1 Lecture 10. Whole Building Heat Loss
California Residential Energy Use - Electricity Electricity by end use category
Lecture 10. Whole Building Heat Loss 1
2
Source: California Energy Commission, "Calif. Statewide Residential Appliance Saturation Study" prepared by KEMA-XENERGY, June 2004.
California Residential Energy Use - Gas
Total Building Envelope Loss
Conduction
Natural gas by end use category
+ Perimeter (slab)
+ Infiltration 3
4
Source: California Energy Commission, "Calif. Statewide Residential Appliance Saturation Study" prepared by KEMA-XENERGY, June 2004.
Review – Conductive Heat Loss Series Heat Transfer - Sum Resistances
Review – Conductive Heat Loss Series Heat Transfer - Sum Resistances
a b
a) ΣR = Ri + Ra + Rb + …+ Ro for each path
(hr-ft2-°F/Btu)
a) ΣR = Ri + Ra + Rb + …+ Ro for each path
(hr-ft2-°F/Btu)
b) U = 1/ΣR for each path always includes air films
(Btu/ hr-ft2-°F)
b) U = 1/ΣR for each path always includes air films
(Btu/ hr-ft2-°F)
(Btu/ hr)
c) Q = U x A x ΔT
•
c) Q = U x A x ΔT
•
a b
(Btu/ hr)
Parallel Heat Transfer - Sum UA values a) U x A for each path
(Btu/hr°F)
b) ΣUA for all paths
(Btu/hr°F)
•
•
•
c) Q = Qa + Qb + … = ΣUA x ΔT 5
a b
6
1
Review - Perimeter (slab) heat loss
Review - Infiltration Unwanted exchange of outside air through “leakage”. Associated heat loss Æ “moving thermal mass” problem
Complicated, radial lines of heat flow Æ simplified method for calculating
•
•
Qslab = F x P x ΔT P = perimeter of floor slab F = perimeter edge loss coefficient (or “slab edge loss factor”)
Qinfil = TM/hr x ΔT = (VHCair x vol/hr) x ΔT = (VHCair x volhouse x AC/hr) x ΔT
(Btu/hr) (ft) (Btu/hr ft °F)
Infiltration Air Exchange Rates Construction Very leaky Loose Medium Tight Very tight
Slab Edge Loss Factor F = 0.55 insulated = 0.81 uninsulated
(Btu/hr ft °F) (Btu/hr ft °F)
7
8
Total Building Envelope Heat Loss
Total Building Envelope Heat Loss •
•
Q env. loss =
Q env. loss =
Conduction + Slab (Perimeter) + Infiltration •
Q cond
ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5
+
∑(UA) ΔT +
•
Q slab (F x P) ΔT
+
Conduction + Slab (Perimeter) + Infiltration
•
•
Q infil
Q cond
+ (VHCair x vol x ac/hr) ΔT
∑(UA)
•
+
Q slab
+
(F x P)
+
•
Q infil
+ (VHCair x vol x ac/hr)
ΔT
Same ΔT = Tindoor - Tout This represents a characteristic of the building only, independent of climate Workbook Handout, p. 16
9
10
Total Building Envelope Heat Loss •
Q env. loss = UAref x ΔT
All paths driven by the same ΔT, a characteristic of the climate
= UAref
Example – Jim Walter House
(Btu/hr)
= UAref x (Tindoor – Tout) (Btu/hr°F)
(°F)
UAref = ∑(UA) + FxP + (VHCair x vol x ac/hr) •
= Q env. loss (Btu/hr°F) ΔT
11
Lecture Handout, p. 16
12
Lecture Handout, example - p. 16
2
Jim Walter home
Who is Jim Walter ?? 1946: Jim Walter starts a homebuilding business in Tampa, Fla., using money borrowed from his father to buy and sell a home for a $300 profit.
Base Case
9 9 9
Conduction Slab edges Infiltration
1960: Mr. Walter's success makes national news, as he is featured in Business Week, the Wall Street Journal, Time magazine and Barron's.
∑ = UAref
Btu/hr F
1964: The company's stock is listed on the NY Stock Exchange. 1969: Jim Walter Corporation ranks No. 287 in the Fortune 500, with sales of $623 million, and has grown into a conglomerate involved in a wide range of businesses from paper to marble to carpet manufacturing. 13
Jim Walter home revisited
Jim Walter home revisited Base Case
Infiltration reduced to 0.5 AC/hr
Base Case
Conduction
Conduction
Slab edges
Slab edges
∑ = UAref
Infiltration reduced to 0.5 AC/hr
Infiltration reduced and insulation increased
Infiltration
Infiltration
15
Lecture Handout, example - p. 16
14
∑ = UAref
Btu/hr F
Lecture Handout, example - p. 16
Btu/hr F
Lecture Handout, example - p. 16
16
Visual break – get ready, rough road ahead!
Example – Whole House Energy Use BASE CASE Walls: prior example (Lec #8) plus 2x4 framing with Rtot=6.7 Ceiling: prior example (Lect #8) Framing: 15% of construction Doors (2): 2” solid core, total 40 ft2 Windows (6): double-pane, 1/2” gap, total 180 ft2 Floor: Slab on grade, insulated Infiltration: Average to loose construction, estimate 1.25 ach. •
Internal gains: Qi = 3500 Btu/hr 17
18
Lecture Handout, example - p. 16
3
Conductive Heat Loss, ΣUA
Calculate areas, perimeter, & volume Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house
= 2 x (30 + 40 ft)
= = = =
= 140 ft
30 ft x 40 ft = 1200 ft2 140 ft x 8 ft = 1120 ft2 gross – doors – windows 1120 ft2 - 40 ft2 - 180 ft2 = 900 ft2
= 30 ft x 40 ft x 8 ft
= 9600 ft3
Lecture Handout, p. 16
19
20
Conductive Heat Loss, ΣUA
Step 1. Calculate areas
STEPS:
Overall dimensions: 30' x 40' house with an 8' ceiling
1. Calculate area for each conductive path using dimensions + % framing
(ft2),
Perimeter, slab Areas ceiling walls (gross) walls (actual)
insulation path (85% of area)
Walls & ceilings
framing path (15% of area)
+ Windows & Doors
Volume, house Lecture Handout, Steps - p. 17
30 ft x 40 ft = 1200 ft2 140 ft x 8 ft = 1120 ft2 gross – doors – windows 1120 ft2 - 40 ft2 - 180 ft2 = 900 ft2
= 30 ft x 40 ft x 8 ft
= 9600 ft3
Lecture Handout, p. 16
22
Table 1 – Conductive Heat Loss, ΣUA
Conductive Heat Loss, ΣUA
A) BASE CASE
23
= = = =
= 140 ft
Given in problem
21
Component
= 2 x (30 + 40 ft)
STEPS:
Area (ft2)
Rtot =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22
.05
51
1. Calculate area for each path (ft2), using dimensions + % framing
Ceiling, framing
1200 * 0.15
11
.09
16
2. Series heat transfer (across each row in table):
Walls, insulation
900 * 0.85
Walls, framing
900 * 0.15
Windows
180
Doors
40
Remember: 13 .08 assume 15% framing 6.7 .15 .49
61
a) ΣR for each path (hr-ft2-°F/Btu) (see earlier examples)
20
b) U = 1/ΣR for each path (Btu/ hr-ft2-°F), includes air films
88
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
24
Lecture Handout, Steps - p. 17
4
Example – Whole House Energy Use
Conduction – Walls & Ceiling Base Case - Given Information
BASE CASE Walls: prior example (Lec #7) plus 2x4 framing with Rtot=6.7
Walls - insulation path: constructed as in prior example problem. See Lecture #8, lecture handout, page 14
Ceiling: prior example (Lect #7) Framing: 15% of construction
Walls - framing path: Given: 2x4 framing with Rtotal = 6.7
Doors (2): 2” solid core, total 40 ft2 Windows (6): double-pane, 1/2” gap, total 180 ft2
Ceiling – insulation & framing paths: constructed as in prior example problem. See Lectures #8+9, lecture handout, page 15
Floor: Slab on grade, insulated Infiltration: Average to loose construction, estimate 1.25 ach. •
Internal gains: Qi = 3500 Btu/hr Lecture Handout, example - p. 16
25
Lecture Handout, example - p. 16
26
Path A
Ceiling
1Path B
2
Previous example: shed roof
Shingles ½” plywood 1-3/4” air gap 5-1/2” batt insul gypsum board
Two sections: 1 – along slope 2 – across slope
R-value (hr ft2 ºF/Btu) Series calculation for each path: Air film Asphalt shingles 1/2" plywood Insulation 1 3/4" air space 2x8 framing Gypsum board Air film
Section 1 (along slope)
Two parallel heat transfer paths:
1
A – across insulation B – across wood frame
Path A
Path B
R Section 2 (across slope)
total
U value = 1/ R total
27
Get ΣR and U-value for each path
A (insulation)
B (wood frame)
0.17 0.44 0.62 19.0 0.95 --0.45 0.62
0.17 0.44 0.62 ----9.06 0.45 0.62
22.25
11.36
0.045
Series
0.088
28
Table 1 – Conductive Heat Loss, ΣUA
Visual break - Lunar eclipse last night!
