Why we need to calculate heating load ?

A) To find how much energy we are going to use?

B) To size the HVAC equipment?

C) To estimate the cost of the building?

D) A and C

E) B and C

F) A, B and C

Objectives

• Finish the heating Load Calculation

• Learn and Practice the calculation of cooling load of buildings

Heat transfer in the building Not only conduction and convection !

Infiltration

• Air transport Sensible energy

Previously defined

• Q = m × cp × ΔT [BTU/hr, W]

• ΔT= T indoor – T outdoor

• or Q = 1.1 BTU/(hr CFM °F) × V × ΔT [BTU/hr]

Latent Infiltration and Ventilation

• Can either track enthalpy and temperature and separate latent and sensible later:• Q total = m × Δh [BTU/hr, W]

• Q latent = Q total - Q sensible = m × Δh - m × cp × ΔT

• Or, track humidity ratio:• Q latent = m × Δw × hfg

Ventilation Example

• Supply 500 CFM of outside air to our classroom• Outside 90 °F 61% RH

• Inside 75 °F 40% RH

• What is the latent load from ventilation?• Q latent = m × hfg × Δw

• Q = ρ × V × hfg × Δw

• Q = 0.076 lbair/ft3 × 500 ft3/min × 1076 BTU/lb × (0.01867 lbH2O/lbair - .00759 lbH2O/lbair) × 60 min/hr

• Q = 26.3 kBTU/hr

What is the difference between ventilation and infiltration?

A) Ventilation refers to the total amount of air entering a space, and infiltration refers only to air that unintentionally enters.

B) Ventilation is intended air entry into a space. Infiltration is unintended air entry.

C) Infiltration is uncontrolled ventilation.

Ground Contact

• Receives less attention:• 3-D conduction problem• Ground temperature is often much closer to indoor air

temperature

• Use F- value for slab floor [BTU/(hr °F ft)] • Note different units from U-value• Multiply by slab edge length• Add to ΣUA• Still need to include basement wall area• Tao and Janis Tables 2.10 and 2.11

More details in ASHRAE handbook -Chapter 29

Ground Contact• 3-D conduction problem

• Ground temperature is often much closer to indoor air temperature

• Use F- value for slab floorMultiply by slab edge length

and Add to ΣUA

Where do you get information about amount of ventilation required?

• ASHRAE Standard 62• Table 2

• Tao and Janis Table 2.9A

Weather Data

• Table 2-2A (Tao and Janis) or• Chapter 28 of ASHRAE Fundamentals

• For heating use the 99% design DB value• 99% of hours during the winter it will be warmer

than this Design Temperature• Elevation, latitude, longitude

• For cooling use the 1% DB and

coincident WB for load calculations

• 1% of hours during the summer will be warmer than this Design Temperature

• Use the 1% design WB for specification of equipment

Weather Data

Cooling Load Calculation

Solar Gain• Affects conductive heat gains because outside surfaces

get hot• Use Q = U·A·ΔT

Replace ΔT with TETD – total equivalent temperature differential

Q = U·A· TETD

• Tables 2-12 – 2-14 in Tao and Janis

Replace ΔT with CLTD (Tables 1 and 2 Chapter 29 of ASHRAE Fundamentals)

Solar Gain

TETD depends on:

- orientation,

- time of day,

- wall properties - surface color- thermal capacity

Glazing

• Q = U·A·ΔT+A×SC×SHGF• Calculate conduction normally Q = U·A·ΔT

• Use U-values from NFRC National Fenestration Rating Council

• ALREADY INCLUDES AIRFILMS• http://cpd.nfrc.org/pubsearch/psMain.asp

• Use the U-value for the actual window that you are going to use

• Only use default values if absolutely necessary• Tao and Janis - no data • Tables 4 and 15, Chapter 31 ASHRAE Fundamentals

Shading Coefficient - SC

• Ratio of how much sunlight passes through relative to a clean 1/8” thick piece of glass

• Depends on• Window coatings• Actually a spectral property• Frame shading, dirt, etc.• Use the SHGC value from NFRC for a particular window

SC=SHGC/0.87• Lower it further for blinds, awnings, shading, dirt

•http://cpd.nfrc.org/pubsearch/psMain.asp

More about Windows

• Spectral coatings (low-e)• Allows visible energy to pass, but limits infrared

radiation• Particularly short wave

• Tints

• Polyester films

• Gas fills

• All improve (lower) the U-value

Low- coatings

Internal gains

• What contributes to internal gains?

