1. Technology in Architecture
Lecture 7
Degree Days
Heating Loads
Annual Fuel Consumption
Simple Payback Analysis

2. Heating Degree Days
Balance Point Temperature (BPT): temperature above which heating is not needed
DDBPT= BPT-TA

3. Sample Calculation January TA=28ºF DD65=65-28= 37 Degree-days/day
x 31 days
= 1,147 degree-days
M: p. 1620, T.C.19

4. Heating Loads

5. Heating Loads Computed for worst case scenario: Do not include:
Pre-dawn at outdoor design dry bulb temperature
Do not include:
Insolation from sun
Heat gain from people, lights, and equipment
Infiltration in nonresidential buildings
Ventilation in residential buildings
SR-3

6. Outdoor Dry Bulb Temperature
M: T.B.1, p. 1571
Outdoor Dry Bulb Temperature
Use Winter Conditions
RESCAN S11: T.B

7. Determine Temperature Difference
Indoor Dry Bulb Temperature (IDBT): 68ºF
Outdoor Dry Bulb Temperature (ODBT): 13ºF
ΔT=IDBT-ODBT=68ºF - 13ºF = 55ºF

8. Determine Envelope U-values
Calculate ΣR and then find U for walls, roofs, floors.
Obtain U values for glazing from manufacturer or other reference

9. Determine Area Quantities
Perform area takeoffs for all building envelope surfaces on each facade:
gross wall area
window area
door area
net wall area
1200 sf -
100’
368 sf -
64 sf
768 sf
12’ 4’ 4’ 8’
Elevation

10. Floor Slabs
For floor slabs at grade, there are two heat loss components:
slab to soil losses
edge losses
M: p. 1680, F.E.1

11. TI=Indoor Air Temperature TGW=Ground Water Temperature
Slab to Soil Losses
Q=Uslab x 0.5 x Aslab x (TI-TGW)
TI=Indoor Air Temperature
TGW=Ground Water Temperature

12. Edge Losses Method I Determine F2 based on heating degree days
M: p. 1680, T.E.11/F. E.1

13. Select F2 based on insulation configuration
Slab Edge Losses
Method II
Select F2 based on insulation configuration
M: 1681, T.E.12

14. TI= Indoor air temperature TO=Outdoor air temperature
Slab Edge Losses
Q=F2 x Slab Perimeter Length x (TI-TO)
where,
TI= Indoor air temperature
TO=Outdoor air temperature

15. Heating Load Example Problem
Building: Office Building
Location: Salt Lake City
ΔT=IDBT-ODBT=68-13=55ºF
Building: 200’ x 100’ (2 stories, 12’-6” each)
Uwall= Btuh/sf-ºF
Uroof= Btuh/sf-ºF
Uwindow= Btuh/sf-ºF
Uslab= Btuh/sf-ºF
Udoor= Btuh/sf-ºF
Slab Edge: R-15 vertical insulation, 2’ deep.
Degree Days = 5,983

16. Heating Load Example Problem
Determine Building Envelope Areas (SF)
Building: 200’ x 100’ (2 stories, 12’-6” each)
N E S W
Gross Wall 5,000 2,500 5,000 2,500
Windows 1, ,
Doors
Net Wall 3,980 1,980 2,950 1,980
Roof/Floor Slab 20,000

17. Heating Loads Insert roof values Insert wall values,
N ,980 55
E ,980 55
S ,950 55
W ,980 55
Insert roof values
Insert wall values
Insert glass values
Insert door values
Insert floor values
N ,000 55 E
S ,000 55 W
N/A N/A N/A N/A
SR-3

18. TI=Indoor Air Temperature TGW=Ground Water Temperature
Slab to Soil Losses
Q=Uslab x 0.5 x Aslab x (TI-TGW)
TI=Indoor Air Temperature
TGW=Ground Water Temperature
Ground Water= 53ºF
ΔT=68ºF-53ºF=15ºF

19. Heating Loads Insert floor values SR-3 0.025 20,000 55,
N ,980 55
E ,980 55
S ,950 55
W ,980 55
Insert floor values
N ,000 55 E
S ,000 55 W
N/A N/A N/A N/A
,000 15
SR-3

20. Select F2 based on insulation configuration
Slab Edge Losses
Method II
Select F2 based on insulation configuration
Slab Edge: R-15 vertical insulation, ’ deep.
F2=0.52
M: 1681, T.E.12

