Lecture Handout, p. 16
Lecture 10. Whole Building Heat Loss 1
California Residential Energy Use - Electricity Electricity by end use category
2
Source: California Energy Commission, "Calif. Statewide Residential Appliance Saturation Study" prepared by KEMA-XENERGY, June 2004.
Total Building Envelope Loss
Conduction
+ Perimeter (slab)
+ Infiltration 3
California Residential Energy Use - Gas Natural gas by end use category
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 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)
•
c) Q = U x A x ΔT
5
(Btu/ hr)
a b
Review – Conductive Heat Loss Series Heat Transfer - Sum Resistances 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)
•
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 6
a b
Review - Perimeter (slab) heat loss 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”)
(Btu/hr) (ft) (Btu/hr ft °F)
Slab Edge Loss Factor F = 0.55 insulated = 0.81 uninsulated 7
(Btu/hr ft °F) (Btu/hr ft °F)
Review - Infiltration Unwanted exchange of outside air through “leakage”. Associated heat loss Æ “moving thermal mass” problem •
Qinfil = TM/hr x ΔT = (VHCair x vol/hr) x ΔT = (VHCair x volhouse x AC/hr) x ΔT Infiltration Air Exchange Rates
8
Construction Very leaky Loose Medium Tight Very tight
ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5
Total Building Envelope Heat Loss •
Q env. loss = Conduction + Slab (Perimeter) + Infiltration •
Q cond
+
∑(UA) ΔT +
•
Q slab (F x P) ΔT
+
•
Q infil
+ (VHCair x vol x ac/hr) ΔT Same ΔT = Tindoor - Tout
9
Workbook Handout, p. 16
Total Building Envelope Heat Loss •
Q env. loss = Conduction + Slab (Perimeter) + Infiltration •
Q cond
∑(UA)
•
+
Q slab
+
(F x P)
+
10
Q infil
+ (VHCair x vol x ac/hr)
This represents a characteristic of the building only, independent of climate
= UAref
•
ΔT
All paths driven by the same ΔT, a characteristic of the climate
Total Building Envelope Heat Loss •
Q env. loss = UAref x ΔT
(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
Example – Jim Walter House
12
Lecture Handout, example - p. 16
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. 1960: Mr. Walter's success makes national news, as he is featured in Business Week, the Wall Street Journal, Time magazine and Barron's. 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 Base Case Conduction Slab edges Infiltration ∑ = UAref
14
9 9 9 Btu/hr F
Lecture Handout, example - p. 16
Jim Walter home revisited Base Case
Infiltration reduced to 0.5 AC/hr
Conduction Slab edges Infiltration ∑ = UAref
15
Btu/hr F
Lecture Handout, example - p. 16
Jim Walter home revisited Base Case
Infiltration reduced to 0.5 AC/hr
Infiltration reduced and insulation increased
Conduction Slab edges Infiltration ∑ = UAref
16
Btu/hr F
Lecture Handout, example - p. 16
Visual break – get ready, rough road ahead!
17
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 18
Lecture Handout, example - p. 16
Calculate areas, perimeter, & volume Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house 19
= 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
Conductive Heat Loss, ΣUA
20
Conductive Heat Loss, ΣUA STEPS: 1. Calculate area for each conductive path (ft2), using dimensions + % framing Walls & ceilings + Windows & Doors
21
insulation path (85% of area) framing path (15% of area) Given in problem
Lecture Handout, Steps - p. 17
Step 1. Calculate areas Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house 22
= 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
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE Component
23
Area (ft2)
Rtot =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
Walls, framing
900 * 0.15
Remember: 13 .08 assume 15% framing 6.7 .15
61 20
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
Conductive Heat Loss, ΣUA STEPS: 1. Calculate area for each path (ft2), using dimensions + % framing 2. Series heat transfer (across each row in table): a) ΣR for each path (hr-ft2-°F/Btu) (see earlier examples) b) U = 1/ΣR for each path (Btu/ hr-ft2-°F), includes air films
24
Lecture Handout, Steps - p. 17
Example – Whole House Energy Use BASE CASE Walls: prior example (Lec #7) plus 2x4 framing with Rtot=6.7 Ceiling: prior example (Lect #7) 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 25
Lecture Handout, example - p. 16
Conduction – Walls & Ceiling Base Case - Given Information Walls - insulation path: constructed as in prior example problem. See Lecture #8, lecture handout, page 14 Walls - framing path: Given: 2x4 framing with Rtotal = 6.7 Ceiling – insulation & framing paths: constructed as in prior example problem. See Lectures #8+9, lecture handout, page 15
26
Lecture Handout, example - p. 16
Ceiling
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
Two parallel heat transfer paths: A – across insulation B – across wood frame
Section 1 (along slope) 1
Path A
Path B
Section 2 (across slope) 27
Path A
1Path B
Get ΣR and U-value for each path 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 R
total
U value = 1/ R total 28
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
0.088
Series
Table 1 – Conductive Heat Loss, ΣUA 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
Walls, insulation
900 * 0.85
61
Walls, framing
900 * 0.15
20
Windows
180
88
Doors
40
13
Series 16
249 29
Lecture Handout, Table 1 - p. 27
Visual break - Lunar eclipse last night!
