Question 10.1: Under a disease condition where the trabeculae become occlud...
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 μL/mmHg and that R is equal to 5 mmHg min/μL. Assume that the inflow flow rate is 2 μL/min.
To solve this problem, we will need to solve the differential equation derived for the Conservation of Mass of the eye:
\frac{d(p_{eye} – p_{norm})}{dt} = -\frac{p_{eye} – p_{norm}}{RC}
\int{\frac{d(p_{eye} – p_{norm})}{p_{eye} – p_{norm}}} = -\int{\frac{dt}{RC}}
\ln(p_{eye} – p_{norm})= \frac{-t}{RC} + K
where K is the indefinite integration constant:
p_{eye} – p_{norm}= K_{0}e^{-t/RC}
where K0 is a different constant from K:
p_{eye}= p_{norm} + K_{0}e^{-t/RC}
To solve for K0, we know that the initial condition is
p_{eye}(t = 0)= p_{norm} + 5 mmHg
p_{norm} + 5 mmHg= p_{norm} + K_{0}e^{0}= p_{norm} + K_{0}
K_{0}= 5 mmHg
p_{eye}= p_{norm} + 5 mmHg(e^{-t/RC})
To solve for the time required to return back to 5% of the steady state normal pressure
1.05p_{norm}= p_{norm} + 5 mmHg(e^{-t/RC})
\frac{0.05p_{norm}}{5 mmHg}= e^{-t/RC}
– \frac{t}{RC}= \ln \left(\frac{0.05p_{norm}}{5 mmHg}\right)= \ln \left(\frac{0.05 \ast Q_{in} \ast R}{5 mmHg}\right)
t = -RC \ln \left(\frac{0.05 \ast Q_{in} \ast R}{5 mmHg}\right)
t= – \left(5\frac{mmHg min}{\mu L}\right) \left(2.8 \frac{\mu L}{mmHg}\right) \ln \left(\frac{0.05 \ast 2 \frac{\mu L}{min} \ast 5 \frac{mmHg min}{\mu L}}{5 mmHg}\right)
t = 32.2 min