Animal Environment & Heat Flow

BSE 2294 Animal Structures and Environments

S. Christian Mariger Ph.D. & Susan W. Gay Ph.D.

Environmental Fundamentals

• Environment is the total of all external conditions that effect the development, response and growth of plants and animals. – Physical factors – Social factors – Thermal factors

• Ventilation is the method of environmental modification for agricultural structures.

Physical Factors

• Space

• Lighting

• Sound

• Gasses

• Equipment

Social Factors

• Number of animals to a pen

• Behavior

Thermal Factors

• Air temperature

• Relative humidity

• Air movement

• Radiation (one type of heat transfer)

Environmental Factors

• Influence:– Animal health– Breeding– Production efficiency – Product quality – Human health – Equipment service life – Building material longevity

Heating and Ventilation Terms

• Heat – the energy transferred from a warmer body to a colder body because of the temperature difference

• Temperature – is a measure of a body’s ability to transfer or receive heat from matter in contact with it.

• Ambient temperature - the temperature of the medium surrounding a body

• British Thermal Unit (Btu) – the quantity of heat required to raise one pound of water one °F

Heating and Ventilation Terms

• Calorie – the quantity of heat required to raise one gram of water one °C

• Specific heat – is the quantity of heat required to raise one pound of material one °F (Units = Btu/lb-°F)

• Sensible heat – is a measure of the energy that accompanies temperature change

• Latent heat – is the heat energy absorbed or released when a material changes phase (ice to water for example)

Sensible and latent heat to change one lb of water from ice to steam

qs = McvΔT

Sensible & latent heat example

• Given a 20 cubic foot water trough that was allowed to freeze to 28° F how many Btu will be required to thaw and warm the water to 40° F.

Sensible & latent heat example

• Find lbs of water.– ρH2O = 62.4 lb/ft3

– 20 ft3 x 62.4 lb/ft3 = 1,248 lbs

Sensible & latent heat example

• Find sensible heat required (Btu) to raise the temp from 28° F to 32° F.– Specific heat of ice = 0.56 Btu/lb - 1° F– 1,248 lbs x 0.56 Btu/lb - 1° F = 699 Btu - 1° F– (32° F – 28° F) x 699 Btu - 1° F = 2,796 Btu

Sensible & latent heat example

• Find the latent heat of fusion for the water.– Latent heat of fusion H2O = 144 Btu/lb

– 1,248 lbs x 144 Btu/lb = 179,712 Btu

Sensible & latent heat example

• Find sensible heat required to raise the temp from 32° F to 40° F.– Specific heat of water = 1.0 Btu/lb - 1° F– 1,248 lbs x 1.0 Btu/lb - 1° F = 1,248 Btu - 1° F– (40° F – 32° F) x 1,248 Btu - 1° F = 9,984 Btu

Sensible & latent heat example

• Sum the Btu’s to find the energy required to raise the temp from 32° F to 40° F.– (32° F – 28° F) = 2,796 Btu– Latent heat of fusion = 179,712 Btu– (40° F – 32° F) = 9,984 Btu– Total = 192,492 Btu

Types of Heat Transfer

Conduction

• Conduction – the exchange of heat between contacting bodies that are at different temperatures or transfer of energy through a material as a result of a temperature gradient.

Conduction is often a heat loss factor as well as a heating factor!

Conduction heat flow

• q = AK (T1 – T2) / L

– A = cross-sectional area of the surface – K = thermal conductivity – L = thickness of the material

– T1 – T2 = ΔT = change in temperature

• q = (A/R) ΔT– R = thermal resistance (L/K)

Conduction example

• Determine the heat transfer through a wall composed of two sheets of ½” plywood (R = 0.62) and 3 ½” of batt insulation (R = 11).

Inside temp = 80° F Outside temp = 20° F

Assume the cross-sectional area “A” is 1ft2

Conduction example

• Find RT for the wall:– Material #1 ½” plywood R = 0.62– Material #2 3 ½” batt insulation R = 11.00 – Material #3 ½” plywood R = 0.62

RT = 12.24

Conduction example

• Find q for the wall:

– q = (A/RT) x (Tinside – Toutside)

– q = (1ft2/12.24) x (80 – 20) = 4.90 Btu/Hour

Heat conduction for a building (qb)

• Calculate the conduction (q) for each building component:

– Ceilings qc - Windows qwi

– Doors qd - Walls qw

– Etc.