30
Walls – insulation path (Example, Lecture #8)
31
Lecture Handout, p. 14
Walls – insulation path (Example, Lecture #8)
R-value (hr ft2 ºF/BTU)
Series
32
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
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 33
Lecture Handout, example - p. 16
Walls – framing path (Example, Lecture #8) Base Case - Given Information 2x4 framing with Rtotal = 6.7 for framing path So…… Uwall framing = 1/Rtotal = 1/6.7 = 0.15
34
(Btu/ hr-ft2-°F)
Lecture Handout, example - p. 16
Table 1 – Conductive Heat Loss, ΣUA 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
Windows
180
88
Doors
40
13 ΣUA=
35
Series
249
Lecture Handout, Table 1 - p. 17
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 36
Lecture Handout, example - p. 16
Windows & Doors 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)
37
Lecture Handout, p. 16
Windows
U = 0.49 (BTU/hr ft2 ºF) 38
Appendix 5, p. 14
Doors
U = 0.33 (BTU/hr ft2 ºF) 39
Appendix 5, p. 14
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE Component
40
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
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
Conductive Heat Loss, ΣUA STEPS: 1. Calculate area for each path (ft2), using dimensions + % framing 2. Series heat transfer (across each row in table):
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), includes air films
c) calculate U x A
41
(Btu/hr°F)
Lecture Handout, Steps - p. 17
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE Component
42
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22 X
.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
=
51 51
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
Table 1 – Conductive Heat Loss, ΣUA 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
Windows
180
.49
88
Doors
40
.33
13
Repeat UxA for each row 43
Lecture Handout, Table 1 - p. 17
Table 1 – Conductive Heat Loss, ΣUA STEPS: 1. Calculate area for each path (ft2) using dimensions + % framing 2. Series heat transfer (across each row in table):
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) 3. Parallel heat transfer (sum last column in table) ΣUA for whole house (Btu/hr°F) 44
Lecture Handout, Steps - p. 17
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE Component
45
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
Parallel
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE Component
Area (ft2)
Ceiling, insulation
1200 * 0.85
Ceiling, framing
1200 * 0.15
Walls, insulation
900 * 0.85
Walls, framing
900 * 0.15
Windows
180
Doors
40
Rtotal U= Contributions 1/Rtotof =ΣR
each component 22
.05 % of total 52 % 27 % 11 .09
51
13 39 % 6.7
.08 32 % .15
61
9%
.49 41 % .33
88
ΣUA= 46
UxA
16
20
13 249
Table 2 – Components of UAref UAref
= ∑(UA) + FxP + (VHCair x vol x ac/hr) •
= Q / ΔT Base Case •
Q / ΔT (Btu/hr°F)
Conduction = ∑ (UxA)
46 %
Slab edges = FxP
14 %
Infiltration = VHC x vol x ac/hr
40 %
∑ = UAref 47
Lecture Handout, Table 2 - p. 18
Table 2 – Components of UAref UAref
= ∑(UA) + FxP + (VHCair x vol x ac/hr) •
= Q / ΔT Base Case •
Q / ΔT (Btu/hr°F) Conduction = ∑ (UxA)
249 249
46 %
Slab edges = FxP
14 %
Infiltration = VHC x vol x ac/hr
40 %
∑ = UAref 48
Lecture Handout, Table 2 - p. 18
Wouldn’t you rather be somewhere else ???
