# 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.

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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 High Temperature Low Temperature Resistance 5

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

7 Conduction Heat Loss

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

9 Convective Heat Loss

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

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

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

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

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

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

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

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

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

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

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 double- glazed 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? 24 R T = R 1 + R 2 + R 3 R 1 R 2 R 3

How do R-values Add? - Example 25 R T 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 R T = 18.50

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

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

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 * A *  T = A *  T R U = 1 R 29

Air Leakage - General Equation q = m * C p *  T – where q = heat flow, Btu/hr m = mass flow of air, lbs/hr C p = specific heat of air, 0.24 Btu/lbs - ° F  T = temperature differential, ° F 30

Air Leakage - General Equation q = m * C p *  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 10x10 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? 2x4 R-value = 4.38 (1.25 per inch) 35

Simple Heat Flow, q, Calculation Typical Wood Framing 36

Minimum Wood Framing Approximately 10 2x4s, 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 Simple Heat Flow, q, Calculation 37

Simple Heat Flow, q, Calculation w/Framing Total Area = 100 ft 2 10x10 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

q insulation = 87.5 * 1 = 6.73 Btuh 13 q framing = 12.5 * 1 = 2.85 Btuh 4.38 q total = q insulation + q framing = 6.73 + 2.85 = 9.58 Btuh 39 Simple Heat Flow, q, Calculation w/Framing

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

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 q total = 9.58 Btuh/ ° F R = A *  T = 100 * 1 = 10.44 q 9.58 TOTAL WALL R 42

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

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 + Window Windows Require Extra Framing Materials 4 extra studs for kings and jacks 2x12 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 10x10 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 q insulation = 64.7 * 1 = 4.98 Btuh 13 q framing = 20.3 * 1 = 4.63 Btuh 4.38 q window = 0.40 *15 * 1 = 6 Btuh q total = 4.98 + 4.63 + 6 = 15.61 Btuh 48

R-Value of the Wall With Framing + Window q total = 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

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56

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

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59 Exterior Finish

R of Cavity Wall Section Inside Air = 0.68 5/8” Gypsum Board = 0.56 3 ½” Cavity Insulation = 13.00 Exterior Finish= 0.94 Outside Air = 0.17 Cavity Wall Section R = 15.35 60

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

R of Framing Wall Section Inside Air =0.68 5/8” Gypsum Board =0.56 3 ½” Framing =4.37 Exterior Finish =0.94 Outside Air =0.17 Framing Wall Section R =6.72 62

63 Framing Factor

64 Framing Factor

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

Assume the Total Wall Area = 100 ft 2 Framing Wall Area = 0.23 * 100 = 23 ft 2 Cavity Wall Area = 100 – 23 = 77 ft 2 Framing q = A*  T = 23 * 1 = 3.42 Btuh R 6.72 Cavity q = A*  T = 77 * 1 = 5.02 Btuh R 15.35 Total q = 3.42 + 5.02 = 8.44 Btuh Total Wall R = A = 100 = 11.85 q 8.44 66 Your Turn - Total Wall R

Assume the Total Wall Area = 100 ft 2 Framing Wall Area = 0.23 * 100 = 23 ft 2 Cavity Wall Area = 100 – 23 = 77 ft 2 Framing = 23% Area but 3.42/8.44 = 41% of Flow 67 Your Turn - Total Wall R

Total Wall U Total Wall R = 11.85 Total Wall U = 1 = 1 = 0.0843 R 11.85 68

69 Total Wall U

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

71 REM/Rate Overall Building UA

72 REM/Rate Annual Energy Cost

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) Adds Moisture – Duct Gain – Solar (Radiation - Windows) – People 76 Losses – ??

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

79 REM/Rate Peak Component Loads 0.57 ACHn 15% Duct Loss to Outside

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

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

86 REM/Rate Annual Component Consumption

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