Viscosity | PhysicsRank

A pipe can carry the same amount each second while a pump keeps doing work. Where does the energy go if the liquid is not speeding up? Follow the velocity across the pipe and the pressure along it, and the missing energy has a clear destination.

We will build the laminar pipe model from a force balance, then use it to compare pumps, branch networks, viscosity measurements and a changing reservoir head. Each step keeps flow rate distinct from speed.

What you will learn to do

Derive a parabolic no-slip velocity profile from pressure and shear forces.
Integrate annular flow to distinguish mean speed, centre speed and volume flow.
Explain a steady pressure drop using mechanical-energy dissipation.
Compare fixed-pressure and fixed-flow changes without swapping the conditions.
Solve linear series/parallel tube networks and audit connector assumptions.
Infer viscosity from data and distinguish the viscous-drain time law from ideal efflux.

Useful before this chapterViscosity, terminal motion and flow regimes →Continuity and energy in flowing fluids →Work, stored energy and power →Exponential change, logarithms and controlled approximations →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Viscous flow through pipes
Learning goals:
- Derive a parabolic no-slip velocity profile from pressure and shear forces.
- Integrate annular flow to distinguish mean speed, centre speed and volume flow.
- Explain a steady pressure drop using mechanical-energy dissipation.
- Compare fixed-pressure and fixed-flow changes without swapping the conditions.
- Solve linear series/parallel tube networks and audit connector assumptions.
- Infer viscosity from data and distinguish the viscous-drain time law from ideal efflux.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

BEFORE YOU READ

Bring back one idea

Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.

Does zero acceleration guarantee that every force on a moving object is zero?

Compare your reasoning

No. A steady velocity requires the net force to vanish, not each individual force. In a fully developed pipe, a forward pressure force balances a retarding shear force even while the fluid keeps moving.

Revisit the explanation →

01 / Viscous flow through pipes

Viscosity measures a response to shear rate

A liquid can flow steadily and still require a continuous push. Adjacent layers exchange momentum, so a faster layer tends to drag a slower one along while being retarded in return. In the Newtonian model, the tangential stress is proportional to the local velocity gradient. It is not proportional to the fluid’s density or simply to its speed relative to the room.

Between a stationary plate and a parallel plate moving at speed U, with gap d and no pressure-driven flow, the steady profile is linear. The shear stress magnitude is ηU/d. In a pipe, the stress varies across the radius, and the resulting speed profile is curved instead. The boundary geometry decides which velocity gradient belongs in the same material law.

The dynamic viscosity η has SI unit Pa·s. A listed value applies at stated conditions, especially temperature. A fluid whose apparent viscosity changes with shear rate needs a more general material model; rigid-pipe Poiseuille flow should not be advertised as a universal description of suspensions, soft tubes or pulsating biological flow.

\tau=\eta\left|\frac{dv}{ds}\right|,\qquad [\eta]=\mathrm{Pa\,s}

Here τ is shear-stress magnitude and s is the coordinate across the velocity gradient. Its direction follows the relative motion and the chosen surface normal.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Viscous flow through pipes
Focus: Viscosity measures a response to shear rate
Learning goals:
- Derive a parabolic no-slip velocity profile from pressure and shear forces.
- Integrate annular flow to distinguish mean speed, centre speed and volume flow.
- Explain a steady pressure drop using mechanical-energy dissipation.
- Compare fixed-pressure and fixed-flow changes without swapping the conditions.
- Solve linear series/parallel tube networks and audit connector assumptions.
- Infer viscosity from data and distinguish the viscous-drain time law from ideal efflux.
Equations in this section (LaTeX):
\tau=\eta\left|\frac{dv}{ds}\right|,\qquad [\eta]=\mathrm{Pa\,s}
Meaning: Here τ is shear-stress magnitude and s is the coordinate across the velocity gradient. Its direction follows the relative motion and the chosen surface normal.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

02 / Viscous flow through pipes

Derive the curved speed profile from a force balance

Choose a cylindrical core of liquid of radius s and length L inside a horizontal pipe. The pressure difference between its end faces pushes it forward with force Δp πs². The surrounding fluid retards it across the curved area 2πsL. In fully developed steady flow, the velocity profile does not change along the pipe, so those forces balance.

