calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diameter of the trachea is 18mm, the air flow velocity out of the trachea is 20cm/s and the density of air is 1.225kg/m3. Also assume that lung volume is decreasing at a rate of 100mL/s.
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Subject:
Physics -
Author:
janefitzgerald -
Created:
1 year ago
Answers 3
Answer:
The time rate of change in air density during expiration is 0.01003kg/m³-s
Explanation:
Given that,
Lung total capacity V = 6000mL = 6 × 10⁻³m³
Air density p = 1.225kg/m³
diameter of the trachea is 18mm = 0.018m
Velocity v = 20cm/s = 0.20m/s
dv /dt = -100mL/s (volume rate decrease)
= 10⁻⁴m³/s
Area for trachea =
\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2
0 - p × Area for trachea =
\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}
-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)
-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)
⇒-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}
\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s
ds/dt = 0.01003kg/m³-s
Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s
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Author:
shyannejohns
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Rate an answer:
9
Answer:
Time rate = 0.010026 kg/m³.s
Explanation:
We are given the following;
Total lung capacity; V = 6000mL = 6 x 10^(-3) m³
diameter of the trachea is 18mm = 0.018m
Air density; ρ = 1.225kg/m³
Velocity; v = 20cm/s = 0.20m/s
From the question, lung volume is decreasing at a rate of 100mL/s.
Thus, dv /dt = -100mL = -0.0001 m³/s
Area of trachea; A = πD²/4
A = (π x 0.018²)/4 = 2.5447 x 10^(-4) m²
Now, let's set up the equation.
-ρAv = (d/dt)(ρV) = V(dρ/dt) + ρ(dv/dt)
Thus,
Plugging in the relevant values to get;
-[1.225 x 2.5447 x 10^(-4) x 0.20] = 6 x 10^(-3)(dρ/dt) + (1.225 x -0.0001)
So,
-0.62345 x 10^(-4) = 6x10^(-3)(dρ/dt) - (1.225 x 10^(-4))
6x10^(-3)(dρ/dt) = (1.225 x 10^(-4)) - 0.62345 x 10^(-4)
6x10^(-3)(dρ/dt) = 0.60155 x 10^(-4)
(dρ/dt) = [0.60155 x 10^(-4)] /(6x10^(-3)) = 0.10026 x 10^(-1) = 0.010026 kg/m³.s
ρ
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Author:
jamesonybhz
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