Exercises — Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations

Q1.1 $==T_0==$ is conserved (the flow across a shock is adiabatic — no heat leaves, and $T_0$ needs only adiabaticity). $==P_0==$ is not conserved: a shock is irreversible, entropy rises, so total pressure drops. Memory hook: temperature = energy = adiabatic; pressure = order = isentropic. See Isentropic flow relations.

Q1.2 $\frac{\gamma }{\gamma -1}=\frac{1.4}{0.4}=3.5$ for pressure; $\frac{1}{\gamma -1}=\frac{1}{0.4}=2.5$ for density.

Q1.3 At $M=0$ the factor $\left(1+\frac{\gamma -1}{2}{M}^2\right)=1$, so all three ratios equal exactly 1. The gas is already at rest, so "static" and "stagnation" coincide — there is no motion to convert.

Q2.1 WHAT: use the adiabatic $T_0$ relation. WHY: only Mach and temperature are needed. $\frac{T_0}{T}=1+\frac{0.4}{2}(1.5{)}^2=1+0.2\times 2.25=1.45$ $T_0=1.45\times 250=362.5\mathrm{K}.$

Q2.2 WHAT: raise the same bracket to the power $3.5$. WHY: stagnation defined by isentropic stop. $\frac{P_0}{P}=(1.45{)}^{3.5}=3.671\dots$ $P_0=3.671\times 40=146.8\mathrm{kPa}.$

Q2.3 WHAT: raise the same bracket to the power $2.5$. WHY: the density ratio is the isentropic ${\rho }_0/\rho$ relation, whose exponent is $\frac{1}{\gamma -1}=2.5$ — we choose it (not the pressure or temperature relation) because the question asks for density. Bracket $=1+0.2(0.8{)}^2=1+0.128=1.128$. $\frac{{\rho }_0}{\rho }=(1.128{)}^{2.5}=\mathrm{1.352.}$ So $\rho /{\rho }_0=1/1.352=0.7396$: the flowing gas has $==26\%==$ less density than at rest ($1-0.7396=0.2604$). Accelerating the gas expands (thins) it — this is why we cannot ignore compressibility once $M\gtrsim 0.3$.

Q3.1 WHAT: invert the pressure relation. WHY: we are given the ratio, want $M$. This is legitimate here because the Pitot is stated subsonic — see the warning box below for why that matters. $\frac{P_0}{P}=1.25={\left(1+0.2M^2\right)}^{3.5}$ Take the $1/3.5$ power of both sides: $1+0.2M^2=1.25^{1/3.5}=1.25^{0.2857}=1.0654$ $0.2M^2=0.0654\Rightarrow M^2=0.3272\Rightarrow M=\mathrm{0.5720.}$ Speed of sound $a=\sqrt{1.4\times 287\times 300}=347.2m/s$, so $V=Ma=0.5720\times 347.2=198.6m/s.$

Q3.2 WHAT: incompressible estimate. $\rho =\frac{P}{RT}=\frac{100000}{287\times 300}=1.1614kg/{\mathrm{m}}^3$. $V_{\text{Bern}}=\sqrt{\frac{2(P_0-P)}{\rho }}=\sqrt{\frac{2\times 25000}{1.1614}}=\sqrt{43055}=207.5m/s.$ Error $=\frac{207.5-198.6}{198.6}=4.5\%$ too high. Bernoulli ignores that the air compresses as it stops, so it over-credits the pressure rise to speed. At $M\approx 0.57$ the error is already several percent — a warning to switch to the full relation.

Q3.3 All three factors are $(1+\text{positive}{)}^{\text{positive}}$, so all increase with $M$ (the curves in the figure only ever climb); at $M=0$ each equals $1$ (the common starting point). The exponents order the steepness: pressure carries $3.5$, density $2.5$, temperature $1$. Hence $==P_0/P==$ climbs fastest (amber), then ${\rho }_0/\rho$ (cyan), then $T_0/T$ (white) — exactly the vertical ordering of the three curves. Limit $M\to \infty$: the bracket $1+\frac{\gamma -1}{2}{M}^2\to \infty$, so all three ratios diverge to $\infty$; because the exponents differ the amber curve outruns the others without bound. Physically: stopping an arbitrarily fast stream dumps arbitrarily large kinetic energy, so total temperature, pressure and density all grow without limit.

Q4.1 WHAT: work backwards from stagnation to static using the same relations, since the reservoir values are the stagnation values everywhere in an isentropic nozzle. Bracket at $M=2$: $1+0.2(2{)}^2=1+0.8=1.8$. $T=\frac{T_0}{1.8}=\frac{500}{1.8}=277.8\mathrm{K}.$ $P=\frac{P_0}{(1.8{)}^{3.5}}=\frac{800}{7.824}=102.3\mathrm{kPa}.$ Speed of sound at the test section: $a=\sqrt{1.4\times 287\times 277.8}=334.1m/s$, so $V=Ma=2.0\times 334.1=668.2m/s.$

Q4.2 WHAT: entropy generated is read from the drop in total pressure alone (total temperature is fixed across an adiabatic shock). $\Delta s=-R\ln \frac{P_{0,2}}{P_{0,1}}=-287\ln \frac{575}{800}=-287\ln (0.71875).$ $\ln (0.71875)=-0.33024,\Delta s=-287\times (-0.33024)=94.8\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}.$ $\Delta s>0$ ✓ — entropy rises, as the Second Law demands for the irreversible shock. Downstream $T_{0,2}=T_{0,1}=500\mathrm{K}$ (the shock is adiabatic; total temperature is untouched even though total pressure fell).

Q5.1 WHAT: form the ratio of the exact dynamic-pressure-normalised over the incompressible one. Exact: $P_0-P=P\left[(1+0.2M^2{)}^{3.5}-1\right]$. With $M=0.5$: bracket $=1+0.2(0.25)=1.05$, and $1.05^{3.5}=1.1783$, so $P_0-P=0.17827P$. Reference $\frac{1}{2}\rho V^2=\frac{\gamma }{2}P{M}^2=0.7\times 0.25P=0.175P$. Ratio $=\frac{0.17827}{0.175}=1.0187$. Series prediction: $1+\frac{M^2}{4}+\frac{2-\gamma }{24}{M}^4+\cdots =1+\frac{0.25}{4}=1+0.0625=1.0625$ to first order — agrees in trend with $1.0187$ once you notice the $M^4$ term $\frac{2-\gamma }{24}{M}^4=\frac{0.6}{24}(0.0625)=0.00156$ pulls it back, and higher terms shave it further. So even at $M=0.5$ Bernoulli under-reads the pressure rise by about $1.9\%$; the $M^2/4$ term is the leading compressibility correction.

Q5.2 Bracket $=1+0.2(0.85{)}^2=1+0.2\times 0.7225=1.1445$. $T_0=1.1445\times 223=255.2\mathrm{K}.$ $P_0=26.5\times (1.1445{)}^{3.5}=26.5\times 1.6026=42.5\mathrm{kPa}.$ The stagnation tip warms from $223\mathrm{K}(-50^{\circ }\mathrm{C})$ to $255\mathrm{K}(-18^{\circ }\mathrm{C})$ — a $+32\mathrm{K}$ ram-heating jump. But $255\mathrm{K}$ is still below the $273\mathrm{K}$ freezing point of water, so super-cooled droplets still freeze on impact: de-icing is required despite the heating. The pressure nearly doubles at the tip (from $26.5$ to $42.5\mathrm{kPa}$), which is what a compressible Pitot must account for.