The resulting shear stress magnitude is sΔp/(2L). It vanishes at the axis and grows toward the wall. Combine it with the Newtonian shear law, using a negative radial velocity gradient because speed decreases outward. Integrating inward from the stationary wall, where no slip makes speed zero, gives a parabola.

Zero velocity at the wall does not mean that the rest of the cross-section is stationary. Conversely, zero shear at the axis does not mean that the centre fluid is stopped: it is where the speed is largest and the profile has zero slope.

\begin{aligned}v(s)&=\frac{\Delta p}{4\eta L}(R^2-s^2)\\ \tau(s)&=\frac{\Delta p\,s}{2L}\end{aligned}

Pressure decreases in the chosen downstream direction. Radius R is fixed; s measures distance from the tube axis between zero and R.

Explain this equation

Read it in words

Axial speed falls quadratically from its maximum at the centre to zero at the stationary wall. The shear stress magnitude grows linearly outward.

Symbols and units

$v(s)$: Downstream velocity at distance s from the tube axis.m/s.
$\Delta p$: Upstream pressure minus downstream pressure across the measuring length; positive for this flow direction.Pascals (Pa).
$\eta$: Constant dynamic viscosity of the Newtonian liquid.Pa·s.
$L$: Distance between the two pressure taps along the uniform tube.Metres (m).
$R$: Internal tube radius.Metres (m).
$s$: Radial distance from the axis, not distance along the tube.Metres (m).
$\tau(s)$: Magnitude of the retarding shear stress on a cylindrical core.Pascals (Pa).

Signs and operations

R² − s² is largest at the centre and zero at the wall. The positive stress shown is a magnitude; the shear force on a forward-moving core acts backward. Pressure force acts forward.

When it applies

Steady, fully developed, axisymmetric laminar flow of an incompressible Newtonian liquid in a straight rigid circular tube; constant viscosity, no slip and negligible body-force component along the horizontal tube. Entrance regions require another profile.

Make it concrete

With R = 1 mm, Δp = 160 Pa, η = 0.002 Pa·s and L = 0.50 m, centre speed is 0.040 m/s. At half-radius it is 0.030 m/s, while wall speed is zero.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \begin{aligned}v(s)&=\frac{\Delta p}{4\eta L}(R^2-s^2)\\ \tau(s)&=\frac{\Delta p\,s}{2L}\end{aligned}
Read it as: Axial speed falls quadratically from its maximum at the centre to zero at the stationary wall. The shear stress magnitude grows linearly outward.
Symbols:
v(s): Downstream velocity at distance s from the tube axis.; m/s.
\Delta p: Upstream pressure minus downstream pressure across the measuring length; positive for this flow direction.; Pascals (Pa).
\eta: Constant dynamic viscosity of the Newtonian liquid.; Pa·s.
L: Distance between the two pressure taps along the uniform tube.; Metres (m).
R: Internal tube radius.; Metres (m).
s: Radial distance from the axis, not distance along the tube.; Metres (m).
\tau(s): Magnitude of the retarding shear stress on a cylindrical core.; Pascals (Pa).
Signs and operations: R² − s² is largest at the centre and zero at the wall. The positive stress shown is a magnitude; the shear force on a forward-moving core acts backward. Pressure force acts forward.
Conditions: Steady, fully developed, axisymmetric laminar flow of an incompressible Newtonian liquid in a straight rigid circular tube; constant viscosity, no slip and negligible body-force component along the horizontal tube. Entrance regions require another profile.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

Locate a speed inside the pipe

A horizontal pipe has internal radius 1.0 mm and a fully developed measuring length of 0.50 m. A Newtonian liquid of viscosity 0.002 Pa·s flows under a 160 Pa pressure drop. Find centre speed, speed at half-radius and wall shear stress, assuming laminar no-slip flow.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Viscous flow through pipes
Focus: Derive the curved speed profile from a force balance
Learning goals:
- Derive a parabolic no-slip velocity profile from pressure and shear forces.
- Integrate annular flow to distinguish mean speed, centre speed and volume flow.
- Explain a steady pressure drop using mechanical-energy dissipation.
- Compare fixed-pressure and fixed-flow changes without swapping the conditions.
- Solve linear series/parallel tube networks and audit connector assumptions.
- Infer viscosity from data and distinguish the viscous-drain time law from ideal efflux.
Equations in this section (LaTeX):
\begin{aligned}v(s)&=\frac{\Delta p}{4\eta L}(R^2-s^2)\\ \tau(s)&=\frac{\Delta p\,s}{2L}\end{aligned}
Meaning: Pressure decreases in the chosen downstream direction. Radius R is fixed; s measures distance from the tube axis between zero and R.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

03 / Viscous flow through pipes

Average over the area, not along a diameter

A thin annulus at radius s carries volume at rate v(s) times its area 2πs ds. Outer annuli have more area than inner ones even when their radial thicknesses match. A simple average of speeds sampled equally along a diameter would give the wrong area-averaged speed.

