Fluid Flow, Continuity, and Bernoulli’s Equation (4B)
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MCAT Chemical and Physical Foundations of Biological Systems › Fluid Flow, Continuity, and Bernoulli’s Equation (4B)
A researcher measures steady, incompressible flow through a horizontal tube and reports $A_1 v_1 = A_2 v_2$ for two points along the tube. Which experimental observation would most directly violate the use of the continuity equation in this form for the segment between points 1 and 2?
A side port between points 1 and 2 withdraws fluid at a constant rate
The tube diameter decreases smoothly between points 1 and 2
The fluid density is approximately constant along the segment
The tube remains horizontal between points 1 and 2
Explanation
This question tests understanding of when the continuity equation applies in its standard form. The continuity equation A₁v₁ = A₂v₂ assumes no fluid is added or removed between the two points—it's based on mass conservation for a closed system. Choice A describes fluid withdrawal between points, which violates this assumption and makes the standard continuity equation invalid, so A is correct. Choice B (diameter change) is exactly when continuity applies. Choice C (horizontal orientation) doesn't affect continuity. Choice D (constant density) is an assumption already built into the incompressible flow model. When applying continuity, always verify that the flow is truly continuous with no sources or sinks between measurement points.
A horizontal pipe carries an incompressible fluid at steady flow. At point 1, the speed is $v_1$ and static pressure is $P_1$. At point 2 downstream, the cross-sectional area is smaller and the speed is measured to be $v_2 > v_1$. Neglecting viscosity, which statement is best explained by Bernoulli’s equation?
$P_2 = P_1$ because pressure depends only on depth in a fluid
$P_2 < P_1$ because some pressure energy converts to kinetic energy
$P_2 > P_1$ because higher speed requires higher static pressure
$P_2 < P_1$ because the fluid’s density decreases in the constriction
Explanation
This question tests conceptual understanding of Bernoulli's equation and energy conservation in fluid flow. Bernoulli's equation represents conservation of mechanical energy: pressure energy plus kinetic energy remains constant (neglecting potential energy for horizontal flow). When velocity increases from v₁ to v₂, kinetic energy increases, so pressure energy must decrease, meaning P₂ < P₁, making choice B correct. Choice A contradicts energy conservation by suggesting both forms of energy increase. Choice C incorrectly applies hydrostatic pressure principles to a flowing fluid. Choice D incorrectly suggests density changes in an incompressible fluid. To understand Bernoulli conceptually, think of it as an energy trade-off: faster flow means more kinetic energy, which must come from somewhere—specifically, from pressure energy.
A syringe pump drives water (density $\rho = 1000,\text{kg/m}^3$) through a rigid, horizontal tube that narrows from radius $r_1 = 2.0,\text{mm}$ to $r_2 = 1.0,\text{mm}$. At the wide section, the static pressure is $P_1 = 120,\text{kPa}$ and the average speed is $v_1 = 0.50,\text{m/s}$. Neglecting viscosity and elevation change, which conclusion is most consistent with Bernoulli’s equation for the static pressure $P_2$ in the narrow section?
$P_2 = P_1$ because the tube is horizontal
$P_2$ is lower only if the flow is turbulent; Bernoulli’s equation requires turbulence
$P_2 < P_1$ because velocity increases in the constriction
$P_2 > P_1$ because continuity requires higher pressure to maintain flow
Explanation
This question tests application of Bernoulli's equation to relate pressure and velocity changes in fluid flow. Bernoulli's equation for horizontal flow states that P₁ + ½ρv₁² = P₂ + ½ρv₂², meaning that as velocity increases, static pressure must decrease to conserve total mechanical energy. Since the tube narrows from radius 2.0 mm to 1.0 mm, the area decreases by a factor of 4, so by continuity, v₂ = 4v₁ = 2.0 m/s. With higher velocity at point 2, Bernoulli's equation requires P₂ < P₁, making choice A correct. Choice C incorrectly suggests pressure increases to maintain flow, confusing the cause-effect relationship—it's the pressure difference that drives flow, not continuity that requires pressure. When applying Bernoulli's equation, always check that kinetic energy increase corresponds to pressure energy decrease.
An IV infusion line carries saline (incompressible) at a volumetric flow rate of $Q = 1.5,\text{mL/s}$. The line transitions from an internal radius of $r_1 = 1.0,\text{mm}$ to $r_2 = 0.50,\text{mm}$ without leakage. Based on conservation of mass (continuity), what is the best prediction for the relationship between average speeds $v_1$ and $v_2$ in the two sections?