A) BASE CASE Component
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22
.05
Ceiling, framing
1200 * 0.15
11
.09
Walls, insulation
900 * 0.85
61
Walls, framing
900 * 0.15
20
Windows
180
88
Doors
40
51
Series 16
13 249
29
Lecture Handout, Table 1 - p. 27
30
5
Walls – insulation path
Walls – insulation path
(Example, Lecture #8)
(Example, Lecture #8)
R-value (hr ft2 ºF/BTU)
Series
Lecture Handout, p. 14
31
air film, inside 1/2” gypsum board 3-1/2” insulation 3/4" plywood air film, outside
0.68 0.45 11.0 0.93 0.17
p.1, first table p.2, upper part p.3, top p.2, top p.1, first table
Rtotal = ΣR U = 1/Rtotal
13.23 0.08
(hr ft2 ºF/BTU) (BTU/hr ft2 ºF)
Lecture Handout, p. 14
32
Example – Whole House Energy Use
Walls – framing path (Example, Lecture #8)
BASE CASE Walls: prior example (Lec #8) plus 2x4 framing with Rtot=6.7
Base Case - Given Information
Ceiling: prior example (Lect #8)
2x4 framing with Rtotal = 6.7 for framing path
Framing: 15% of construction Doors (2): 2” solid core, total 40 ft2
So…… Uwall framing = 1/Rtotal = 1/6.7 = 0.15
Windows (6): double-pane, 1/2” gap, total 180 ft2
(Btu/ hr-ft2-°F)
Floor: Slab on grade, insulated Infiltration: Average to loose construction, estimate 1.25 ach. •
Internal gains: Qi = 3500 Btu/hr Lecture Handout, example - p. 16
33
Table 1 – Conductive Heat Loss, ΣUA
Example – Whole House Energy Use
A) BASE CASE Component
BASE CASE
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22
.05
51
Ceiling, framing
1200 * 0.15
11
.09
16
Walls, insulation
900 * 0.85
13
.08
Walls, framing
900 * 0.15
6.7
.15
Walls: prior example (Lec #8) plus 2x4 framing with Rtot=6.7 Ceiling: prior example (Lect #8) Framing: 15% of construction
61
Doors (2): 2” solid core, total 40 ft2
20
Windows (6): double-pane, 1/2” gap, total 180 ft2
Series
Windows
180
88
Floor: Slab on grade, insulated
Doors
40
13
Infiltration: Average to loose construction, estimate 1.25 ach.
249
Internal gains: Qi = 3500 Btu/hr
ΣUA= 35
Lecture Handout, example - p. 16
34
Lecture Handout, Table 1 - p. 17
•
36
Lecture Handout, example - p. 16
6
Windows & Doors
Windows
Base Case - Given Information Windows (6) double-pane, 1/2” gap, total 180 Ft2 (see Appendix 5, p.14) Doors (2) 2” solid core, total 40 ft2 (see Appendix 5, p.14)
U = 0.49 (BTU/hr ft2 ºF) Lecture Handout, p. 16
37
Appendix 5, p. 14
38
Table 1 – Conductive Heat Loss, ΣUA
Doors
A) BASE CASE Component
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22
.05
51
Ceiling, framing
1200 * 0.15
11
.09
16
Walls, insulation
900 * 0.85
13
.08
61
Walls, framing
900 * 0.15
6.7
.15
20 88
Windows
180
.49
Doors
40
.33
13
ΣUA=
249
U = 0.33 (BTU/hr ft2 ºF) Appendix 5, p. 14
39
Lecture Handout, Table 1 - p. 17
40
Conductive Heat Loss, ΣUA
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE
STEPS: Component
Rtotal =ΣR
U= 1/Rtot
UxA
1. Calculate area for each path (ft2), using dimensions + % framing
Ceiling, insulation
1200 * 0.85
22 X
.05
2. Series heat transfer (across each row in table):
Ceiling, framing
1200 * 0.15
11
.09
16
9 a) ΣR for each path (hr-ft2-°F/Btu) (see earlier examples)
Walls, insulation
900 * 0.85
13
.08
61
9 b) U = 1/ΣR for each path (Btu/ hr-ft2-°F),
Walls, framing
900 * 0.15
6.7
.15
20
.49
88
includes air films
c) calculate U x A
41
Area (ft2)
(Btu/hr°F)
Lecture Handout, Steps - p. 17
42
Windows
180
Doors
40
=
51 51
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
7
Table 1 – Conductive Heat Loss, ΣUA
Table 1 – Conductive Heat Loss, ΣUA
A) BASE CASE Component
STEPS:
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22
.05
51
1. Calculate area for each path (ft2) using dimensions + % framing
Ceiling, framing
1200 * 0.15
11
.09
16
2. Series heat transfer (across each row in table):
Walls, insulation
900 * 0.85
13
.08
61
Walls, framing
900 * 0.15
6.7
.15
20
9 a) ΣR for each path (hr-ft2-°F/Btu) (see earlier examples) 9 b) U = 1/ΣR for each path (Btu/ hr-ft2-°F) 9 c) U x A for each path (Btu/hr-°F)
Windows
180
.49
88
Doors
40
.33
13
3. Parallel heat transfer (sum last column in table) ΣUA for whole house (Btu/hr°F)
Repeat UxA for each row Lecture Handout, Table 1 - p. 17
43
Lecture Handout, Steps - p. 17
44
Table 1 – Conductive Heat Loss, ΣUA
Table 1 – Conductive Heat Loss, ΣUA
A) BASE CASE
A) BASE CASE
Component Ceiling, insulation
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
1200 * 0.85
22
.05
51
Parallel
Component
Area (ft2)
Ceiling, insulation
1200 * 0.85
Ceiling, framing
1200 * 0.15
11
.09
16
Ceiling, framing
1200 * 0.15
Walls, insulation
900 * 0.85
13
.08
61
Walls, insulation
900 * 0.85
Walls, framing
900 * 0.15
6.7
.15
20
Walls, framing
900 * 0.15
Windows
180
.49
88
Windows
180
Doors
40
.33
13
Doors
40
ΣUA=
249
Lecture Handout, Table 1 - p. 17
45
22
UxA
.05 % of total 27 % .09
51
52 % 11 13 39 % 6.7
.08 32 % .15
61
9%
.49 41 % .33
88
16
20
13 249
46
Table 2 – Components of UAref
= ∑(UA) + FxP + (VHCair x vol x ac/hr)
UAref
•
= Q / ΔT
= ∑(UA) + FxP + (VHCair x vol x ac/hr) •
= Q / ΔT Base Case
Base Case •
•
Q / ΔT (Btu/hr°F)
Q / ΔT (Btu/hr°F)
Conduction = ∑ (UxA)
46 %
Conduction = ∑ (UxA)
Slab edges = FxP
14 %
Slab edges = FxP
14 %
Infiltration = VHC x vol x ac/hr
40 %
Infiltration = VHC x vol x ac/hr
40 %
∑ = UAref 47
U=
ΣUA=
Table 2 – Components of UAref UAref
Rtotal
Contributions 1/Rtotof =ΣR each component
Lecture Handout, Table 2 - p. 18
249 249
46 %
∑ = UAref 48
Lecture Handout, Table 2 - p. 18
8
Wouldn’t you rather be somewhere else ???
