An engineer says 'the stagnation pressure at this point in the flow is 150 kPa.' What does this actually mean, physically?
AA pressure gauge installed at that point in the flow reads 150 kPa
BIf the fluid at that point were brought to rest isentropically (without losses), it would have a pressure of 150 kPa — this value encodes the total mechanical energy the fluid carries
CThe velocity at that point is zero and the static pressure is 150 kPa
DThe flow is everywhere at 150 kPa pressure, and velocity only increases at narrowed sections
Stagnation pressure is not a 'real' pressure that a gauge at every point would read — it is a thermodynamic bookkeeping quantity representing the total mechanical energy per unit volume in the flow. Only at an actual stagnation point (like the tip of a Pitot tube, where velocity genuinely equals zero) does a pressure gauge read the stagnation pressure directly. Everywhere else, the gauge reads the lower static pressure P, and V > 0. The stagnation pressure is best understood as: 'if I could stop this parcel isentropically, how much pressure would it have?' That hypothetical value encodes its kinetic + pressure energy combined.
Question 2 Multiple Choice
A Pitot tube in an airflow reads stagnation pressure P₀ = 120 kPa at its stagnation port. A static port nearby reads P = 100 kPa. The air density is ρ = 1.2 kg/m³. Using the incompressible Bernoulli relation, what is the flow velocity?
AV = √(2 × 20,000 / 1.2) ≈ 183 m/s
BV = (120,000 − 100,000) / 1.2 ≈ 16,700 m/s
CV = √(120,000 / 1.2) ≈ 316 m/s
DV = 2 × (120,000 − 100,000) / 1.2 ≈ 33,333 m/s
From P₀ = P + ½ρV², we get V = √(2(P₀ − P)/ρ) = √(2 × 20,000 / 1.2) = √(33,333) ≈ 183 m/s. The pressure difference ΔP = 20,000 Pa (not 20 Pa), so unit consistency is critical. Option B divides rather than takes the square root — a common error. Option C uses P₀ directly rather than the pressure difference, which is wrong because P₀ includes both static and dynamic contributions; only the difference (P₀ − P) represents the kinetic energy converted to pressure at the stagnation point.
Question 3 True / False
In steady, inviscid, incompressible flow, stagnation pressure P₀ = P + ½ρV² is constant everywhere in the flow field.
TTrue
FFalse
Answer: False
Bernoulli's equation guarantees that P₀ is constant *along a streamline* in irrotational, inviscid, incompressible, steady flow. Different streamlines can — and generally do — have different stagnation pressures, especially if they originate from different upstream conditions. Only in a uniform, irrotational flow field does every streamline carry the same stagnation pressure, making P₀ constant throughout the field. In a viscous flow or across a shock wave, stagnation pressure drops due to irreversible energy losses.
Question 4 True / False
In compressible flow at Mach 0.8, the static pressure equals the stagnation pressure minus the incompressible dynamic pressure: P = P₀ − ½ρV².
TTrue
FFalse
Answer: False
The formula P₀ = P + ½ρV² is the *incompressible* approximation, valid only at low Mach numbers where density variations are negligible. At M = 0.8, compressibility effects are significant and the correct relation is the isentropic formula: P₀/P = (1 + (γ−1)/2 × M²)^(γ/(γ−1)). For air (γ = 1.4) at M = 0.8, this gives P₀/P ≈ 1.524, meaning P ≈ 0.656 P₀ — substantially different from what the incompressible formula predicts. Using the incompressible formula at transonic Mach numbers introduces significant error and would predict incorrect velocities in aircraft applications.
Question 5 Short Answer
Explain why stagnation conditions (P₀, T₀) are the natural reference state for analyzing compressible nozzle flows, and what physical processes would change them.
Think about your answer, then reveal below.
Model answer: In an isentropic (adiabatic and frictionless) process, stagnation pressure P₀ and stagnation temperature T₀ remain constant along streamlines even as static pressure and temperature vary dramatically. As fluid accelerates through a converging nozzle, static pressure and temperature fall sharply while velocity increases — but P₀ and T₀ stay fixed, encoding the total energy content of the flow. This makes them ideal reference quantities: all local flow conditions can be expressed as fractions of these fixed reference values using isentropic ratios, and the entire flow field is characterized by specifying the inlet stagnation conditions alone. What changes them: frictional losses (viscosity, boundary layers) irreversibly reduce P₀ by converting ordered kinetic energy to thermal energy (entropy production); a normal shock wave dramatically drops P₀; heat addition increases T₀ and reduces P₀ (Rayleigh flow). Any process that generates entropy reduces P₀ — so a drop in stagnation pressure along a duct is a direct thermodynamic measure of irreversibility.