14. air intro statics

7/29/2019 14. Air Intro Statics

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Why do we need to understand the impactof air on buildings?

Air continually bathes our buildings interior and exterior oftenmoving, creatingpressure variations across the building enclosure.

Thesepressure variations cause air leakage through the enclosure,resulting in heat losses (both sensible and latent) and in turn increasethe buildings energy requirements to maintain comfort conditions.

Air movement through the enclosure results in the transport of watervapour, and as a result, the potential for condensation and waterdamage to the envelope components increases.

The moving air introduces draughts within the interior of the buildingand in turn potentially reduces the level of human comfort

Control of air movement offers potential for natural ventilation.


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2. Dynamic Air Characteristics (aerodynamics)

The study of the properties of air in motion, or its dynamic properties

the characteristics of moving air, and especially of the interaction

between the air and solid stationary objects

also the interaction of solid objects moving through still air


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Units of Pressure

The basic SI metric unit ofpressure is the Pascal.

Force:

Newton (N) = kg.m/s2

Pressure (force /unit area):

Pascal (Pa) = N/m2 = (kg.m/s2)/m2

= kg/m.s2

Units of Pressure

When quantifying the pressure of air, it isoften convenient to express it as the heightof another fluid which it can support. Oftenmercuryor waterare used as standard

reference fluids.

P1 P2

hIn the U-tube manometer the differencebetween pressures P1 and P2 is the pressurerequired to support height hof the fluid:

P1 P2 = P = g h

where: = density of the fluid (kg/m2)g = gravitational constant (9.8 m/sec2)

h = height difference of the fluid (m)

Units of Pressure

1 standard atmosphere of air pressure is slightly over 100 kPa, whichis the pressure which will support about 10 metres of water.

Therefore:

100 kPa will support about . . . . . . . . . . . . . . . . . . . 10 m of water

10 kPa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 m of water

1 kPa or 1,000 Pa . . . . . . . . . . . . . . . . . . . . . . . . . . 100 mm of water

100 Pa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 mm of water

10 Pa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 mm of water

1 Pa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0.1 mm of water

Static Properties of Air

One of the characteristics of fluids(liquids or gasses), which distinguishthem from solids, is that their shape

and volume change significantly as aresult of the forces or pressureswhich are applied to them.


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Di

ff

erence in air densities

at room temperature, air

density is about 1.2 kg/m

3

Di

ff

erence in air densities

Because of gravitational forces, at any point in a

static fluid, the pressure is given by:

P = g h

P

The negative sign indicates that the pressure increasesas one moves down in the fluid.

The denser the fluid the higher the value of the

more rapidly the pressure changes.

where:

P = static pressure (Pa)

h = height of fluid exerting a pressure

= mass density (kg/m3)

g = gravitational constant (9.8 m/s2)

If atmospheric pressure at sea level is 101.325 kPa, what

is the pressure at 10m below sea level and 10m abovesea level?

density of water = 1,000 kg/m3

density of air = 1.18 kg/m3

Example: Fluid pressures

P+10

P10

P0

10m

10m


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P+10

P10

P0

10m

10m

At 10m above sea level:P+10 = P0 gh

= 101,325 Pa

(1.18 kg/m3) (9.81 m/s2) (+10 m)= 101,325 Pa 116 Pa

= 101,209 Pa = 101.21 kPa






P+10

P10

P0

10m

10m

At 10m belowsea level:P10 = P0 gh

= 101,325 Pa

(1,000 kg/m3) (9.81 m/s2) (10 m)= 101,325 Pa + 98,100 Pa

= 199,425 Pa = 199.43 kPa






P+10

P10

P0

10m

10m

101.21 kPa

199.43 kPa

101.325 kPa

P = + 98.1 kPa

P = 0.12 kPa

Example: Fluid pressure in container

If a cylinder with a closed end is pushed 3 metres below the water level

with closed end down, what is the pressure exerted on the bottom ofthe cylinder if the atmospheric pressure at sea level is 101.325 kPa?

density of sea water = 1,000 kg/m3


Pa

Pw

3m

P0


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If a cylinder with a closed end is pushed 3 metres below the water level

with closed end down, what is the pressure exerted on the bottom ofthe cylinder if the atmospheric pressure at sea level is 101.325 kPa?



