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BME 365R Quantitative Physiology
Homework 2 Solution
Due September 20, 2011; 6:00PM
1.
(20) Airway resistance is important in respiratory medicine.
The following questions are
related to the paper entitled “Extrathoracic airway resistance in man.”
Download the
paper from Blackboard to answer the following questions:
A.
(6) Consider Figure 2 showing plots of air flow vs. pressure difference in the upper
airway.
Identify which plotted quantities correspond to an effort and which correspond to
a flow.
From Figure 2A estimate the upper airway resistance during resting ventilation
when the pressure difference (
Δ
p
) is 1 cmH
2
O.
During hyperventilation does the upper
airway resistance (at a pressure difference
p
= 1 cmH
2
O) increase or decrease relative to
resting ventilation?
Estimate the upper airway resistance at a pressure difference
p
= 1
cmH
2
O during hyperventilation.
x axis
p is effort, y axis exp flow and insp flow is flow
since the plot is not linear, we need to draw a line on the partial linear part on
p
= 1
cmH
2
O point(flow is approximately 0.75), the intercept of this line on y axis
approximately 0.4 is the flow when
p
= 0 cmH
2
O
Rr=
p/flow=1/(0.750.4)=2.86
cmH
2
O/(l/secs)
Similar for hyperventilation
Rh=
p/flow=1/(1.250.5)=1.33
cmH
2
O/(l/secs)
During hyperventilation the upper airway resistance (at a pressure difference
Δ
p = 1
cmH2O) decrease relative to resting ventilation.
B.
(6) Considering the upper airway as a discrete element with a flow and pressure
difference, use the values determined in Part A to compute the power in Watts input into
the upper airway for both resting ventilation and hyperventilation (take a pressure
difference of
p
= 1 cmH
2
O).
Using the computed powers into the upper airway and
considering a 1
Ω
linear resistor, determine the current into and voltage across the resistor
so that the thermal power generated by the resistor is equivalent to the power input into
the upper airway during resting ventilation and hyperventilation.
For resting ventilation
The power is
Pr=
=
p*flow=1*0.75=0.75
cm
H
2
O
*L/secs =750cm
H
2
O
*ml/secs*133Pa/1.36
cm
H
2
O*m
3
/10
6
ml=0.073Watts
I=(P/R)
1/2
=(0.073/1)
½
=0.27A
V=IR=0.27V
For hyperventilation
The power is
Pr=
=
p*flow=1*1.25=1.25
cm
H
2
O
*L/secs =1250cm
H
2
O
*ml/secs*133Pa/1.36
cm
H
2
O*m
3
/10
6
ml=0.12Watts
I=(P/R)
1/2
=(0.12/1)
½
=0.35A
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 Spring '09
 Rylander

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