1) Why do we calculate heating and cooling loads?

A) To estimate amount of energy used for heating and cooling by a building

B) To size heating and cooling equipment for a building

C) Because Dr. Siegel tells us to

A note about units

• Quiz #6 question asked for the humidity ratio• No one included units• Every quantity that has dimensions needs to have

those units included• Failure to included units will be counted as a

wrong answer in the future

Objectives

• Review 1-D conduction

• Use knowledge of heat transfer to calculate heating and cooling loads• Conduction• Ventilation• Ground contact• Solar gains• Internal gains

1-D Conduction

90 °F

70 °F TUATl

kAq

q heat transfer rate [W, BTU/hr]q heat flux [W/m2, BTU/(hr ft2)]

TR

Aq

k conductivity [W/(m °C), BTU/(hr ft °F)]l length [m, cm, ft, in]ΔT temperature difference [°C, K, R, °F]A surface area [m2, ft2]

lk

A

U = k/l = 1/R U-Value [W/(m2 °C), BTU/(hr ft2 °F)]R = l/k = 1/U R-Value [m2 °C/W, hr ft2 °F/BTU]

Material k Values

Material k W/(m K)1

Steel 64 - 41

Soil 0.52

Wood 0.16 - 0.12

Fiberglass 0.046 - 0.035

Polystyrene 0.029

1At 300 KIncropera and DeWitt (2002) Appendix A

1-D Conduction

q heat transfer rate [W]

TR

Aq

RU ,

1

k conductivity [W/(m °C)]l length [m]

90 °F

70 °F

lk

AU = k/l

ΔT temperature difference [°C]A surface area [m2]

U U-Value [W/(m2 °C)]

q = UAΔT

90 °F

70 °F

l1k1 k2

l2• R = l/k

• q = (A/Rtotal)ΔT

• Add resistances in series• Add U-values in parallel

2

2

1

1

2121

111

k

l

k

lR

UUURRR

total

totaltotal

R1/A R2/A

Tout TinTmid

2) To find a total U value for a wall, which of the following formulas is not

acceptable?A) U total = 1/R total

B) U total = U1+U2+U3

C) U total = 1/R1+1/R2+1/R3

D) U total = 1/(1/U2+1/U2+1/U3)

E) B and C

Tout

Tin

R1/A R2/ARo/A

Tout

Ri/A

Tin

•Air film (Table 25-1)•Surface conductance

•Convection heat transfer coefficient•Use ε = 0.9

•Direction/orientation•Air speed

•Air space (Table 25-3)•Orientation•Thickness

The Drape Defense

• http://utwired.engr.utexas.edu/conservationmyths/

• Do drapes limit heat loss?

• Do drapes limit heat gain?

l1k1, A1 k2, A2

l2

l3

k3, A3

A2 = A1

(l1/k1)/A1

R1/A1

ToutTin

(l2/k2)/A2

R2/A2

(l3/k3)/A3

R3/A3

1. Add resistances for series

2. Add U-Values for parallel

R1/A1

ToutTin

R2/A2

R3/A3

1. R1/A1 + R2/A2 = (R1 + R2) /A1 = R12 /A1=1/(U12A1)

2. R3 /A3=1/(U3A3)

3. U3A3 + U12A1

4. q = (U3A3 + U12A1)ΔT

A1=A2

U1A1

U2(A2+A4)

U3A3

U5A5

Relationship between temperature and heat loss

3) Which of the following statements about a material is true?

A) A high U-value is a good insulator, and a high R-value is a good conductor.

B) A high U-value is a good conductor, and a high R-value is a good insulator.

C) A high U-value is a good insulator, and a high R-value is a good insulator.

D) A high U-value is a good conductor, and a high R-value is a good conductor.

Example

• Consider a 1 ft × 1 ft × 1 ft box

• Two of the sides are 1” thick extruded expanded polystyrene foam

• The other four sides are 1” thick plywood

• The inside of the box needs to be maintained at 40 °F

• The air around the box is still and at 80 °F

• How much cooling do you need?

4)What is the R-value of 1” of plywood?

