C/P 01 · Fluid flow
A Narrowed Tube
Use continuity, Bernoulli and Poiseuille to read two flow experiments.
Passage
Three relationships describe a fluid moving through a tube. The first is continuity: for an incompressible fluid, the volume flow rate Q is the same at every cross-section, so Q = A × v, where A is the cross-sectional area and v is the mean speed. The second is Bernoulli’s principle: for an ideal fluid (no viscosity) moving through a horizontal tube, the quantity P + ½ρv² is the same at every point, where P is the pressure and ρ is the density. The third is Poiseuille’s law, which applies to a viscous fluid in steady flow through a long, narrow tube:
Here ΔP is the pressure difference between the two ends of the tube, r is its radius, L is its length and η is the viscosity of the fluid. Real fluids follow all three relationships at once, but in a given experiment one of them usually dominates.
Part 1. A student pumps a water–glycerol mixture (ρ = 1.0 × 10³ kg/m³) through a horizontal rigid tube that has a short narrow section between two wide sections. Pressure gauges at each section read gauge pressure. Speed was measured with a flow meter in the wide sections only. Table 1 shows the results for one pumping rate.
Table 1. Measurements along the tube at one flow rate.
| Section | Radius (mm) | Mean speed (m/s) | Gauge pressure (kPa) |
|---|---|---|---|
| Upstream, wide | 4 | 1 | 10.0 |
| Narrow | 2 | not measured | 2.5 |
| Downstream, wide | 4 | 1 | 9.9 |
Part 2. The student then measures how the flow rate through straight tubes depends on the tube and the fluid. Each tube is connected between the same two reservoirs, so the pressure difference ΔP is the same for every tube. η₀ is the viscosity of the original mixture, and a second mixture with twice the viscosity is also tested. Table 2 shows the flow rate for each tube.
Table 2. Steady flow rate through four tubes at the same ΔP.
| Tube | Radius (mm) | Length (cm) | Viscosity | Flow (mL/s) |
|---|---|---|---|---|
| A | 2 | 10 | η₀ | 8.0 |
| B | 1 | 10 | η₀ | 0.5 |
| C | 2 | 20 | η₀ | 4.0 |
| D | 2 | 10 | 2η₀ | 4.0 |
Questions
Question 1 of 6
What is the mean speed of the fluid in the narrow section in Part 1?
This would be right if the speed were proportional to 1/r. Continuity ties the speed to the area, and the area depends on r².
This would need the area to fall by a factor of 8. Halving the radius reduces the area by a factor of 4.
This treats the speed as proportional to 1/r⁴, which is the radius dependence in Poiseuille’s law, not in continuity.
Why the answer is what it is
By continuity, Q = A × v stays constant. Area is proportional to r², so halving the radius cuts the area to one fourth and raises the speed by a factor of 4, from 1 m/s to 4 m/s.
Review: Fluids · Equations: Flow
Question 2 of 6
In Table 1, the pressure falls from 10.0 kPa to 2.5 kPa in the narrow section and is 9.9 kPa again downstream. Which statement best explains this pattern?
If friction had removed the pressure, it would not come back downstream. The pressure returns almost fully, so most of the drop was not a permanent loss.
The tube is horizontal, so the height term in Bernoulli’s equation does not change.
The fluid is treated as incompressible. The pressure change comes from the change in speed, not from compression.
Why the answer is what it is
Where the fluid speeds up, the pressure falls because P + ½ρv² stays constant. When the tube widens again, the fluid slows and the pressure comes back. The 0.1 kPa that does not return is the energy lost to viscosity, which is small compared with the 7.5 kPa exchange.
Review: Pressure–Flow Relationships · Equations: Flow
Question 3 of 6
The student doubles the pumping rate, so the speed in the wide sections becomes 2 m/s. By about what factor does the pressure difference between a wide section and the narrow section change?
The pressure difference depends on v², not on v. Doubling the speed does not just double the difference.
This would need the pressure difference to depend on v³. It depends on v².
The radii set the ratio of the speeds, but the pressure difference also depends on how fast the fluid is moving.
Why the answer is what it is
The pressure difference is ½ρ(v_narrow² − v_wide²). The speeds are 1 and 4 m/s at first, then 2 and 8 m/s: ½(1000)(16 − 1) = 7,500 Pa becomes ½(1000)(64 − 4) = 30,000 Pa. Because the difference depends on v², doubling every speed multiplies it by 4.
Review: Pressure–Flow Relationships · Equations: Flow
Question 4 of 6
The student tests a new tube, E, with a radius of 1 mm and a length of 20 cm, using the original mixture at the same ΔP. What flow rate is expected?
This is the flow for a 1 mm tube that is 10 cm long (Tube B). Tube E is also twice as long, which cuts the flow in half again.
This treats the flow as proportional to r², not r⁴. Poiseuille’s law has r⁴.
This treats the flow as proportional to r, not r⁴: halving the radius and doubling the length would give 8.0 ÷ 2 ÷ 2 = 2.0. Poiseuille’s law has r⁴.
Why the answer is what it is
Compared with Tube A (8.0 mL/s), Tube E has half the radius, which cuts the flow by 2⁴ = 16, and twice the length, which cuts it by another 2. The flow is 8.0 ÷ 16 ÷ 2 = 0.25 mL/s.
Review: Equations: Flow · Fluids, Pressure, and Flow
Question 5 of 6
Which pair of tubes in Table 2 best isolates the effect of viscosity on the flow rate?
These tubes differ in length, not in viscosity.
These tubes differ in both radius and length, and neither differs in viscosity.
These tubes differ in radius, not in viscosity.
Why the answer is what it is
A good comparison changes one variable and holds everything else the same. Tubes A and D have the same radius, length and ΔP, and differ only in viscosity. Doubling the viscosity halves the flow, from 8.0 to 4.0 mL/s.
Review: Equations: Flow
Question 6 of 6
In the body, the pressure difference across a tissue’s blood supply is unchanged. Which change would reduce blood flow to the tissue the most?
Flow is inversely proportional to length, so this only halves the flow.
Flow is inversely proportional to viscosity, so this only halves the flow.
This cuts the flow to one fourth, which is a large effect but smaller than a change in radius.
Why the answer is what it is
Flow depends on r⁴, so halving the radius cuts the flow to 1/16. This is why small changes in the diameter of arterioles are the body’s main way of directing blood flow, while length and viscosity matter much less.
Review: Vascular Regulation · Cardiovascular Physiology · Equations: Flow