑UA + FP)(ti-to)}/(1006ρ)(ti-to))
= {40,000w – (468 W/k +165 W/k) (20K)}/(1006)(1.14 kg/m3)(20k)
= 1.2 m3/s

20. Assignment 1
1) Calculate the sensible, latent and total heat production from a flock of 30,000 white leghorn laying hens with average weight of 1.9 kg/bird keep in house at 24oC.
2) A poultry house used to raise 10,000 broilers is operating in INSAT, Sungai Chuchoh. The latent and sensible heat of broilers is 6 W/kg. The growth rate is estimated at kg/day. Calculate the latent and sensible heat produced by the broilers at day 40.
3) A poultry building is lighted by fluorescent lights generated 40 w/m2 of floor area. The building has floor area of 1200 m2. Estimate the sensible heat generated from this source.

21. 4) Determine the ventilation rate (m3/s) required to maintain a barn at 18oC, given following condition:
No of cow = 1000 , Av. wt = 500 kg
Wall area = 274m2, R-value = 2.5 m2K/W
Ceiling area = 520m2, R-value = 1.7m2K/W
Window area = 12m2, R-value = 0.35m2K/W
Door area = 15m2, R-value = 0.59m2K/W
Perimeter length = 120m, heat loss factor = 1.7 W/m K
to = -10oC

22. REFERENCESS
Albright, L. D. (1990). Environment control for animals and plants. ASAE Textbook.