A Pitot tube is inserted into a flow. The stagnation port and static port register identical pressures. What does this tell you about the flow at that location?
AThe fluid density is too low to generate a measurable pressure difference
BThe flow velocity at that point is zero or negligible
CThere is a blockage in the stagnation port preventing pressure buildup
DThe tube is misaligned with the flow, so stagnation is incomplete
Dynamic pressure = stagnation pressure − static pressure = ½ρV². If both ports read the same pressure, the difference is zero, meaning ½ρV² = 0, hence V = 0. This is the direct consequence of the operating principle: the entire measurement relies on the pressure rise that results from stagnating the moving fluid and converting its kinetic energy into pressure. No velocity means no kinetic energy to convert, so no pressure difference. A misaligned tube would reduce (not zero out) the stagnation pressure.
Question 2 Multiple Choice
In a water-filled manometer connected to a Pitot tube immersed in water flow, the velocity formula simplifies to V = √(2gΔh), where Δh is the height difference. Why does fluid density not appear in this result?
AWater is incompressible, so density does not affect pressure in any liquid measurement
BDynamic pressure ½ρV² and the manometer pressure ρgΔh both contain ρ, which cancels when they are set equal
CThis formula is a special approximation valid only at low flow speeds where density effects are small
DGravity acts equally on both sides of the manometer, eliminating the density term
Setting the dynamic pressure equal to the manometer pressure rise: ½ρV² = ρgΔh. Since ρ appears on both sides and is the same fluid (manometer fluid = flowing fluid), it cancels, giving V = √(2gΔh). This elegant result means velocity can be read directly from a ruler measurement of liquid height, with no need to know the fluid density. If the manometer uses a different fluid (e.g., mercury), the two densities are unequal and both must be accounted for.
Question 3 True / False
A Pitot tube directly measures the velocity of the fluid flowing past it.
TTrue
FFalse
Answer: False
A Pitot tube measures pressure — specifically the difference between stagnation (total) pressure and static pressure. Velocity is then derived from this measurement using V = √(2ΔP/ρ), which requires knowing the fluid density. The device converts kinetic energy into a measurable pressure rise; the velocity is inferred, not directly sensed. This distinction matters: errors in the assumed fluid density propagate directly into the velocity calculation.
Question 4 True / False
The stagnation point at the Pitot tube's front face creates a pressure higher than the surrounding static pressure because the fluid's kinetic energy is converted into pressure energy there.
TTrue
FFalse
Answer: True
This is the core physical principle. Bernoulli's equation along the stagnation streamline gives P_static + ½ρV² = P_stagnation (with elevation terms equal at both points). At the stagnation point, V = 0, so all kinetic energy ½ρV² has been converted into additional pressure. The stagnation pressure therefore exceeds the static pressure by exactly the dynamic pressure ½ρV², which is what the instrument measures. This is an energy trade-off: kinetic energy → pressure energy.
Question 5 Short Answer
What physical principle does a Pitot tube exploit to measure flow velocity, and what two pressures must it measure to do so?
Think about your answer, then reveal below.
Model answer: A Pitot tube exploits Bernoulli's equation, which states that pressure and kinetic energy trade off along a streamline: P + ½ρV² = constant. By facing a port directly into the flow, fluid is brought to rest (stagnated), converting all kinetic energy into a pressure rise. The stagnation port measures total pressure P_total = P_static + ½ρV². A flush static port measures P_static (the ambient pressure undisturbed by flow). The difference ΔP = P_total − P_static = ½ρV² is the dynamic pressure, from which velocity is calculated as V = √(2ΔP/ρ).
The key insight is that the Pitot tube doesn't sense velocity directly — it senses the pressure consequence of stopping the flow. The two-port design (stagnation + static) isolates exactly the dynamic pressure term from Bernoulli's equation. Knowing fluid density then completes the velocity calculation. This simplicity (one pressure difference measurement) is why Pitot tubes are widely used from aircraft cockpits to introductory fluid mechanics labs.