Question

A pressure difference of 24400 Pa is used to inject a sample onto a 58.4-cm-long capillary that has a diameter of 60.0 μm. Assuming the viscosity of the sample is 0.00102 kg/(m·s), how much time is required to inject a sample that is 2.35% of the length of the capillary?

Answer 1

By using "Poiseuille equation" we can determine the viscosity, ,of liquid passing through a capilllary .

The  "Poiseuille equation" is   = (Pr⁴t)/(8VL) -------------->(i)

where  = viscosity of the liquid , P = difference of pressure , r = radius of capillary tube , L = length of capillary , V = volume of liquid flow through the tube , t = time required , =3.14

for the given question , = 0.00102 kg/(m·s) , P=24400 Pa =24400 kg/(m.s²) r=60.0 m =6*10^-5 m ,L =58.4 cm = 0.584 m ,

Now 2.35% of the length of the capillary = (0.584*2.35)/100 = 0.0137 m , so the 0.0137 m length of capillary contain the sample liquid.  

The volume of capillary tube = r²h , here h = 0.0137 m

so the volume of sample liquid,V = {3.14*(6*10^-5)² *(0.0137)} = 1.55*10^-10 m³

from (i) equation we get ,

t = (8VL )/(Pr⁴) = [8*(1.55*10^-10 m³)*(0.584 m)*(0.00102 kg.m^-1.s^-1)] / [(24400 kg.m^-1.s^-2)*3.14*(6*10^-5 m)⁴]

or, t = 0.74 s

Therefore , 0.74 s time required to inject the given sample .