49
Perimeter (Slab) Heat Loss, F x P
50
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 51
Lecture Handout, example - p. 16
Perimeter (Slab) Heat Loss, F x P Slab-on-grade with 2” polystyrene rigid insulation Different places to put insulation
Slab Edge Loss Factor F = 0.55 Btu/hr ft °F insulated = 0.81 Btu/hr ft °F uninsulated 52
Calculate areas, perimeter, & volume Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house 53
= 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
Perimeter (Slab) Heat Loss, F x P F x P = 0.55 (Btu/hr-ft-°F) x 140 (ft) = 77 (Btu/hr°F)
54
Lecture Handout, Table 2 - p. 18
Perimeter (Slab) Heat Loss, F x P 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 55
Lecture Handout, Table 2 - p. 18
Infiltration = VHC x vol x AC/hr
56
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 57
Lecture Handout, example - p. 16
Infiltration, VHC x vol x AC/hr Base Case: Average to loose construction GIVEN: Estimate 1.25 ach
Infiltration Air Construction Very leaky Loose Medium Tight Very tight
58
Exchange Rates ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5 Lecture Handout, Example - p. 16
Calculate areas, perimeter, & volume 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
= 900 ft2 = 9600 ft3
Lecture Handout, p. 16
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)
60
Lecture Handout, Table 2 - p. 18
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)
Insert F x P value 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 61
Lecture Handout, Table 2 - p. 18
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)
Insert F x P value into Table 2 Base Case •
Q / ΔT (Btu/hr°F)
62
Conduction = ∑ (UxA)
249
46 %
Slab edges = FxP
14 %
Infiltration = VHC x vol x ac/hr
77 77 216
∑ = UAref
542
40 %
Lecture Handout, Table 2 - p. 18
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)
63
Conduction = ∑ (UxA)
249
Slab edges = FxP Infiltration = VHC x vol x ac/hr
77 77 216
∑ = UAref
542
Lecture Handout, Table 2 - p. 18
UAref characterizes the house, but how much energy are you using?
64
Example – Hourly Energy Loss
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
= UAref x ΔT = 542 (Btu /hr-°F) x (68 – 30 °F) = 20,406 Btu/hr
Table 2 – Calculate UAref, Base Case Contributions of each component WHAT CAN BE IMPROVED ???
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)
66
40 %
7th inning stretch
67
Case B. Reduce Infiltration Rate Now, back to Jim Walter and improvements to the base case.
68
Case B. Reduce Infiltration Rate
69
Case B. Reduce Infiltration Rate
70
Case B. Reduce Infiltration Rate 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
71
Exchange Rates ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5 Lecture Handout, Example - p. 16
Case B. Infiltration
È
- 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)
START with Æ
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
Case B. Infiltration
È
- 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) A. Base Case
B. Infil È
•
Q / ΔT (Btu/hr°F)
73
Conduction = ∑ (UxA)
249
249
Slab edges = FxP
77
77
Infiltration = VHC x vol x ac/hr
216
86
∑ = UAref
542
412 24% reduction
Case B. Infiltration
È
- 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
•
Q / ΔT (Btu/hr°F)
74
Conduction = ∑ (UxA)
249
60%
Slab edges = FxP
77
19%
Infiltration = VHC x vol x ac/hr
86
21%
∑ = UAref
412 24% reduction
Case C. Add insulation Rigid insulation
75
Case C. Add Insulation Add 2” polystyrene board to entire surface of walls & ceiling. This new insulation covers both original insulation & framing paths
76
Workbook Handout, Example - p. 16
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
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
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
Series
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249
78
Workbook Handout, Table 1 - p. 17
Case C. Add Insulation – new R-values 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
.09 .053
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
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249 191
79
Workbook Handout, Table 1 - p. 17
Case C. Infil.È + Insul. Ç - New UAref ?
Conduction
A. Base Case 249
B. Infil È 249
Slab edges
77
77
Infiltration
216
86
∑ = UAref
542
412 24% reduction
80
C. Infil È + Insul. Ç 191
Case C. Infil.È + Insul. Ç - New UAref ?
