Assignment 4 CHEM 233
April 8, 2024; due April 15, 2024
Questions
1. In blood, leukocytes attack viruses and neutralize them. Leukocytes have antibodies on their surface,
which recognize a specific antigen on the surface of a virus. After a virus infection has been successfully
eliminated, only a few leukocytes with the specific antibody remain in the blood, thus allowing a
reasonably quick response in case the body gets re-infected by the same virus. Let us call these remaining
leukocytes “sleeping” leukocytes. Assume there is one sleeping leukocyte per 50 mL of blood left when
reinfection occurs. Virus particles multiply quickly before they are recognized and attacked.
a) How many virus particles will be present in 50 mL of blood when the rate at which they are detected
has reached 1 virus particle/day? (Afterwards, the body produces more leukocytes with the right antibodies
and the rate goes up quickly).
Hint: consider the heterocoagulation between leukocytes and virus particles. The rate at which
viruses collide with a single leukocyte is given by
Here αo is the capture efficiency (assumed to be one), G the effective shear rate in blood (assumed to be
100 s-1,) n is the number of virus particles per unit volume, a1 the radius of a leukocyte (assumed to be 3
μm) and a2 the radius of the virus (10 nm).
Answer:
J = 1/day = 1/24x3600
Substituting values in the equation for J yield n = 3.2x109/m3
Hence in 50 mL, there are 1.6x105 virus particles.
b). Assuming the virus has a density of 1 g/cm3, what is the corresponding concentration of virus
particles (g/L)?
Answer:
n = 3.2x109/m3 = 3.2x106/L
c = n x Vvirus x density = n x 4/3 πa3 x 1000 g/L
Hence c = 1.3 x 10-11 g/L
2. Consider an aqueous suspension of spherical particles with a radius of 50 nm and a volume
fraction of 10-4.
a) What is the perikinetic coagulation rate and the coagulation half-time, assuming fast coagulation?
4
Answer: 3 𝜋𝑎3 𝑛 = 10−4
𝑘𝑇
1.4∙10−21
𝑚2
Hence n = 1.9 x 1017/m3 Also 𝐷 = 6𝜋𝜂𝑎 = 6𝜋10−3∙5∙10−8 = 1.5 ∙ 10−13 𝑠 and αp = 1
The coagulation rate is
f = 8π aDn2
𝑓 = 8𝜋 ∙ 5 ∙ 10−81.5 ∙ 10−13 ∙ 1.92 ∙ 1034 = 7.0 ∙ 1015
𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛𝑠
𝑚3 𝑠
b) The onset of fast perikinetic coagulation was found to occur at a NaCl concentration of 0.1 M.
At 0.05 M, the coagulation rate was half the fast rate. Draw a (perikinetic) stability diagram for this
suspension.
Answer:
3. For the same suspension as in question 2
a) What is the orthokinetic coagulation rate and coagulation half-time at G = 10 s-1, assuming the
orthokinetic coagulation efficiency αo = 0.8? What is the orthokinetic coagulation half-time at 10,000 s-1,
assuming αo = 0.2?
Answer:
16
f = a oGa 3 n 2
3
16
For G =10 s-1: 𝑓 = 3 0.8 ∙ 10 ∙ 53 ∙ 10−24 ∙ 1.92 ∙ 1034= 1.9∙ 1013 m-3 s-1
16
For G = 10,000 s-1: 𝑓 = 3 0.2 ∙ 10,000 ∙ 53 ∙ 10−24 ∙ 1.92 ∙ 1034= 4.8∙ 1016 m-3 s-1
b) What is the Peclet number at G = 10 s-1 and 10,000 s-1?
Answer:
Pe = Ga2/D
For G =10 s-1: 10∙25∙10-16/1.5 ∙ 10−13 = 0.17
For G = 10,000 s-1: 104∙25∙10-16/1.5 ∙ 10−13 = 1.7∙103