Intraocular Fluid Flow - Biofluidmechanics: An Introduction to Fluid Mechanics Macrocirculation and Microcirculation - page 286

where R is the resistance to aqueous humor flow. Using the two previous relationships to
solve for the Conservation of Mass, we get
dV eye
dt 5 Q in

2 Q out

C dp eye

p norm
R 2

p eye
R dt 5

dp eye
dt 1

p eye 2 p norm

RC

0 ð 10 : 7 Þ

5

where p norm is the steady-state pressure that normally acts and regulates the inflow of
aqueous humor.

Example
Under a disease condition where the trabeculae become occluded, the intraocular pressure
transiently increases by 5 mmHg. If the trabeculae occlusion is removed, which brings the flow
resistance back to normal levels, what is the time required to bring the intraocular pressure back
to within 5% of the steady state value? Assume that C is equal to 2.8

μ L/mmHg and that R is

equal to 5 mmHg min/ μ L. Assume that the inflow flow rate is 2 μ L/min.

Solution
To solve this problem, we will need to solve the differential equation derived for the
Conservation of Mass of the eye:

d ð p eye 2 p norm Þ
dt

p eye 2 p norm
RC

52

ð d ð p eye 2 p norm Þ
p eye 2 p norm 52

ð dt
RC

ð p eye 2 p norm Þ 5 2 t ln

RC 1 K

where K is the indefinite integration constant:
p eye 2 p norm 5 K 0 e 2 t = RC
where K 0 is a different constant from K:
p eye 5 p norm 1 K 0 e 2 t = RC
To solve for K 0 , we know that the initial condition is
p eye ð t 5 0 Þ 5 p norm 1 5 mmHg

5 p norm 1 K 0 e 0

p norm 1 5 mmHg

5 p norm 1 K 0

K 0 5 5 mmHg

ð e 2 t = RC Þ

p eye 5 p norm 1 5 mmHg