Integrating over the circular cross-section gives the volume flow Q. Its mean speed is Q/(πR²), exactly half the centre speed for this parabolic profile. This is where the fourth power of radius enters: a larger radius increases the available area and also increases the velocity produced by the same pressure gradient.

\begin{aligned}Q&=\int_0^R v(s)\,2\pi s\,ds\\ &=\frac{\pi R^4\Delta p}{8\eta L}\\ \bar v&=\frac{Q}{\pi R^2}=\frac{v(0)}2\end{aligned}

Q is volume per unit time, not a speed. This is Poiseuille’s law for the stated measuring segment.

Explain this equation

Read it in words

Add the flow through concentric rings. The result depends on radius to the fourth power, and its corresponding mean speed is half the centre value.

Symbols and units

$Q$: Volumetric flow through any cross-section of this incompressible steady segment.m³/s.
$v(s)$: Axial speed in the annulus at radial distance s from the tube axis.m/s.
$2\pi s\,ds$: Area of a thin annulus at radius s with radial thickness ds.m².
$R$: Internal tube radius; halve a stated diameter before using it.Metres (m).
$\Delta p$: Pressure loss from upstream to downstream across L.Pascals (Pa).
$\eta,\ L$: Dynamic viscosity and measuring length, respectively.Pa·s and m.
$\bar v,\ v(0)$: Area-averaged speed and centre speed.m/s.

Signs and operations

The integral sums v times area. A fourth power magnifies radius errors: using diameter in place of radius would multiply predicted flow by sixteen. The bar indicates an average over area.

When it applies

Use the fully developed laminar Newtonian, no-slip circular-pipe model. The fourth-power comparison assumes pressure drop, length and viscosity are fixed. If the pump instead fixes flow, pressure must adjust.

Make it concrete

For R = 1 mm, Δp = 80 Pa, η = 0.002 Pa·s and L = 0.50 m, Q = π × 10⁻⁸ m³/s, about 1.885 mL/min. The mean speed is 0.010 m/s.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \begin{aligned}Q&=\int_0^R v(s)\,2\pi s\,ds\\ &=\frac{\pi R^4\Delta p}{8\eta L}\\ \bar v&=\frac{Q}{\pi R^2}=\frac{v(0)}2\end{aligned}
Read it as: Add the flow through concentric rings. The result depends on radius to the fourth power, and its corresponding mean speed is half the centre value.
Symbols:
Q: Volumetric flow through any cross-section of this incompressible steady segment.; m³/s.
v(s): Axial speed in the annulus at radial distance s from the tube axis.; m/s.
2\pi s\,ds: Area of a thin annulus at radius s with radial thickness ds.; m².
R: Internal tube radius; halve a stated diameter before using it.; Metres (m).
\Delta p: Pressure loss from upstream to downstream across L.; Pascals (Pa).
\eta,\ L: Dynamic viscosity and measuring length, respectively.; Pa·s and m.
\bar v,\ v(0): Area-averaged speed and centre speed.; m/s.
Signs and operations: The integral sums v times area. A fourth power magnifies radius errors: using diameter in place of radius would multiply predicted flow by sixteen. The bar indicates an average over area.
Conditions: Use the fully developed laminar Newtonian, no-slip circular-pipe model. The fourth-power comparison assumes pressure drop, length and viscosity are fixed. If the pump instead fixes flow, pressure must adjust.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

Translate a profile into a collected volume

A liquid of viscosity 0.002 Pa·s flows through a 1.0 mm radius circular tube over a 0.50 m measuring length with an 80 Pa pressure drop. Under the fully developed laminar no-slip model, how many millilitres pass a section in one minute?

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

INTERACTIVE LAB

Explore the speed profile and flow

Predict what halving radius will do at fixed pressure drop. Then change one control at a time. Compare the centre speed, mean speed, volume flow and Reynolds number; the dashed curve retains the initial setting.

