Load CalculationsHeat Balance Method - Example

Prof. Jeffrey D. SpitlerSchool of Mechanical and Aerospace Engineering, Oklahoma State University

Weather

50'-0"

10'-0

"

20'-0"

20'-0

"20

'-0"

15'-0'' 15'-0"

Office Office

OfficeConferenceRoom

Storage

Corridor

Lobby

Toilet

1 2 3

4 5

6

7

8

50'-0

"

Building Layout

Area: 2500 square feetDivided into zones for convenience:

North Zone:Office 1, Office 3, Storage, Corridor, Toilet

South Zone:Conference Room, Office 2, Lobby

Basement

Construction: North Wall

MPS 159, Fig. 5-9Layers: outside-inside:

Facing Brick 4”Sheathing; R-1Fiberglass insulation R-11Gypsum Wall Board0.5”

Construction: North Wall

Construction: W,S,E Walls

MPS 143Facing Brick 4”Cement Mortar 0.5”Concrete Block Cinder Aggregate 8”Air Gap / Furring strips 0.75”Gypsum Wall Board 0.5”

Construction: W,S,E Walls

Items 5 & 6 are two materials in parallel: 1x3 furring, reflective airspaceTreat as a composite material with same resistance and capacitance

16” o.c.

Representative area

9”

Composite Material: Equiv. k

Ffthrin.-Btu 0.38 ofty conductivi

a with material a toequivalent is this,0.75" thicknessIf

)Ffthr

Btu0.507(also

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BtuFfthr0.94

0.208ft

BtuFfthr2.77

0.792ft

111

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−−

=

−−=

−=

−−+

−−=

+=′

+′

=′ furring

furring

air

air

furringaire RA

RA

RRR

Composite material: Equiv. ρcpFor composite material to have equivalent heat storage, (ρcpV)equivalent = Σ (ρcpV)

(ρcpV) air approximately zero

( )

33

33

ftlbm72.4

0.0625ftFlbm

Btu0.39F

Btu115.0compositeEquivalent

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FBtu115.0)(

FBtu115.0ft0.0098

FlbmBtu0.39

ftlbm30)(

=

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=

−=

=

=

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=

ρ

ρ

ρ

p

compositep

pinep

c

Vc

Vc

Composite Material: Entry

But, was it important?

Note: This approach is “conservative”; particularly so with steel studs

Construction: W,S,E Walls

Construction: Roof

MPS 144, Table 5-4b(“insulated” version)Built Up Roofing 0.4”R-4 insulationConcrete Slab 2”Air Gap 4”Light Weight Aggregate Plaster 0.75”

Construction: Interior partitions

Assume insulated for sound attenuation; ignore stud.

Construction: Floor

Remember, layers are defined from “outside-to-inside”,

Or bottom to top for this case.

Construction: Slab-on-Grade Floorand Basement Walls

12” dirt partly arbitrary.

Construction: Windows

MPS 148, e=0.6 on one pane

Construction: Doors

Door Info:Area = 20 square feetLW Emissivity in and out = 0.9U- Factor = 0.39

Added as a window, with SHGC=0.0

Internal Heat Gains: Lighting

Best to use lighting plan.We assumed 1 W/ft2

Internal Heat Gains: Equip.

We assumed 1 W/ft2

Internal Heat Gains: People

Office 1: 3 peopleOffices 2,3: 2 peopleConf. rm.: 20 people

InfiltrationAssume 0.5 ACH0.1 (?)ACH basement

Ventilation SummaryChicago 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.

Room # NameFloor Area (sq. ft.)

ODA CFM/sq.. Ft.

ODA (CFM) ODA (ACH)

ODA Fraction

1 Office 400 0.2 80 1.2 18%2 Storage 300 0.1666667 50 1 24%3 Office 300 0.2 60 1.2 22%4 Conference Room 400 0.5 200 3 33%5 Office 300 0.2 60 1.2 22%6 Corridor 400 0 0 0 0%7 Lobby 250 0 0 0 0%8 Toilet 150 0 0 0 0%9 Basement 2500 0 0 0 0%

Sum 5000 450

Airflow Rates

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

Results of Interest

HeatingRoom LoadsRoom AirflowsZone Loads/ Airflows /Coil LoadsBuilding Coil Loads

CoolingRoom LoadsRoom AirflowsZone Loads/ Airflows /Coil LoadsBuilding Coil Loads

Reports

Cooling Load (w/o ventilation)

Airflow Rates Required

Room-by-Room Summary

Room # Name

Sens. Cooling Coil Load (Btu/hr)

Supply Flow Rate (CFM)

1 Office 9626 4462 Storage 4435 2063 Office 5881 2734 Conference Room 12938 6005 Office 5765 2676 Corridor 5925 2757 Lobby 12166 5648 Toilet 3408 1589 Basement 33085 1533

Sum of peaks 93229 4322

Total Building Coil Load (w/ ventilation)

Building Heating Load

Questions?

Appendix: Step-by-step procedure

Step1: Open a new file by clicking either the new file icon or menu itemStep2: Specify weather cooling or heating load calculation is desiredStep3: 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

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 toolbarStep6: enter the zone information that is supply air temperature , the interior set point temperature and the orientation with respect to North and press okStep7: 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.

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.

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

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

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