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, ft2
R = resistance, ft2-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
R1
R2
R3
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, ft2
U = conductance, Btu/ft2-hr-°F
T = temperature differential, °F
26
Equations - Conduction
• Where does U come from?
– Table values
• How do they add?
1 =
1 + 1
UT
U1
U2
• 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 * 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 ft3 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
• ft3/hr = ACHnat * Volume (ft3)
– where
• ACHnat = Natural Air Changes per hour
• Volume = volume of the conditioned space
• q = 0.018* ACHnat * Volume (ft3) * T
33
Simple Heat Flow, q, Calculation
Assume 10x10 wall A = 100 ft2
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
Simple Heat Flow, q, Calculation
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 in2 / 144 in2 per ft2 = 12.5 ft2
37
Simple Heat Flow, q, Calculation w/Framing
Total Area = 100 ft2 10x10 wall
Cavity Insulation R-value = 13
Framing R = 4.38
Framing Area = 12.5 ft2
Cavity Insulation Area = 100 – 12.5 = 87.5 ft2
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________
_A1_ + _A2_
R1
R2
43
Simple Heat Flow, q, Calculation
What if there is a window in the wall?
Window:
Size 3 ft x 5 ft = 15 ft2
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 ft2 of extra framing
Total framing = 12.5 + 7.8 = 20.3 ft2
46
Simple Heat Flow, q, Calculation
With Framing + Window
Total Area = 100 ft2 10x10 wall
Cavity Insulation R-value = 13
Framing R-value = 4.38
Framing Area = 20.3 ft2
Window U-factor = 0.40
Window Area = 15 ft2
Cavity Insulation Area = 100 – 20.3 - 15 = 64.7 ft2
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
55
56
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
58
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 =
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