49
Perimeter (Slab) Heat Loss, F x P
50
Example – Whole House Energy Use
Perimeter (Slab) Heat Loss, F x P
BASE CASE
Slab-on-grade with 2” polystyrene rigid insulation
Walls: prior example (Lec #8) plus 2x4 framing with Rtot=6.7
Different places to put insulation
Ceiling: prior example (Lect #8) Framing: 15% of construction Doors (2): 2” solid core, total 40 ft2 Windows (6): double-pane, 1/2” gap, total 180 ft2 Floor: Slab on grade, insulated Infiltration: Average to loose construction, estimate 1.25 ach.
Slab Edge Loss Factor F = 0.55 Btu/hr ft °F insulated = 0.81 Btu/hr ft °F uninsulated
•
Internal gains: Qi = 3500 Btu/hr Lecture Handout, example - p. 16
51
52
Calculate areas, perimeter, & volume
Perimeter (Slab) Heat Loss, F x P
Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house 53
F x P = 0.55 (Btu/hr-ft-°F) x 140 (ft) = 77 (Btu/hr°F)
= 2 x (30 + 40 ft)
= 140 ft
= = = =
= 1200 ft2 = 1120 ft2
30 ft x 40 ft 140 ft x 8 ft gross – doors – windows 1120 ft2 - 40 ft2 - 180 ft2
= 30 ft x 40 ft x 8 ft
= 900 ft2 = 9600 ft3
Lecture Handout, p. 16
54
Lecture Handout, Table 2 - p. 18
9
Perimeter (Slab) Heat Loss, F x P
Infiltration = VHC x vol x AC/hr
F x P = 0.55 (Btu/hr-ft-°F) x 140 (ft) = 77 (Btu/hr°F) Insert F x P # into Table 2 Base Case •
Q / ΔT (Btu/hr°F) Conduction = ∑ (UxA)
249
46 %
Slab edges = FxP
77 77
14 %
Infiltration = VHC x vol x ac/hr
40 %
∑ = UAref Lecture Handout, Table 2 - p. 18
55
56
Example – Whole House Energy Use
Infiltration, VHC x vol x AC/hr
BASE CASE Walls: prior example (Lec #8) plus 2x4 framing with Rtot=6.7
Base Case: Average to loose construction GIVEN: Estimate 1.25 ach
Ceiling: prior example (Lect #8) Framing: 15% of construction
Infiltration Air Construction Very leaky Loose Medium Tight Very tight
Doors (2): 2” solid core, total 40 ft2 Windows (6): double-pane, 1/2” gap, total 180 ft2 Floor: Slab on grade, insulated Infiltration: Average to loose construction, estimate 1.25 ach. •
Exchange Rates ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5
Internal gains: Qi = 3500 Btu/hr Lecture Handout, example - p. 16
57
58
Calculate areas, perimeter, & volume
Lecture Handout, Example - p. 16
Infiltration = VHC x vol x AC/hr
Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house 59
= 2 x (30 + 40 ft)
= 140 ft
= = = =
= 1200 ft2 = 1120 ft2
30 ft x 40 ft 140 ft x 8 ft gross – doors – windows 1120 ft2 - 40 ft2 - 180 ft2
= 30 ft x 40 ft x 8 ft
= VHCair x vol x AC/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 1.25 (house vol/hr) = 216 (Btu/hr °F)
= 900 ft2 = 9600 ft3
Lecture Handout, p. 16
60
Lecture Handout, Table 2 - p. 18
10
Infiltration = VHC x vol x AC/hr
Infiltration = VHC x vol x AC/hr
VHCair x vol x ac/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 1.25 (house vol/hr) = 216 (Btu/hr°F)
VHCair x vol x ac/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 1.25 (house vol/hr) = 216 (Btu/hr°F)
Insert F x P value into Table 2
Insert F x P value into Table 2
Base Case
Base Case
•
•
Q / ΔT (Btu/hr°F)
Q / ΔT (Btu/hr°F)
Conduction = ∑ (UxA)
249
46 %
Conduction = ∑ (UxA)
249
46 %
Slab edges = FxP
77 77
14 %
Slab edges = FxP
14 %
40 %
Infiltration = VHC x vol x ac/hr
77 77 216
∑ = UAref
542
Infiltration = VHC x vol x ac/hr ∑ = UAref
Lecture Handout, Table 2 - p. 18
61
40 %
Lecture Handout, Table 2 - p. 18
62
UAref characterizes the house, but how much energy are you using?
Table 2 – Calculate UAref, Base Case Finally, sum the conductive, slab edge, and infiltration loss rates to find UAref, the overall rate of heat loss UAref = ∑(UA) + FxP + (VHCair x vol x ac/hr) Base Case •
Q / ΔT (Btu/hr°F) Conduction = ∑ (UxA)
249
Slab edges = FxP Infiltration = VHC x vol x ac/hr
77 77 216
∑ = UAref
542
Lecture Handout, Table 2 - p. 18
63
64
Example – Hourly Energy Loss
Table 2 – Calculate UAref, Base Case Contributions of each component WHAT CAN BE IMPROVED ???
