Question 12.2: A patient is currently connected to an extracorporeal device...
A patient is currently connected to an extracorporeal device to remove the excess salt from his or her blood (see Figure 12.7). Imagine that the inflow blood contains 5 grams of salt and the blood inflow flow rate is 5 mL/min. The blood outflow rate is also 5 mL/min. The inflow rate of the dialysate is 2 mL/min, and the outflow flow rate is also 2 mL/min. There is a flow between the two compartments of 1 mL/min in each direction. The initial concentration of salt within the dialysate is 0 grams, and the initial concentration of salt within the blood is 150 g. The volume of the blood compartment is 100 mL, and the volume of the dialysate compartment is 100 mL. Determine the salt quantity as a function of time for both compartments of this system.
First, set up the differential equations and the initial conditions:
\frac{dc_{1}}{dt}= – \frac{1 ml/min}{100 mL}c_{1} + \frac{1 mL/min}{100 mL}c_{2} + 5\frac{grams}{min}
\frac{dc_{2}}{dt}= \frac{1 ml/min}{100 mL}c_{1} – \frac{1 mL/min}{100 mL}c_{2}
c1(0) = 150 g
c2 (0) = 0 g
Using the Laplace Transform, this system becomes
sC1 – c1(0) = – 0.01C1 + 0.01C2 + \frac{5}{s}
sC2 – c2(0) = 0.01C1 – 0.01C2
C1(s + 0.01) – 0.01C2 = \frac{5}{s} + 150= \frac{150s + 5}{s}
0.01C1 = C2(s + 0.01)
C1 = 100C2(s + 0.01)
[100C2(s + 0.01)](s + 0.01) – 0.01C2 = \frac{150s + 5}{s}
100C2(s² + 0.02s) = \frac{150s + 5}{s}
C_{2}= \frac{1.5s + 0.05}{s^{2}(s + 0.02)}= \frac{-5}{s} + \frac{2.5}{s^{2}} + \frac{5}{s + 0.02}
C_{1}= \frac{150s^{2} + 6.5s + 0.05}{s^{2}(s + 0.02)}= \frac{200}{s} – \frac{2.5}{s^{2}} – \frac{50}{s + 0.02}
c_{1}(t)= 200 – 2.5t – 50e^{-0.02t}
c_{2}(t)= -5 + 2.5t + 5e^{-0.02t}