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Cakulate the time rate of change of air density during expiration. Assume that the lu ig 3.11) has a total volume of 6000 ml, the diameter of the trachea is 18 mm, the airflo locity out of the trachea is 20 cm/s, and the density of air is 1.225 kg/m. Also assume th ng volume is decreasing at a rate of 100 mL/s.

Cakulate the time rate of change of air density during expiration. Assume that the lu ig 3.11) has a total volume of 6000 ml, the diameter of the trachea is 18 mm, the airflo locity out of the trachea is 20 cm/s, and the density of air is 1.225 kg/m. Also assume th ng volume is decreasing at a rate of 100 mL/s.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Cakulate the time rate of change of air density during expiration Assume that the lung (Fig. 3.11) has a total volume of 6000 ml, the diameter of the trachea is 18 mm, the airflow velocity out of the trachea is 20 cm/s, and the density of air is 1.225 kg/m. Also assume that lung volume is decreasing at a rate of 100 mL/s. Hello sir, I want the same solution, but in a detailed way and mention his data, a question, and a solution in detailing mathematics without words. Solution We will start from Eq. (3.24) because we are asked for the time rate of change of density. We are asked to find the time rate of change of air density; this suggests that Example 3.5 condis tions are representing a nonsteady flow scenario. In addition, we were told what the rate of change in the lung volume is during this procedure, further supporting the use of Eq. (3.24). pdV+ (3.24 ams Assume that at the instant in time that we are measuring the system, density is uniform within the volume of interest. This allows us to remove density from within the first integral Since no information was provided about how density varies within the volume, we would be unable to solve this problem without making the assumption. = 0 amu The volume integral of the volume of interest is just the volume of interest. Therefore, Eq (3.24) can be represented as o = v2d+ (19 if we also assume that the density and velocity are not functions of the outflow area of interest (the second integral obtains a positive solution because we are representing an outflow velocity and this velocity aligns with the area vector). Using the product rule av at +p=-pA Solving this equation with the known values -vgd- dt (18 mm |125 (20)(지| ml -100- cm 1.225- 0.01 m's 2 6000 mL
Transcribed Image Text:Cakulate the time rate of change of air density during expiration Assume that the lung (Fig. 3.11) has a total volume of 6000 ml, the diameter of the trachea is 18 mm, the airflow velocity out of the trachea is 20 cm/s, and the density of air is 1.225 kg/m. Also assume that lung volume is decreasing at a rate of 100 mL/s. Hello sir, I want the same solution, but in a detailed way and mention his data, a question, and a solution in detailing mathematics without words. Solution We will start from Eq. (3.24) because we are asked for the time rate of change of density. We are asked to find the time rate of change of air density; this suggests that Example 3.5 condis tions are representing a nonsteady flow scenario. In addition, we were told what the rate of change in the lung volume is during this procedure, further supporting the use of Eq. (3.24). pdV+ (3.24 ams Assume that at the instant in time that we are measuring the system, density is uniform within the volume of interest. This allows us to remove density from within the first integral Since no information was provided about how density varies within the volume, we would be unable to solve this problem without making the assumption. = 0 amu The volume integral of the volume of interest is just the volume of interest. Therefore, Eq (3.24) can be represented as o = v2d+ (19 if we also assume that the density and velocity are not functions of the outflow area of interest (the second integral obtains a positive solution because we are representing an outflow velocity and this velocity aligns with the area vector). Using the product rule av at +p=-pA Solving this equation with the known values -vgd- dt (18 mm |125 (20)(지| ml -100- cm 1.225- 0.01 m's 2 6000 mL
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