UAref = 542 (Btu/hr-°F) When the indoor temperature is 68 °F and the outdoor temperature is 30 °F ……. how much energy is lost through the envelope? •
Q env. loss
65
Base Case
Percent of total
Conduction = ∑ (UxA)
249
46 %
Slab edges = FxP
14 %
Infiltration = VHC x vol x ac/hr
77 77 216
∑ = UAref
542
•
Q / ΔT (Btu/hr°F)
= UAref x ΔT = 542 (Btu /hr-°F) x (68 – 30 °F) = 20,406 Btu/hr
40 %
66
11
7th inning stretch
Case B. Reduce Infiltration Rate Now, back to Jim Walter and improvements to the base case.
67
68
Case B. Reduce Infiltration Rate
69
Case B. Reduce Infiltration Rate
70
Case B. Reduce Infiltration Rate
Case B. Infiltration
71
START with Æ
Exchange Rates ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5 Lecture Handout, Example - p. 16
- New UAref?
VHCair x vol x ac/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 0.5 (house vol/hr) = 86 (Btu/hr°F)
Tightening efforts around the house have lowered the rate of infiltration. Assume the new rate is in the tight construction range at 0.5 ac/hr Infiltration Air Construction Very leaky Loose Medium Tight Very tight
È
A. Base Case
B. Infil È
•
Q / ΔT (Btu/hr°F)
72
Conduction = ∑ (UxA)
249
249
Slab edges = FxP
77
77
Infiltration = VHC x vol x ac/hr
216
86
∑ = UAref
542
412
Workbook Handout, Table 2 - p. 18
12
Case B. Infiltration
È
- New UAref?
Case B. Infiltration
VHCair x vol x ac/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 0.5 (house vol/hr) = 86 (Btu/hr°F) A. Base Case
- New UAref?
VHCair x vol x ac/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 0.5 (house vol/hr) = 86 (Btu/hr°F)
B. Infil È
Percent of total
B. Infil È
•
•
Q / ΔT (Btu/hr°F)
Q / ΔT (Btu/hr°F)
Conduction = ∑ (UxA)
249
249
Conduction = ∑ (UxA)
249
60%
Slab edges = FxP
77
77
Slab edges = FxP
77
19%
Infiltration = VHC x vol x ac/hr
216
86
Infiltration = VHC x vol x ac/hr
86
21%
∑ = UAref
542
412
∑ = UAref
412
24% reduction
73
24% reduction
74
Case C. Add insulation
Case C. Add Insulation
Rigid insulation
Add 2” polystyrene board to entire surface of walls & ceiling. This new insulation covers both original insulation & framing paths
75
Workbook Handout, Example - p. 16
76
Case C. Add Insulation • R-value of new insulation Æ Series heat transfer • 2 “ polystyrene, k= 0.25 (Btu-in/ hr-ft2-°F) • R = d/k = 2 (in) / 0.25 (Btu-in/ hr-ft2-°F) = 8 (hr-ft2-°F/Btu) • Add R-8 in series to wall and ceiling paths
77
È
Workbook Handout, Steps - p. 17
Case C. Add Insulation – new R-values A) BASE CASE
START with Æ Component
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
Ceiling, insulation
1200 * 0.85
22
.05
Ceiling, framing
1200 * 0.15
11
.09
16
Walls, insulation
900 * 0.85
13
.08
61
Walls, framing
900 * 0.15
6.7
.15
20
.49
88
Windows
180
Doors
40
78
UxA 51
Series
.33
13
ΣUA=
249
Workbook Handout, Table 1 - p. 17
13
Case C. Add Insulation – new R-values
Case C. Infil.È + Insul. Ç - New UAref ?
C) ADD INSULATION Component
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
2 22+8=30
.05 .033
51 34
Ceiling, framing
1200 * 0.15
11+8=19 11
.053 .09
16 10
Walls, insulation
900 * 0.85
13 13+8=21
.08 .048
61 37
Walls, framing
900 * 0.15
6.7+8=14.7 6.7
.15 .068
20 9
.49
88
Windows
180
Doors
40
.33
13
ΣUA=
249 191
Workbook Handout, Table 1 - p. 17
79
Conduction
A. Base Case 249
B. Infil È 249
Slab edges
77
77
Infiltration
216
86
∑ = UAref
542
412 24% reduction
80
Jim Walter home
Case C. Infil.È + Insul. Ç - New UAref ?
Base Case
Conduction
A. Base Case 249
B. Infil È 249
C. Infil È + Insul. Ç 191
Slab edges
77
77
77
Infiltration
216
86
86
∑ = UAref
542
412
354
24% reduction
35% reduction
81
82
Infiltration reduced to 0.5 AC/hr
Infiltration reduced and insulation increased
Conduction
249 (46%)
249 (60%)
191 (54%)
Slab edges
77 (14%)
77 (19%)
77 (22%)
Infiltration
216 (40%)
86 (19%)
∑ = UAref
542 Btu/hr F
86 (24%)
412 Btu/hr F
354 Btu/hr F
24% reduction
35% reduction
Lecture Handout, Example - p. 18
Hourly vs. Annual Energy Consumption
Hourly vs. Annual Energy Consumption
1. Relate envelope losses to furnace use Æ Balance Point Temperature, Tbp
• What did we actually calculate ? • Will annual energy savings from energy conservation measures also be 35% ?
2. Add up hourly furnace energy use over a period of time Æ Degree Days, DD
• How does envelope heat loss relate to furnace energy use (and $$)? • How can we “add up” all the hourly energy loss calculations for a longer period?
83
C. Infil È + Insul. Ç 191
84
14
Balance Point Game
1. Balance Point Temperature (Tbp) Defined as …..
This week in section you play with the Balance Point Temperature variables as part of a game. Make sure you read Lab #4 and Reader Chapter #8 before you go.
The outdoor temperature at which heat losses from • the building (Qenv loss) exactly match internal heat • gains (Qi), and below which the furnace turns on.
•
Q env loss Tout
Tin
•
Qi
85
86
Influence of Loads on Balance Point Temperature
Balance Point Temperature (Tbp)
•
Tbp = Ttherm – Qi / UAref
Mathematically, start with heat balance…. •
•
Qgain = Qenv loss •
•
Qi + Qfurnace = UAref x (Tindoor - Tout) =0
= Tbp = Ttherm (thermostat)
Rearranging: •
Tbp = Ttherm – Qi / UAref (°F)
•
Workbook Handout, p. 19
87
88
Example - Table 2 – Calculate Tbp
• Why does Balance Point Temperature make Cris Benton giddy ?
Assume internal gains of 3,500 Btu/hr and alternate thermostat settings (68 °F and 72 °F)
UAref Ttherm = 68 °F
Ttherm = 72 °F 89
B. Infil È 412
C. Infil È + Insul. Ç 354
68 – 3500/542 = 61.5 72 – 3500/542 = 65.5
68 – 3500/412 = 59.5 72 – 3500/412 = 63.5
68 – 3500/354 = 58.1 72 – 3500/354 = 62.1
•
Qi = high UAref = low Tbp = low
To be continued …..
Tbp = Ttherm – Qi / UAref = Ttherm – 3500 / UAref
A. Base Case 542
Qi = low UAref = high Tbp = high
• Enough about the envelope – we pay for furnace energy use. How do we calculate that? • But this was all instantaneous Btu/hr. What about energy use over a period of time, such as a season? Or a year? • Show me the money. What is it going to cost to heat the house? • And tell me again why is electricity so evil? 90
15
Lecture 10. Whole Building Heat Loss 1
2
Source: California Energy Commission, "Calif. Statewide Residential Appliance Saturation Study" prepared by KEMA-XENERGY, June 2004.