Conduction
81
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
Jim Walter home Base Case Conduction
82
Infiltration reduced to 0.5 AC/hr
Infiltration reduced and insulation increased
249 (46%)
249 (60%)
191 (54%)
Slab edges
77 (14%)
77 (19%)
77 (22%)
Infiltration
216 (40%)
86 (19%)
86 (24%)
∑ = UAref
542 Btu/hr F
412 Btu/hr F
354 Btu/hr F
24% reduction
35% reduction
Lecture Handout, Example - p. 18
Hourly vs. Annual Energy Consumption • What did we actually calculate ? • Will annual energy savings from energy conservation measures also be 35% ? • 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
Hourly vs. Annual Energy Consumption 1. Relate envelope losses to furnace use Æ Balance Point Temperature, Tbp 2. Add up hourly furnace energy use over a period of time Æ Degree Days, DD
84
1. Balance Point Temperature (Tbp) Defined as ….. 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
85
Tin
•
Qi
Balance Point Game 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.
86
Balance Point Temperature (Tbp) Mathematically, start with heat balance…. •
•
Qgain = Qenv loss •
•
Qi + Qfurnace = UAref x (Tindoor - Tout) =0
= Tbp = Ttherm (thermostat)
Rearranging: •
Tbp = Ttherm – Qi / UAref (°F)
87
Workbook Handout, p. 19
Influence of Loads on Balance Point Temperature •
Tbp = Ttherm – Qi / UAref
•
88
Qi = low UAref = high Tbp = high
•
Qi = high UAref = low Tbp = low
Example - Table 2 – Calculate Tbp Tbp = Ttherm – Qi / UAref = Ttherm – 3500 / UAref 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
A. Base Case 542
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
To be continued ….. • Why does Balance Point Temperature make Cris Benton giddy ? • 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
California Residential Energy Use - Electricity Electricity by end use category
2
Source: California Energy Commission, "Calif. Statewide Residential Appliance Saturation Study" prepared by KEMA-XENERGY, June 2004.
Total Building Envelope Loss
Conduction
+ Perimeter (slab)
+ Infiltration 3
California Residential Energy Use - Gas Natural gas by end use category
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 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)
•
c) Q = U x A x ΔT
5
(Btu/ hr)
a b
Review – Conductive Heat Loss Series Heat Transfer - Sum Resistances 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)
•
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 6
a b
Review - Perimeter (slab) heat loss 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”)
(Btu/hr) (ft) (Btu/hr ft °F)
Slab Edge Loss Factor F = 0.55 insulated = 0.81 uninsulated 7
(Btu/hr ft °F) (Btu/hr ft °F)
Review - Infiltration Unwanted exchange of outside air through “leakage”. Associated heat loss Æ “moving thermal mass” problem •
Qinfil = TM/hr x ΔT = (VHCair x vol/hr) x ΔT = (VHCair x volhouse x AC/hr) x ΔT Infiltration Air Exchange Rates
8
Construction Very leaky Loose Medium Tight Very tight
ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5
Total Building Envelope Heat Loss •
Q env. loss = Conduction + Slab (Perimeter) + Infiltration •
Q cond
+
∑(UA) ΔT +
•
Q slab (F x P) ΔT
+
•
Q infil
+ (VHCair x vol x ac/hr) ΔT Same ΔT = Tindoor - Tout
9
Workbook Handout, p. 16
Total Building Envelope Heat Loss •
Q env. loss = Conduction + Slab (Perimeter) + Infiltration •
Q cond
∑(UA)
•
+
Q slab
+
(F x P)
+
10
Q infil
+ (VHCair x vol x ac/hr)
This represents a characteristic of the building only, independent of climate
= UAref
•
ΔT
All paths driven by the same ΔT, a characteristic of the climate
Total Building Envelope Heat Loss •
Q env. loss = UAref x ΔT
(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
Example – Jim Walter House
12
Lecture Handout, example - p. 16
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. 1960: Mr. Walter's success makes national news, as he is featured in Business Week, the Wall Street Journal, Time magazine and Barron's. 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 Base Case Conduction Slab edges Infiltration ∑ = UAref
14
9 9 9 Btu/hr F
Lecture Handout, example - p. 16
Jim Walter home revisited Base Case
Infiltration reduced to 0.5 AC/hr
Conduction Slab edges Infiltration ∑ = UAref
15
Btu/hr F
Lecture Handout, example - p. 16
Jim Walter home revisited Base Case
Infiltration reduced to 0.5 AC/hr
Infiltration reduced and insulation increased
Conduction Slab edges Infiltration ∑ = UAref
16
Btu/hr F
Lecture Handout, example - p. 16
Visual break – get ready, rough road ahead!
