Viscosity and Poiseuille Flow (4B)
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MCAT Chemical and Physical Foundations of Biological Systems › Viscosity and Poiseuille Flow (4B)
A lab compares two Newtonian fluids flowing through the same rigid cylindrical capillary (same $r$ and $L$) under the same pressure drop $\Delta P$. Fluid X has viscosity $\eta_X = 1.0\ \text{mPa}\cdot\text{s}$ and Fluid Y has viscosity $\eta_Y = 4.0\ \text{mPa}\cdot\text{s}$ at the measurement temperature. Flow is laminar for both. Based on Poiseuille flow, what is the expected ratio $Q_X/Q_Y$?
$Q_X/Q_Y = 1$ because both experience the same $\Delta P$.
$Q_X/Q_Y = 4$ because $Q \propto 1/\eta$ with other variables fixed.
$Q_X/Q_Y = 16$ because $Q \propto 1/\eta^2$.
$Q_X/Q_Y = 1/4$ because higher viscosity increases flow resistance.
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law demonstrates that Q = (πr⁴ΔP)/(8ηL), establishing an inverse proportionality between flow rate and viscosity. In this comparison, Fluid Y has four times the viscosity of Fluid X (4.0 vs 1.0 mPa·s) while using the same capillary and pressure drop, resulting in Fluid Y having one-fourth the flow rate. Choice B is correct because Q_X/Q_Y = η_Y/η_X = 4.0/1.0 = 4, accurately reflecting the Q ∝ 1/η relationship. Choice A reverses the ratio, choice C incorrectly squares the viscosity effect, and choice D ignores viscosity's crucial role. When comparing fluids in Poiseuille flow, the flow rate ratio equals the inverse of the viscosity ratio.
A researcher evaluates whether a capillary-flow assay is sensitive to small manufacturing variation in tube radius. Two nominally identical rigid cylindrical capillaries have the same length $L$ and are used with the same Newtonian fluid at the same temperature and the same applied pressure drop $\Delta P$. Capillary A has radius $r$, while Capillary B has radius $0.90r$. Assuming laminar Poiseuille flow, which outcome is most consistent with the expected change in volumetric flow rate?
$Q_B \approx 0.81Q_A$ because flow rate is proportional to cross-sectional area.
$Q_B \approx 0.66Q_A$ because flow rate scales as $r^4$.
$Q_B \approx 1.11Q_A$ because a smaller radius increases speed (Bernoulli principle).
$Q_B \approx 0.90Q_A$ because flow rate is directly proportional to radius.
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law reveals that Q = (πr⁴ΔP)/(8ηL), demonstrating the critical r⁴ dependence that makes flow extremely sensitive to radius variations. When Capillary B has radius 0.90r compared to Capillary A's radius r, the flow rate ratio is Q_B/Q_A = (0.90)⁴ = 0.6561 ≈ 0.66. Choice C is correct because it accurately calculates that Q_B ≈ 0.66Q_A based on the fourth-power radius relationship in Poiseuille flow. Choices A and B underestimate the effect by assuming linear or quadratic relationships, while choice D contradicts physics with incorrect Bernoulli reasoning. This extreme sensitivity to radius (10% decrease causes 34% flow reduction) explains why precise manufacturing tolerances are crucial for microfluidic devices.
To compare two capillaries, a student drives the same Newtonian fluid (viscosity $\eta$) through each under identical pressure drop $\Delta P = 1.0\ \text{kPa}$. Capillary 1 has length $L$ and radius $r$. Capillary 2 has length $2L$ and radius $r$. The flow is laminar in both. Based on Poiseuille’s equation, which outcome is consistent with these conditions?
Both capillaries have the same flow rate because $\Delta P$ is the same.
Capillary 2 has one-fourth the flow rate because $Q \propto 1/L^2$.
Capillary 2 has twice the flow rate because a longer tube sustains flow longer.
Capillary 2 has half the flow rate of Capillary 1 because $Q \propto 1/L$.
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law shows that Q = (πr⁴ΔP)/(8ηL), indicating flow rate is inversely proportional to tube length. In this comparison, Capillary 2 has twice the length (2L) of Capillary 1 while all other parameters remain constant, resulting in half the flow rate. Choice A is correct because it accurately states that Capillary 2 has half the flow rate of Capillary 1 due to the Q ∝ 1/L relationship. Choice B incorrectly suggests longer tubes increase flow, choice C ignores the length effect, and choice D incorrectly proposes a quadratic relationship. When analyzing Poiseuille flow, verify that each parameter's effect matches the law's predictions: linear relationships for ΔP, η, and L, but fourth power for r.
In a microfluidics experiment, a Newtonian buffer is driven through a straight cylindrical glass capillary of length $L = 10\ \text{cm}$ and radius $r = 0.50\ \text{mm}$ under a constant pressure drop $\Delta P = 2.0\ \text{kPa}$. The flow is verified to be laminar and fully developed. The same capillary is then used with a second buffer at the same temperature, identical except its dynamic viscosity is doubled (from $\eta$ to $2\eta$). Based on Poiseuille flow, what change in volumetric flow rate $Q$ is expected under the same $\Delta P$, $L$, and $r$?