MEMBERSThe interactive lab is part of membership.Unlock →

STOP & THINK

A fully developed laminar circular-pipe profile has centre speed 0.060 m/s. What is its area-averaged speed in m/s?

Answer in m/s.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Viscous flow through pipes
Focus: Average over the area, not along a diameter
Learning goals:
- Derive a parabolic no-slip velocity profile from pressure and shear forces.
- Integrate annular flow to distinguish mean speed, centre speed and volume flow.
- Explain a steady pressure drop using mechanical-energy dissipation.
- Compare fixed-pressure and fixed-flow changes without swapping the conditions.
- Solve linear series/parallel tube networks and audit connector assumptions.
- Infer viscosity from data and distinguish the viscous-drain time law from ideal efflux.
Equations in this section (LaTeX):
\begin{aligned}Q&=\int_0^R v(s)\,2\pi s\,ds\\ &=\frac{\pi R^4\Delta p}{8\eta L}\\ \bar v&=\frac{Q}{\pi R^2}=\frac{v(0)}2\end{aligned}
Meaning: Q is volume per unit time, not a speed. This is Poiseuille’s law for the stated measuring segment.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

04 / Viscous flow through pipes

A pressure drop can pay for dissipation without changing speed

In a uniform fully developed pipe segment, the same velocity profile enters and leaves. There is no net gain in kinetic-energy flow between its ends, and a horizontal segment has no gravitational change. Yet the inlet pressure is higher. The pressure work is transferred to internal energy through viscous shearing.

The pressure power supplied to a volume flow is ΔpQ. Ideal, lossless Bernoulli would predict equal pressures at equal heights and equal speeds. Applying that equation unchanged here would erase the very mechanism sustaining the flow. A mechanical-energy balance must include the viscous loss.

This energy statement concerns the measuring segment. A complete pumping system may also need power for height changes, acceleration, entrance losses and connectors. Real pump input can exceed the power delivered to the liquid.

P_{\mathrm{diss}}=\Delta p\,Q,\qquad [P]=\mathrm{Pa}\,\mathrm{m^3/s}=\mathrm W

For this uniform horizontal fully developed segment, pressure-work loss equals viscous dissipation.

WORKED EXAMPLE

The speed can stay steady while energy is lost

Across a horizontal uniform pipe segment, the measured pressure drop is 120 Pa and steady volume flow is 2.0 mL/s. The entering and leaving profiles are identical. What mechanical power is dissipated within this segment?

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Viscous flow through pipes
Focus: A pressure drop can pay for dissipation without changing speed
Learning goals:
- Derive a parabolic no-slip velocity profile from pressure and shear forces.
- Integrate annular flow to distinguish mean speed, centre speed and volume flow.
- Explain a steady pressure drop using mechanical-energy dissipation.
- Compare fixed-pressure and fixed-flow changes without swapping the conditions.
- Solve linear series/parallel tube networks and audit connector assumptions.
- Infer viscosity from data and distinguish the viscous-drain time law from ideal efflux.
Equations in this section (LaTeX):
P_{\mathrm{diss}}=\Delta p\,Q,\qquad [P]=\mathrm{Pa}\,\mathrm{m^3/s}=\mathrm W
Meaning: For this uniform horizontal fully developed segment, pressure-work loss equals viscous dissipation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

05 / Viscous flow through pipes

Ask what the pump holds fixed before using radius scaling

“Narrowing reduces flow by the fourth power” is incomplete unless the pressure drop, viscosity and length are held fixed. A pump that actively maintains the same flow produces a different comparison: the required pressure drop increases as inverse radius to the fourth power.

The mean speed and Reynolds number also respond differently. At fixed pressure drop, mean speed scales as R² and Reynolds number as R³. At fixed flow, mean speed scales as inverse R² and Reynolds number as inverse R. A narrowing that reduces Reynolds number in one experiment increases it in the other.

Always state which experiment the comparison describes. After calculating the new parameters, check that both states still justify the laminar model. A mathematically valid scaling is not a licence to extrapolate beyond it.

WORKED EXAMPLE

Two pumps answer the same narrowing differently

A tube initially carries 4.0 mL/min under laminar no-slip conditions. Its internal radius is halved while viscosity and measuring length stay fixed. Compare the flow if pressure drop stays fixed with the required pressure if flow stays fixed. Assume both final states remain within the model.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

STOP & THINK

A pump maintains constant volume flow while a tube’s radius decreases. Under the same laminar model, what must happen to the pressure drop?