California Residential Energy Use - Gas
Total Building Envelope Loss
Conduction
Natural gas by end use category
+ Perimeter (slab)
+ Infiltration 3
4
Source: California Energy Commission, "Calif. Statewide Residential Appliance Saturation Study" prepared by KEMA-XENERGY, June 2004.
Review – Conductive Heat Loss Series Heat Transfer - Sum Resistances
Review – Conductive Heat Loss Series Heat Transfer - Sum Resistances
a b
a) ΣR = Ri + Ra + Rb + …+ Ro for each path
(hr-ft2-°F/Btu)
a) ΣR = Ri + Ra + Rb + …+ Ro for each path
(hr-ft2-°F/Btu)
b) U = 1/ΣR for each path always includes air films
(Btu/ hr-ft2-°F)
b) U = 1/ΣR for each path always includes air films
(Btu/ hr-ft2-°F)
(Btu/ hr)
c) Q = U x A x ΔT
•
c) Q = U x A x ΔT
•
a b
(Btu/ hr)
Parallel Heat Transfer - Sum UA values a) U x A for each path
(Btu/hr°F)
b) ΣUA for all paths
(Btu/hr°F)
•
•
•
c) Q = Qa + Qb + … = ΣUA x ΔT 5
a b
6
1
Review - Perimeter (slab) heat loss
Review - Infiltration Unwanted exchange of outside air through “leakage”. Associated heat loss Æ “moving thermal mass” problem
Complicated, radial lines of heat flow Æ simplified method for calculating
•
•
Qslab = F x P x ΔT P = perimeter of floor slab F = perimeter edge loss coefficient (or “slab edge loss factor”)
Qinfil = TM/hr x ΔT = (VHCair x vol/hr) x ΔT = (VHCair x volhouse x AC/hr) x ΔT
(Btu/hr) (ft) (Btu/hr ft °F)
Infiltration Air Exchange Rates Construction Very leaky Loose Medium Tight Very tight
Slab Edge Loss Factor F = 0.55 insulated = 0.81 uninsulated
(Btu/hr ft °F) (Btu/hr ft °F)
7
8
Total Building Envelope Heat Loss
Total Building Envelope Heat Loss •
•
Q env. loss =
Q env. loss =
Conduction + Slab (Perimeter) + Infiltration •
Q cond
ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5
+
∑(UA) ΔT +
•
Q slab (F x P) ΔT
+
Conduction + Slab (Perimeter) + Infiltration
•
•
Q infil
Q cond
+ (VHCair x vol x ac/hr) ΔT
∑(UA)
•
+
Q slab
+
(F x P)
+
•
Q infil
+ (VHCair x vol x ac/hr)
ΔT
Same ΔT = Tindoor - Tout This represents a characteristic of the building only, independent of climate Workbook Handout, p. 16
9
10
Total Building Envelope Heat Loss •
Q env. loss = UAref x ΔT
All paths driven by the same ΔT, a characteristic of the climate
= UAref
Example – Jim Walter House
(Btu/hr)
= UAref x (Tindoor – Tout) (Btu/hr°F)
(°F)
UAref = ∑(UA) + FxP + (VHCair x vol x ac/hr) •
= Q env. loss (Btu/hr°F) ΔT
11
Lecture Handout, p. 16
12
Lecture Handout, example - p. 16
2
Jim Walter home
Who is Jim Walter ?? 1946: Jim Walter starts a homebuilding business in Tampa, Fla., using money borrowed from his father to buy and sell a home for a $300 profit.
Base Case
9 9 9
Conduction Slab edges Infiltration
1960: Mr. Walter's success makes national news, as he is featured in Business Week, the Wall Street Journal, Time magazine and Barron's.
∑ = UAref
Btu/hr F
1964: The company's stock is listed on the NY Stock Exchange. 1969: Jim Walter Corporation ranks No. 287 in the Fortune 500, with sales of $623 million, and has grown into a conglomerate involved in a wide range of businesses from paper to marble to carpet manufacturing. 13
Jim Walter home revisited
Jim Walter home revisited Base Case
Infiltration reduced to 0.5 AC/hr
Base Case
Conduction
Conduction
Slab edges
Slab edges
∑ = UAref
Infiltration reduced to 0.5 AC/hr
Infiltration reduced and insulation increased
Infiltration
Infiltration
15
Lecture Handout, example - p. 16
14
∑ = UAref
Btu/hr F
Lecture Handout, example - p. 16
Btu/hr F
Lecture Handout, example - p. 16
16
Visual break – get ready, rough road ahead!
Example – Whole House Energy Use BASE CASE Walls: prior example (Lec #8) plus 2x4 framing with Rtot=6.7 Ceiling: prior example (Lect #8) Framing: 15% of construction Doors (2): 2” solid core, total 40 ft2 Windows (6): double-pane, 1/2” gap, total 180 ft2 Floor: Slab on grade, insulated Infiltration: Average to loose construction, estimate 1.25 ach. •
Internal gains: Qi = 3500 Btu/hr 17
18
Lecture Handout, example - p. 16
3
Conductive Heat Loss, ΣUA
Calculate areas, perimeter, & volume Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house
= 2 x (30 + 40 ft)
= = = =
= 140 ft
30 ft x 40 ft = 1200 ft2 140 ft x 8 ft = 1120 ft2 gross – doors – windows 1120 ft2 - 40 ft2 - 180 ft2 = 900 ft2
= 30 ft x 40 ft x 8 ft
= 9600 ft3
Lecture Handout, p. 16
19
20
Conductive Heat Loss, ΣUA
Step 1. Calculate areas
STEPS:
Overall dimensions: 30' x 40' house with an 8' ceiling
1. Calculate area for each conductive path using dimensions + % framing
(ft2),
Perimeter, slab Areas ceiling walls (gross) walls (actual)
insulation path (85% of area)
Walls & ceilings
framing path (15% of area)
+ Windows & Doors
Volume, house Lecture Handout, Steps - p. 17
30 ft x 40 ft = 1200 ft2 140 ft x 8 ft = 1120 ft2 gross – doors – windows 1120 ft2 - 40 ft2 - 180 ft2 = 900 ft2
= 30 ft x 40 ft x 8 ft
= 9600 ft3
Lecture Handout, p. 16
22
Table 1 – Conductive Heat Loss, ΣUA
Conductive Heat Loss, ΣUA
A) BASE CASE
23
= = = =
= 140 ft
Given in problem
21
Component
= 2 x (30 + 40 ft)
STEPS:
Area (ft2)
Rtot =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22
.05
51
1. Calculate area for each path (ft2), using dimensions + % framing
Ceiling, framing
1200 * 0.15
11
.09
16
2. Series heat transfer (across each row in table):
Walls, insulation
900 * 0.85
Walls, framing
900 * 0.15
Windows
180
Doors
40
Remember: 13 .08 assume 15% framing 6.7 .15 .49
61
a) ΣR for each path (hr-ft2-°F/Btu) (see earlier examples)
20
b) U = 1/ΣR for each path (Btu/ hr-ft2-°F), includes air films
88
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
24
Lecture Handout, Steps - p. 17
4
Example – Whole House Energy Use
Conduction – Walls & Ceiling Base Case - Given Information
BASE CASE Walls: prior example (Lec #7) plus 2x4 framing with Rtot=6.7
Walls - insulation path: constructed as in prior example problem. See Lecture #8, lecture handout, page 14
Ceiling: prior example (Lect #7) Framing: 15% of construction
Walls - framing path: Given: 2x4 framing with Rtotal = 6.7
Doors (2): 2” solid core, total 40 ft2 Windows (6): double-pane, 1/2” gap, total 180 ft2
Ceiling – insulation & framing paths: constructed as in prior example problem. See Lectures #8+9, lecture handout, page 15
Floor: Slab on grade, insulated Infiltration: Average to loose construction, estimate 1.25 ach. •
Internal gains: Qi = 3500 Btu/hr Lecture Handout, example - p. 16
25
Lecture Handout, example - p. 16
26
Path A
Ceiling
1Path B
2
Previous example: shed roof
Shingles ½” plywood 1-3/4” air gap 5-1/2” batt insul gypsum board
Two sections: 1 – along slope 2 – across slope
R-value (hr ft2 ºF/Btu) Series calculation for each path: Air film Asphalt shingles 1/2" plywood Insulation 1 3/4" air space 2x8 framing Gypsum board Air film
Section 1 (along slope)
Two parallel heat transfer paths:
1
A – across insulation B – across wood frame
Path A
Path B
R Section 2 (across slope)
total
U value = 1/ R total
27
Get ΣR and U-value for each path
A (insulation)
B (wood frame)
0.17 0.44 0.62 19.0 0.95 --0.45 0.62
0.17 0.44 0.62 ----9.06 0.45 0.62
22.25
11.36
0.045
Series
0.088
28
Table 1 – Conductive Heat Loss, ΣUA
Visual break - Lunar eclipse last night!
