# Chapter 3 :Hydraulic circuit analysis

The previous chapters have introduced the governing equations for fluid flow and provided solutions for simple unidirectional flows. While we can only rarely use these simple solutions to exactly describe real flows, these unidirectional flows provide a framework for generating engineering estimates for many flows, and these estimation techniques reduce complex systems with large numbers of components into a relatively simple approximate linear description. This is relevant for flow in microdevices both (1) because microdevices can straightforwardly be made with large numbers of microchannels and (2) because those microchannels often have a geometry that leads to flows that are well approximated by the unidirectional solutions presented in Chapter 2. Our approach, called hydraulic circuit analysis, involves assuming that the Poiseuille flow derived in Chapter 2 provides a sound engineering estimate of the pressure drops and flow rates through long, mostly straight channels, even if the cross section is not exactly circular and the channels are neither perfectly straight nor infinite in extent. We thus write an approximate linear relation between pressure drops and flow rates through these channels—the Hagen-Poiseuille law. This law, combined with conservation of mass, approximately prescribes fluid flow through complex networks by solution of sets of algebraic equations.

### 3.1 Hydraulic circuit analysis

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### 3.2 Hydraulic circuit equivalents for fluid flow in microchannels

•   3.2.1 Analytic representation of sinusoidal pressures and flowrates
•   3.2.2 Hydraulic impedance
•   3.2.3 Hydraulic circuit relations
•   3.2.4 Series and parallel component rules

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### 3.3 Solution techniques

Hydraulic circuit systems lead to algebraic systems of equations that can be solved directly using symbolic or matrix manipulation software. Given the hydraulic impedances of the channels, the unknowns are the pressures at the N nodes, and the flowrates in the M channels. The N equations from conservation of mass at each node and M equations from theHagen-Poiseuille law for each channel define the system. Thus, a N +M×N +M matrix equation can be written and inverted to obtain the solution.

VIDEO: Matrix formalism for solving the Hagen-Poiseuille law for channel networks.

If the channels are all purely resistive, then the pressure and flowrate phasors are real. If the channels include compliance, then the pressure and flowrate phasors are complex and the angle of the complex numbers describes the phase lag of the response.

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EXAMPLE PROBLEM 3.2: Calculate the hydraulic radius of

1. a glass channel made using photoresist lithographically patterned to width w = 240 μm and etched isotropically in HF to a depth of d = 30 μm.
2. a channel cast in polydimethylsiloxane made by patterning SU-8 molds of width w = 40 μm and depth d = 30 μm and casting the PDMS over the SU-8 mold.

Note that isotropic etches lead to microchannel cross sections consisting of circular side walls and flat top and bottom surfaces, while casting over SU-8 makes roughly rectangular channel cross-sections (see Figure 3.9). Figure 3.9: Shapes of channels based on (a) wet etching of glass using HF and (b) casting over an SU-8 template.

Solution: Note that for circular channels, rh = r. For infinitely flat plates separated by d, rh = d. The rectangular section has a hydraulic radius approximately equal to half the length of the sides, while the isotropically etched channel, because it is wide, starts to approach rh = d.

HF etch: SU-8 mold: .

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EXAMPLE PROBLEM 3.3:

Consider the fluid circuit shown in Figure. Write the 7 equations in the unknowns p1, p2, p3, p4, Q1, Q2, Q3. Three of these equations are given by the problem statement, one is conservation of mass, and three are Hagen-Poiseuille relations. Write these equations as a matrix equation. Invert the matrix to solve the system for the 7 unknowns. Figure 3.10: A simple fluid circuit. All channels are circular in cross-section with radius R = 20 μm. L1 = 0.5 cm, L2 = 1 cm, and L3 = 2 cm. Pressures at the ports are p1 = 0.11 MPa, p3 = 0.1 MPa, and p4 = 0.1 MPa. The fluid is water, with viscosity 1×10-3 Pa s.

Solution: First we must solve for the hydraulic resistance of the three channels, using Rh = 8ηL∕πR4. This gives hydraulic resistances of 7.96×1013, 1.59×1014, and 3.18×1014 N∕s m5 for the three channels.