It decreases as radius to the fourth power
It increases as inverse radius to the fourth power
It remains constant by definition

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Viscous flow through pipes
Focus: Ask what the pump holds fixed before using radius scaling
Learning goals:
- Derive a parabolic no-slip velocity profile from pressure and shear forces.
- Integrate annular flow to distinguish mean speed, centre speed and volume flow.
- Explain a steady pressure drop using mechanical-energy dissipation.
- Compare fixed-pressure and fixed-flow changes without swapping the conditions.
- Solve linear series/parallel tube networks and audit connector assumptions.
- Infer viscosity from data and distinguish the viscous-drain time law from ideal efflux.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

06 / Viscous flow through pipes

Combine pressure losses and flow rates consistently

Define a hydraulic resistance as pressure drop divided by volume flow. For this linear laminar model it depends on viscosity and geometry, not on the chosen pressure drop. Sections in series carry the same incompressible steady flow; their pressure drops add. Parallel branches between common pressure nodes have the same pressure drop; their flows add.

The algebra resembles a resistor network, but pressure is not electric potential and volume flow is not electric current. Keep their units and physical assumptions explicit. Neglecting the loss at a connector is a stated approximation, especially when tube sizes change abruptly.

\begin{aligned}\mathcal R&=\frac{8\eta L}{\pi R^4}\\ \Delta p&=\mathcal R Q\\ \mathcal R_{\mathrm{series}}&=\sum_i\mathcal R_i\\ \frac1{\mathcal R_{\mathrm{parallel}}}&=\sum_i\frac1{\mathcal R_i}\end{aligned}

These resistance combinations apply to linear laminar segments between well-defined pressure nodes, with connector losses excluded.

Explain this equation

Read it in words

Hydraulic resistance converts a flow into its required pressure drop. Series adds resistances; parallel adds conductances, which are reciprocal resistances.

Symbols and units

$\mathcal R$: Hydraulic resistance, distinct from the tube radius R.Pa·s/m³.
$\eta$: Dynamic viscosity of the liquid.Pa·s.
$L, R$: Tube length and internal radius.Metres (m).
$\Delta p$: Driving pressure difference between the segment’s nodes.Pascals (Pa).
$Q$: Volume flow rate through the segment.m³/s.
$\sum_i$: Add contributions from all the specified segments or branches.An operation, not another physical quantity.
$\mathcal R_i$: Resistance of individual tube i. The index labels each tube; series and parallel subscripts label the equivalent resistance of the whole arrangement.Pa·s/m³.

Signs and operations

The reciprocal in the parallel relation expresses the addition of flow per unit pressure difference. The fourth power acts on tube radius; the calligraphic resistance symbol is a different quantity.

When it applies

Use steady incompressible, Newtonian, fully developed laminar segments with no slip and negligible connector/end losses. Pressure nodes must correspond to the same elevation, or use the appropriate pressure-plus-gravity driving head.

Make it concrete

Two equal-length tubes of radii r and 2r have resistances in the ratio 16:1. In series the smaller tube takes 16/17 of the total pressure loss; in parallel the larger tube carries 16/17 of the total flow.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Equation (LaTeX): \begin{aligned}\mathcal R&=\frac{8\eta L}{\pi R^4}\\ \Delta p&=\mathcal R Q\\ \mathcal R_{\mathrm{series}}&=\sum_i\mathcal R_i\\ \frac1{\mathcal R_{\mathrm{parallel}}}&=\sum_i\frac1{\mathcal R_i}\end{aligned}
Read it as: Hydraulic resistance converts a flow into its required pressure drop. Series adds resistances; parallel adds conductances, which are reciprocal resistances.
Symbols:
\mathcal R: Hydraulic resistance, distinct from the tube radius R.; Pa·s/m³.
\eta: Dynamic viscosity of the liquid.; Pa·s.
L, R: Tube length and internal radius.; Metres (m).
\Delta p: Driving pressure difference between the segment’s nodes.; Pascals (Pa).
Q: Volume flow rate through the segment.; m³/s.
\sum_i: Add contributions from all the specified segments or branches.; An operation, not another physical quantity.
\mathcal R_i: Resistance of individual tube i. The index labels each tube; series and parallel subscripts label the equivalent resistance of the whole arrangement.; Pa·s/m³.
Signs and operations: The reciprocal in the parallel relation expresses the addition of flow per unit pressure difference. The fourth power acts on tube radius; the calligraphic resistance symbol is a different quantity.
Conditions: Use steady incompressible, Newtonian, fully developed laminar segments with no slip and negligible connector/end losses. Pressure nodes must correspond to the same elevation, or use the appropriate pressure-plus-gravity driving head.

Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

WORKED EXAMPLE

Locate the pressure loss in a series pair

Two long horizontal tube segments carry the same Newtonian liquid in series. Their lengths are equal and their radii are r and 2r. Total pressure drop is 170 Pa. Assuming fully developed laminar flow and negligible junction loss, find the pressure drop in each.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

WORKED EXAMPLE

Locate the flow in a parallel pair

The same equal-length tubes of radii r and 2r are instead connected in parallel between common pressure nodes. Total flow is 3.4 mL/min. Under the same linear laminar assumptions with negligible connector losses, find each branch flow.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Viscous flow through pipes
Focus: Combine pressure losses and flow rates consistently
Learning goals:
- Derive a parabolic no-slip velocity profile from pressure and shear forces.
- Integrate annular flow to distinguish mean speed, centre speed and volume flow.
- Explain a steady pressure drop using mechanical-energy dissipation.
- Compare fixed-pressure and fixed-flow changes without swapping the conditions.
- Solve linear series/parallel tube networks and audit connector assumptions.
- Infer viscosity from data and distinguish the viscous-drain time law from ideal efflux.
Equations in this section (LaTeX):
\begin{aligned}\mathcal R&=\frac{8\eta L}{\pi R^4}\\ \Delta p&=\mathcal R Q\\ \mathcal R_{\mathrm{series}}&=\sum_i\mathcal R_i\\ \frac1{\mathcal R_{\mathrm{parallel}}}&=\sum_i\frac1{\mathcal R_i}\end{aligned}
Meaning: These resistance combinations apply to linear laminar segments between well-defined pressure nodes, with connector losses excluded.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

07 / Viscous flow through pipes

Fit the relation and audit its limits

A pressure-versus-flow measurement can estimate viscosity if the tube geometry is known and the model is valid. A constant viscosity predicts a straight line through zero after the pressure zero has been corrected. A nonzero intercept needs investigation; it is not itself a measurement of a different viscosity.

The radius deserves careful measurement because resistance contains its inverse fourth power. A small fractional radius error contributes roughly four times that fractional error to a viscosity inferred from this experiment. Temperature drift, a noncircular tube, entrance development and connector losses can also bias the inference.

Reynolds number is one diagnostic, not a complete certificate. Specify diameter as the length scale for a circular pipe and mean speed as the speed scale. Transition depends on inlet disturbances and the apparatus, so a single threshold should not be treated as a universal switch. Low Reynolds number does not remove the need for a fully developed measuring section or a Newtonian material response.

WORKED EXAMPLE

Separate a fitted slope from a pressure zero

A 0.50 m measuring length of a 1.0 mm radius tube gives measured pressure differences 210 Pa at 1.8 mL/min and 330 Pa at 3.0 mL/min. Suppose the flow follows the linear laminar model but the pressure instrument may have a constant zero offset. Estimate that offset and the viscosity. Both points alone do not establish why an offset exists.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Viscous flow through pipes
Focus: Fit the relation and audit its limits
Learning goals:
- Derive a parabolic no-slip velocity profile from pressure and shear forces.
- Integrate annular flow to distinguish mean speed, centre speed and volume flow.
- Explain a steady pressure drop using mechanical-energy dissipation.
- Compare fixed-pressure and fixed-flow changes without swapping the conditions.
- Solve linear series/parallel tube networks and audit connector assumptions.
- Infer viscosity from data and distinguish the viscous-drain time law from ideal efflux.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

08 / Viscous flow through pipes

A viscous drain has a different time law

If a large reservoir drains through a long narrow tube, the pressure driving the flow can decrease with the liquid height. In a quasi-steady viscous-dominated model, the instantaneous flow is proportional to the hydrostatic head h. Volume conservation then makes the height’s rate of decrease proportional to h itself, producing an exponential.

This differs from a large-reservoir ideal orifice, where exit speed is proportional to the square root of height. The two time laws are not competing algebraic tricks: they represent different dominant energy balances. A short outlet dominated by acceleration cannot be modelled as a long viscous tube simply because both let liquid out.