21. Heating Loads Insert floor values SR-3 0.025 20,000 55,
N ,980 55
E ,980 55
S ,950 55
W ,980 55
Insert floor values
N ,000 55 E
S ,000 55 W
N/A N/A N/A N/A
,000 15
SR-3

22. Infiltration
Residential buildings use infiltration to provide fresh air
“Air change/hour (ACH) method” or
“Crack length method”
Prone to subjective interpretation
Vulnerable to construction defects
Provides a relatively approximate result

23. ASHRAE Standard 62.1-2016 (M: p. 1696, T.F.1)
Ventilation Analysis
Non-residential buildings use ventilation to provide fresh air and to offset infiltration effects.
ASHRAE Standard (M: p. 1696, T.F.1)
Estimates the number of people/1000 sf of usage type
Prescribes minimum ventilation/person for usage type

24. ASHRAE 62.1-2016 Defines space occupancy and ventilation loads
M: p. 1697, T.F.1

25. Ventilation Load — Sensible
40,000 sf x 5people/1,000sf = 200 people
200 people x 17 cfm/person = 3,400 cfm
3,400 cfm x 60min/hr = 204,000cfh

26. Heating Loads Input Ventilation Load—Sensible SR-3 0.025 20,000 55,
N ,980 55
E ,980 55
S ,950 55
W ,980 55
Input Ventilation Load—Sensible
N ,000 55 E
S ,000 55 W
N/A N/A N/A N/A
,000 15
204,
SR-3

27. Ventilation Load — Latent
Determine ΔW
WI= #H2O/#dry air
-WO= #H2O/#dry air
ΔW= #H2O/#dry air

28. Heating Loads Input Ventilation Load — Latent SR-3
, ,500
N , ,583
E ,
S , ,762
W , ,880
Input Ventilation Load — Latent
N , ,050
E ,525
S , ,100
W ,525
,210
N/A N/A N/A N/A
, ,000
,160
204, ,960
204, ,308
SR-3

29. Heating Load Total Load 469,435 Btuh 5.9 6.8 8.8 SR-3 14.5 0.3 63.8
, , ,500
5.9
N , ,583
E , ,880
S , ,762
W , , ,105
Total Load
469,435 Btuh
6.8
N , ,050
E ,525
S , ,100
W , ,200
14.5
,210 1,210
0.3
N/A N/A N/A N/A
, ,000
,160 41,160
8.8
204, ,960
204, , ,268
63.8
SR-3
469,435

30. Annual Fuel Consumption

31. Annual Fuel Usage (E) E= UA x DDBPT x 24 AFUE x V where:
UA: heating load/ºF
DDBPT: degree days for given balance point
AFUE: annual fuel utilization efficiency
V: fuel heating value

32. Calculating UA QTotal= UA x ΔT UA= QTotal/ΔT From earlier example:
QTotal=469,435 Btuh
ΔT= 55ºF
UA=469,435/55=8,535 Btuh/ºF

33. Determine AFUE
Annual Fuel Utilization Efficiency of an electric heating system is 100%
M: p. 311, T.9.4

34. Determine Heat Content (V)
Heat content is the quantity of Btu/unit
Note: Natural Gas is sold in therms (100 cf)
M: p. 309, T.9.2

35. Annual Fuel Usage Example
What is the expected annual fuel usage for a house in Salt Lake City if its peak heating load is 35,750 Btuh?
UA=Q/ΔT
UA=35,750/55= 650 Btuh/ºF

36. Determine AFUE
Annual Fuel Utilization Efficiency of an electric heating system is 100%
M: p.311, T.9.4

37. Determine Heat Content (V)
Heat content is the quantity of Btu/unit
M: p. 309, T.9.2

38. Annual Fuel Usage — Electricity
E= UA x DD x 24
AFUE x V
EELEC =(650)(5,983)(24)/(1.0)(3,413)
=27,347 kwh/yr
If electricity is $0.0735/kwh, then
annual cost = $2,010

39. Annual Fuel Usage — Gas E= UA x DDBPT x 24 AFUE x V
=1,111 therms/yr
If gas is $0.41/therm, then
annual cost = $456

40. Simple Payback Analysis

41. Simple Payback Heating System Cost Comparison First Electricity 6,000
($)
Electricity 6,000
Oil 8,000
Gas 8,900

42. Simple Payback Heating System Cost Comparison
First Annual Incremental Incremental Simple
Cost Fuel Cost First Cost Annual Savings Payback
($) ($/yr) ($) ($/yr) (yrs)
Electricity 6,000 2,
Oil 8,000 1,152 2,
Gas 8, ,900 1,
If money is available, select gas furnace system