$Q$ remains unchanged because the pressure drop is held constant.
$Q$ decreases to one-fourth of its original value because viscous resistance scales with $\eta^2$.
$Q$ decreases to one-half of its original value because $Q \propto 1/\eta$ for laminar capillary flow.
$Q$ increases because higher viscosity increases the pressure transmitted to the fluid (Bernoulli effect).
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law states that volumetric flow rate Q = (πr⁴ΔP)/(8ηL), showing that flow rate is inversely proportional to viscosity. In this scenario, doubling the viscosity from η to 2η while keeping all other parameters constant will halve the flow rate. Choice C is correct because it accurately reflects that Q ∝ 1/η for laminar capillary flow, resulting in Q decreasing to one-half its original value. Choice A incorrectly suggests Q ∝ 1/η², while choices B and D contradict the fundamental inverse relationship between flow rate and viscosity. When solving Poiseuille flow problems, always check that the relationship between Q and each variable matches the law's predictions.
A physiologist approximates flow through a small arteriole as steady laminar flow in a rigid cylindrical tube. A vasodilator increases the arteriole radius by 10% (from $r$ to $1.10r$) without changing $\Delta P$ across the segment, its length $L$, or blood viscosity $\eta$. Based on Poiseuille’s equation, which change in flow rate is most consistent with this model?
Flow rate increases by about 10% because $Q \propto r$.
Flow rate increases by about 46% because $Q \propto r^4$.
Flow rate decreases because a larger radius lowers velocity and thus lowers $Q$ (Bernoulli reasoning).
Flow rate increases by about 21% because $Q \propto r^2$.
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law establishes that Q = (πr⁴ΔP)/(8ηL), showing flow rate scales with the fourth power of radius. When the arteriole radius increases by 10% (from r to 1.10r), the flow rate increases by (1.10)⁴ = 1.4641, representing a 46.41% increase. Choice C is correct because it accurately calculates the 46% increase resulting from the Q ∝ r⁴ relationship in Poiseuille flow. Choices A and B underestimate by assuming linear or quadratic relationships, while choice D contradicts physics by suggesting larger radii decrease flow. In physiological applications, this r⁴ dependence explains why small vessel diameter changes dramatically affect blood flow and why vasoregulation is so effective.
A device uses laminar flow through a cylindrical capillary to deliver a drug solution. The designer can change only one parameter while keeping the others constant: pressure drop $\Delta P$, tube length $L$, and fluid viscosity $\eta$ remain fixed. Which modification is most effective for increasing the volumetric flow rate $Q$ by approximately an order of magnitude (about $10\times$) while remaining within the Poiseuille-flow model?
Increase the tube radius by a factor of 1.8 (since $Q \propto r^2$).
Decrease the tube length by a factor of 2 (since $Q \propto 1/L$).
Increase the pressure drop by a factor of 2 (since $Q \propto 1/\Delta P$).
Increase the tube radius by a factor of 2 (since $Q \propto r^4$).
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law shows Q = (πr⁴ΔP)/(8ηL), revealing that flow rate depends on r⁴, making radius changes most effective for large flow increases. To achieve a 10-fold increase in Q, the radius must increase by ⁴√10 ≈ 1.78, so doubling the radius yields 2⁴ = 16-fold increase, exceeding the target. Choice A is correct because increasing radius by factor of 2 produces the desired order-of-magnitude increase through the r⁴ dependence. Choice B only doubles flow rate, choice C incorrectly assumes r² dependence, and choice D incorrectly inverts the pressure relationship. For optimizing Poiseuille flow systems, radius adjustments provide the most dramatic effects due to the fourth-power relationship.
A physiology lab models blood flow through a small artery as steady, laminar Poiseuille flow in a rigid cylindrical vessel. During a cold-pressor test, sympathetic activation causes the artery radius to decrease from $r$ to $0.90r$ while mean arterial pressure and vessel length remain approximately constant over the short interval. Viscosity is assumed unchanged. Based on Poiseuille’s relationship $Q \propto r^4$, what is the expected change in flow rate through that artery?
(You may use: $(0.90)^4 \approx 0.66$.)
Flow decreases to about $0.66Q$ because of the $r^4$ dependence.
Flow is unchanged because pressure is the only determinant of flow in a tube.
Flow decreases to about $0.90Q$ because flow is directly proportional to radius.
Flow increases because constriction increases fluid speed by Bernoulli’s principle.
