epgep heating infiltration 2013

INFILTRATION

Tams Tirpk

Building Service Engineering Department

Room 131, 1st floor, building D

[email protected]

Lecture structure

Define natural ventilation and

infiltration

Calculation of infiltration Calculation of infiltration

Methods

Examples

(Measurement of infiltration)

Ventilation and Infiltration

Air exchange of outdoor air with the

air already in a building

Ventilation

Infiltration

Why are they important in HVAC?

provide comfortable and healthy

indoor environment

Ventilation

Intentional

3 types

Natural ventilation

Mechanical ventilationMechanical ventilation

Hybrid ventilation

Natural ventilation

Windows, doors, skylights, roof ventilatiors, stacks connecting to registers, specially designed inlet and outlet openings

Infiltration

Unintentional

Driven by natural and/or artificial

pressure differences

Mostly present in

tall, leaky or partially pressurized

buildings and lobby areas

residential buildings

Window and door frames, cracks etc.

Location of Common Air

Leakage Paths

Source: www.njenergyhomes.com

Driving mechanisms for natural

ventilation and infiltration

Pressure differences caused by

Stack effect: air density differences

due to temperature differences

between indoor air and outdoor airbetween indoor air and outdoor air

Wind

Depends on the characteristic of the

openings

Stack pressure

Hydrostatic pressure caused by the

weight of a column of air located inside

or outside a building

For a single column of air the stack For a single column of air the stack

pressure can be calculated as:

Where:

ps = stack pressure, Pa pr = stack pressure at reference height, Pa

g = gravitational constant, 9.81 m/s2 = indoor or outdoor air density, kg/m3

r = indoor or outdoor air density at reference temperature Tr, kg/m3

H = height above reference plane, m T = temperature, K

H

T

TgpHgpp r

rrrs ==

Stack pressure

Positive when the

building is

pressurized relative

to outdoorsto outdoors

Neutral Pressure

Level (NPL) is not

necessarily located

at the mid-height of

the building.

Wind pressure

The wind pressure is given

by the Bernoulli equation:

Where:

pw = wind surface pressure relative to outdoor static pressure in undisturbed flow, Pa

= outside air density, kg/m3 (about 1.2)

cp = wind surface pressure coefficient, dimensionless

U = wind speed, m/s

Combined driving forces

The pressure differences due to wind pressure, stack pressure and mechanical system are considered in considered in

combination by addingthem together and thendetermining the airflow rate through each opening due to this total pressure difference.

Calculation of natural


1. Airflow through large intentional openings:

(ASHRAE Fundamentals 2005, Ch 27.10)

/2 pACQ = /2 pACQD

=

Where:

Q = airflow rate, m3/s

CD = discharge coefficient for opening, CD = 0.61 for sharp-edge orifice

A = cross-sectional area of opening, m2

p = pressure difference across opening, Pa

= air density, kg/m3



2. Airflow through small openings:

p

bhQ =

3

pL

Q =12

Where:

Q = flow rate, m3/s

b = length of crack, m

h = height of crack, m

L = depth of crack in flow direction, m


= absolute viscosity of air, Pa s



3. Power-law equation crack flow equation:

(ASHRAE Fundamentals 2005, Ch 27.12)

npkLQ =Where:

Q = flow rate, m3/s

k = flow coefficient, m3 s-1 m-1 Pa-n

L = length of crack, m


n = flow exponent, n=0.51, typically between 0.6 and 0.7, the

most often used value n=2/3=0.67

npkLQ =

Simplified models of residential


1. Single-zone models:

basic model: The basic model uses the

effective air leakage area AL

at 4 Pa, which

can be obtained from a whole-building can be obtained from a whole-building

pressurization test.

enhanced model: The enhanced model uses

pressurization test results to characterize

house air leakage through the leakage

coefficient c and the pressure exponent n.

Basic Model

The basic model uses the effective air leakage

area ALat 4 Pa, which can be obtained from a

whole-building pressurization test.

