Question

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.

Answer 1

The time rate of change in air density during expiration is 0.01003kg/m³-s

Step-by-step 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 =

`(\pi )/(4) d^2\\= 0.785* 0.018^2\\= 2.5434 *10^-^4m^2`

0 - p × Area for trachea =

`(d)/(dt) (pv)=v(ds)/(dt) + p(dv)/(dt)`

`-1.225*2.5434*10^-^4*0.20=6*10^-^3(ds)/(dt) +1.225(-1*10^-^4)`

`-1.225*2.5434*10^-^4*0.20=6*10^-^3(ds)/(dt) +1.225(-1*10^-^4)`

⇒
`-0.623133*10^-^4+1.225*10^-^4=6*10^-^3(ds)/(dt)`

`(ds)/(dt) = (0.6018*10^-^4)/(6*10^-^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

Answer 2

Time rate = 0.010026 kg/m³.s

Step-by-step 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