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law shows that flow rate depends on the fourth power of radius: Q ∝ r⁴ when other parameters are constant. In this scenario, the artery radius decreases from r to 0.90r during sympathetic activation, which dramatically reduces flow due to the r⁴ dependence. Choice A is correct because (0.90)⁴ ≈ 0.66, meaning flow decreases to about 66% of its original value. Choice B fails because it assumes a linear relationship between radius and flow, missing the critical fourth-power dependence. In similar questions, always remember the strong r⁴ dependence and calculate the effect of radius changes by raising the ratio to the fourth power.
In a capillary viscometry setup, a constant pressure source applies $\Delta P$ across a rigid tube of radius $r$ and length $L$. The fluid is Newtonian and the flow is verified to be laminar. The investigator accidentally records the tube diameter $d$ as if it were the radius when predicting the effect of changing tube size. If the true radius is doubled (from $r$ to $2r$) while $\Delta P$, $L$, and $\eta$ are held constant, which prediction for the change in $Q$ is consistent with Poiseuille flow?
$Q$ increases by a factor of 4 because $Q \propto r^2$.
$Q$ increases by a factor of 16 because $Q \propto r^4$.
$Q$ increases by a factor of 2 because $Q \propto r$.
$Q$ increases by a factor of 8 because $Q \propto r^3$.
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law clearly states that volumetric flow rate depends on the fourth power of radius: Q ∝ r⁴ when other parameters are constant. In this scenario, doubling the radius from r to 2r increases flow by a factor of 2⁴ = 16, regardless of any confusion about diameter versus radius in the setup description. Choice D is correct because it accurately reflects the r⁴ dependence that is fundamental to laminar flow in tubes. Choices A, B, and C fail because they suggest incorrect power relationships (r¹, r², and r³ respectively) that do not match Poiseuille's law. In similar questions, always remember the strong fourth-power radius dependence, which makes radius changes the most dramatic factor affecting flow rate.
A lab uses the same capillary (fixed $r$ and $L$) to measure flow of a Newtonian fluid at two temperatures. At $20^\circ\text{C}$, the viscosity is $\eta_{20}$ and the measured flow rate is $Q_{20}$. At $37^\circ\text{C}$, the viscosity decreases to $0.75\eta_{20}$ while the applied pressure drop $\Delta P$ is kept constant and the flow remains laminar. Based on Poiseuille flow, what is the expected $Q_{37}$ relative to $Q_{20}$?
$Q_{37} = 1.33Q_{20}$ because flow is inversely proportional to viscosity.
$Q_{37} = 1.75Q_{20}$ because warmer fluids always flow faster by Bernoulli’s principle.
$Q_{37} = 0.75Q_{20}$ because flow is proportional to viscosity.
$Q_{37} = Q_{20}$ because viscosity changes do not affect laminar flow.
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law states that flow rate is inversely proportional to viscosity: Q ∝ 1/η when other parameters remain constant. In this scenario, warming the fluid from 20°C to 37°C reduces viscosity to 0.75η₂₀, which increases flow rate by the reciprocal factor. Choice B is correct because Q₃₇ = Q₂₀ × (η₂₀/0.75η₂₀) = Q₂₀ × (1/0.75) = 1.33Q₂₀, properly applying the inverse viscosity relationship. Choice A fails because it incorrectly assumes flow is proportional to viscosity rather than inversely proportional. In similar questions, remember that lower viscosity means easier flow, and calculate the flow ratio as the inverse of the viscosity ratio.
In a microfluidics study, researchers drive an incompressible Newtonian fluid through a straight glass capillary under steady, laminar conditions. The capillary length is held constant at $L = 10\ \text{cm}$ and the pressure drop is fixed at $\Delta P = 20\ \text{kPa}$. The temperature is kept constant so viscosity does not change during a run. The team replaces Fluid 1 (viscosity $\eta_1 = 1.0\ \text{mPa}\cdot\text{s}$) with Fluid 2 (viscosity $\eta_2 = 2.0\ \text{mPa}\cdot\text{s}$) while keeping the same capillary radius. Based on Poiseuille flow, what change in volumetric flow rate $Q$ is expected?
(Use Poiseuille proportionality: $Q \propto \frac{\Delta P, r^4}{\eta L}$.)
$Q$ decreases by a factor of 2 because $Q \propto 1/\eta$.
$Q$ doubles because higher viscosity increases the pressure-driven flux.
$Q$ is unchanged because $\Delta P$ and $L$ are unchanged.
$Q$ decreases by a factor of 16 because $Q \propto 1/r^4$ and viscosity increased.
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law states that volumetric flow rate is inversely proportional to viscosity: Q ∝ ΔPr⁴/(ηL). In this scenario, doubling the viscosity from 1.0 to 2.0 mPa·s while keeping all other parameters constant will halve the flow rate. Choice C is correct because it accurately reflects this inverse relationship: when viscosity doubles, flow rate decreases by a factor of 2. Choice A fails because it incorrectly suggests that higher viscosity increases flow, which contradicts the physics of viscous resistance. In similar questions, always check that flow rate decreases with increasing viscosity and verify the proportionality relationships are applied correctly.