A. 0.62 BTU/(hr∙°F∙ft2)

B. 1.24 BTU/(hr∙°F∙ft2)

C. 0.81 hr∙°F∙ft2/BTU

D. 0.62 hr∙°F∙ft2/BTU

E. 1.24 hr∙°F∙ft2/BTU

5)What is the U-value of the plywood walls?

A. 0.81 BTU/(hr∙°F∙ft2)

B. 0.38 BTU/(hr∙°F∙ft2)

C. 1.24 BTU/(hr∙°F∙ft2)

D. 0.48 BTU/(hr∙°F∙ft2)

The Moral of the Story

1. Calculate R-values for each series path

2. Convert them to U-values

3. Find the appropriate area for each U-value

4. Multiply U-valuei by Areai

5. Sum UAi

6. Calculate q = UAtotalΔT

Infiltration (Convection)

• Air carries sensible energy• q = M × C × ΔT [BTU/hr, W]

• M mass flow rate = ρ × Q [lb/hr, kg/s]• ρ air density (0.076 lb/ft3, 1.2 kg/m3 @ STP)

• Q volumetric flow rate [CFM, m3/s]

• C specific heat of air• 0.24 BTU/(lb °F), 1007 kJ/(kg K) @ STP

• For similar indoor and outdoor conditions• ρ and C are often combined

• q = 1.08 BTU min/(ft3 °F hr ) × Q × ΔT

Latent Infiltration and Ventilation

• Can either track enthalpy and temperature and separate latent and sensible later• q = M × ΔH [BTU/hr, W]

• Or, track humidity ratio• q = M × hfg × ΔW

• hfg = ~1076 BTU/lb, 2.5 kJ/kg

• M = ρ × Q [lb/hr, kg/s]

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 = M × hfg × ΔW

• q = ρ × Q × 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

6) 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) The terms can be used interchangeably.

Where do you get information about amount of ventilation required?

• ASHRAE Standard 62• Table 2• Available on website from library• Hotly debated – many addenda and changes

Ground Contact

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

temperature

• Use F- value (from simulations) [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

• WA State Energy Code heat loss tables

Weather Data

• Chapter 27 of ASHRAE Fundamentals For heating use the 99% DB value• 99% of hours during the winter it will be warmer

than this Design Temperature• Elevation, latitude, longitude

• Heating dry-bulb– 99.6% and 99% values

• 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• Facing page

• 0.4%, 1%, 2% cooling DB and MWB

• 0.4%, 1%, 2% cooling WB and MDB

Solar Gain

• Increased conduction because outside surfaces got hot

• Use q = UAΔT1. Replace ΔT with TETD

• Tables 2-11 – 2-13 in Tao and Janis (2001)• 4 pm for a dark colored surface

2. Replace ΔT with CLTD (Tables 1 and 2 Chapter 28 of ASHRAE Fundamentals)

3. Sol-air temperature• Table 29-15 (example)

Glazing

• q = UAΔT+A×SC×SHGF• Calculate conduction normally q = UAΔT

• Use U-values from NFRC Certified Products Directory• ALREADY INCLUDES AIRFILMS

• http://www.nfrc.org/nfrcpd.html

• Use the U-value for the actual window that you are going to use• Only use default values if absolutely necessary (Tables 4

and 15, Chapter 30 ASHRAE Fundamentals)

Solar Gain Through Windows

• Add to conduction A× SHGF × SC• SHGF = solar heat gain factor

• Measure of how much energy comes through an average “perfect” window

• Depends on– Latitude

– Orientation

– Time of Day

– Time of Year

• Tabulated in ASHRAE Fundamentals 1997 Chapter 29 Table 15

• Tao and Janis Table 2-15 for 40° latitude (July 21 @ 8 am)

Shading Coefficient

• 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• Lower it further for dirt, blinds, awnings, shading

•http://www.nfrc.org/nfrcpd.html

More about Windows

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

radiation• Particularly short wave• Can see it with a match/lighter in older windows

• 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?

Conclusions

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

• Convection principles used to account for infiltration and ventilation

• Radiation for solar gain and increased conduction

• Include sensible and internal gains