$v_2 = 2 v_1$ because radius halves
$v_2 = v_1/4$ because smaller radius increases resistance
$v_2 = 4 v_1$ because area scales as $r^2$
$v_2 = v_1$ because $Q$ is given in mL/s
Explanation
This question tests understanding of how the continuity equation relates to circular pipe geometry. For a circular pipe, area A = πr², so when radius changes from r₁ to r₂ = 0.50r₁, the area changes by a factor of (r₂/r₁)² = 0.25. By the continuity equation Q = A₁v₁ = A₂v₂, if A₂ = 0.25A₁, then v₂ = 4v₁, making choice B correct. Choice A incorrectly assumes velocity scales linearly with radius rather than inversely with area. Choice C reverses the relationship, suggesting smaller pipes have lower velocity, which contradicts continuity for steady flow. To avoid confusion, always remember that area scales as radius squared, and velocity is inversely proportional to area for constant flow rate.
Blood (assume incompressible, $\rho = 1060\ \text{kg/m}^3$) flows steadily through a horizontal artery segment that has a localized stenosis. Upstream area is $A_1 = 4.0\ \text{mm}^2$ and at the stenosis is $A_2 = 1.0\ \text{mm}^2$. If the upstream speed is $v_1 = 0.25\ \text{m/s}$ and viscous losses are neglected, what would be expected for the pressure at the stenosis compared with upstream, based on continuity and Bernoulli?
Lower pressure upstream because higher pressure is required to maintain constant volumetric flow
Higher pressure at the stenosis because smaller area increases resistance
Same pressure at the stenosis because the artery is horizontal ($\Delta h=0$)
Lower pressure at the stenosis because speed is higher in the smaller area
Explanation
This question tests understanding of pressure changes in arterial stenosis using Bernoulli's principle. The continuity equation shows that when area decreases from 4.0 mm² to 1.0 mm² (factor of 4), velocity increases from 0.25 m/s to 1.0 m/s (factor of 4). According to Bernoulli's equation for horizontal flow, this velocity increase requires a pressure decrease to conserve total mechanical energy. The pressure at the stenosis must be lower than upstream pressure because the fluid's kinetic energy has increased. Answer choice A incorrectly associates smaller area with higher resistance and pressure, confusing steady-state flow principles with viscous effects. In medical contexts, this pressure drop at stenoses can affect blood flow patterns and is why severe stenoses can compromise tissue perfusion despite maintaining flow continuity.
A clinician models blood flow through a carotid artery segment as steady, incompressible flow along a streamline. At point 1 (lower elevation), $v_1 = 0.8\ \text{m/s}$ and $P_1 = 13\ \text{kPa}$ (gauge). At point 2, the vessel is at a higher elevation by $\Delta h = 0.20\ \text{m}$ and the speed is $v_2 = 0.8\ \text{m/s}$. Take $\rho = 1060\ \text{kg/m}^3$ and $g = 9.8\ \text{m/s}^2$, neglect viscosity. Based on Bernoulli’s Equation, what is most consistent for $P_2$ relative to $P_1$?
$P_2 > P_1$ because pressure increases with height in flowing fluids
$P_2 < P_1$ by approximately $\rho g\Delta h$
$P_2 < P_1$ by approximately $\tfrac{1}{2}\rho(v_2^2-v_1^2)$
$P_2 = P_1$ because speeds are equal
Explanation
This question tests application of Bernoulli's equation including gravitational potential energy changes. The complete Bernoulli equation is P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂. Since v₁ = v₂ = 0.8 m/s, the kinetic energy terms cancel, leaving P₁ + ρgh₁ = P₂ + ρgh₂. Rearranging: P₂ = P₁ - ρg(h₂ - h₁) = P₁ - ρgΔh. With Δh = 0.20 m positive (point 2 is higher), P₂ < P₁ by approximately ρgΔh = (1060 kg/m³)(9.8 m/s²)(0.20 m) ≈ 2080 Pa ≈ 2.1 kPa. Answer choice A incorrectly suggests pressure increases with height in flowing fluids, confusing static fluid behavior with dynamic flow. In medical contexts, this hydrostatic pressure difference explains why blood pressure varies with measurement height relative to the heart.
A saline solution (density $\rho = 1000\ \text{kg/m}^3$) flows steadily through a horizontal catheter that narrows from cross-sectional area $A_1 = 4.0\ \text{mm}^2$ to $A_2 = 1.0\ \text{mm}^2$. The average speed in the wide section is measured as $v_1 = 0.50\ \text{m/s}$. Assuming incompressible flow and no leakage, what would be expected for the average speed $v_2$ in the narrow section?
