1. Animal Environment & Heat Flow BSE 2294 Animal Structures and Environments S. Christian Mariger Ph.D. & Susan W. Gay Ph.D.

2. 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.

3. Physical Factors Space Lighting Sound Gasses Equipment

4. Social Factors Number of animals to a pen Behavior

5. Thermal Factors Air temperature Relative humidity Air movement Radiation (one type of heat transfer)

6. Environmental Factors Influence: –Animal health –Breeding –Production efficiency –Product quality –Human health –Equipment service life –Building material longevity

7. 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

8. 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)

9. Sensible and latent heat to change one lb of water from ice to steam q s = Mc v ΔT

10. 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.

11. Sensible & latent heat example Find lbs of water. – ρ H2O = 62.4 lb/ft 3 –20 ft 3 x 62.4 lb/ft 3 = 1,248 lbs

12. 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

13. Sensible & latent heat example Find the latent heat of fusion for the water. – Latent heat of fusion H 2 O = 144 Btu/lb – 1,248 lbs x 144 Btu/lb = 179,712 Btu

14. 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

15. 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

16. Types of Heat Transfer

17. 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!

18. Conduction heat flow q = AK (T 1 – T 2 ) / L –A = cross-sectional area of the surface –K = thermal conductivity –L = thickness of the material –T 1 – T 2 = ΔT = change in temperature q = (A/R) ΔT –R = thermal resistance (L/K)

19. 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° FOutside temp = 20° F Assume the cross-sectional area “A” is 1ft 2

20. Conduction example Find R T 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 R T = 12.24

21. Conduction example Find q for the wall: –q = (A/R T ) x (T inside – T outside ) –q = (1ft 2 /12.24) x (80 – 20) = 4.90 Btu/Hour

22. Heat conduction for a building (q b ) Calculate the conduction (q) for each building component: –Ceilings q c - Windows q wi –Doors q d - Walls q w –Etc. Add all the conductions to find the conduction for the building (q b ) q b = q c + q wi + q d + q w + q...... (in Btu/hr)

23. 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 T 1 – T 2 = (q/A) R T 1 = 80° FT 4 = 20° FT 2 = ?T 3 = ? R 2 = 11 R 1 = 0.62R 3 = 0.62

24. Conduction temperature change Temp at point 2 T 2 = T 1 – (q/A) R 1 T 2 = 80° F – (4.9/1) x 0.62 = 77° F Temp at point 3 T 3 = T 2 – (q/A) R 2. T 3 = 77° F – (4.9/1) x 11.0 = 23° F

25. Convection Heat transferred to or from a body by mass movement of either a liquid or a gas

26. Convection Convection is often used for interior heating

27. 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.

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

29. Typical Environmental Effects (dairy cattle example)

30. Heat stress occurs in animals when their heat gain is greater than their heat loss. Body heat Metabolism Physical activity Performance Environment Radiation (sun) Convection (air) Conduction (resting surface)

31. Heat stress has a severe impact on cow performance and health. Increases Respiration rate Sweating Water intake Decreases Dry matter intake Feed passage rate Blood flow to internal organs Milk production Reproduction performance

32. 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

33. Animals can lose heat by sensible or latent heat losses. Sensible heat Conduction (direct contact) Convection (air movement) Radiation (line of sight) Latent heat Evaporation (phase change)

34. As air temperatures increase, animals cannot lose as much sensible heat, so they pant and sweat (evaporation). Indirect radiation Digestive heat Indirect radiation Convecti on Conductio n Direct radiation

35. As relative humidity rises, an animal losses less heat by evaporation. Evaporation (from skin) Evaporation (respiratory tract)

36. No Stress Mild Stress Heat Stress Severe Stress Dead Cows 72 80 90 100 110 120 Temperature (F) Relative Humidity (%) 0 20 40 60 80 100

37. How can you tell if a cow is suffering from heat stress? Rectal temperatures Above 102.5 ° F Respiration rates > 80 breaths per minute Decreases in Dry matter intake Milk production