• Add all the conductions to find the conduction for the building (qb)

qb = qc + qwi + qd + qw + q...... (in Btu/hr)

Conduction temperature change

• We can also calculate the temperature from one side to the next for each layer in the wall.

• Determine the temperatures at points 2 and 3. – Where T1 – T2 = (q/A) R

T1 = 80° F T4 = 20° FT2 = ? T3 = ?

R2 = 11

R1 = 0.62 R3 = 0.62

Conduction temperature change

• Temp at point 2• T2 = T1 – (q/A) R1

• T2 = 80° F – (4.9/1) x 0.62 = 77° F

• Temp at point 3• T3 = T2 – (q/A) R2 . • T3 = 77° F – (4.9/1) x 11.0 = 23° F

Convection

• Heat transferred to or from a body by mass movement of either a liquid or a gas

Convection

• Convection is often used for interior heating

Radiation

• The exchange of thermal energy between objects by electromagnetic waves.

• Radiant energy is transferred between two bodies in both directions, not just from warmer to cooler.

Radiation

• Here is an example of infra red (IR) radiation being used in an interior heating application

Typical Environmental Effects (dairy cattle example)

Heat stress occurs in animals when their

heat gain is greater than their heat loss.

Body heatMetabolism

Physical activity

Performance

Environment Radiation (sun)

Convection (air)

Conduction (resting surface)

Heat stress has a severe impact on cow performance and health.

IncreasesRespiration rate

Sweating

Water intake

DecreasesDry matter intake

Feed passage rate

Blood flow to internal organs

Milk production

Reproduction performance

Cows are much more comfortable at cooler

temperatures than humans.

Thermal comfort zone

41 – 77 °F

Lower critical temperature

Neonatal calves 55 °F

Mature cows 13 °F

Upper critical temperature

77 – 78 °F

Animals can lose heat by sensible or latent

heat losses.

Sensible heatConduction (direct contact)

Convection (air movement)

Radiation (line of sight)

Latent heatEvaporation (phase change)

As air temperatures increase, animals cannot lose as much sensible heat, so they

pant and sweat (evaporation).

Indirect radiation

Digestive heat

Indirect radiation

Convection

Conduction

Direct radiation

As relative humidity rises, an animal losses

less heat by evaporation.

Evaporation (from skin)

Evaporation (respiratory tract)

NoStress

MildStress

HeatStress

SevereStress

DeadCows

72

80

90

100

110

120

Te

mp

era

ture

(F

)Relative Humidity (%)

0 20 40 60 80 100

How can you tell if a cow is suffering from

heat stress?

Rectal temperaturesAbove 102.5 °F

Respiration rates> 80 breaths per minute

Decreases inDry matter intake

Milk production

How can heat stress be managed?

Shade

Air exchange

Air velocity

Water

Shade lowers the solar heat load from direct

and, sometimes, indirect radiation.

Good air exchange or ventilation of confinement housing is essential to animal comfort.

RemovesHot, moist air

IncreasesConvective heat loss

Recommended1000 cfm per cow

Cows’ cooling ability is improved by increasing the air velocity over the animal’s skin.

RemovesHot, moist air in contact with

the animal

TurbulenceDisrupt the boundary layer

Recommended220 to 440 fpm (2.5 to 5 mph)

Water improves animal cooling through

evaporation.

Watering locationsIncrease in hot weather

Sprinkling systemsWet cow’s hide

Increase direct evaporation

Evaporative cooling padsCools air directly

Cows cooled by convection

Thermal effects on other species

Heat Balance

Heat balance

• To maintain constant room temperature, heat produced by the animals and heaters must equal the heat lost through the building structure and by ventilation.

• Heat gain (Qh) = Heat loss (QT): – Qf + Qs = Qvent + Qb

Heat removed by ventilation (Qvent)

• Ventilation removes heat by replacing warm in side air with cold outside air.