• How much?

• What about latent internal gains?

Internal gains

• ASHRAE Fundamentals ch. 29 or handouts• Table 1 – people

• Table 2 – lighting, Table 3 – motors

• Table 5 – cooking appliances

• Table 6 -10 Medical, laboratory, office

• Tao and Janis - People only - Table 2.17

The latent load of a building is needed to calculate __________.

A) Heating only

B) Cooling only

C) Heating and cooling

Summary:Heating and cooling loads

• Heating - Everything gets converted to a UA, UF, mcp • Sum and multiply it by the design temperature difference

• Cooling loads have additional components• Internal gains

• Solar gain

• Increased gain through opaque surfaces

• Also need to calculate latent cooling load

Heating and Cooling Load Procedures

• Handout• Calculate heating load• Calculate cooling load• Need to also calculate latent cooling load

Conclusions

• Conduction and convection principles can be used to calculate heat loss for individual components

• Air transport principles used to account for infiltration and ventilation

• Radiation for solar gain and increased conduction

• Include sensible and internal gains

Reading Assignment

• Readings: Tao and Janis

Chapter 2

Example problem• Calculate the cooling load for the building in Pittsburgh PA with the geometry shown on

figure. On east north and west sides are buildings which create shade on the whole wall.

• Windows: Horizontal slider, Manufacturer:  American Window Alliance, Inc, CDP number AMW-K-3-00028 ttp://cpd.nfrc.org/pubsearch/psMain.asp• Walls: 4” face brick + 2” insulation + 4” concrete block, Uvalue = 0.1, Dark color• Roof: 2” internal insulation + 4” concrete , Uvalue = 0.120 , Dark color• Below the building is basement wit temperature of 75 F.

• Internal design parameters:• air temperature 75 F• Relative humidity 50%

• Find the amount of fresh air that needs to be supplied by ventilation system.

Example problem• Internal loads:

• 10 occupants, who are there from 8:00 A.M. to 5:00 P.M.doing moderately active office work

• 1 W/ft2 heat gain from computers and other office equipment from 8:00 A.M. to 5:00 P.M.

• 0.2 W/ft2 heat gain from computers and other office equipment from 5:00 P.M. to 8:00 A.M.

• 1.5 W/ft2 heat gain from suspended fluorescent lights from 8:00 A.M. to 5:00 P.M.

• 0.3 W/ft2 heat gain from suspended fluorescent lights from 5:00 P.M. to 8:00 A.M.

• Infiltration:• 0.5 ACH per hour

Example solution

• SOLUTION steps (see handouts):

• 1. Calculate cooling load from conduction through opaque surfaces using TETD.

• 2. Calculate conduction and solar transmission through windows.

• 3. Add sensible internal gains and infiltration. • 4. The result is your raw sensible cooling load. • 5. Calculate latent internal gains. • 6. Calculate latent gains due to infiltration. • 7. The sum of 5 and 6 is your raw latent cooling load.

Example solution

• SOLUTION:

• For which hour to do the calculation ?

• With computer calculation for all and select the largest.

Example solutionFor which hour to do the calculation when you do manual calculation?

• Identify the major single contributor to the cooling load and do the calculation for the hour when the maximum cooling load for this contributor appear.

• For example problem major heat gains are through the roof or solar through windows!

Roof: maximum TETD=61F at 6 pm (Table 2.12)South windows: max. SHGF=109 Btu/hft2 at 12 am (July 21st Table 2.15 A)

If you are not sure, do the calculation for both hours: at 6 pm

Roof gains = A x U x TETD = 900 ft2 x 0.12 Btu/hFft2 x 61 F = 6.6 kBtu/hWindow solar gains = A x SC x SHGF =80 ft2 x 0.71 x 10 Btu/hft2 = 0.6 kBtu/h total = 7.2 kBtu/h

at 12 am Roof gains = A x U x TETD = 900 ft2 x 0.12 Btu/hFft2 x 30 F = 3.2 kBtu/hWindow solar gains = A x SC x SHGF =80 ft2 x 0.71 x 109 Btu/hft2 = 6.2 kBtu/h total= 9.4 kBtu/h

For the example critical hour is July 12 AM.

Solution

• On the board

Example 2 How to calculate Cooling Load for

HVAC design

• If the room with no outdoor influence has 4 lighting fixtures with 100 W each and 10 students,

what is the needed relative humidity and temperature of supply air if only required amount of fresh air is supplied

and room temperature is 75 F and RH 50%?