Viscous flow: connect pressure loss, radius, and resistance

MCAT content library / Chemistry & physics / Viscous flow: connect pressure loss, radius, and resistance

A pressure drop can maintain steady flow without speeding the fluid up. Read two tube experiments, then test what a narrowed radius or changed fluid would actually predict.

About 20 minutes, including the worked examples and checks.

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Useful foundations

Ideal fluid flow: use continuity before Bernoulli

On this page

A steady flow can require a continuing pressure drop
State the tube model before using the fourth power
Worked records: read resistance from the axes
Specify what is held fixed during narrowing
Connect pressure loss to energy and model limits
Carry the physical relation into circulation with its boundaries intact
Check your reasoning
Continue learning
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A steady flow can require a continuing pressure drop

In a horizontal tube of constant area, a steady incompressible flow has the same average speed at its ends. A real viscous fluid can still require higher pressure upstream. The pressure difference supplies the work needed to overcome resistance; it need not produce a continuing increase in average speed. AAMC includes viscosity, Poiseuille flow, turbulence, and circulatory applications within its fluid scope.¹

Dynamic viscosity, η, describes a fluid’s resistance to shear, with SI units Pa·s. Layers moving at different speeds exert viscous forces on one another. With the no-slip condition at a stationary tube wall, fluid at the wall is stationary relative to it, while fluid farther inward can move faster. Fully developed laminar flow in the ideal circular-tube model has a parabolic velocity profile; Q/A gives its cross-sectional average speed, not its centerline speed.²⁶

State the tube model before using the fourth power

For steady, fully developed laminar flow of an incompressible Newtonian fluid through a straight, rigid circular tube of uniform internal radius r and length L, the horizontal-tube result is Q = πr⁴Δp/(8ηL). Newtonian here means a constant viscosity relates shear stress to the velocity gradient under the specified conditions. The result also assumes no slip and negligible entrance effects over the section being analyzed.²⁶

Define hydraulic resistance as R = Δp/Q. Under these conditions, R = 8ηL/(πr⁴). At fixed geometry and viscosity, Q increases linearly with Δp. Increasing length or viscosity raises resistance; increasing internal radius lowers it strongly. The SI resistance unit is Pa·s/m³.³

Use the actual internal radius in the formula. An internal diameter of 0.80 mm means r = 0.40 mm = 4.0 × 10⁻⁴ m. Substituting the diameter as though it were the radius would overpredict flow by 2⁴ = 16. In a ratio comparison, radius and diameter have the same fractional change, so a diameter ratio can be raised to the fourth power if it is used consistently.

Worked records: read resistance from the axes

The records below are original illustrative data for two separate horizontal tubes. Their lengths are equal, they contain the same fluid at the same controlled temperature, and the stated laminar-flow model applies throughout the displayed range. Their internal radii can differ. Each tube’s pressure difference and steady flow are measured separately.

Original illustrative laminar-flow records. Pressure drop is plotted against volume flow, so each line’s slope is the hydraulic resistance. Data values are supplied in the table.

Q (mL/s)	Tube A Δp (kPa)	Tube B Δp (kPa)
0	0	0
0.10	0.80	0.20
0.20	1.60	0.40
0.30	2.40	0.60

Illustrative pressure drops at matched volume flows

Because pressure drop is on the vertical axis, slope equals Δp/Q, or resistance. Tube A’s slope is 8.0 kPa/(mL/s), and B’s is 2.0 kPa/(mL/s). In SI units these are 8.0 × 10⁹ and 2.0 × 10⁹ Pa·s/m³. Reversing the axes would make the slope Q/Δp, the conductance rather than the resistance.

At the same 0.40 kPa pressure difference, the model predicts Q_A = 0.050 mL/s and Q_B = 0.20 mL/s. Since length and viscosity are equal, R_A/R_B = 4 = (r_B/r_A)⁴. Thus B’s radius is √2, about 1.41, times A’s radius. Four times the flow does not require four times the radius.

Specify what is held fixed during narrowing

Replace tube A with a tube of the same length but an internal radius uniformly reduced to 0.80 of its former value. Keep the same fluid, temperature, and pressure difference, and retain the laminar assumptions. The flow ratio is Q_new/Q_old = 0.80⁴ = 0.4096. At 0.80 kPa, the former 0.100 mL/s becomes about 0.0410 mL/s.

If the goal is instead to restore 0.100 mL/s through the narrower tube, pressure must increase by 1/0.4096, about 2.44. The required pressure difference becomes about 1.95 kPa. “Narrowing changes the flow” is therefore incomplete without specifying whether pressure or flow is controlled.

This calculation changes the radius along the entire modeled tube. A short local constriction has its own length and can introduce additional entrance, exit, or turbulent losses. It cannot automatically be treated as though the whole original tube had that smaller radius.