Pa

Pw

3m

P0

At 3m below the sea level, the pressureof the air in the cylinder is:

Pa = P0 gh

= 101,325 Pa

(1.18 kg/m3)(9.81 m/s2)(3 m)

= 101,360 Pa = 101.36 kPa

The pressure difference acting on the

bottom of the cylinder is:

P = Pw Pa

= 130.755 kPa 101.360 kPa

= 29.40 kPa (inwards)


Pa

Pw

3m

P0

and the pressure of the water outside the cylinder is:

Pw = P0 gh

= 101,325 Pa (1,000 kg/m3)(9.81 m/s2)(3 m)

= 130,755 Pa = 130.76 kPa


Pa

Pw

3m

P0

P = 29.40 kPa130.76 kPa

101.325 kPa

101.36 kPa


If the same cylinder is pushed 3m below the water level with the

closed end up, what is the air pressure within the cylinder if theatmospheric pressure at sea level is 101.325 kPa?



Pw

3m

P0

Pa, bottom

Pa, top


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If the same cylinder is pushed 3m below the water level with the

closed end up, what is the air pressure within the cylinder if theatmospheric pressure at sea level is 101.325 kPa?



Pw

3m

P0

Pa, bottom

Pa, top

At 3m below the sea level, the pressure of the

water is the same as before:Pw = P0 gh

= 101,325 Pa

(1,000 kg/m3)(9.81 m/s2)(3 m)

= 130,755 Pa = 130.76 kPa


This will be the same as the air pressure just above the water surface in

the cylinder. Near the top of the cylinder, the air pressure will bereduced by the influence of the fluid height:

Pa, top = Pa, bottom gh

= 130,755 Pa (1.18 kg/m3)(9.81 m/s2)(+3 m)

= 130,720 Pa = 130.72 kPa

The pressure difference acting on thetop of the cylinder is:

P = P0 Pa, top

= 101.33 kPa 130.72 kPa

= 29.39 kPa (outwards)

Pw

3m

Pa, bottom

Pa, top

P0


130.72 kPa

130.76 kPa

101.33 kPaP = 29.39 kPa

130.76 kPa Pw

3m

Pa, bottom

Pa, top

P0


3m

P0 P

w, bottom

Pw, top

Pa, top

If the same cylinder is pulled 3m above water level with the

closed end up, what is the air pressure within the cylinder ifthe atmospheric pressure at sea level is 101.325 kPa?



At 3m above the sea level, the pressureof the air is:

Pw = P0 gh

= 101,325 Pa (1.18 kg/m3)(9.81 m/s2)(+3 m)

= 101,290 Pa = 101.29 kPa


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3m

P0 P

w, bottom

Pw, top

Pa, top

The pressure of the water at the bottom of the cylinder is equal to P0

(sea level air pressure). The pressure of the water inside at the top of thecylinder is:

The difference in pressure across the top of

the cylinder is:

101.29 kPa 71.90 kPa = 29.39 kPa

Pw = P0 gh

= 101,325 Pa (1,000 kg/m3) (9.81 m/s2) (+3 m)

= 71,895 Pa = 71.90 kPa


3m

P0 P

w, bottom

Pw, top

Pa, top

71.90 kPa

101.33 kPa

P = 29.39 kPa101.29 kPa

Example: Air pressure in balloon

An analogous situation to that of the previous example can be is

found in the case of a hot air balloon, though the difference in densityis due to temperature differences of air rather than differing fluids.

Assuming a balloon diameter of 15m, what isthe buoyancy (lift) if the air within the balloon isheated to 30C above the outside temperature

of 10C?

Assume an atmospheric pressure of 101.325 kPa.

From the table, air density is:

at 10C . . . . . . 1.25 kg/m3

at 40C . . . . . . 1.13 kg/m3

15 m

20m

20 m above the opening inside the balloon the air pressure is:


Pint = P0 intgh = 101,325 Pa (1.13 kg/m3)(9.81 m/s2)(20 m)

= 101,103 Pa

Pext = P0 extgh

= 101,325 Pa (1.25 kg/m3)(9.81 m/s2)(20 m)

= 101,080 Pa

Similarly, 20 m above the opening outsidethe balloon the air pressure is:

15 m

20m

The difference is therefore 23 Pa.


