Question

Blood takes about 1.55 s to pass through a 2.00 mm long capillary. If the diameter of the capillary is 5.00 μm and the pressure drop is 2.45 kPa, calculate the viscosity ???? of blood. Assume laminar flow.

Answer 1

The viscosity
`\mathbf{\eta = 7.416 * 10^(-4) \ N.s/m^2}`

Step-by-step explanation:

From the given information:

Time t = 1.55 s

The radius of capillary = 5.00 μm /2

The pressure drop ΔP = 2.45 kPa

The length of the capillary = 2.00 mm

∴

The viscosity of the blood flow can be calculated by using the formula:

`\eta = (r^2 \Delta P )/(8Lv)`

where;

v = L/t

Then;

`\eta = (r^2 \Delta P )/(8L((L)/(t)))`

`\eta = (((5 * 10^(-6) \ m)/(2))^2(2.45 * 10^3 \ Pa) )/(8(2.0 * 10^(-3) \ m ) ((2.0 * 10^(-3) \ m )/(1.55 \ s )))`

`\eta = 7.416 * 10^(-4) \ Pa.s`

To (N.s/m²)

`\mathbf{\eta = 7.416 * 10^(-4) \ N.s/m^2}`