Some speech aerodyamics basics
1. relationship between volume and pressure
2. relationship between pressure and flow
3. relationship between particle velocity and volume
velocity
4. Reynolds Number – types of flow
Volume – three dimensional space occupied by any
solid liquid or gas.
Lung volume
4500 – 2000 cm3
breathing = change volume by 500-1000 cm3
Vocal tract volume (glottis to constriction)
/b/ ~ 140 cm3
/d/ ~ 85 cm3
/g/ ~ 40 cm3
Pressure – force per unit of area
measured in cm H2O
i.e. the height of a column of water can can be supported
by the pressure.
Atmospheric pressure – pressure at which atmosphere
presses on us.
1030 cm H2O at sea level
850 cm H2O in Denver (elevation 5200 ft)
765 cm H2O in Mexico City (elevation 8000 ft)
Pressure – force per unit of area
Relative air pressure is the difference between pressure
in an enclosed cavity and atmospheric pressure.
measure intraoral pressure with a tube in the mouth
leading to a calibrated pressure transducer.
1. The relationship between volume and pressure.
Boyle's Law. “The pressure of a given quantity of gas
varies inversely with its volume.”
As volume decreases, pressure increases
- e.g. exhaling
As volume increases, pressure decreases
- e.g. inhaling
p2 = (V1 x p1)/V2
V1 = volume at time 1
V2 = volume at time 2
p1 = pressure at time 1
p2 = pressure at time 2
1. The relationship between volume and pressure.
Boyle's Law.
p2 = (V1 x p1)/V2
Example 1: pressure build up during /p/.
p1 = 1030 cm H2O (atmospheric pressure)
V1 = 3140 cm3 (3000 – lung volume, 140 mouth volume)
V2 = 3040 cm3 (reduced lung volume during exhale)
p2 = (3140 x 1030) / 3040 = 1064 cm H2O
The relative pressure increase is 34 cm H2O
1. The relationship between volume and pressure.
Boyle's Law.
p2 = (V1 x p1)/V2
Example 2: pressure build up during /k'/.
p1 = 1030 cm H2O (atmospheric pressure)
V1 = 40 cm3 (mouth volume at the start of /k'/)
V2 = 39 cm3 (mouth volume after larynx is raised in /k'/)
p2 = (40 x 1030) / 39 = 1056 cm H2O
The relative pressure increase is 26 cm H2O
2. The relationship between presssure and flow
particle velocity = the rate of change of position in
a specified direction.
no airflow when pressure inside
equals pressure outside
pin = pout
pin
pout
2. The relationship between presssure and flow
particle velocity = the rate of change of position in
a specified direction.
outgoing airflow when
pin > pout
pin
pout
2. The relationship between presssure and flow
There are some air particles moving from inside
the container to out.
How fast are these air particles moving?
u = 1303 x √(pin – pout)
pin
, in cm/second
pout
2. The relationship between presssure and flow
particle velocity = the rate of change of position in a
specified direction.
Example 1: particle velocity in /p/
pin = 1064 cm H2O
pout = 1030 cm H2O
u = 1303 x √(pin – pout) = 1303 x √(34) = 7598 cm/s
pin
u
pout
2. The relationship between presssure and flow
Example 2: particle velocity through the glottis in
voicing
pin = 1042 cm H2O
pout = 1038 cm H2O
u = 1303 x √(pin – pout) = 1303 x √(4) = 2606 cm/s
pin
u
pout
3. Particle velocity and volume velocity
volume velocity = the volume of fluid/gas passing
through a system per unit of time.
pin
u
pout
each particle is traveling at a speed of u cm/s
How many particles travel through the opening per
second? Depends on the size of the opening.
3. Particle velocity and volume velocity
volume velocity = the volume of fluid/gas passing
through a system per unit of time. (This quantity is
important in speech research partly because it is
easy to measure.)
Vv = a x ū
ū = average particle velocity (cm/s)
a = cross-sectional area of the channel (cm2)
Vv = volume velocity (cm3/s)
3. Particle velocity and volume velocity
volume velocity
Vv = a x ū
ū = average particle velocity (cm/s)
a = cross-sectional area of the channel (cm2)
Vv = volume velocity (cm3/s)
Example 1. Volume velocity in modal voicing
p = 3 cm H2O, a = 0.15 cm2
u = 1303 √p = 1303 x 1.73 = 2257 cm/s
Vv = 0.15 * 2257 = 339 cm3/s
3. Particle velocity and volume velocity
volume velocity
Vv = a x ū
Example 2. Volume velocity in breathy sounds
----- same as example 1 -----p = 3 cm H2O
u = 1303 √p = 1303 x 1.73 = 2257 cm/s
------ unique to example 2 ----a = 0.3 cm2
Vv = 0.3* 2257 = 677 cm3/s
(in modal voicing it was 339 cm3/s)
3. Particle velocity and volume velocity
For a give pressure, the volume velocity (how
much air is escaping the vocal tract) is proportional
to the size of the constriction.
modal voice – less flow than breathy voice
voiced sounds – less flow than voiceless sounds
4. Laminar and turbulent flow
Reynold’s number – ratio of inertial forces (fluid
density, velocity, channel characteristics) to viscosity.
Re = (d x ū)/0.19
critical value = 1800
Re > 1800 will result in turbulent flow
d = √(4/pi x a) , diameter of tube in cm
a
, cross-sectional area of tube in cm2
4. Laminar and turbulent flow
Reynold’s number – Re = (d x ū)/0.19
example 1. A voiceless approximant [ ! ].
p = 0.8 cm H2O
a = 0.4 cm2, d = 0.71 cm
ū = 1303 x √p = 1165 cm/s
Re = 4377 which is greater than 1800, therefore
we expect this sound to have turbulent flow.
4. Laminar and turbulent flow
example 2. A voiced approximant [ l ].
The oral air pressure is less than in the voiceless
approximant because most of the pressure is trapped
behind the glottis.
p = 0.13 cm H2O
a = 0.4 cm2, d = 0.71 cm
ū = 1303 x √p = 470 cm/s
Re = 1764 which is less than 1800, therefore
we expect this sound to have laminar flow.