The exponential’s arbitrarily late tail is a property of this approximation. At sufficiently small heads, capillary pressure, remaining geometry, leaking air or other omitted effects can matter. Do not turn an asymptotic idealization into a claim about how an actual vessel empties.

A\frac{dh}{dt}=-Q=-\frac{\rho gh}{\mathcal R},\qquad h(t)=h_0e^{-t/\tau},\qquad\tau=\frac{A\mathcal R}{\rho g}

A is the reservoir’s constant horizontal area. The quasi-steady model neglects entrance/exit kinetic losses, capillary effects and changes in tube resistance.

WORKED EXAMPLE

Each halving takes the same time in this model

A constant-area reservoir drains through a long narrow tube in the quasi-steady viscous-dominated model. Its head falls from 8.0 cm to 4.0 cm in 30 s. How much longer does the model predict for the fall from 4.0 cm to 2.0 cm, and what is its time constant?

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Viscous flow through pipes
Focus: A viscous drain has a different time law
Learning goals:
- Derive a parabolic no-slip velocity profile from pressure and shear forces.
- Integrate annular flow to distinguish mean speed, centre speed and volume flow.
- Explain a steady pressure drop using mechanical-energy dissipation.
- Compare fixed-pressure and fixed-flow changes without swapping the conditions.
- Solve linear series/parallel tube networks and audit connector assumptions.
- Infer viscosity from data and distinguish the viscous-drain time law from ideal efflux.
Equations in this section (LaTeX):
A\frac{dh}{dt}=-Q=-\frac{\rho gh}{\mathcal R},\qquad h(t)=h_0e^{-t/\tau},\qquad\tau=\frac{A\mathcal R}{\rho g}
Meaning: A is the reservoir’s constant horizontal area. The quasi-steady model neglects entrance/exit kinetic losses, capillary effects and changes in tube resistance.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

BRING THE IDEAS TOGETHER

What should stay with you

Viscosity relates shear stress to velocity gradient in the stated Newtonian model.
Poiseuille flow is an area integral; its mean speed is half its centre value.
The radius-to-fourth-power rule needs a stated controlled pressure or flow.
A pressure drop can pay for dissipation while a velocity profile stays steady.
Check material, geometry, flow regime and entrance conditions before extrapolating.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

In a fully developed laminar circular tube, the centre speed is 0.080 m/s. What is the speed at a radial distance 0.5R, in m/s?

Answer in m/s.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 2

A horizontal uniform fully developed pipe segment has pressure drop 250 Pa and volume flow 4.0 mL/s. Its entering and leaving velocity profiles are identical. What power is dissipated, in mW?

Answer in mW.

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 3

Within the same Poiseuille model, radius doubles while pressure drop, viscosity and measuring length remain fixed. The new volume flow is…

Four times the original
Eight times the original
Sixteen times the original

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

SELF CHECK 4

The shear-stress magnitude is zero on the axis of a fully developed circular-pipe profile. What follows?

The centre fluid is stationary
The velocity gradient is zero there, while speed is maximal
There can be no pressure drop

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

Weight the speed by areaIntermediate · Hints and guided reasoning A flow controller pays a pressure costAdvanced · Hints and guided reasoning A wider branch can also be longerAdvanced · Hints and guided reasoning The shorter narrow segment can dominateAdvanced · Hints and guided reasoning Use the slope when the zero is uncertainAdvanced · Hints and guided reasoning Recognize a constant fractional decreaseAdvanced · Hints and guided reasoning A larger sphere falls through syrupIntermediate · Hints and guided reasoning

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

A tube radius is halved. Can you immediately say that its volume flow falls by sixteen?

Compare your reasoning

Only if pressure drop, viscosity and length stay fixed and the laminar model remains valid. If the source instead fixes flow, the required pressure drop rises sixteenfold. Decide what is controlled before taking a ratio.

Revisit the explanation →

A few days later: change the situation

Why does averaging the plotted speeds at equally spaced points along a diameter fail to give the volume flow?

Compare your reasoning

Those sample points do not represent equal cross-sectional areas. An outer annulus covers more area than an inner annulus of the same thickness. Weight each velocity by 2πs ds; the parabolic profile’s area mean is half its centre speed.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

Back to chapter contents ↑