Viscosity and Poiseuille Flow (4B)
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MCAT Chemical and Physical Foundations of Biological Systems › Viscosity and Poiseuille Flow (4B)
A student uses a syringe pump to generate steady laminar flow of a Newtonian fluid through a capillary. For a given trial, $\Delta P$, $r$, and $\eta$ are held constant while the student compares two capillary lengths: $L_1 = 4\ \text{cm}$ and $L_2 = 12\ \text{cm}$. Which statement is consistent with Poiseuille flow for the ratio of flow rates $Q_1/Q_2$?
$Q_1/Q_2 = 1$ because viscosity sets the flow rate, not length.
$Q_1/Q_2 = 9$ because $Q \propto 1/L^2$.
$Q_1/Q_2 = 1/3$ because longer tubes increase velocity via reduced wall interactions.
$Q_1/Q_2 = 3$ because $Q \propto 1/L$.
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 linear relationship between flow rate and tube length. Comparing capillaries with L₁ = 4 cm and L₂ = 12 cm while keeping other parameters constant, the flow rate ratio is Q₁/Q₂ = L₂/L₁ = 12/4 = 3. Choice A is correct because it accurately identifies that Q ∝ 1/L, resulting in the shorter capillary having three times the flow rate of the longer one. Choice B reverses the relationship, choice C incorrectly squares the length effect, and choice D ignores length's influence entirely. For Poiseuille flow comparisons, remember that doubling length halves flow rate, tripling length reduces flow to one-third, and so on.
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 $r$?
$Q$ 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).
$Q$ decreases to one-fourth of its original value because viscous resistance scales with $\eta^2$.
$Q$ remains unchanged because the pressure drop is held constant.
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 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 $Q_X/Q_Y$?
$Q_X/Q_Y = 4$ because $Q \propto 1/\eta$ with other variables fixed.
$Q_X/Q_Y = 1$ because both experience the same $\Delta P$.
$Q_X/Q_Y = 16$ 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.
An investigator models blood flow through a straight artery segment as laminar Poiseuille flow. During mild hypothermia, plasma viscosity increases by 25% (from $\eta$ to $1.25\eta$) while arterial radius, length, and the pressure drop across the segment are assumed unchanged. Under these assumptions, which statement best describes the influence of viscosity in this system?
Volumetric flow rate is unchanged because viscosity only affects turbulent flow.
Volumetric flow rate decreases to $0.80$ of baseline because $Q \propto 1/\eta$.
Volumetric flow rate increases to $1.25$ of baseline because $Q \propto \eta$.
Volumetric flow rate decreases to $0.64$ of baseline because $Q \propto 1/\eta^2$.
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 is inversely proportional to viscosity. In this hypothermia scenario, increasing viscosity by 25% (from η to 1.25η) while keeping radius, length, and pressure drop constant will reduce flow rate to Q/1.25 = 0.80Q. Choice A is correct because it accurately calculates that flow rate decreases to 0.80 of baseline, reflecting the Q ∝ 1/η relationship. Choice B incorrectly suggests a direct proportionality, choice C wrongly limits viscosity effects to turbulent flow, and choice D proposes an incorrect quadratic relationship. For medical applications of Poiseuille flow, remember that even modest viscosity changes can significantly impact perfusion.
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 twice the flow rate because a longer tube sustains flow longer.
Capillary 2 has one-fourth the flow rate because $Q \propto 1/L^2$.
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.
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 2 (since $Q \propto r^4$).
Increase the tube radius by a factor of 1.8 (since $Q \propto r^2$).
Increase the pressure drop by a factor of 2 (since $Q \propto 1/\Delta P$).
Decrease the tube length by a factor of 2 (since $Q \propto 1/L$).
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 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 46% because $Q \propto r^4$.
Flow rate increases by about 10% because $Q \propto r$.
Flow rate increases by about 21% because $Q \propto r^2$.
Flow rate decreases because a larger radius lowers velocity and thus lowers $Q$ (Bernoulli reasoning).
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.
In an experiment, laminar flow of a Newtonian fluid is established through a cylindrical capillary. The operator accidentally increases the pressure drop from $\Delta P$ to $2\Delta P$ while simultaneously switching to a fluid with twice the viscosity (from $\eta$ to $2\eta$). The tube radius $r$ and length $L$ are unchanged. Under Poiseuille flow, what is the expected net effect on the volumetric flow rate $Q$?
$Q$ is halved because $Q \propto \eta/\Delta P$.
$Q$ increases by a factor of 4 because $Q \propto(\Delta P)^2$.
$Q$ is unchanged because the factor of 2 in $\Delta P$ cancels the factor of 2 in $\eta$.
$Q$ doubles because pressure dominates viscosity in laminar flow.
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law states Q = (πr⁴ΔP)/(8ηL), showing flow rate is directly proportional to pressure drop and inversely proportional to viscosity. When pressure doubles (ΔP to 2ΔP) and viscosity doubles (η to 2η) simultaneously, these effects exactly cancel: Q_new = (πr⁴·2ΔP)/(8·2η·L) = Q_original. Choice B is correct because it recognizes that the factor of 2 increase in pressure drop perfectly compensates for the factor of 2 increase in viscosity, leaving flow rate unchanged. Choices A and C incorrectly weight one parameter over the other, while choice D proposes an incorrect quadratic pressure dependence. When multiple parameters change in Poiseuille flow, multiply their individual effects to find the net result.
A researcher measures laminar flow of a glycerol–water mixture through a rigid cylindrical capillary at constant temperature. The capillary length is held fixed at $L = 5.0\ \text{cm}$ and the applied pressure drop is held fixed at $\Delta P = 1.5\ \text{kPa}$. The capillary radius is increased from $r$ to $2r$ by switching to a wider tube of the same length. Assuming Poiseuille flow applies, what change in volumetric flow rate $Q$ is expected?
$Q$ increases by a factor of 2 because $Q \propto r$.
$Q$ increases by a factor of 4 because $Q \propto r^2$.
$Q$ increases by a factor of 16 because $Q \propto r^4$.
$Q$ decreases by a factor of 2 because wider tubes reduce shear rate and slow the flow.
Explanation
This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law reveals that volumetric flow rate Q = (πr⁴ΔP)/(8ηL), demonstrating that flow rate is proportional to the fourth power of radius. In this experiment, doubling the radius from r to 2r while maintaining constant pressure drop, length, and viscosity will increase flow rate by a factor of 2⁴ = 16. Choice D is correct because it accurately identifies that Q ∝ r⁴, resulting in a 16-fold increase in flow rate. Choices A and B incorrectly assume linear or quadratic relationships, while choice C contradicts the physics by suggesting wider tubes decrease flow. For Poiseuille flow problems, remember that the r⁴ dependence makes radius changes the most dramatic factor affecting flow rate.
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?
$Q_B \approx 1.11Q_A$ because a smaller radius increases speed (Bernoulli principle).
$Q_B \approx 0.90Q_A$ because flow rate is directly proportional to radius.
$Q_B \approx 0.81Q_A$ because flow rate is proportional to cross-sectional area.
$Q_B \approx 0.66Q_A$ because flow rate scales as $r^4$.
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.