Connect pressure loss to energy and model limits

The net pressure work supplied to a volume ΔV moving through the section is ΔpΔV; the corresponding rate is ΔpQ. In the horizontal constant-area steady case, the fluid’s endpoint kinetic energies per volume match. The supplied mechanical energy is dissipated, ultimately increasing internal thermal energy. This is why a pressure-loss term is needed beyond the loss-free Bernoulli account.³

At higher speeds, or near disturbances, the simple laminar model may fail. The Reynolds number Re = ρvD/η compares inertial and viscous effects, using average speed v and internal diameter D. Increasing speed increases Re when the other terms are fixed. A transition threshold depends on the flow setting, so neither one high speed nor a curved pressure–flow plot alone establishes a unique mechanism.⁴

If a slope changes, examine fluid temperature, internal geometry, length, pressure calibration, and whether the flow is steady and developed. A changed fluid can alter the tube as well as viscosity. Repeating measurements is useful, but repeating the same uncontrolled comparison does not isolate which factor changed.

Carry the physical relation into circulation with its boundaries intact

In circulation, pressure differences drive flow and vessel geometry affects resistance. Blood vessels also expand, recoil, branch, and experience pulsatile pressures. A single rigid-tube prediction is therefore a specified approximation, not a complete account of the arterial or venous network.⁵

For a branching bed, compare total cross-sectional area and total flow before inferring average speed. One small capillary does not carry the entire incoming arterial flow; the many parallel branches share it. The large combined capillary area helps explain low average capillary speeds even though each individual vessel is narrow.⁵

The useful transferable sequence is to identify the controlled pressure or flow, verify the geometry and flow regime, and then apply the appropriate resistance relation. A strong radius dependence makes geometry especially valuable to measure before attributing a changed flow to the fluid itself.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

A long rigid circular tube is replaced by one of the same length whose internal diameter is 0.40 mm instead of 0.80 mm. The fluid, temperature, and pressure difference are unchanged, and steady fully developed laminar conditions remain valid. What percentage of the original volume flow does the new tube carry?

A. 6.25%

Correct. The diameter and radius both decrease by a factor of two. At fixed pressure, viscosity and length, Q scales with r⁴, so the flow is (1/2)⁴ = 1/16 = 6.25% of its former value.

B. 25%

Review this choice. This applies the area ratio rather than the Poiseuille flow ratio. Area scales with radius squared, while pressure-driven laminar flow in this model scales with radius to the fourth power.

C. 50%

Review this choice. This scales flow directly with diameter. The fourth-power relationship produces a substantially larger reduction.

D. 200%

Review this choice. This reverses the effect of narrowing at fixed pressure difference. Resistance increases, so volume flow decreases.

The principle to carry forward

Use the internal dimension and the correct exponent. A diameter ratio is valid in a radius-ratio formula because the factors of two cancel.

Check 2

At a controlled temperature, replacing one Newtonian fluid with another changes a tube’s straight Δp-versus-Q slope from 8 to 16 kPa/(mL/s). The tube’s length and internal radius are verified unchanged, pressure readings are calibrated, and the same laminar model applies. Which change accounts for the slope increase?

A. Doubling the tested flow rate while leaving viscosity unchanged.

Review this choice. That moves the operating point along the same line. With unchanged geometry and viscosity, it does not double the line’s slope.

B. Doubling the tested pressure difference while leaving viscosity unchanged.

Review this choice. A larger pressure difference produces a larger flow on the original line. It changes an operating point, not the resistance under the stated linear model.

C. Doubling the fluid’s dynamic viscosity.

Correct. The slope is R = 8ηL/(πr⁴). With length and radius fixed, doubling R requires doubling η within the stated model.

D. Halving the fluid’s dynamic viscosity.

Review this choice. This would halve the resistance and the Δp-versus-Q slope, the opposite of the observed change.

The principle to carry forward

Distinguish a changed operating point from a changed relationship. Axis choice determines whether the fitted slope is resistance or conductance.

Check 3

A new fluid gives half the old fluid’s steady flow through the same nominal tube at the same calibrated pressure difference and controlled temperature. The tube material may swell or contract when exposed to the replacement fluid. Which added measurement most directly tests an alternative to “the new fluid is more viscous”?

A. Measure the tube’s external diameter before either fluid is introduced.

Review this choice. The external diameter before exposure does not determine the internal radius under each flowing condition. It leaves the proposed geometry change unresolved.

B. Measure the tube’s internal radius with each fluid under the operating conditions.

Correct. Radius has a fourth-power effect in the model, and the material may change after fluid exposure. Measuring the relevant internal geometry addresses a competing explanation for the lower flow.

C. Measure the total fluid remaining in each supply bottle after the trial.

Review this choice. Supply volume may help document usage, but it does not directly distinguish a viscosity change from a changed internal flow radius under the stated conditions.

D. Measure the peak flow immediately after switching on the pressure source.

Review this choice. The startup peak is a transient measurement. It does not resolve the internal geometry during the steady comparison and introduces a condition outside the stated steady model.

The principle to carry forward

A shared label such as “the same tube” does not guarantee the same operating geometry. Measure the variable that supports the competing explanation.