ERT 460 CONTROLLED ENVIRONMENT DESIGN II CONTROLLED ENVIRONMENT

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

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

1. 0: Temperature in plant/animal/ aquaculture structures Environment parameters to be consider: – Temperature – Light intensity & Solar radiation – RH – Wind speed – Gases (CO 2 & NH 3)

• 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. 4

Greenhouse for crop

Animals Barns need good in-house environment (temp, RH, light, O 2)

Aquaculture Building

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.

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

2. 2: Steady state SEB 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

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.

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.

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

Example 1: Determine the sensible heat production from 100 dairy cow each at 590 kg wt housed in a barn at 12 o. C. Solution: refer to Appendix 5 -1: • At 10 o. C 500 kg cow lost SHP = 1. 5 W/kg • At 15 o. C 500 kg cow lost SHP = 1. 2 W/kg • Interpolating linearly for 12 o. C, (qs )500 kg = 1. 2 + (2/5)(1. 5 -1. 2) = 1. 32 W/kg • 590 kg cow lose heat >500 kg cow by the ratio: (590/500)0. 734 (qs )590 kg = 1. 32 W/kg (500) (590/500)0. 734 = 745 W/cow • If 100 cow, total sensible heat : 745 W/cow X 100 cow = 74. 5 k. W

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

Example 2: A poultry building is lighted by fluorescent lights generated 40 w/m 2 of floor area. The barn is 12 m wide by 40 m long. Estimate the sensible heat generated from this source. Solution: Floor area = 12 m X 40 m = 480 m 2 Total lighting capacity = 480 m 2 X 40 w/m 2 = 19, 200 w = 19. 2 kw Assume additional 20% heat generated by ballast use: 1. 2 X 19. 2 Kw = 23 k. W

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/m 3 V = ventilation rate Animal House: • i) qs & qe are combined into 1 = q s • 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))

Example 3 Determine the ventilation rate (m 3/s) required to maintain a dairy barn at 15 oc, given following condition: • No of cow = 60 , Av. wt = 580 kg • Wall area = 274 m 2, R-value = 2. 05 m 2 K/W • Ceiling area = 520 m 2, R-value = 1. 97 m 2 K/W • Window area = 12 m 2, R-value = 0. 30 m 2 K/W • Door area = 15 m 2, R-value = 0. 49 m 2 K/W • Perimeter length = 110 m, heat loss factor = 1. 5 W/m K • to = -5 o. C • Assume no heat from light & motor, little solar heating

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)0. 734 X (60 cows) = 40, 000 W • Air density , ρ = 1. 14 kg/m 3 • FP = (110 m) 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, 000 w – (468 W/k +165 W/k) (20 K)}/(1006)(1. 14 kg/m 3)(20 k) = 1. 2 m 3/s

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 24 o. C. 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 0. 058 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/m 2 of floor area. The building has floor area of 1200 m 2. Estimate the sensible heat generated from this source.

4) Determine the ventilation rate (m 3/s) required to maintain a barn at 18 o. C, given following condition: • No of cow = 1000 , Av. wt = 500 kg • Wall area = 274 m 2, R-value = 2. 5 m 2 K/W • Ceiling area = 520 m 2, R-value = 1. 7 m 2 K/W • Window area = 12 m 2, R-value = 0. 35 m 2 K/W • Door area = 15 m 2, R-value = 0. 59 m 2 K/W • Perimeter length = 120 m, heat loss factor = 1. 7 W/m K • to = -10 o. C

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