Viscosity and Poiseuille Flow (4B) - Chem & Physics

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?
Answer: Increase the tube radius by a factor of 2 (since $Q \propto r^4$ ). 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?
Answer: Flow rate increases by about 46% because $Q \propto r^4$ . 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 research group models blood flow through a small artery as laminar flow in a rigid cylindrical tube. At rest, the artery has radius $r$ . During sympathetic activation, smooth muscle constriction reduces the radius to $0.75r$ while the pressure drop across the segment and blood viscosity are assumed constant over the short interval. According to Poiseuille's relationship, what happens to the volumetric flow rate $Q$ ?
Answer: $Q$ decreases to about $0.32Q_0$ because $Q \propto r^4$ . This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, emphasizing the strong dependence on radius to the fourth power. In this scenario, constricting the artery radius to 0.75r while keeping other factors constant significantly reduces the flow rate. Choice A is correct because it accurately reflects the relationship Q ∝ $r^4$ , resulting in Q ≈ 0.32 Q0. Choice C fails because smaller radius decreases flow rate, not increases velocity in a way that boosts Q. In similar questions, always calculate the radius factor raised to the fourth power and verify assumptions of constant viscosity and pressure.
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 $r$ ?
Answer: $Q$ decreases to one-half of its original value because $Q \propto 1/\eta$ for laminar capillary flow. 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 researcher uses the same capillary (fixed $r$ and $L$ ) and increases the pressure drop until the flow transitions from laminar to turbulent. While still in the laminar regime, which observation is most consistent with Poiseuille flow as $\Delta P$ is increased?
Answer: $Q$ increases linearly with $\Delta P$ until turbulence occurs. This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with linear dependence on ΔP in laminar regime. In this scenario, increasing ΔP drives higher flow until turbulence begins. Choice A is correct because it accurately reflects the linear increase Q ∝ ΔP while laminar. Choice B fails because Q is linear with ΔP, not sqrt(ΔP), in viscous-dominated laminar flow. In similar questions, distinguish laminar from turbulent regimes and confirm the linear pressure-flow relationship.
A capillary perfusion assay uses a constant-flow pump that maintains a fixed volumetric flow rate $Q$ through a rigid tube (radius $r$ , length $L$ ). The fluid is switched to one with double the viscosity while $Q$ , $r$ , and $L$ are held constant and flow remains laminar. According to Poiseuille's relationship, what happens to the required pressure drop $\Delta P$ ?
Answer: $\Delta P$ doubles because $\Delta P \propto \eta$ for fixed $Q$ . This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with ΔP ∝ η for fixed Q. In this scenario, doubling viscosity at constant Q requires adjusting the pressure drop. Choice A is correct because it accurately reflects that higher viscosity doubles the required ΔP to maintain Q. Choice B fails because higher viscosity increases, not reduces, shear stress and required pressure. In similar questions, rearrange Poiseuille's equation for the target variable and verify constant flow conditions.
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 $Q$ ?
Answer: $Q$ is unchanged because the factor of 2 in $\Delta P$ cancels the factor of 2 in $\eta$ . 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.
In a controlled perfusion experiment, a rigid capillary carries a Newtonian fluid at laminar flow. The investigator keeps $Q$ constant using a syringe pump but then reduces the tube radius from $r$ to $0.80r$ (same $L$ and $\eta$ ). According to Poiseuille flow, how must $\Delta P$ change to maintain the same $Q$ ?
Answer: $\Delta P$ increases by about $1/(0.80)^4 \approx 2.44$ . This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with ΔP ∝ 1/ $r^4$ for fixed Q. In this scenario, reducing radius to 0.80r requires adjusting ΔP to maintain Q. Choice B is correct because it accurately reflects the factor 1/ $(0.80)^4$ ≈ 2.44. Choice C fails because it uses 1/ $r^2$ instead of 1/ $r^4$ . In similar questions, rearrange for ΔP and apply the inverse fourth-power scaling for radius changes.
A student predicts that narrowing a capillary will always increase volumetric flow because the fluid speed increases in the smaller section. In a setup with constant $\Delta P$ across a single uniform capillary segment (laminar, Newtonian fluid), the radius is reduced from $r$ to $0.50r$ with $L$ and $\eta$ unchanged. Which result is consistent with Poiseuille flow?
Answer: $Q$ decreases to $\tfrac{1}{16}Q_0$ because $Q \propto r^4$ . This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with strong dependence on $r^4$ . In this scenario, halving the radius significantly impacts the flow resistance. Choice A is correct because it accurately reflects Q ∝ $r^4$ , leading to Q = $(1/2)^4$ Q0 = 1/16 Q0. Choice C fails because while velocity increases, volumetric flow decreases due to smaller cross-section. In similar questions, always apply the $r^4$ term and distinguish between velocity and flow rate.
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?
Answer: $Q$ increases by a factor of 16 because $Q \propto r^4$ . 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 laminar flow experiment uses a capillary with fixed $L$ and $r$ . The investigator reports that switching to a fluid with higher viscosity increased the measured flow rate at the same pressure drop. Which statement best describes the influence of viscosity in this system under Poiseuille flow assumptions?
Answer: Higher viscosity should decrease flow rate because viscous resistance increases. This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with Q inversely proportional to η. In this scenario, the reported increase in Q with higher η contradicts the law. Choice A is correct because it accurately states that higher viscosity should decrease Q due to increased resistance. Choice B fails because higher viscosity decreases, not increases, flow rate. In similar questions, always recall the inverse relationship and identify inconsistencies with experimental reports.