17
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 18
Lecture Handout, example - p. 16
Calculate areas, perimeter, & volume Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house 19
= 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
Conductive Heat Loss, ΣUA
20
Conductive Heat Loss, ΣUA STEPS: 1. Calculate area for each conductive path (ft2), using dimensions + % framing Walls & ceilings + Windows & Doors
21
insulation path (85% of area) framing path (15% of area) Given in problem
Lecture Handout, Steps - p. 17
Step 1. Calculate areas Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house 22
= 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
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE Component
23
Area (ft2)
Rtot =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
Walls, framing
900 * 0.15
Remember: 13 .08 assume 15% framing 6.7 .15
61 20
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
Conductive Heat Loss, ΣUA STEPS: 1. Calculate area for each path (ft2), using dimensions + % framing 2. Series heat transfer (across each row in table): a) ΣR for each path (hr-ft2-°F/Btu) (see earlier examples) b) U = 1/ΣR for each path (Btu/ hr-ft2-°F), includes air films
24
Lecture Handout, Steps - p. 17
Example – Whole House Energy Use BASE CASE Walls: prior example (Lec #7) plus 2x4 framing with Rtot=6.7 Ceiling: prior example (Lect #7) 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 25
Lecture Handout, example - p. 16
Conduction – Walls & Ceiling Base Case - Given Information Walls - insulation path: constructed as in prior example problem. See Lecture #8, lecture handout, page 14 Walls - framing path: Given: 2x4 framing with Rtotal = 6.7 Ceiling – insulation & framing paths: constructed as in prior example problem. See Lectures #8+9, lecture handout, page 15
26
Lecture Handout, example - p. 16
Ceiling
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
Two parallel heat transfer paths: A – across insulation B – across wood frame
Section 1 (along slope) 1
Path A
Path B
Section 2 (across slope) 27
Path A
1Path B
Get ΣR and U-value for each path 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 R
total
U value = 1/ R total 28
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
0.088
Series
Table 1 – Conductive Heat Loss, ΣUA 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
Walls, insulation
900 * 0.85
61
Walls, framing
900 * 0.15
20
Windows
180
88
Doors
40
13
Series 16
249 29
Lecture Handout, Table 1 - p. 27
Visual break - Lunar eclipse last night!
30
Walls – insulation path (Example, Lecture #8)
31
Lecture Handout, p. 14
Walls – insulation path (Example, Lecture #8)
R-value (hr ft2 ºF/BTU)
Series
32
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
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 33
Lecture Handout, example - p. 16
Walls – framing path (Example, Lecture #8) Base Case - Given Information 2x4 framing with Rtotal = 6.7 for framing path So…… Uwall framing = 1/Rtotal = 1/6.7 = 0.15
34
(Btu/ hr-ft2-°F)
Lecture Handout, example - p. 16
Table 1 – Conductive Heat Loss, ΣUA 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
Windows
180
88
Doors
40
13 ΣUA=
35
Series
249
Lecture Handout, Table 1 - p. 17
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 36
Lecture Handout, example - p. 16
Windows & Doors 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)
37
Lecture Handout, p. 16
Windows
U = 0.49 (BTU/hr ft2 ºF) 38
Appendix 5, p. 14
Doors
U = 0.33 (BTU/hr ft2 ºF) 39
Appendix 5, p. 14
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE Component
40
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
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
Conductive Heat Loss, ΣUA STEPS: 1. Calculate area for each path (ft2), using dimensions + % framing 2. Series heat transfer (across each row in table):
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), includes air films
c) calculate U x A
41
(Btu/hr°F)
Lecture Handout, Steps - p. 17
