Question

The viscosity of water at 25C is about .00089 Pa s. The inner diameter of a PE160 polyethylene tubingis 1.14mm. Assume steady state laminar flow.

A) What pressure is necessary to get a flow of 5 mL min^-1 through a 20 cm length of this tubing?

B) What is the velocity of the flow?

C) What pressure is necessary to get a flow of 5 mL min^-1 through the same PE160 tubing if plasma is used, with viscosity of .002 Pa s?

D) What pressure would be needed for the same flow of water through a 20cm length of PE60 tubing with i.d.=.76mm?

Answer 1

We know, Hagen POiseuille's equation is :

Given flow rate, Q = 5 mL/min = 0.083*10^-6 m³/s

Pipe length, L = 20 cm = 0.2 m

Inner diameter = 1.14 mm = 1.14*10^-3 m

So, inner radius, r = 0.57*10^-3 m

Viscosity = 0.00089 Pa s

Putting these values in the above equation, we get :

delta P = 356.400 Pa

(b)

Velocity of flow, v = volumetric flow rate/Cross sectional area = Q/(pi*r²) = 0.083*10^-6/(pi*(0.57*10^-3)²) = 0.0813 m/s

(c) For plasma, viscosity = 0.002 Pa.s

Putting the same values in the equation except for this new value of viscosity, we get :

delta P = 800.89 Pa

(d)

When internal diameter = 0.76 mm = 0.76*10^-3 m , internal radius, r = 0.76/2*10^-3 m = 0.38*10^-3 m

Putting the same values in the equation as in part (a) except for this new value of r , we get :

delta P = 1804.275 Pa

Answer 2

a)P=128uLQ/piD^4=128*0.00089*0.2*0.005*0.001/(pi*60*1.14^4*10^-12)=357.8Pa

b)v=4Q/piD^2=4*0.005*0.001/(pi*60*1.14^2*10^-6)=8.16cm/s

c)P is directly proportional to u

P1/u1=P2/u2

P2=P1u2/u1=357.8*0.002/0.00089=804Pa

d)P=128*0.00089*0.2*0.005*0.001/(pi*60*0.76^4*10^-12)=1811.5Pa