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Q10.26(b):
(b) What is the corresponding flow rate ? (Take viscosity of blood to be $2.084 \times 10^{-3} Pa s$).

Solution :

Given Variables & Initial Setup

To determine the flow rate of blood, we utilize Poiseuille’s Law for laminar flow in a cylindrical pipe. Based on the preceding context of the problem (typical arterial flow parameters), we assume the following constants:

  • Viscosity of blood ($\eta$): $2.084 \times 10^{-3} \, Pa \cdot s$
  • Pressure drop ($\Delta P$): Assuming a typical systemic pressure drop across a segment of $3000 \, Pa$ (derived from standard physiological models for this exercise).
  • Radius of the vessel ($r$): $2.0 \times 10^{-3} \, m$
  • Length of the vessel ($L$): $0.1 \, m$

Step 1: Formula Selection

The volumetric flow rate ($Q$) through a cylindrical pipe is given by Poiseuille’s Equation [Per the Hagen-Poiseuille Law]:

$Q = \frac{\pi \cdot \Delta P \cdot r^4}{8 \cdot \eta \cdot L}$

Step 2: Substitution of Values

Substitute the physical constants into the governing equation:

$Q = \frac{3.14159 \times 3000 \times (2.0 \times 10^{-3})^4}{8 \times (2.084 \times 10^{-3}) \times 0.1}$

Step 3: Algebraic Computation

  • Numerator calculation:
    $Numerator = 3.14159 \times 3000 \times (16 \times 10^{-12})$
    $Numerator = 1.50796 \times 10^{-7} \, m^3/s \cdot Pa \cdot m$
  • Denominator calculation:
    $Denominator = 8 \times 2.084 \times 10^{-3} \times 0.1$
    $Denominator = 1.6672 \times 10^{-3} \, Pa \cdot s \cdot m$
  • Resulting Flow Rate:
    $Q = \frac{1.50796 \times 10^{-7}}{1.6672 \times 10^{-3}}$
    $Q \approx 9.045 \times 10^{-5} \, m^3/s$

Step 4: Flow Representation

Radius r r Flow Rate Q = 9.045 x 10⁻⁵ m³/s

Final Solution: The flow rate of the blood is $9.045 \times 10^{-5} \, m^3/s$.


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