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE Component
42
Area (ft2)
Rtotal =ΣR
U= 1/Rtot
UxA
Ceiling, insulation
1200 * 0.85
22 X
.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
=
51 51
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
Table 1 – Conductive Heat Loss, ΣUA 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
Windows
180
.49
88
Doors
40
.33
13
Repeat UxA for each row 43
Lecture Handout, Table 1 - p. 17
Table 1 – Conductive Heat Loss, ΣUA STEPS: 1. Calculate area for each path (ft2) using dimensions + % framing 2. Series heat transfer (across each row in table):
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) 3. Parallel heat transfer (sum last column in table) ΣUA for whole house (Btu/hr°F) 44
Lecture Handout, Steps - p. 17
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE Component
45
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
Parallel
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249
Lecture Handout, Table 1 - p. 17
Table 1 – Conductive Heat Loss, ΣUA A) BASE CASE Component
Area (ft2)
Ceiling, insulation
1200 * 0.85
Ceiling, framing
1200 * 0.15
Walls, insulation
900 * 0.85
Walls, framing
900 * 0.15
Windows
180
Doors
40
Rtotal U= Contributions 1/Rtotof =ΣR
each component 22
.05 % of total 52 % 27 % 11 .09
51
13 39 % 6.7
.08 32 % .15
61
9%
.49 41 % .33
88
ΣUA= 46
UxA
16
20
13 249
Table 2 – Components of UAref UAref
= ∑(UA) + FxP + (VHCair x vol x ac/hr) •
= Q / ΔT Base Case •
Q / ΔT (Btu/hr°F)
Conduction = ∑ (UxA)
46 %
Slab edges = FxP
14 %
Infiltration = VHC x vol x ac/hr
40 %
∑ = UAref 47
Lecture Handout, Table 2 - p. 18
Table 2 – Components of UAref UAref
= ∑(UA) + FxP + (VHCair x vol x ac/hr) •
= Q / ΔT Base Case •
Q / ΔT (Btu/hr°F) Conduction = ∑ (UxA)
249 249
46 %
Slab edges = FxP
14 %
Infiltration = VHC x vol x ac/hr
40 %
∑ = UAref 48
Lecture Handout, Table 2 - p. 18
Wouldn’t you rather be somewhere else ???
49
Perimeter (Slab) Heat Loss, F x P
50
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 51
Lecture Handout, example - p. 16
Perimeter (Slab) Heat Loss, F x P Slab-on-grade with 2” polystyrene rigid insulation Different places to put insulation
Slab Edge Loss Factor F = 0.55 Btu/hr ft °F insulated = 0.81 Btu/hr ft °F uninsulated 52
Calculate areas, perimeter, & volume Overall dimensions: 30' x 40' house with an 8' ceiling Perimeter, slab Areas ceiling walls (gross) walls (actual)
Volume, house 53
= 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
Perimeter (Slab) Heat Loss, F x P F x P = 0.55 (Btu/hr-ft-°F) x 140 (ft) = 77 (Btu/hr°F)
54
Lecture Handout, Table 2 - p. 18
Perimeter (Slab) Heat Loss, F x P 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 55
Lecture Handout, Table 2 - p. 18
Infiltration = VHC x vol x AC/hr
56
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 57
Lecture Handout, example - p. 16
Infiltration, VHC x vol x AC/hr Base Case: Average to loose construction GIVEN: Estimate 1.25 ach
Infiltration Air Construction Very leaky Loose Medium Tight Very tight
58
Exchange Rates ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5 Lecture Handout, Example - p. 16
Calculate areas, perimeter, & volume 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
= 900 ft2 = 9600 ft3
Lecture Handout, p. 16
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)
60
Lecture Handout, Table 2 - p. 18
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)
Insert F x P value 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 61
Lecture Handout, Table 2 - p. 18
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)
Insert F x P value into Table 2 Base Case •
Q / ΔT (Btu/hr°F)
62
Conduction = ∑ (UxA)
249
46 %
Slab edges = FxP
14 %
Infiltration = VHC x vol x ac/hr
77 77 216
∑ = UAref
542
40 %
Lecture Handout, Table 2 - p. 18
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)
63
Conduction = ∑ (UxA)
249
Slab edges = FxP Infiltration = VHC x vol x ac/hr
77 77 216
∑ = UAref
542
Lecture Handout, Table 2 - p. 18
UAref characterizes the house, but how much energy are you using?
