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