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Buoyancy force = pressure x area = (23 Pa) (7.5m)2 = 4,064 N

Since F=ma

this force would support a mass of:

m = (4,064 N)/(9.8 m/s2) = 415 kg

15 m

20m

where: p = pressure difference across enclosure

o = outside density (kg/m3) i = inside density (kg/m3)

h = distance from neutral plane (m)

g = gravitational constant (9.81 m/s2)

Stack effect

i

o

neutral

plane

h

The stack effect is the result of the buoyancy caused by thedifference in densities between interior and exterior air.

Thepressure difference across the building enclosure at adistance hfrom the neutral plane is:

p = og h i g h = g h (oi )

where:

pt = barometric pressure (Pa)Ra = gas constant for air = 287.1 J/kg

To and Ti = outside and inside temperatures (K)

o=

pt

RaTo

i=pt

RaTi

ps=ghpt

Ra

1

To

1

Ti

=0.0342hpt1

To

1

Ti

=0.0342hptT

TiTo

To avoid the need to calculate air densities at the different

temperatures, this equation can be reduced by incorporating Boyles lawequation, rearranged for density:

PV = w R T

substituting into:

Stack effect

ps = g h (oi )

ps=0.0342hptT

TiTo

This represents the pressure difference across any point on a buildingenclosure at a vertical distance h from the neutral plane, given interior

and exterior temperatures.

Note that the temperatures must be in absolute (Kelvin) degrees.

Stack effect

ps=ghptRa

1

To

1

Ti

=0.0342hpt

1

To

1

Ti

=0.0342hpt

T

TiTo


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Example: Stack effect

A 100 m high building has an average interior air temperature of 22C

and the exterior temperature is 20C.

i

o

Density of Dry Air as a Function of Temperature(at Sea Level)

1.00

1.05

1.10

1.15

1.20

1.25

1.30

1.35

1.40

1.45

1.50

1.55

1.60

- 50 -4 5 - 40 - 35 -3 0 - 25 -2 0 - 15 -1 0 - 5 0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 0 4 5 5 0 5 5 6 0 6 5 7 0

Temperature (C)

Density(kg/m3)


A 100m high building has an average interior air temperature of 22C


How does the pressure differential

between exterior and interior acrossthe enclosure vary with height

a) if openings are only near the

bottom of the building?

density of 22C air = 1.18 kg/m3


i

o

Example:Stack effect

Since openings are only at the bottom of the building, assume

the neutral plane is at ground level, then the pressuredifference across the building enclosure at any height is:

i

o

h

neutral

plane

ps = g h (oi )

Therefore, at 100 m height (top of building):

p = (1.96 kg/s2 m2) 100 m

= 196 Pa

= (9.81 m/s2) h (1.38 1.18 kg/m3)

= (1.96 kg/s2 m2) h


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i

o

h

neutral

plane

Alternate method using temperatures directly:

Ti and To must be in absolute degrees (K):

22C = 295K and 20C = 253K

ps=0.0342hptT

TiTo

Therefore:

= 196 Pa

ps =0.0342(100m)(101.32kPa)

42C

(295K)(253K)


A 100 m high building has an average interior air temperature of 22C


How does the pressure differentialbetween exterior and interior across

the enclosure vary with height



i

o

b) if openings are evenly distributedthroughout the height of the building?

Assume the neutral plane is at mid-height where the pressure differencemust be zero, increasing in both directions from that plane.


i

o

neutral

plane

h

Similarly, at ground level:

p = 98 Pa

At top of the building (+50 m from neutral plane):p = (9.81 m/s2)(+50 m)(1.38 1.18 kg/m3)

= + 98 Pa


i

o

neutral

plane

h

Alternate method:

= 98 Pa

(positive or negative depending on

direction from neutral axis)

ps=0.0342(50m)(101.32Pa)42C

(295K)(253K)


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Air pressures inside and outside a heated building with a single

opening at the bottom and no internal partitions

Air pressures inside and outside a heated building with equal

openings at top and bottom and no internal partitions

Air pressures inside and outside a heated building with each storey being

completely isolated and having equal openings top and bottom

Air pressures inside and outside a heated building with

uniform distribution of openings through the enclosure,

the floors and the walls of the elevator shaft


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14. air intro statics

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