$v_2 = 0.50\ \text{m/s}$
$v_2 = 0.125\ \text{m/s}$
$v_2 = 2.0\ \text{m/s}$
$v_2 = 1.0\ \text{m/s}$
Explanation
This question tests understanding of the continuity equation for incompressible fluid flow. The continuity equation states that for steady flow of an incompressible fluid, the product of cross-sectional area and velocity must remain constant: A₁v₁ = A₂v₂. In this catheter scenario, we have A₁ = 4.0 mm², v₁ = 0.50 m/s, and A₂ = 1.0 mm². Solving for v₂: v₂ = (A₁/A₂)v₁ = (4.0/1.0) × 0.50 = 2.0 m/s. Choice A incorrectly divides velocities instead of multiplying, a common error when students confuse the inverse relationship between area and velocity. To verify continuity problems, always check that flow rate Q = Av remains constant at both points.
A contrast agent (assume incompressible) flows through a bifurcating vessel where the parent vessel has cross-sectional area $A_0 = 6\ \text{mm}^2$ and average speed $v_0 = 0.40\ \text{m/s}$. It splits into two daughter branches with areas $A_1 = 2\ \text{mm}^2$ and $A_2 = 4\ \text{mm}^2$. If flow divides such that average speeds in both branches are equal ($v_1 = v_2$), what is the expected common branch speed?
$0.20\ \text{m/s}$
$0.80\ \text{m/s}$
$0.60\ \text{m/s}$
$0.40\ \text{m/s}$
Explanation
This question tests continuity equation application in bifurcating vessels. The continuity equation requires that flow rate entering equals flow rate exiting: Q₀ = Q₁ + Q₂, or A₀v₀ = A₁v₁ + A₂v₂. Given A₀ = 6 mm², v₀ = 0.40 m/s, A₁ = 2 mm², A₂ = 4 mm², and v₁ = v₂ = v, we solve: 6(0.40) = 2v + 4v = 6v, giving v = 0.40 m/s. Choice C incorrectly assumes flow splits proportionally to area ratios without considering the constraint of equal velocities. When branches have equal velocities, the parent velocity equals the branch velocity only when total branch area equals parent area, which is true here (2 + 4 = 6 mm²).
A lab measures volumetric flow rate through an IV line as $Q = 2.0\ \text{mL/s}$. The tubing has an inner radius $r = 0.50\ \text{mm}$ at a straight section. Assuming steady incompressible flow, what is the best estimate of the average fluid speed in this section? (Use $1\ \text{mL} = 1\ \text{cm}^3$.)
$v \approx 2.5\ \text{m/s}$
$v \approx 25\ \text{m/s}$
$v \approx 0.25\ \text{m/s}$
$v \approx 0.025\ \text{m/s}$
Explanation
This question tests calculation of flow velocity from volumetric flow rate using the continuity equation. The relationship between volumetric flow rate Q and average velocity v is Q = Av, where A is cross-sectional area. Given Q = 2.0 mL/s = 2.0 cm³/s and r = 0.50 mm = 0.050 cm, the area is A = πr² = π(0.050)² = 0.00785 cm². Therefore, v = Q/A = 2.0/0.00785 = 255 cm/s = 2.55 m/s ≈ 2.5 m/s. Choice B incorrectly uses diameter instead of radius or makes a unit conversion error, a common mistake when working with small medical tubing. Always verify units: when Q is in cm³/s and A in cm², velocity comes out in cm/s, which must be converted to m/s.
A fluid with density $\rho = 1000\ \text{kg/m}^3$ flows steadily through a vertical pipe from point 1 (lower) to point 2 (higher). The pipe diameter is constant, so $v_1 = v_2$. The height increase is $\Delta h = h_2 - h_1 = 0.50\ \text{m}$. Neglecting viscosity, what pressure change $P_2 - P_1$ is most consistent with Bernoulli’s equation? (Use $g = 9.8\ \text{m/s}^2$.)
$P_2 - P_1 \approx +0.20\ \text{kPa}$
$P_2 - P_1 \approx +4.9\ \text{kPa}$
$P_2 - P_1 \approx -4.9\ \text{kPa}$
$P_2 - P_1 \approx -0.20\ \text{kPa}$
Explanation
This question tests Bernoulli's equation with gravitational potential energy changes. The complete Bernoulli equation is P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂. Since the pipe has constant diameter, v₁ = v₂, so the kinetic energy terms cancel. This gives P₂ - P₁ = ρg(h₁ - h₂) = -ρgΔh = -1000(9.8)(0.50) = -4900 Pa = -4.9 kPa. The negative sign indicates pressure decreases with height, as expected since the fluid must overcome gravity. Choice C might result from using incorrect units or forgetting the density term. Remember: when fluid flows upward against gravity at constant velocity, pressure must decrease by ρgh to provide the driving force.