• BME 200

# Solving for flow in parallel vessels the circuits way

When I wrote my notes I had assumed that fractions would be an easy way to explain this problem. However, I messed up the example and it is incorrect, so please strike it from your notes. Here is another way to solve the problem that may make more sense.

Given the vessel network below, find $$Q_1, Q_2,$$ and $$Q_3$$.

Note that I have drawn a little more information in the diagram above to make it clear that you can treat the vessels as parallel. Add them in parallel:

$$\frac{1}{R_{parallel}}=\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3}$$

Solving for $$R_{parallel}$$ gives:

$$R_{parallel} = 0.545\,\textrm{PRU}$$

New circuit:

We can use $$\Delta P = QR$$ to figure out the pressure drop across all three resistors combined. So,

$$\Delta P = 3(0.545) = 1.635\,\textrm{mmHg}$$

Since the three resistors are in parallel, the $$\Delta P$$ across each is the same. We didn’t conver this in class because it is something you learn in circuits. So, knowing that $$\Delta P$$ is the same across each vessel, you can calculate the $$Q$$ for each vessel:

$$Q_1=\frac{\Delta P}{R_1} = \frac{1.635}{1} = 1.635\,\textrm{mL/sec}$$

$$Q_2=\frac{\Delta P}{R_2} = \frac{1.635}{2} = 0.818\,\textrm{mL/sec}$$

$$Q_3=\frac{\Delta P}{R_3} = \frac{1.635}{3} = 0.545\,\textrm{mL/sec}$$

# If you’re really interested, keep reading

Okay, if you’re not scared off by now, here’s the proper fraction way to do it. Multiply the main volumetric flowrate, $$Q$$, by the inverse of the vessel resistance and divide by all the resistances in parallel. So, for each flow rate that looks like this:

$$Q_1 = Q\left ( \frac{\frac{1}{R_1}}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}\right ) = 1.635\,\textrm{mL/sec}$$

$$Q_2 = Q\left ( \frac{\frac{1}{R_2}}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}\right ) = 0.818\,\textrm{mL/sec}$$

$$Q_3 = Q\left ( \frac{\frac{1}{R_3}}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}\right ) = 0.545\,\textrm{mL/sec}$$

Last updated:
September 6, 2018