38. How can heat stress be managed? Shade Air exchange Air velocity Water

39. Shade lowers the solar heat load from direct and, sometimes, indirect radiation.

40. Good air exchange or ventilation of confinement housing is essential to animal comfort. Removes Hot, moist air Increases Convective heat loss Recommended 1000 cfm per cow

41. Cows’ cooling ability is improved by increasing the air velocity over the animal’s skin. Removes Hot, moist air in contact with the animal Turbulence Disrupt the boundary layer Recommended 220 to 440 fpm (2.5 to 5 mph)

42. Water improves animal cooling through evaporation. Watering locations Increase in hot weather Sprinkling systems Wet cow’s hide Increase direct evaporation Evaporative cooling pads Cools air directly Cows cooled by convection

43. Thermal effects on other species

44. Heat Balance

45. 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 (Q h ) = Heat loss (Q T ): –Q f + Q s = Q vent + Q b

46. Heat removed by ventilation (Q vent ) 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.

47. Heat removed by ventilation (Q vent ) Q vent = (1.1)(Fan rate cfm)(Δt)

48. Ventilation (Q vent ) 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.

49. Ventilation (Q vent ) example (Q vent ) = (1.1) (fan rate) (T i – T o ) (Q vent ) = (1.1) (1,200) (65 – 15) (Q vent ) = 66,000 Btu/hour

50. Heat lost through the structure (Q b ) We have discussed heat lost through structure in terms of thermal resistance (R) and thermal conductivity (K). Q = AK (T 1 – T 2 ) / L –A = cross-sectional area of the surface –K = thermal conductivity –L = thickness of the material –T 1 – T 2 = Δt = change in temperature Q = (A/R) Δt –R = thermal resistance (L/K) Q b = q c + q wi + q d + q w + q......

51. Heat gain (Q h ) Heat gain in an animal structure comes from two major sources: –Supplemental heat (Q f ) – the heat provided by various heaters. –Animal sensible heat (Q s ) – the heat the animals give up to the environment. Conduction Convection Radiation Evaporation (latent heat of vaporization)

52. Animal sensible heat (Q s ) Assumptions: 1.air velocity 20- 30 fpm 2.humidity 50% 3.surface temp of walls are equal to air temp

53. 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 (T o = 20° F) is 7 cfm for each animal. The total heat loss for the structure Q B = 14,000 Btu/hour and the animal sensible heat Q s = 375 Btu / hour / head. Will supplemental heat (Q f ) be required for this structure, if so how much?

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

55. Forced ventilation example Find the heat removed by ventilation (Q vent ) –Q vent = (1.1) (fan rate) (Δt) –Q vent = (1.1) (350) (60 – 20) –Q vent = 15,400 Btu/hour

56. Forced ventilation example Find the total heat loss (Q T ) –(Q T ) = Q vent + Q B –(Q T ) = 15,400 Btu/hour + 14,000 Btu/hour –(Q T ) = 29,400 Btu/hour

57. Forced ventilation example Find animal sensible heat (Q s ) –(Q s ) = (# animals) (Btu/hour – head) –(Q s ) = (50) (375) = 18,750 Btu/hour

58. Forced ventilation example If Q f + Q s = Q vent + Q B then: Find the supplemental heat (Q f ) –(Q f ) = Q vent + Q B – Q s –OR –(Q f ) = Q T – Q s –(Q f ) = 29,400 Btu/hour - 18,750 Btu/hour –(Q f ) = 10,650 Btu/hour

59. Moisture Balance

60. 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%

61. 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!

62. Where does the moisture come from? Incoming air Animal waste Animal respiration Feed and water

63. Swine ventilation rates

64. Air tempering systems Tempering warms or cools air before it enters the animal housing portion of a structure.

65. Air tempering systems Tempering systems include: –air make-up systems –air blending systems –heat exchangers –solar collectors –earth tubes –evaporative coolers

66. Air pre-heating

67. Air blending

68. Heat exchangers

69. Solar collectors

70. Earth tubes

71. Evaporative cooling