2

1000UCtC

AQ

ws

L+=

Where:

Q = airflow rate, m3/s

AL = effective air leakage area, cm2

Cs = stack coefficient, (L/s)2/(cm4K)

t = average indoor-outdoor temperature difference for time interval of calculation, K

Cw = wind coefficient, (L/s)2/[cm4 (m/s)2]

U = average wind speed measured at local weather station for

time interval of calculation, m/s

Problem for the basic model

Estimate the infiltration for:

two-story house

effective leakage area AL=500 cm2

and a volume of 340 m3L

and a volume of 340 m3

predominant wind perpendicular to the street (shelter class 3), v=6.7 m/s

outdoor temperature te=-19C

indoor temperature ti=20C

Problem for the basic model

2

1000UCtC

AQ

ws

L+=

27.6000231.0)39000290.0(500

+27.6000231.0)39000290.0(

1000

500+

h

m

s

m33

2650736.0 =

ASHRAE Fundamentals 2005

Air exchange rate:

hm

hm

V

Qn

178.0

340

265

3

3

===

Simplified models of residential


2. Infiltration air flow rate calculation according

to EN 12831:2003 (Heating systems in

buildings Method for calculation of the

design heat load)design heat load)

iiiienVV =

50inf,2&

Where:

n50 = air exchange rate per hour (1/h), resulting from a pressure difference of 50 Pa

between the inside and the outside of the building, including the effects of air inlets;

ei = shielding coefficient

i = height correction factor, which takes into account the increase in wind velocity

with the height of the space from ground level.

EN 12831:2003

D 5.2. Air exchange rate n50

Default values for the air exchange rate, n50

for the whole building resulting from pressure difference of 50 Pa between inside and outside:

n50 (h-1)

Degree of air-tightness of the building envelope (quality of

Construction

Degree of air-tightness of the building envelope (quality of

the window seal)

high (high quality

sealed windows

and doors)

medium (double

glaze windows,

normal seal)

low (single glaze

windows, no

sealant)

single family

dwellings< 4 4-10 > 10

other dwellings or

buildings< 2 2-5 > 5

The whole building air exchange rates may be expressed for other pressure

differences than 50 Pa, but these results should be adapted to suit the equation above.

EN 12831:2003

D 5.3 Shielding coefficient e Default values for the shielding coefficient, e, are:

Shielding Class

e

Heated space

without

exposed

openings

Heated space

with one

exposed

opening

Heated space with

more than one

exposed

openingopenings opening opening

No shielding (building in

windy areas, high rise

buildings in city centres)

0 0.03 0.05

Moderate shielding (buildings

in the country with trees

or other buildings around

them, suburbs)

0 0.02 0.03

Heavy shielding (average

height buildings in city

centres, buildings in

forests)

0 0.01 0.02

EN 12831:2003

D 5.4 Height correction factor -

Default values for the height correction

factor, , are:

Height of heated space above ground-Height of heated space above ground-

level (centre of room height to

ground level)

0 10 m 1.0

> 10 30 m 1.2

> 30 m 1.5

Problem for EN 12831:2003

Infiltration air flow rate calculation:

Estimate the infiltration for a flat on the

second floor of a two-story house situated

in the centre of Copenhagen. Two windows

(which are double glazed windows and

have PVC spacers meaning some

tightness) face a walking street. The flat

has a volume of 100 m3.

Problem for EN 12831:2003

iiiienVV = 50inf, 2

&

hmhm 313 12102.031002 =

Measurement of infiltration

airflows

Measurement of infiltration airflows

can be done with

TRACER GAS METHODS:

Constant concentration

Constant emission method

Decay method

or Passive tracer gas method

References

ASHRAE Fundamentals 2001, Ch 26

ASHRAE Fundamentals 2005, Ch 27 ASHRAE Fundamentals 2005, Ch 27

European Standard EN 12831:2003

epgep heating infiltration 2013

Documents

wind pressure

stack pressure positive

outdoor static pressure

outdoor air density

air density differences

total pressure difference

wind surface pressure

outside air density