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Some measurements on a blood sample at 37 ˚C (98.6 ˚F) indicate a shearing stress of 0.540 N/m2 for a corresponding rate of shearing strain of 198 s-1. Determine (a) the apparent viscosity of the blood and (b) its ratio to the viscosity of water at the same temperature.

Answer :

boffeemadrid

Answer:

[tex]0.00273\ \text{Pa s}[/tex]

3.944

Explanation:

[tex]\tau[/tex] = Shearing stress = [tex]0.54\ \text{N/m}^2[/tex]

[tex]\dot{\gamma}[/tex] = Rate of shearing strain = [tex]198\ \text{s}^{-1}[/tex]

(a) Viscosity is given by

[tex]\mu_b=\dfrac{\tau}{\dot{\gamma}}\\\Rightarrow \mu_b=\dfrac{0.54}{198}\\\Rightarrow \mu_b=0.00273\ \text{Pa s}[/tex]

The apparent viscosity of the blood is [tex]0.00273\ \text{Pa s}[/tex]

(b) Viscosity of water at 37 ˚C = 0.0006922 Pa s = [tex]\mu_w[/tex]

The ratio is

[tex]\dfrac{\mu_b}{\mu_w}=\dfrac{0.00273}{0.0006922}\\\Rightarrow \dfrac{\mu_b}{\mu_w}=3.944[/tex]

The required ratio is 3.944

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