Load Calculations Heat Balance Method - Example. Prof. Jeffrey D. Spitler School of Mechanical and Aerospace Engineering, Oklahoma State University

Transcription

1 Load Calculations Heat Balance Method - Example Prof. Jeffrey D. Spitler School of Mechanical and Aerospace Engineering, Oklahoma State University

2 Weather

3 20'-0" 15'-0'' 15'-0" '-0" Office Storage Office 6 10'-0" Corridor 8 50'-0" Toilet '-0" Conference Room Office 7 Lobby 50'-0"

4 Building Layout Area: 2500 square feet Divided into zones for convenience: North Zone: Office 1, Office 3, Storage, Corridor, Toilet South Zone: Conference Room, Office 2, Lobby Basement

5 Construction: North Wall MPS 159, Fig. 5-9 Layers: outside-inside: Facing Brick 4 Sheathing; R-1 Fiberglass insulation R-11 Gypsum Wall Board0.5

6 Construction: North Wall

7 Construction: W,S,E Walls MPS 143 Facing Brick 4 Cement Mortar 0.5 Concrete Block Cinder Aggregate 8 Air Gap / Furring strips 0.75 Gypsum Wall Board 0.5

8 Construction: W,S,E Walls Items 5 & 6 are two materials in parallel: 1x3 furring, reflective airspace Treat as a composite material with same resistance and capacitance 16 o.c. Representative area 9

9 Composite Material: Equiv. k 1 R 1 R 1 R air furring = + = + e air furring Rair R furring A A = 0.792ft 2 hr ft F 2.77 Btu ft 2 hr ft F 0.94 Btu 2 = Btu hr F (also = hr Btu 2 ft ) F If thickness = 0.75", this is equivalent Btu -in. conductivity of hr ft F to a material with a

10 Composite material: Equiv. ρc p For composite material to have equivalent heat storage, (ρc p V) equivalent = Σ (ρc p V) (ρc p V) air approximately zero 3 ( ft ) lbm Btu Btu ( ρc pv ) pine = = ft lbm F F Btu ( ρc pv ) composite = F Btu Leaving c p = 0.39, lbm F Btu F lbm Equivalent composite ρ = = Btu ft ft lbm F

11 Composite Material: Entry But, was it important? Note: This approach is conservative ; particularly so with steel studs

12 Construction: W,S,E Walls

13 Construction: Roof MPS 144, Table 5-4b ( insulated version) Built Up Roofing 0.4 R-4 insulation Concrete Slab 2 Air Gap 4 Light Weight Aggregate Plaster 0.75

14 Construction: Interior partitions Assume insulated for sound attenuation; ignore stud.

15 Construction: Floor Remember, layers are defined from outside-to-inside, Or bottom to top for this case.

16 Construction: Slab-on-Grade Floor and Basement Walls 12 dirt partly arbitrary.

17 Construction: Windows MPS 148, e=0.6 on one pane

18 Construction: Doors Door Info: Area = 20 square feet LW Emissivity in and out = 0.9 U- Factor = 0.39 Added as a window, with SHGC=0.0

19 Internal Heat Gains: Lighting Best to use lighting plan. We assumed 1 W/ft 2

20 Internal Heat Gains: Equip. We assumed 1 W/ft 2

21 Internal Heat Gains: People Office 1: 3 people Offices 2,3: 2 people Conf. rm.: 20 people

22 Infiltration Assume 0.5 ACH 0.1 (?)ACH basement

23 Ventilation Summary Chicago ordinances specify required ventilation; outdoor air (ODA) may be 1/3 of the required ventilation rate. This requirement may be met by bringing in 450 CFM ODA. The ODA (ACH) values given below were entered as ventilation heat gains for each of the corresponding rooms. ODA Floor Area CFM/sq.. ODA ODA Room # Name (sq. ft.) Ft. (CFM) ODA (ACH) Fraction 1 Office % 2 Storage % 3 Office % 4 Conference Room % 5 Office % 6 Corridor % 7 Lobby % 8 Toilet % 9 Basement % Sum

24 Airflow Rates Airflow rates are calculated to meet sensible loads in space, given hot deck / cold deck temperature.

25 Results of Interest Heating Room Loads Room Airflows Zone Loads/ Airflows /Coil Loads Building Coil Loads Cooling Room Loads Room Airflows Zone Loads/ Airflows /Coil Loads Building Coil Loads

26 Reports

27 Cooling Load (w/o ventilation)

28 Airflow Rates Required

29 Room-by-Room Summary Room # Sens. Cooling Coil Load (Btu/hr) Supply Flow Rate (CFM) Name 1 Office Storage Office Conference Room Office Corridor Lobby Toilet Basement Sum of peaks

30 Total Building Coil Load (w/ ventilation)

31 Building Heating Load

32 Questions?

33 Appendix: Step-by-step procedure Step1: Open a new file by clicking either the new file icon or menu item Step2: Specify weather cooling or heating load calculation is desired Step3: Click the location button and specify the location of the building Step4: specify the outside air temperature by clicking the special outside conditions button, then press ok

34 Appendix: Step-by-step procedure Step5: on the tree structure select the building and under the operations menu click add or click the add icon on the toolbar Step6: enter the zone information that is supply air temperature, the interior set point temperature and the orientation with respect to North and press ok Step7: now select the zone and do the add procedure; this time the software adds a room. Enter the room height. Step8: Select the room and once again doing the add procedure we get the option of adding different heat gain types such as people, equipment, infiltration, lights, the wall elements etc.

35 Appendix: Step-by-step procedure Step9: enter the wall layer and window type information in the wall element. Step10: subsequently enter the light, infiltration, equipment, etc. information. Step11: when all the information is entered under the calculate menu item, execute for building or execute for room, whichever is desirable. Step12: to view the results under the calculate menu item, click view output for building.

36 Importing room-by-room results into Excel (1)

37 Importing room-by-room results into Excel (2)

38 Importing room-by-room results into Excel (3)