64
Example – Hourly Energy Loss
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
= UAref x ΔT = 542 (Btu /hr-°F) x (68 – 30 °F) = 20,406 Btu/hr
Table 2 – Calculate UAref, Base Case Contributions of each component WHAT CAN BE IMPROVED ???
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)
66
40 %
7th inning stretch
67
Case B. Reduce Infiltration Rate Now, back to Jim Walter and improvements to the base case.
68
Case B. Reduce Infiltration Rate
69
Case B. Reduce Infiltration Rate
70
Case B. Reduce Infiltration Rate 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
71
Exchange Rates ac/hr 2-6 1-2 0.75 - 1 0.5 – 0.75 0.05 – 0.5 Lecture Handout, Example - p. 16
Case B. Infiltration
È
- 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)
START with Æ
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
Case B. Infiltration
È
- 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) A. Base Case
B. Infil È
•
Q / ΔT (Btu/hr°F)
73
Conduction = ∑ (UxA)
249
249
Slab edges = FxP
77
77
Infiltration = VHC x vol x ac/hr
216
86
∑ = UAref
542
412 24% reduction
Case B. Infiltration
È
- 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
•
Q / ΔT (Btu/hr°F)
74
Conduction = ∑ (UxA)
249
60%
Slab edges = FxP
77
19%
Infiltration = VHC x vol x ac/hr
86
21%
∑ = UAref
412 24% reduction
Case C. Add insulation Rigid insulation
75
Case C. Add Insulation Add 2” polystyrene board to entire surface of walls & ceiling. This new insulation covers both original insulation & framing paths
76
Workbook Handout, Example - p. 16
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
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
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
Series
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249
78
Workbook Handout, Table 1 - p. 17
Case C. Add Insulation – new R-values 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
.09 .053
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
Windows
180
.49
88
Doors
40
.33
13
ΣUA=
249 191
79
Workbook Handout, Table 1 - p. 17
Case C. Infil.È + Insul. Ç - New UAref ?
Conduction
A. Base Case 249
B. Infil È 249
Slab edges
77
77
Infiltration
216
86
∑ = UAref
542
412 24% reduction
80
C. Infil È + Insul. Ç 191
Case C. Infil.È + Insul. Ç - New UAref ?
Conduction
81
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
Jim Walter home Base Case Conduction
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Infiltration reduced to 0.5 AC/hr
Infiltration reduced and insulation increased
249 (46%)
249 (60%)
191 (54%)
Slab edges
77 (14%)
77 (19%)
77 (22%)
Infiltration
216 (40%)
86 (19%)
86 (24%)
∑ = UAref
542 Btu/hr F
412 Btu/hr F
354 Btu/hr F
24% reduction
35% reduction
Lecture Handout, Example - p. 18
Hourly vs. Annual Energy Consumption • What did we actually calculate ? • Will annual energy savings from energy conservation measures also be 35% ? • 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
Hourly vs. Annual Energy Consumption 1. Relate envelope losses to furnace use Æ Balance Point Temperature, Tbp 2. Add up hourly furnace energy use over a period of time Æ Degree Days, DD
84
1. Balance Point Temperature (Tbp) Defined as ….. 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
85
Tin
•
Qi
Balance Point Game 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.
86
Balance Point Temperature (Tbp) Mathematically, start with heat balance…. •
•
Qgain = Qenv loss •
•
Qi + Qfurnace = UAref x (Tindoor - Tout) =0
= Tbp = Ttherm (thermostat)
Rearranging: •
Tbp = Ttherm – Qi / UAref (°F)
87
Workbook Handout, p. 19
Influence of Loads on Balance Point Temperature •
Tbp = Ttherm – Qi / UAref
•
88
Qi = low UAref = high Tbp = high
•
Qi = high UAref = low Tbp = low
Example - Table 2 – Calculate Tbp Tbp = Ttherm – Qi / UAref = Ttherm – 3500 / UAref 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
A. Base Case 542
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
To be continued ….. • Why does Balance Point Temperature make Cris Benton giddy ? • 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