A) BASE CASE Component
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22
.05
Ceiling, framing
1200 * 0.15
11
.09
Walls, insulation
900 * 0.85
61
Walls, framing
900 * 0.15
20
Windows
180
88
Doors
40
51
Series 16
13 249
29
Lecture Handout, Table 1 - p. 27
30
5
Walls – insulation path
Walls – insulation path
(Example, Lecture #8)
(Example, Lecture #8)
R-value (hr ft2 ºF/BTU)
Series
Lecture Handout, p. 14
31
air film, inside 1/2” gypsum board 3-1/2” insulation 3/4" plywood air film, outside
0.68 0.45 11.0 0.93 0.17
p.1, first table p.2, upper part p.3, top p.2, top p.1, first table
Rtotal = ΣR U = 1/Rtotal
13.23 0.08
(hr ft2 ºF/BTU) (BTU/hr ft2 ºF)
Lecture Handout, p. 14
32
Example – Whole House Energy Use
Walls – framing path (Example, Lecture #8)
BASE CASE Walls: prior example (Lec #8) plus 2x4 framing with Rtot=6.7
Base Case - Given Information
Ceiling: prior example (Lect #8)
2x4 framing with Rtotal = 6.7 for framing path
Framing: 15% of construction Doors (2): 2” solid core, total 40 ft2
So…… Uwall framing = 1/Rtotal = 1/6.7 = 0.15
Windows (6): double-pane, 1/2” gap, total 180 ft2
(Btu/ hr-ft2-°F)
Floor: Slab on grade, insulated Infiltration: Average to loose construction, estimate 1.25 ach. •
Internal gains: Qi = 3500 Btu/hr Lecture Handout, example - p. 16
33
Table 1 – Conductive Heat Loss, ΣUA
Example – Whole House Energy Use
A) BASE CASE Component
BASE CASE
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22
.05
51
Ceiling, framing
1200 * 0.15
11
.09
16
Walls, insulation
900 * 0.85
13
.08
Walls, framing
900 * 0.15
6.7
.15
Walls: prior example (Lec #8) plus 2x4 framing with Rtot=6.7 Ceiling: prior example (Lect #8) Framing: 15% of construction
61
Doors (2): 2” solid core, total 40 ft2
20
Windows (6): double-pane, 1/2” gap, total 180 ft2
Series
Windows
180
88
Floor: Slab on grade, insulated
Doors
40
13
Infiltration: Average to loose construction, estimate 1.25 ach.
249
Internal gains: Qi = 3500 Btu/hr
ΣUA= 35
Lecture Handout, example - p. 16
34
Lecture Handout, Table 1 - p. 17
•
36
Lecture Handout, example - p. 16
6
Windows & Doors
Windows
Base Case - Given Information Windows (6) double-pane, 1/2” gap, total 180 Ft2 (see Appendix 5, p.14) Doors (2) 2” solid core, total 40 ft2 (see Appendix 5, p.14)
U = 0.49 (BTU/hr ft2 ºF) Lecture Handout, p. 16
37
Appendix 5, p. 14
38
Table 1 – Conductive Heat Loss, ΣUA
Doors
A) BASE CASE Component
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22
.05
51
Ceiling, framing
1200 * 0.15
11
.09
16
Walls, insulation
900 * 0.85
13
.08
61
Walls, framing
900 * 0.15
6.7
.15
20 88
Windows
180
.49
Doors
40
.33
13
ΣUA=
249
U = 0.33 (BTU/hr ft2 ºF) Appendix 5, p. 14
39
Lecture Handout, Table 1 - p. 17
40
Conductive Heat Loss, ΣUA
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE
STEPS: Component
Rtotal =ΣR
U= 1/Rtot
UxA
1. Calculate area for each path (ft2), using dimensions + % framing
Ceiling, insulation
1200 * 0.85
22 X
.05
2. Series heat transfer (across each row in table):
Ceiling, framing
1200 * 0.15
11
.09
16
9 a) ΣR for each path (hr-ft2-°F/Btu) (see earlier examples)
Walls, insulation
900 * 0.85
13
.08
61
9 b) U = 1/ΣR for each path (Btu/ hr-ft2-°F),
Walls, framing
900 * 0.15
6.7
.15
20
.49
88
includes air films
c) calculate U x A
41
Area (ft2)
(Btu/hr°F)
Lecture Handout, Steps - p. 17
42
Windows
180
Doors
40
=
51 51
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
7
Table 1 – Conductive Heat Loss, ΣUA
Table 1 – Conductive Heat Loss, ΣUA
A) BASE CASE Component
STEPS:
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22
.05
51
1. Calculate area for each path (ft2) using dimensions + % framing
Ceiling, framing
1200 * 0.15
11
.09
16
2. Series heat transfer (across each row in table):
Walls, insulation
900 * 0.85
13
.08
61
Walls, framing
900 * 0.15
6.7
.15
20
9 a) ΣR for each path (hr-ft2-°F/Btu) (see earlier examples) 9 b) U = 1/ΣR for each path (Btu/ hr-ft2-°F) 9 c) U x A for each path (Btu/hr-°F)
Windows
180
.49
88
Doors
40
.33
13
3. Parallel heat transfer (sum last column in table) ΣUA for whole house (Btu/hr°F)
Repeat UxA for each row Lecture Handout, Table 1 - p. 17
43
Lecture Handout, Steps - p. 17
44
Table 1 – Conductive Heat Loss, ΣUA
Table 1 – Conductive Heat Loss, ΣUA
A) BASE CASE
A) BASE CASE
Component Ceiling, insulation
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
1200 * 0.85
22
.05
51
Parallel
Component
Area (ft2)
Ceiling, insulation
1200 * 0.85
Ceiling, framing
1200 * 0.15
11
.09
16
Ceiling, framing
1200 * 0.15
Walls, insulation
900 * 0.85
13
.08
61
Walls, insulation
900 * 0.85
Walls, framing
900 * 0.15
6.7
.15
20
Walls, framing
900 * 0.15
Windows
180
.49
88
Windows
180
Doors
40
.33
13
Doors
40
ΣUA=
249
Lecture Handout, Table 1 - p. 17
45
22
UxA
.05 % of total 27 % .09
51
52 % 11 13 39 % 6.7
.08 32 % .15
61
9%
.49 41 % .33
88
16
20
13 249
46
Table 2 – Components of UAref
= ∑(UA) + FxP + (VHCair x vol x ac/hr)
UAref
•
= Q / ΔT
= ∑(UA) + FxP + (VHCair x vol x ac/hr) •
= Q / ΔT Base Case
Base Case •
•
Q / ΔT (Btu/hr°F)
Q / ΔT (Btu/hr°F)
Conduction = ∑ (UxA)
46 %
Conduction = ∑ (UxA)
Slab edges = FxP
14 %
Slab edges = FxP
14 %
Infiltration = VHC x vol x ac/hr
40 %
Infiltration = VHC x vol x ac/hr
40 %
∑ = UAref 47
U=
ΣUA=
Table 2 – Components of UAref UAref
Rtotal
Contributions 1/Rtotof =ΣR each component
Lecture Handout, Table 2 - p. 18
249 249
46 %
∑ = UAref 48
Lecture Handout, Table 2 - p. 18
8
Wouldn’t you rather be somewhere else ???
