22-Agric-A1 Applied Plant, Animal or Human Physiology · May 2018
Question 6 of 6: Chicken Calorimetry and Scaled Barn Heat Production
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 04-Agric-A1 Applied Plant, Animal or Human
Physiology, National Exams May 2018 — a three-hour closed-book
examination; one of two approved calculator models (Casio or Sharp) is permitted. The rubric
states that five (5) questions constitute a complete exam paper and that the
first five questions appearing in the answer book are marked (worth 20 marks each, 100 marks
total); all six (6) printed questions are worked here as a complete study resource.
Reference texts. M.K. Yousef (ed.), Stress Physiology in
Livestock, Vol. I — Basic Principles, CRC Press (thermoregulation, thermoneutral
zone, piloerection, endotherm/ectotherm physiology); K. Schmidt-Nielsen, Animal Physiology:
Adaptation and Environment, 5th ed. (metabolic body-size scaling, Kleiber's law,
thermoconformers, calorimetry); P. McDonald et al., Animal Nutrition, 7th ed.
(gross/digestible/metabolizable/net energy, feed-energy partition, growth efficiency);
R.L. Curtis, Environmental Management in Animal Agriculture, Iowa State University
Press (thermoneutral zone, animal housing microclimate); D.M. Lewis & T.R. Morris,
Poultry Lighting: the Theory and Practice (photoperiodism); ASABE Standards (American
Society of Agricultural and Biological Engineers), Livestock Energetics and Thermal
Environmental Management (design sensible heat production, calorimetry).
Question 6: Chicken Calorimetry and Scaled Barn Heat Production (20 marks)
Direct vs. indirect calorimetry. This problem uses a direct
calorimeter: the birds are housed in a sealed chamber and their heat output is captured and
measured directly from the temperature and humidity rise of a metered air stream passed through
the chamber (a ventilated, open-circuit variant of a heat-sink calorimeter) — in contrast
to an indirect calorimeter, which infers heat production from the birds' oxygen
consumption and carbon dioxide production via a fixed gas-exchange-to-heat conversion (e.g.
Brouwer's equation) rather than by capturing the heat itself.
Check: the source lists an "entropy" value at inlet (17.5 kJ/kg) and
outlet (43.6 kJ/kg); given the units (kJ/kg) this almost certainly mislabels moist-air specific
enthalpy. Neither value is required once dry-bulb temperature, humidity ratio, and the
stated specific heat/latent heat constants are given directly, so the calculation below (as
instructed by the question) uses only those quantities, not the "entropy" figures.
Given.
Quantity
Value
Airflow rate (at inlet)
1.5×10−3 m³/s
Inlet: dry-bulb T / humidity ratio / specific volume
10°C / 0.00294 kg/kg / 0.81 m³/kg
Outlet: dry-bulb T / humidity ratio
24°C / 0.00763 kg/kg
Specific heat of air, $c_p$
1.0 kJ/kg·K
Latent heat of vaporization, $h_{fg}$
2257 kJ/kg
Calorimeter: chicken count / average mass
5 / 2.0 kg
Barn: chicken count / average mass
3000 / 2.5 kg
Find. The specific (W/kg) sensible and latent heat production rates in the
calorimeter, and the total sensible heat production of the 3,000-chicken barn.
Approach. Convert the inlet volumetric airflow to a dry-air mass flow rate
using the inlet specific volume, then use that mass flow with the sensible-heat ($c_p\,\Delta T$)
and latent-heat ($h_{fg}\,\Delta W$) relations across the inlet-to-outlet rise to get the total
heat rates, divide by the calorimeter birds' total mass for the specific (W/kg) rates, then
rescale the sensible-heat result from the 2.0-kg reference bird to the barn's 2.5-kg bird using
the metabolic body-size law (heat production ∝ $M^{0.75}$, not linearly with mass) before
multiplying by the barn's 3,000-bird count.
Dry-air mass flow rate. The airflow is measured at the inlet, so divide by
the inlet specific volume:
$$\dot{m}_{air} = \frac{\dot{V}}{v_{in}} = \frac{1.5\times10^{-3}}{0.81} =
\boxed{1.8519\times10^{-3}\ \text{kg/s}}$$
Sensible and latent heat rates in the calorimeter. Applying $c_p\,\Delta T$
and $h_{fg}\,\Delta W$ across the measured inlet-to-outlet rise:
$$\dot{Q}_{sens} = \dot{m}_{air}\,c_p\,(T_{out}-T_{in}) =
1.8519\times10^{-3}\times1.0\times(24-10) = \boxed{25.93\ \text{W}}$$
$$\dot{Q}_{lat} = \dot{m}_{air}\,h_{fg}\,(W_{out}-W_{in}) =
1.8519\times10^{-3}\times2257\times(0.00763-0.00294) = \boxed{19.60\ \text{W}}$$
Specific rates per kg of calorimeter bird. The five calorimeter chickens
total $5\times2.0 = 10$ kg:
$$\dot{q}_{sens} = \frac{25.93}{10} = \boxed{2.593\ \text{W/kg}}, \qquad
\dot{q}_{lat} = \frac{19.60}{10} = \boxed{1.960\ \text{W/kg}}$$
Rescale sensible heat from the 2.0-kg reference bird to the barn's 2.5-kg
bird. Per-bird sensible heat in the calorimeter is $25.93/5 = 5.185$ W for a 2.0-kg
bird. Total metabolic (and hence sensible) heat production scales with body mass to the power
0.75 (Kleiber's law), so:
$$\left(\frac{M_{barn}}{M_{cal}}\right)^{0.75} = \left(\frac{2.5}{2.0}\right)^{0.75} =
\boxed{1.1822}$$
$$\dot{Q}_{sens,\,barn\ bird} = 5.185 \times 1.1822 = \boxed{6.130\ \text{W per bird}}$$
This is below the value a naive linear (per-kg) scaling would give
($2.593 \times 2.5 = 6.482$ W/bird), which is the trap this two-step approach is designed to
avoid.
Total sensible heat production, 3,000-chicken barn.
$$\dot{Q}_{sens,\,barn} = 6.130 \times 3000 = \boxed{18{,}390\ \text{W} \approx 18.4\ \text{kW}}$$
This design sensible-heat load (radiative + convective + conductive losses to the barn air and
surfaces) is what a ventilation engineer uses to size the barn's winter ventilation rate (or
supplemental heat) so as to remove excess sensible heat and hold the barn near its own
thermoneutral design temperature.