14. Air Intro Statics
Transcript of 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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Example: Fluid pressures
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
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
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
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
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
density of air = 1.18 kg/m3
Pa
Pw
3m
P0
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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
density of air = 1.18 kg/m3
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)
Example: Fluid pressure in container
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
Example: Fluid pressure in container
Pa
Pw
3m
P0
P = 29.40 kPa130.76 kPa
101.325 kPa
101.36 kPa
Example: Fluid pressure in container
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?
density of sea water = 1,000 kg/m3
density of air = 1.18 kg/m3
Pw
3m
P0
Pa, bottom
Pa, top
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Example: Fluid pressure in container
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?
density of sea water = 1,000 kg/m3
density of air = 1.18 kg/m3
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
Example: Fluid pressure in container
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
Example: Fluid pressure in container
130.72 kPa
130.76 kPa
101.33 kPaP = 29.39 kPa
130.76 kPa Pw
3m
Pa, bottom
Pa, top
P0
Example: Fluid pressure in container
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?
density of sea water = 1,000 kg/m3
density of air = 1.18 kg/m3
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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Example: Fluid pressure in container
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
Example: Fluid pressure in container
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:
Example: Air pressure in balloon
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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Example: Air pressure in balloon
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)
Example: Stack effect
A 100m high building has an average interior air temperature of 22C
and the exterior temperature is 20C.
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
density of 20C air = 1.38 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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Example: Stack effect
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)
Example:Stack effect
A 100 m high building has an average interior air temperature of 22C
and the exterior temperature is 20C.
How does the pressure differentialbetween exterior and interior across
the enclosure vary with height
density of 22C air = 1.18 kg/m3
density of 20C air = 1.38 kg/m3
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.
Example: Stack effect
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
Example:Stack effect
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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