49
Perimeter (Slab) Heat Loss, F x P
50
Example – Whole House Energy Use
Perimeter (Slab) Heat Loss, F x P
BASE CASE
Slab-on-grade with 2” polystyrene rigid insulation
Walls: prior example (Lec #8) plus 2x4 framing with Rtot=6.7
Different places to put insulation
Ceiling: prior example (Lect #8) Framing: 15% of construction Doors (2): 2” solid core, total 40 ft2 Windows (6): double-pane, 1/2” gap, total 180 ft2 Floor: Slab on grade, insulated Infiltration: Average to loose construction, estimate 1.25 ach.
Slab Edge Loss Factor F = 0.55 Btu/hr ft °F insulated = 0.81 Btu/hr ft °F uninsulated
•
Internal gains: Qi = 3500 Btu/hr Lecture Handout, example - p. 16
51
52
Calculate areas, perimeter, & volume
Perimeter (Slab) Heat Loss, F x P
Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house 53
F x P = 0.55 (Btu/hr-ft-°F) x 140 (ft) = 77 (Btu/hr°F)
= 2 x (30 + 40 ft)
= 140 ft
= = = =
= 1200 ft2 = 1120 ft2
30 ft x 40 ft 140 ft x 8 ft gross – doors – windows 1120 ft2 - 40 ft2 - 180 ft2
= 30 ft x 40 ft x 8 ft
= 900 ft2 = 9600 ft3
Lecture Handout, p. 16
54
Lecture Handout, Table 2 - p. 18
9
Perimeter (Slab) Heat Loss, F x P
Infiltration = VHC x vol x AC/hr
F x P = 0.55 (Btu/hr-ft-°F) x 140 (ft) = 77 (Btu/hr°F) Insert F x P # into Table 2 Base Case •
Q / ΔT (Btu/hr°F) Conduction = ∑ (UxA)
249
46 %
Slab edges = FxP
77 77
14 %
Infiltration = VHC x vol x ac/hr
40 %
∑ = UAref Lecture Handout, Table 2 - p. 18
55
56
Example – Whole House Energy Use
Infiltration, VHC x vol x AC/hr
BASE CASE Walls: prior example (Lec #8) plus 2x4 framing with Rtot=6.7
Base Case: Average to loose construction GIVEN: Estimate 1.25 ach
Ceiling: prior example (Lect #8) Framing: 15% of construction
Infiltration Air Construction Very leaky Loose Medium Tight Very tight
Doors (2): 2” solid core, total 40 ft2 Windows (6): double-pane, 1/2” gap, total 180 ft2 Floor: Slab on grade, insulated Infiltration: Average to loose construction, estimate 1.25 ach. •
Exchange Rates ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5
Internal gains: Qi = 3500 Btu/hr Lecture Handout, example - p. 16
57
58
Calculate areas, perimeter, & volume
Lecture Handout, Example - p. 16
Infiltration = VHC x vol x AC/hr
Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house 59
= 2 x (30 + 40 ft)
= 140 ft
= = = =
= 1200 ft2 = 1120 ft2
30 ft x 40 ft 140 ft x 8 ft gross – doors – windows 1120 ft2 - 40 ft2 - 180 ft2
= 30 ft x 40 ft x 8 ft
= VHCair x vol x AC/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 1.25 (house vol/hr) = 216 (Btu/hr °F)
= 900 ft2 = 9600 ft3
Lecture Handout, p. 16
60
Lecture Handout, Table 2 - p. 18
10
Infiltration = VHC x vol x AC/hr
Infiltration = VHC x vol x AC/hr
VHCair x vol x ac/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 1.25 (house vol/hr) = 216 (Btu/hr°F)
VHCair x vol x ac/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 1.25 (house vol/hr) = 216 (Btu/hr°F)
Insert F x P value into Table 2
Insert F x P value into Table 2
Base Case
Base Case
•
•
Q / ΔT (Btu/hr°F)
Q / ΔT (Btu/hr°F)
Conduction = ∑ (UxA)
249
46 %
Conduction = ∑ (UxA)
249
46 %
Slab edges = FxP
77 77
14 %
Slab edges = FxP
14 %
40 %
Infiltration = VHC x vol x ac/hr
77 77 216
∑ = UAref
542
Infiltration = VHC x vol x ac/hr ∑ = UAref
Lecture Handout, Table 2 - p. 18
61
40 %
Lecture Handout, Table 2 - p. 18
62
UAref characterizes the house, but how much energy are you using?
Table 2 – Calculate UAref, Base Case Finally, sum the conductive, slab edge, and infiltration loss rates to find UAref, the overall rate of heat loss UAref = ∑(UA) + FxP + (VHCair x vol x ac/hr) Base Case •
Q / ΔT (Btu/hr°F) Conduction = ∑ (UxA)
249
Slab edges = FxP Infiltration = VHC x vol x ac/hr
77 77 216
∑ = UAref
542
Lecture Handout, Table 2 - p. 18
63
64
Example – Hourly Energy Loss
Table 2 – Calculate UAref, Base Case Contributions of each component WHAT CAN BE IMPROVED ???
