The trachea has a diameter of 18mm; air flows through it at alinear velocity of

Question

The trachea has a diameter of 18mm; air flows through it at alinear velocity of 80cm/s. Each small bronchus has a diameter of1.3mm;air flows through the small bronchi at a linear velocity of15cm/s.

Calculate the volumetric flow rate, mass flow rate, molar flowrate of air through each of these regions of the respiratorysystem. Also, calulate the Reynolds number for each compartment,given he formula:

Re= ( Dv ) /

where D is diameter, v is linear velocity, is density,and is viscosity.The viscosity of air is 1.84 x 10^-4g/(cm*s)

Explanation / Answer

The trachea has a diameter of 18mm; air flows through it at alinear velocity of 80cm/s. Each small bronchus has a diameter of1.3mm;air flows through the small bronchi at a linear velocity of15cm/s.

Calculate the volumetric flow rate, mass flow rate, molar flowrate of air through each of these regions of the respiratorysystem. Also, calulate the Reynolds number for each compartment,given he formula:

Re= ( Dv ) /

trachea

volumetric flow rate = linear velocity*cross sectionalarea

cross sectional area = d2/4

volumetric flow rate = (80 cm/s)*(*[1.8 cm]2)/4= 203.575204 ~ 204 cm3/s

PV = nRT

dn/dt = (P/RT)*dV/dt

dn/dt = (1 atm/ [0.0821 L*atm/mol/K *298K *1000 cm3/L] )*204cm3/s

dn/dt = 0.00832080717 mol/s = 8.32 x 10-3mol/s      MOLAR FLOW RATE

molar mass of air = 29 g/mol

mass flow rate = 8.32 x 10-3 mol/s   *(29 g/mol) = 2.41 x 10-1 g/s

Re = D*v/u

= [n/V]*molar mass = molar mass*P/RT

Re = D*v*molar mass*P/[RT*]

Re =

(0.018 m)*(0.8 m/s)*(0.029 kg/mol)*(101325 Pa)/[8.314 J/mol/K *298 K * 1.8e-4 g/cm-s *1kg/1000g *100 cm/1m]

Re = 948.807946 ~ 9.49 x 102 (noteReynolds Number is dimensionless)

======================

bronchus, v = 15 cm/s, d = 1.3 mm

volumetric flow rate = (15 cm/s)*(*[1.3 cm]2)/4= 19.9098434 ~ 19.9 cm3/s

PV = nRT

dn/dt = (P/RT)*dV/dt

dn/dt = (1 atm/ [0.0821 L*atm/mol/K *298K *1000 cm3/L] )*19.9cm3/s

dn/dt = 0.000813782644 mol/s = 8.14 x 10-4mol/s      MOLAR FLOW RATE

molar mass of air = 29 g/mol

mass flow rate = 8.14 x 10-4 mol/s *(29 g/mol) = 2.36 x 10-2 g/s

Re = D*v/u

= [n/V]*molar mass = molar mass*P/RT

Re = D*v*molar mass*P/[RT*]

Re =

(0.0013m)*(0.15m/s)*(0.029kg/mol)*(101325Pa)/[8.314 J/mol/K *298 K * 1.8e-4 g/cm-s *1kg/1000g *100 cm/1m]

Re = 12.8484409 ~ 1.28 x 101 (noteReynolds Number is dimensionless)

Note: You did not specify the temperature. Perhaps T = 310 K isbetter than T = 298K. Substitute accordingly

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