Heat Loss Gain Calculations 1 How

Heat Loss & Gain Calculations 1

How Heat Moves in Homes Conduction is the transfer of heat through solid objects, such as the ceilings, walls, and floors of a home. Insulation (and multiple layers of glass in windows) reduces conduction losses. The direction of heat flow is from hot to cold, so this illustration shows conduction from a warm interior to a cooler outdoors. 2

Conduction Heat Loss High Temperature Low Temperature 3

Conduction Heat Loss High Temperature Low Temperature As Temperature Differences Increase, Heat Loss Increases 4

Conduction Heat Loss Low Temperature High Temperature Resistance 5

Conduction Heat Loss High Temperature Low Temperature As Resistance Increases, Heat Loss Decreases 6

Conduction Heat Loss 7

How Heat Moves in Homes Convection is the flow of heat by currents of air. Air currents are caused by pressure differences, stirring fans, and air density changes as it heats and cools. As air becomes heated, it becomes less dense and rises; as air cools, it becomes more dense and sinks. 8

Convective Heat Loss 9

Convective Heat Loss - Air Leakage High Pressure Low Pressure 10

Convective Heat Loss - Air Leakage High Pressure Low Pressure As Pressure Differences Increase, Heat Loss Increases 11

Convective Heat Loss - Air Leakage High Pressure Low Pressure As Leakage Area Decreases, Heat Loss Decreases 12

What Causes Pressure? Natural High Pressure Low Pressure Windward Side of House 13

What Causes Pressure? Natural High Pressure Low Pressure Leeward Side of House 14

What Causes Pressure? Mechanical High Pressure Low Pressure Tight Supply Duct & No Return Duct 15

What Causes Pressure? Mechanical High Pressure Low Pressure Tight Supply Duct & Return Duct Leaks 16

What Causes Pressure? Mechanical High Pressure Low Pressure Supply Duct Leaks & Tight Return Ducts 17

What Causes Pressure? Stack Effect Higher Pressure Hot Air Rises Lower Pressure 18

What Causes Pressure? Stack Effect Higher Pressure Hot Air Rises Lower Pressure Effected by Height and Temperature Gradient 19

What Causes Pressure? Stack Effect Higher Pressure Hot Air Rises Neutral Pressure Plane Lower Pressure 20

How Heat Moves in Homes Radiation is the movement of energy in waves from warm to cooler objects across empty spaces, such as radiant heat traveling from the inner panes of glass to outer panes in doubleglazed windows in winter. 21

Equations - Conduction q = A * T R – where • • q = heat flow, Btu/hr A = area, ft 2 R = resistance, ft 2 -hr-°F/Btu T = temperature differential, °F Higher temperature – Lower temperature 22

Where Do You Get R? • Table of R-values for various materials • Some values are for entire thickness – Brick – Plywood – Gypsum Board • Some values are per inch of thickness – Wood (framing) – Insulation 23

How do R-values Add? RT = R 1 + R 2 + R 3 R 1 R 2 R 3 24

How do R-values Add? - Example RT for a Structurally Insulated Panel (SIP) ½ inch plywood = 1. 25 4 inch Rigid Foam Center = 4 per inch = 16. 00 ½ inch plywood = 1. 25 RT = 18. 50 25

Equations - Conduction q = U * A * T – where • • q = heat flow, Btu/hr A = area, ft 2 U = conductance, Btu/ft 2 -hr-°F T = temperature differential, °F 26

Equations - Conduction • Where does U come from? – Table values • How do they add? 1 = 1 + 1 UT U 1 U 2 • Commonly provided for the entire assembly 27

U-factor A U-factor is used to describe an area that is composed of several materials. Example: Window U-factor includes the glass, frame, and sash. 28

Relationship Between R and U q = U * A * T q = A* T R U * A * T = A * T R U=1 R 29

Air Leakage - General Equation q = m * Cp * T – where • • q = heat flow, Btu/hr m = mass flow of air, lbs/hr Cp = specific heat of air, 0. 24 Btu/lbs -°F T = temperature differential, °F 30

Air Leakage - General Equation q = m * Cp * T – where does m come from? m = mass flow of air, lbs/hr Under normal conditions in a home: Density of Air = 13. 5 ft 3 per lb air Cubic Feet of Air = m 13. 5 31

Air Leakage For Ducts q = 1. 08 * cfm * T (ducts) – where • cfm = duct leakage rate to the outside – where does the 1. 08 come from? cfm * 0. 24 * 60 min/hr = cfm * 1. 08 13. 5 ft³/lb air 32

