Atherosclerotic plaque reduces the effective radius of a coronary artery by 50%. Assuming laminar flow conditions hold, by what factor does flow resistance increase?
A2× — resistance doubles because the cross-sectional area is halved
B4× — resistance quadruples because area depends on radius squared
C8× — resistance increases with the cube of radius reduction
D16× — resistance increases by the fourth power of the radius ratio
From Q = πR⁴ΔP/(8μL), flow resistance (ΔP/Q) ∝ 1/R⁴. Reducing R by half means R⁴ decreases by (1/2)⁴ = 1/16, so resistance increases 16-fold. This is why a 50% stenosis is medically critical — the heart cannot compensate for a 16-fold increase in resistance. The R⁴ relationship governs pipe flow because both the available flow area and the velocity profile (shaped by the no-slip condition and viscous resistance) change with radius.
Question 2 Multiple Choice
Water flows through a long straight pipe at Re = 800 (well within the laminar regime). If the pipe radius is doubled while keeping pressure drop ΔP, fluid viscosity μ, and pipe length L constant, by what factor does volumetric flow rate increase?
A2× — flow rate is proportional to radius
B4× — flow rate is proportional to cross-sectional area (R²)
C8× — flow rate is proportional to R³
D16× — flow rate is proportional to R⁴
Q = πR⁴ΔP/(8μL) shows Q ∝ R⁴ when ΔP, μ, and L are held constant. Doubling R gives Q increases by 2⁴ = 16. This fourth-power sensitivity makes Hagen-Poiseuille flow very responsive to small changes in pipe radius — the reason it dominates design considerations in microfluidics, biomedical devices, and any viscous-flow application where pipe geometry must be precisely controlled.
Question 3 True / False
In fully developed Hagen-Poiseuille flow, the fluid velocity at the pipe centerline equals the average (mean) velocity across the entire cross-section.
TTrue
FFalse
Answer: False
The parabolic velocity profile has its maximum at the centerline (r = 0): u_max = ΔPR²/(4μL). Integrating over the cross-section gives the mean velocity V_avg = u_max/2. The centerline velocity is exactly twice the mean velocity, not equal to it. This factor of two has practical consequences: in chemical reactors, the fastest fluid elements spend half as much time in the reactor as the average, leading to non-uniform conversion and residence time distributions.
Question 4 True / False
The Hagen-Poiseuille equation accurately predicts pressure drop in the region immediately downstream of a pipe entrance, before the velocity profile has fully developed.
TTrue
FFalse
Answer: False
The Hagen-Poiseuille solution assumes fully developed flow — the velocity profile shape is unchanged along the pipe length (∂u/∂x = 0). Near the pipe entrance, the flow transitions from a flat entry profile to the parabolic profile, and pressure drop per unit length is higher than the fully-developed prediction during this entrance region. The entrance length L_e ≈ 0.06 Re·D can extend many pipe diameters. Applying Hagen-Poiseuille to a short pipe or entrance region overestimates flow rate for a given pressure drop.
Question 5 Short Answer
Why does volumetric flow rate in a pipe depend on the fourth power of radius rather than the second power (cross-sectional area), and what physical mechanism explains this?
Think about your answer, then reveal below.
Model answer: The fourth-power dependence comes from two contributing R² factors. First, a larger radius provides more cross-sectional area for flow (∝ R²). Second, a larger radius means the parabolic velocity profile spans a wider range — the centerline (maximum) velocity increases with R² because viscous resistance over a longer radial distance is less, allowing faster fluid. Integrating this wider, faster parabolic profile over the larger cross-section gives Q ∝ R² × R² = R⁴.
This is why the R⁴ law is so dramatic in practice: a modest change in radius changes both how much fluid the pipe can hold and how fast that fluid moves, compounding the effect. In turbulent flow the relationship is weaker (closer to R^2.5) because turbulent mixing flattens the velocity profile, eliminating the velocity-amplification component of the R⁴ scaling.