UAref = 542 (Btu/hr-°F) When the indoor temperature is 68 °F and the outdoor temperature is 30 °F ……. how much energy is lost through the envelope? •
Q env. loss
65
Base Case
Percent of total
Conduction = ∑ (UxA)
249
46 %
Slab edges = FxP
14 %
Infiltration = VHC x vol x ac/hr
77 77 216
∑ = UAref
542
•
Q / ΔT (Btu/hr°F)
= UAref x ΔT = 542 (Btu /hr-°F) x (68 – 30 °F) = 20,406 Btu/hr
40 %
66
11
7th inning stretch
Case B. Reduce Infiltration Rate Now, back to Jim Walter and improvements to the base case.
67
68
Case B. Reduce Infiltration Rate
69
Case B. Reduce Infiltration Rate
70
Case B. Reduce Infiltration Rate
Case B. Infiltration
71
START with Æ
Exchange Rates ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5 Lecture Handout, Example - p. 16
- New UAref?
VHCair x vol x ac/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 0.5 (house vol/hr) = 86 (Btu/hr°F)
Tightening efforts around the house have lowered the rate of infiltration. Assume the new rate is in the tight construction range at 0.5 ac/hr Infiltration Air Construction Very leaky Loose Medium Tight Very tight
È
A. Base Case
B. Infil È
•
Q / ΔT (Btu/hr°F)
72
Conduction = ∑ (UxA)
249
249
Slab edges = FxP
77
77
Infiltration = VHC x vol x ac/hr
216
86
∑ = UAref
542
412
Workbook Handout, Table 2 - p. 18
12
Case B. Infiltration
È
- New UAref?
Case B. Infiltration
VHCair x vol x ac/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 0.5 (house vol/hr) = 86 (Btu/hr°F) A. Base Case
- New UAref?
VHCair x vol x ac/hr = 0.018 (Btu/ft3 °F) x 9600 (ft3/house vol) x 0.5 (house vol/hr) = 86 (Btu/hr°F)
B. Infil È
Percent of total
B. Infil È
•
•
Q / ΔT (Btu/hr°F)
Q / ΔT (Btu/hr°F)
Conduction = ∑ (UxA)
249
249
Conduction = ∑ (UxA)
249
60%
Slab edges = FxP
77
77
Slab edges = FxP
77
19%
Infiltration = VHC x vol x ac/hr
216
86
Infiltration = VHC x vol x ac/hr
86
21%
∑ = UAref
542
412
∑ = UAref
412
24% reduction
73
24% reduction
74
Case C. Add insulation
Case C. Add Insulation
Rigid insulation
Add 2” polystyrene board to entire surface of walls & ceiling. This new insulation covers both original insulation & framing paths
75
Workbook Handout, Example - p. 16
76
Case C. Add Insulation • R-value of new insulation Æ Series heat transfer • 2 “ polystyrene, k= 0.25 (Btu-in/ hr-ft2-°F) • R = d/k = 2 (in) / 0.25 (Btu-in/ hr-ft2-°F) = 8 (hr-ft2-°F/Btu) • Add R-8 in series to wall and ceiling paths
77
È
Workbook Handout, Steps - p. 17
Case C. Add Insulation – new R-values A) BASE CASE
START with Æ Component
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
Ceiling, insulation
1200 * 0.85
22
.05
Ceiling, framing
1200 * 0.15
11
.09
16
Walls, insulation
900 * 0.85
13
.08
61
Walls, framing
900 * 0.15
6.7
.15
20
.49
88
Windows
180
Doors
40
78
UxA 51
Series
.33
13
ΣUA=
249
Workbook Handout, Table 1 - p. 17
13
Case C. Add Insulation – new R-values
Case C. Infil.È + Insul. Ç - New UAref ?
C) ADD INSULATION Component
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
2 22+8=30
.05 .033
51 34
Ceiling, framing
1200 * 0.15
11+8=19 11
.053 .09
16 10
Walls, insulation
900 * 0.85
13 13+8=21
.08 .048
61 37
Walls, framing
900 * 0.15
6.7+8=14.7 6.7
.15 .068
20 9
.49
88
Windows
180
Doors
40
.33
13
ΣUA=
249 191
Workbook Handout, Table 1 - p. 17
79
Conduction
A. Base Case 249
B. Infil È 249
Slab edges
77
77
Infiltration
216
86
∑ = UAref
542
412 24% reduction
80
Jim Walter home
Case C. Infil.È + Insul. Ç - New UAref ?
Base Case
Conduction
A. Base Case 249
B. Infil È 249
C. Infil È + Insul. Ç 191
Slab edges
77
77
77
Infiltration
216
86
86
∑ = UAref
542
412
354
24% reduction
35% reduction
81
82
Infiltration reduced to 0.5 AC/hr
Infiltration reduced and insulation increased
Conduction
249 (46%)
249 (60%)
191 (54%)
Slab edges
77 (14%)
77 (19%)
77 (22%)
Infiltration
216 (40%)
86 (19%)
∑ = UAref
542 Btu/hr F
86 (24%)
412 Btu/hr F
354 Btu/hr F
24% reduction
35% reduction
Lecture Handout, Example - p. 18
Hourly vs. Annual Energy Consumption
Hourly vs. Annual Energy Consumption
1. Relate envelope losses to furnace use Æ Balance Point Temperature, Tbp
• What did we actually calculate ? • Will annual energy savings from energy conservation measures also be 35% ?
2. Add up hourly furnace energy use over a period of time Æ Degree Days, DD
• How does envelope heat loss relate to furnace energy use (and $$)? • How can we “add up” all the hourly energy loss calculations for a longer period?
83
C. Infil È + Insul. Ç 191
84
14
Balance Point Game
1. Balance Point Temperature (Tbp) Defined as …..
This week in section you play with the Balance Point Temperature variables as part of a game. Make sure you read Lab #4 and Reader Chapter #8 before you go.
The outdoor temperature at which heat losses from • the building (Qenv loss) exactly match internal heat • gains (Qi), and below which the furnace turns on.
•
Q env loss Tout
Tin
•
Qi
85
86
Influence of Loads on Balance Point Temperature
Balance Point Temperature (Tbp)
•
Tbp = Ttherm – Qi / UAref
Mathematically, start with heat balance…. •
•
Qgain = Qenv loss •
•
Qi + Qfurnace = UAref x (Tindoor - Tout) =0
= Tbp = Ttherm (thermostat)
Rearranging: •
Tbp = Ttherm – Qi / UAref (°F)
•
Workbook Handout, p. 19
87
88
Example - Table 2 – Calculate Tbp
• Why does Balance Point Temperature make Cris Benton giddy ?
Assume internal gains of 3,500 Btu/hr and alternate thermostat settings (68 °F and 72 °F)
UAref Ttherm = 68 °F
Ttherm = 72 °F 89
B. Infil È 412
C. Infil È + Insul. Ç 354
68 – 3500/542 = 61.5 72 – 3500/542 = 65.5
68 – 3500/412 = 59.5 72 – 3500/412 = 63.5
68 – 3500/354 = 58.1 72 – 3500/354 = 62.1
•
Qi = high UAref = low Tbp = low
To be continued …..
Tbp = Ttherm – Qi / UAref = Ttherm – 3500 / UAref
A. Base Case 542
Qi = low UAref = high Tbp = high
• Enough about the envelope – we pay for furnace energy use. How do we calculate that? • But this was all instantaneous Btu/hr. What about energy use over a period of time, such as a season? Or a year? • Show me the money. What is it going to cost to heat the house? • And tell me again why is electricity so evil? 90
15
