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?
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
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
The viscosity of water at 25C is about .00089 Pa s. The inner diameter of a...
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