• If humidity is constant we know the specific heat of air.

• If we also know the difference between the outside temp and the inside temp (Δt)

• If we also know how much air is being exchanged in Cubic Feet/Minute (cfm)

• Then we can calculate the heat removed by ventilation.

Heat removed by ventilation (Qvent)

Qvent = (1.1)(Fan rate cfm)(Δt)

Ventilation (Qvent) example

• A building is ventilated at 1,200 cfm. The inside temperature is 65° F and the outside temperature is 15° F. Determine the rate of heat removal.

Ventilation (Qvent) example

• (Qvent) = (1.1) (fan rate) (Ti – To)

• (Qvent) = (1.1) (1,200) (65 – 15)

• (Qvent) = 66,000 Btu/hour

Heat lost through the structure (Qb)

• We have discussed heat lost through structure in terms of thermal resistance (R) and thermal conductivity (K).

• Q = AK (T1 – T2) / L– A = cross-sectional area of the surface – K = thermal conductivity – L = thickness of the material – T1 – T2 = Δt = change in temperature

• Q = (A/R) Δt– R = thermal resistance (L/K)

• Qb = qc + qwi + qd + qw + q......

Heat gain (Qh)

• Heat gain in an animal structure comes from two major sources:– Supplemental heat (Qf) – the heat provided by

various heaters. – Animal sensible heat (Qs) – the heat the

animals give up to the environment. • Conduction • Convection • Radiation • Evaporation (latent heat of vaporization)

Animal sensible heat (Qs)

Assumptions: 1. air velocity 20-

30 fpm

2. humidity 50%

3. surface temp of walls are equal to air temp

Forced ventilation example

• Fifty (50) pigs in the growing stage (100 lbs) are housed at a temperature of 60° F. The cold weather ventilation rate (To = 20° F) is 7 cfm for each animal. The total heat loss for the structure QB = 14,000 Btu/hour and the animal sensible heat Qs = 375 Btu / hour / head. Will supplemental heat (Qf) be required for this structure, if so how much?

Forced ventilation example

• Find the required ventilation:– (# animals) x (cfm/animal) = fan rate – fan rate = (50) x (7 cfm) = 350 cfm

Forced ventilation example

• Find the heat removed by ventilation (Qvent)

– Qvent = (1.1) (fan rate) (Δt)

– Qvent = (1.1) (350) (60 – 20)

– Qvent = 15,400 Btu/hour

Forced ventilation example

• Find the total heat loss (QT)

– (QT) = Qvent + QB

– (QT) = 15,400 Btu/hour + 14,000 Btu/hour

– (QT) = 29,400 Btu/hour

Forced ventilation example

• Find animal sensible heat (Qs)

– (Qs) = (# animals) (Btu/hour – head)

– (Qs) = (50) (375) = 18,750 Btu/hour

Forced ventilation example

• If Qf + Qs = Qvent + QB then:

• Find the supplemental heat (Qf)

– (Qf) = Qvent + QB – Qs

– OR

– (Qf) = QT – Qs

– (Qf) = 29,400 Btu/hour - 18,750 Btu/hour

– (Qf) = 10,650 Btu/hour

Moisture Balance

What about moisture?

• As ventilating air moves through a structure it evaporates moisture from the floor, pits and other wet surfaces.

• As animals breath, moisture is lost from their respiratory system to the air.

• To maintain a desirable temperature, enough moisture must be removed to keep the relative humidity below 70%

Moisture balance

• To maintain a constant rate of moisture:

• Moisture Loss = Moisture production

• The moisture holding capacity of air nearly doubles with each 20° F increase in temperature!

Where does the moisture come from? • Incoming air

• Animal waste

• Animal respiration

• Feed and water

Swine ventilation rates

Air tempering systems

• Tempering warms or cools air before it enters the animal housing portion of a structure.

Air tempering systems

• Tempering systems include: – air make-up systems – air blending systems – heat exchangers – solar collectors – earth tubes – evaporative coolers

Air pre-heating

Air blending

Heat exchangers

Solar collectors

Earth tubes

Evaporative cooling