Air Leakage for an Entire House • q = 0. 018* ft³/hr * T – where • ft³/hr = air leakage rate for the entire house • Where does the 0. 018 come from? ft³/hr * 0. 24 = ft³/hr * 0. 018 13. 5 ft³/lb air • ft 3/hr = ACHnat * Volume (ft 3) – where • ACHnat = Natural Air Changes per hour • Volume = volume of the conditioned space • q = 0. 018* ACHnat * Volume (ft 3) * T 33

Simple Heat Flow, q, Calculation Assume 10 x 10 wall A = 100 ft 2 Cavity Insulation R value = 13 T = 1 degree q = 100 * 1 = 7. 69 Btuh 13 What is missing? 34

Simple Heat Flow, q, Calculation What about the wood framing? 2 x 4 R-value = 4. 38 (1. 25 per inch) 35

Simple Heat Flow, q, Calculation Typical Wood Framing 36

Simple Heat Flow, q, Calculation Minimum Wood Framing Approximately 10 2 x 4 s, 10 ft long Each stud: 1. 5 inches wide 10 ft * 12 inches/ft = 120 inches long 10 studs * 1. 5 in * 120 in = 1800 square inches 1800 in 2 / 144 in 2 per ft 2 = 12. 5 ft 2 37

Simple Heat Flow, q, Calculation w/Framing Total Area = 100 ft 2 10 x 10 wall Cavity Insulation R-value = 13 Framing R = 4. 38 Framing Area = 12. 5 ft 2 Cavity Insulation Area = 100 – 12. 5 = 87. 5 ft 2 T = 1 degree 38

Simple Heat Flow, q, Calculation w/Framing qinsulation = 87. 5 * 1 = 6. 73 Btuh 13 qframing = 12. 5 * 1 = 2. 85 Btuh 4. 38 qtotal = qinsulation + qframing = 6. 73 + 2. 85 = 9. 58 Btuh 39

Calculating R when q is Known q = A * T R multiply both sides by R R * q = R *A * T R R * q = A * T 40

Calculating R when q is Known R * q = A * T Divide both sides by q: R * q = A * T q q R = A * T q 41

R-Value of the Entire Wall w/Framing qtotal = 9. 58 Btuh/°F R = A * T = 100 * 1 = 10. 44 q 9. 58 TOTAL WALL R 42

R-Value of the Entire Wall w/Framing Another Equation to Calculate Total Wall R RT = _______AT____ _A 1_ + _A 2_ R 1 R 2 43

Simple Heat Flow, q, Calculation What if there is a window in the wall? Window: Size 3 ft x 5 ft = 15 ft 2 U-factor = 0. 40 44

Framing + Window 45

Simple Heat Flow, q, Calculation With Framing + Windows Require Extra Framing Materials 4 extra studs for kings and jacks 2 x 12 36 inch long for the header Approximately 7. 8 ft 2 of extra framing Total framing = 12. 5 + 7. 8 = 20. 3 ft 2 46

Simple Heat Flow, q, Calculation With Framing + Window Total Area = 100 ft 2 10 x 10 wall Cavity Insulation R-value = 13 Framing R-value = 4. 38 Framing Area = 20. 3 ft 2 Window U-factor = 0. 40 Window Area = 15 ft 2 Cavity Insulation Area = 100 – 20. 3 - 15 = 64. 7 ft 2 T = 1 degree 47

Simple Heat Flow, q, Calculation With Framing + Window qinsulation = 64. 7 * 1 = 4. 98 Btuh 13 qframing = 20. 3 * 1 = 4. 63 Btuh 4. 38 qwindow = 0. 40 *15 * 1 = 6 Btuh qtotal = 4. 98 + 4. 63 + 6 = 15. 61 Btuh 48

R-Value of the Wall With Framing + Window qtotal = 15. 61 Btuh/°F q = A * T R R = A * T = 100 * 1 = 6. 41 q 15. 61 49

R-Value Comparison Cavity Insulation Only R = 13 Cavity Insulation + Framing R = 10. 44 Cavity Insulation + Framing + Window R = 6. 41 50

Your Turn Total Area = 1000 ft² Ceiling R = 38 Pull Down Stairs Area = 15 ft² R=2 What is the R value of the total ceiling? 51

Your Turn Ceiling q = (1000 – 15) = 25. 92 38 Pull Down Stairs q = 15 = 7. 5 2 Total q = 25. 92 + 7. 5 = 33. 42 R = _1000_ = 29. 92 33. 42 52

