The effect of pump rate and channel height on Urea clearance (COMSOL modeling)

ByDean Johnson

Comsol modeling was done to find the ideal channel height and volumetric flow rate. What is clear from the simulation is that ever increasing flow rates in the model lead to increased clearance and decreasing channel heights give some benefit but not on the order previously expected. In addition, (Depner 2005) shows that increasing the flow rate improves the clearance only up to a point, with this ‘point’ being dependent on the device. There is a formula for the maximum clearance for counter flow systems but not for our in-beaker set up. Testing at higher flow rates with the stronger membranes will let us explore this. We will also need to understand the minimum and maximum flow rates for the device. The minimum rate will be determined from the clearance K and the maximum treatment time, remember KT/V > 1.2 for 75% URR.

Figure 1: Concentration as a function of volumetric flow rate. Initial Concentration is shown as the straight line. Shallow channels have lower concentrations.
Figure 2: Extraction Ratio, ER, as a function of volumetric flow rate. This is the percentage of the flowing fluid volume cleared of urea. Again, shallow channels clear a higher percentage of the urea (for a smaller volume as will be seen in the next graph)
Figure 3: Instantaneous clearance, K, versus volumetric flow rate. For very high rates, the clearance is nearly identical. In practice this probably won’t be the case. For lower flow rates, in the regime of interest, the shallow channels do have a slight advantage over the deeper channels, but not as much as would be indicated by the 2D geometric analysis with which we are all familiar.
Figure 3: Clearance of Urea for three channel heights, 100 µm, 300 µm, and 500 µm versus flow rate. For each channel, the point at which the Peclet number, Pe, is one is shown with a red marker.

E.G. For the 100-µm channels (500 µm x 10 cm) at ~35 µL/min blood flow rate we get 4 µL/min of clearance. K = 4 µL/min per, V = 5 L, so to get KT/V = 1.2 requires T = 1.2V/K = 1.2*5/4×10-6 = 1.5×106 min for a single channel. For 20 channels per chip, 80 µL/min, and 100 chips per device, 8000 µL/min. that gives us 750 min of dialysis, 12.5 hours. We still need to do better. It is still not certain that we can pump at that rate, 35 µL/min, through the channels, which would be a total rate of 70 mL/min for a total KTot of 8 mL/min.

 

References:

Depner, T. A. (2005) ‘Hemodialysis adequacy: Basic essentials and practical points for the nephrologist in training’, Hemodialysis International, 9(3), 241-254.

 

 

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2 Comments

  1. What is to prevent us from pumping at 35 ul/min through a channel? What pressure is required?

  2. I think we need to plan on lengthening the membranes, adding channels, and flowing faster. I would like to get this down to 20 chips and 5-6 hours.

    Also if the device can operate at blood pressure 2-3 PSI, we can avoid adding a pump.

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