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 $Q_{20}$ ?
Answer: $Q_{37} = 1.33Q_{20}$ because flow is inversely proportional to viscosity. 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.
A capillary flow assay uses the same pressure drop and tube length, but compares two radii: $r_1 = 0.40\ \text{mm}$ and $r_2 = 0.60\ \text{mm}$ . The fluid is Newtonian and flow is laminar. What is the best estimate of $Q_2/Q_1$ under Poiseuille flow?
Answer: $(0.60/0.40)^4 \approx 5.1$ . This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with Q ∝ $r^4$ . In this scenario, comparing two radii under identical conditions highlights the $r^4$ dependence. Choice B is correct because it accurately calculates $(0.60/0.40)^4$ ≈ $(1.5)^4$ ≈ 5.1. Choice C fails because it inverts the ratio incorrectly. In similar questions, always raise the radius ratio to the fourth power and ensure other parameters are constant.
In a capillary-flow experiment, a Newtonian fluid is driven through a rigid tube at low Reynolds number (laminar). The investigator wants to increase the volumetric flow rate $Q$ without changing the tube geometry ( $r$ and $L$ fixed). Two interventions are considered: 1) Increase the pressure drop from $\Delta P$ to $2\Delta P$ . 2) Add a solute that doubles viscosity from $\eta$ to $2\eta$ . If both interventions are applied simultaneously and the system remains laminar, what is the net effect on $Q$ according to Poiseuille's law $Q \propto \Delta P/\eta$ (with geometry fixed)?
Answer: $Q$ is unchanged because the factor of 2 in $\Delta P$ cancels the factor of 2 in $\eta$ . This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law shows that flow rate is proportional to pressure drop and inversely proportional to viscosity: Q ∝ ΔP/η. In this scenario, doubling the pressure drop increases flow by factor of 2, while doubling the viscosity decreases flow by factor of 2, resulting in no net change. Choice B is correct because these two effects exactly cancel: Q_new = Q_original × (2ΔP)/(2η) = Q_original. Choice A fails because it incorrectly assumes pressure changes dominate over viscosity changes, when in fact they have equal but opposite effects. In similar questions, remember that Q depends linearly on both ΔP (direct) and η (inverse), so multiplicative changes in these parameters combine accordingly.
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: $Q \propto \frac{\Delta P\, r^4}{\eta L}$ .)
Answer: $Q$ decreases by a factor of 2 because $Q \propto 1/\eta$ . 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.
A lab compares flow through two parallel capillaries connected to the same pressure source, each experiencing the same $\Delta P$ . Capillary A has radius $r$ and length $L$ . Capillary B has radius $r$ and length $2L$ . Same Newtonian fluid, laminar flow. Which statement is consistent with Poiseuille flow for the ratio $Q_A/Q_B$ ?
Answer: $Q_A/Q_B = 2$ because halving length doubles flow rate. This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with Q ∝ 1/L. In this scenario, comparing lengths with same radius and ΔP shows the length effect. Choice A is correct because it accurately reflects QA / QB = 2 due to half the resistance in A. Choice C fails because resistance scales linearly with L, not $L^2$ . In similar questions, use the inverse proportionality to length for equal radii and confirm parallel setup.
A clinician compares flow through two parallel arterioles supplying the same tissue bed. Both vessels experience the same pressure drop $\Delta P$ and contain blood of the same viscosity $\eta$ over the short time window. Arteriole 1 has radius $r$ and length $L$ . Arteriole 2 has radius $0.80r$ and length $0.50L$ . Assuming laminar Poiseuille flow, what is the ratio $Q_2/Q_1$ ? (Use $Q \propto r^4/L$ and $(0.80)^4 \approx 0.41$ .)
Answer: $Q_2/Q_1 \approx 0.82$ because radius decreases flow to 0.41 and halving length doubles it. This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law shows that flow rate scales as Q ∝ r⁴/L when pressure drop and viscosity are constant. In this scenario, Arteriole 2 has radius 0.80r (reducing flow by (0.80)⁴ = 0.41) and length 0.50L (increasing flow by factor of 2), giving Q₂/Q₁ = 0.41 × 2 = 0.82. Choice A is correct because it properly combines the r⁴ effect with the inverse length dependence to get the ratio of approximately 0.82. Choice D fails because it incorrectly assumes flow is proportional to r/L instead of r⁴/L, missing the critical fourth-power radius dependence. In similar questions, always use the correct power relationships and combine effects multiplicatively.
A lab measures laminar flow of a Newtonian fluid through a capillary at constant $\Delta P$ , $L$ , and $r$ . Temperature is increased from $20^\circ\text{C}$ to $40^\circ\text{C}$ , and the fluid's dynamic viscosity decreases by 30% (to $0.70\eta_0$ ). All other parameters are unchanged. What change in $Q$ is expected from Poiseuille flow?
Answer: $Q$ increases to about $1.43Q_0$ because $Q \propto 1/\eta$ . This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with Q inversely proportional to η. In this scenario, decreasing viscosity to 0.70 η0 through temperature increase alters the flow rate. Choice B is correct because it accurately reflects Q ∝ 1/η, leading to Q ≈ 1.43 Q0. Choice A fails because viscosity and flow rate are inversely, not directly, proportional. In similar questions, always recall the inverse relationship with viscosity and verify if other parameters remain constant.
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?
Answer: $Q$ increases by a factor of 16 because $Q \propto r^4$ . 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.
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?
Answer: Capillary 2 has half the flow rate of Capillary 1 because $Q \propto 1/L$ . 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.

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