The seven equations are

### 3.1 Hydraulic circuit analysis

*** Note: Sorry, this page is currently unavailable in the online version.

### 3.2 Hydraulic circuit equivalents for fluid flow in microchannels

•   3.2.1 Analytic representation of sinusoidal pressures and flowrates
•   3.2.2 Hydraulic impedance
•   3.2.3 Hydraulic circuit relations
•   3.2.4 Series and parallel component rules

*** Note: Sorry, this page is currently unavailable in the online version.

### 3.3 Solution techniques

Hydraulic circuit systems lead to algebraic systems of equations that can be solved directly using symbolic or matrix manipulation software. Given the hydraulic impedances of the channels, the unknowns are the pressures at the N nodes, and the flowrates in the M channels. The N equations from conservation of mass at each node and M equations from theHagen-Poiseuille law for each channel define the system. Thus, a N +M×N +M matrix equation can be written and inverted to obtain the solution.

VIDEO: Matrix formalism for solving the Hagen-Poiseuille law for channel networks.

If the channels are all purely resistive, then the pressure and flowrate phasors are real. If the channels include compliance, then the pressure and flowrate phasors are complex and the angle of the complex numbers describes the phase lag of the response.

__________________________________________________________________________________________________________________________________________________________
EXAMPLE PROBLEM 3.2: Calculate the hydraulic radius of

1. a glass channel made using photoresist lithographically patterned to width w = 240 μm and etched isotropically in HF to a depth of d = 30 μm.
2. a channel cast in polydimethylsiloxane made by patterning SU-8 molds of width w = 40 μm and depth d = 30 μm and casting the PDMS over the SU-8 mold.

Note that isotropic etches lead to microchannel cross sections consisting of circular side walls and flat top and bottom surfaces, while casting over SU-8 makes roughly rectangular channel cross-sections (see Figure 3.9). Figure 3.9: Shapes of channels based on (a) wet etching of glass using HF and (b) casting over an SU-8 template.

Solution: Note that for circular channels, rh = r. For infinitely flat plates separated by d, rh = d. The rectangular section has a hydraulic radius approximately equal to half the length of the sides, while the isotropically etched channel, because it is wide, starts to approach rh = d.

HF etch: SU-8 mold: .

__________________________________________________________________________________________________________________________________________________________

__________________________________________________________________________________________________________________________________________________________
EXAMPLE PROBLEM 3.3:

Consider the fluid circuit shown in Figure. Write the 7 equations in the unknowns p1, p2, p3, p4, Q1, Q2, Q3. Three of these equations are given by the problem statement, one is conservation of mass, and three are Hagen-Poiseuille relations. Write these equations as a matrix equation. Invert the matrix to solve the system for the 7 unknowns. Figure 3.10: A simple fluid circuit. All channels are circular in cross-section with radius R = 20 μm. L1 = 0.5 cm, L2 = 1 cm, and L3 = 2 cm. Pressures at the ports are p1 = 0.11 MPa, p3 = 0.1 MPa, and p4 = 0.1 MPa. The fluid is water, with viscosity 1×10-3 Pa s.

Solution: First we must solve for the hydraulic resistance of the three channels, using Rh = 8ηL∕πR4. This gives hydraulic resistances of 7.96×1013, 1.59×1014, and 3.18×1014 N∕s m5 for the three channels.

The seven equations are (3.31) (3.32) (3.33) (3.34) (3.35) (3.36)

and (3.37)

In matrix form, this is (3.38)

The solution is p2 = 0.1057 MPa, Q1 = 54×10-12 m3∕s, Q2 = 36×10-12 m3∕s, and Q3 = 18×10-12 m3∕s.

### 3.4 Summary

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### 3.6 Exercises

*** Note: Sorry, this page is currently unavailable in the online version. (3.32) (3.33) (3.34) (3.35) (3.36)

and (3.37)

In matrix form, this is (3.38)

The solution is p2 = 0.1057 MPa, Q1 = 54×10-12 m3∕s, Q2 = 36×10-12 m3∕s, and Q3 = 18×10-12 m3∕s.

### 3.4 Summary

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