HERS Rating Software Examples Must know: • Areas • R / U values • Temperature Differential – Indirectly by knowing what is on the other side of the surface 53

Above Grade Wall Properties 54

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Conduction Heat Loss Typical Resistances in a Wall Cavity Insulation Gypsum Board Exterior Finish Inside Air Film Outside Air Film High Temperature Low Temperature 57

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Exterior Finish 59

R of Cavity Wall Section Inside Air = 0. 68 5/8” Gypsum Board = 3 ½” Cavity Insulation = Exterior Finish= Outside Air = Cavity Wall Section R = 0. 56 13. 00 0. 94 0. 17 15. 35 60

Conduction Heat Loss Typical Resistances in a Wall Gypsum Board Inside Air Film High Temperature Framing Exterior Finish Outside Air Film Low Temperature 61

R of Framing Wall Section Inside Air = 0. 68 5/8” Gypsum Board = 3 ½” Framing = Exterior Finish = Outside Air = Framing Wall Section R = 0. 56 4. 37 0. 94 0. 17 6. 72 62

Framing Factor 63

Framing Factor 64

Your Turn - Total Wall R Cavity Wall Section R = 15. 35 Framing Wall Section R = 6. 72 Framing Factor = 0. 23 (23% of the wall is framing) Remember - the objective is to calculate “q” correctly 65

Total Wall U Total Wall R = 11. 85 Total Wall U = 1 = 0. 0843 R 11. 85 68

Total Wall U 69

Total UA for a House 2006 IECC Compliance (2006 IRC, Chapter 11, Energy Efficiency) • Prescriptive • Overall Building UA • Annual Energy Cost 70

REM/Rate Overall Building UA 71

REM/Rate Annual Energy Cost 72

HVAC Design Peak Loads • Heating – What is T? • Winter Design Temperature • Lexington = 6°F • Inside Temperature? Typical 68°F – T = 68 – 6 = 62°F 73

HVAC Design Peak Loads Heating • Losses (q’s) – Shell (UA for House) – Infiltration (ACHnat) – Duct Loss (cfm) • Gains – ? ? (People are not considered) 74

HVAC Design Peak Loads • Cooling – What is T? • Summer Design Temperature • Lexington = 91°F • Inside Temperature? Typical 76°F – T = 91 – 76 = 15°F 75

HVAC Design Peak Loads Cooling • Gains (q’s) - Complex – Shell (UA for House) – Infiltration (ACHnat) • Losses – ? ? • Adds Moisture – Duct Gain – Solar (Radiation - Windows) – People 76

HVAC Design Peak Loads Is T the same for all surfaces? 77

HVAC Design Peak Loads Is T the same for all surfaces? Basement Walls to the Ground Ceiling to the Attic Wall to the Garage Floor to the Crawl Space 78

REM/Rate Peak Component Loads 0. 57 ACHn 15% Duct Loss to Outside 79

HVAC Annual Loads • Heating – What is an annual T? • Heating Degree Days 65°F - Average daily temperature Add them for one year • Lexington = 4683 HDD • q = U * A * T T = Heating Degree Days * 24 • Close but More Complex 80

HVAC Annual Loads • Cooling – What is an annual T? • Cooling Degree Days Average daily temperature – 65°F Add them for one year • Lexington = 1175 CDD • More Complex Calculation – Solar Radiation – Dehumidification 81

REM/Rate Annual Component Loads 82

HVAC Annual Consumption • Heating Equipment Efficiency – Heat Pump • Heating Season Performance Factor (HSPF) – Btu/Watt-hr – Geothermal Heat Pump • Coefficient of Performance (COP) – Watt-hr output / Watt-hr input – Gas (Combustion) • Annual Fuel Utilization Efficiency (AFUE) – Btu output / Btu input 83

HVAC Annual Consumption • Cooling Equipment Efficiency – Heat Pump / Air Conditioner • Seasonal Energy Efficiency Ratio (SEER) – Btuh/Watt – Geothermal Heat Pump • Energy Efficiency Ratio (EER) – Btuh/Watt 84

HVAC Annual Consumption Equipment Efficiency Adjustment in REM/Rate Formula Created by Florida Solar Center • Cooling – Reduced for Hotter Climates • Lexington: Label SEER = 13, Reduced SEER = 12. 2 • Heating – Heat Pump – Reduced for Cooler Climates • Lexington: Label HSPF = 7. 7, Reduced HSPF = 5. 7 85

REM/Rate Annual Component Consumption 86