Exercises — Mass flow rate ṁ and its relation to throat area

WHAT: we count kilograms crossing the line each second. WHY: this is the raw definition $\dot{m}=\rho Av$ — no choking, no thermodynamics, just "stuff sweeping past." $\dot{m}=0.8\times 0.05\times 400=16\text{kg/s}$ Units check: $\frac{\text{kg}}{{\text{m}}^3}\cdot {\text{m}}^2\cdot \frac{\text{m}}{\text{s}}=\frac{\text{kg}}{\text{s}}$ ✔.

Depends on: throat area $A^{\ast }$, chamber pressure $p_0$, chamber temperature $T_0$, and gas type through $\gamma$ and $R$. Does NOT depend on: the downstream (exit / back) pressure, once the throat is choked. WHY: at $M=1$ pressure signals cannot travel upstream past the sonic throat, so the chamber "never hears" a lower back-pressure. See Choked Flow and Sonic Conditions.

Step A — coefficient (gas only). The exponent is $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.3}{0.6}=3.8333$: $\Gamma =\sqrt{1.3}{\left(\frac{2}{2.3}\right)}^{3.8333}=1.14018\times (0.86957{)}^{3.8333}=1.14018\times 0.59454=0.67793$ Step B — the $\sqrt{R{T}_0}$ denominator: $\sqrt{R{T}_0}=\sqrt{300\times 2500}=\sqrt{750000}=866.03$ Step C — assemble $\dot{m}=\frac{A^{\ast }{p}_0}{\sqrt{R{T}_0}}\Gamma$: $\dot{m}=\frac{0.015\times 4\times 10^6}{866.03}\times 0.67793=\frac{60000}{866.03}\times 0.67793=69.281\times 0.67793=46.97\text{kg/s}$ WHY split it up? Separating the gas-only part ($\Gamma$) from the engine part ($A^{\ast }{p}_0/\sqrt{R{T}_0}$) makes errors visible and reuse easy.

WHAT: invert the choked formula for $A^{\ast }$. WHY: $\dot{m}$ is linear in $A^{\ast }$, so this is a clean division. $A^{\ast }=\frac{\dot{m}\sqrt{R{T}_0}}{p_0\Gamma }$ $\Gamma$: $\sqrt{1.2}{\left(\frac{2}{2.2}\right)}^{\frac{2.2}{0.4}}=1.09545\times (0.90909{)}^{5.5}=1.09545\times 0.58471=0.64051$. $\sqrt{R{T}_0}:$ $\sqrt{340\times 3200}=\sqrt{1088000}=1043.07$. $A^{\ast }=\frac{250\times 1043.07}{7\times 10^6\times 0.64051}=\frac{260767}{4483570}=0.05816{\text{m}}^2$ So about $A^{\ast }\approx 0.0582{\text{m}}^2$ (a throat radius of $\sqrt{A^{\ast }/\pi }\approx 0.136\text{m}$).

WHAT: use the proportionalities, not the full numbers. $\dot{m}\propto \frac{A^{\ast }{p}_0}{\sqrt{T_0}}$

• $p_0$: $\times 3$
• $A^{\ast }$: $\times \frac{1}{2}$
• $T_0$: $\times 4\Rightarrow \sqrt{T_0}\times 2\Rightarrow$ contributes $\times \frac{1}{2}$ $\text{factor}=3\times \frac{1}{2}\times \frac{1}{2}=\frac{3}{4}=0.75$ Result: $\dot{m}$ drops to 75% of its original value. WHY $\sqrt{T_0}$ in the denominator? Hotter gas at fixed pressure is less dense; the mass you can shove through falls even though the gas moves faster. See the Isentropic Flow Relations.

Only the gas-dependent group changes: $\dot{m}\propto \frac{\Gamma (\gamma )}{\sqrt{R}}$. Gas P: ${\Gamma }_P=\sqrt{1.4}{\left(\frac{2}{2.4}\right)}^{3.0}=1.18322\times (0.83333{)}^3=1.18322\times 0.57870=0.68475$. Then ${\Gamma }_P/\sqrt{R_P}=0.68475/\sqrt{287}=0.68475/16.9411=0.040419$. Gas Q: ${\Gamma }_Q=\sqrt{1.2}{\left(\frac{2}{2.2}\right)}^{5.5}=1.09545\times 0.58471=0.64051$. Then ${\Gamma }_Q/\sqrt{R_Q}=0.64051/\sqrt{320}=0.64051/17.8885=0.035805$. Ratio P:Q $=0.040419/0.035805=1.1289$. Result: Gas P flows about 12.9% more mass under identical chamber & throat conditions. WHY $1/\sqrt{R}$? Bigger $R$ (lighter molecules) means lower density at the same $p,T$, so less mass per second — even though light gas is great for exhaust speed and Specific Impulse.

WHAT: combine mass flow with exhaust velocity. WHY: with pressure-matched exit, thrust is simply $F=\dot{m}c$ — see Thrust Equation and Effective Exhaust Velocity. $F=46.97\times 2600=122122\text{N}\approx 122\text{kN}$ Units: $\frac{\text{kg}}{\text{s}}\cdot \frac{\text{m}}{\text{s}}=\frac{\text{kg⋅m}}{{\text{s}}^2}=\text{N}$ ✔.

Step 1 — required mass flow from $F=\dot{m}c$: $\dot{m}=\frac{F}{c}=\frac{500000}{3000}=166.67\text{kg/s}$ Step 2 — the choking coefficient (exponent $\frac{2.25}{0.5}=4.5$): $\Gamma =\sqrt{1.25}{\left(\frac{2}{2.25}\right)}^{4.5}=1.11803\times (0.88889{)}^{4.5}=1.11803\times 0.58848=0.65792$ Step 3 — denominator $\sqrt{R{T}_0}=\sqrt{360\times 3000}=\sqrt{1080000}=1039.23$. Step 4 — invert for $A^{\ast }$: $A^{\ast }=\frac{\dot{m}\sqrt{R{T}_0}}{p_0\Gamma }=\frac{166.67\times 1039.23}{8\times 10^6\times 0.65792}=\frac{173205}{5263360}=0.03291{\text{m}}^2$ So $A^{\ast }\approx 0.0329{\text{m}}^2$ (throat radius $\approx 0.102\text{m}$). WHY this order? Thrust fixes $\dot{m}$; then the choked formula (inverted) fixes geometry. Two clean steps, two different physics laws chained.

WHAT: build the normalised flux $G(M)$, differentiate, solve $G^{\prime }(M)=0$. WHY: the location of the peak is exactly the physical choke point — proving it needs calculus, not three sample points. Recall from the definitions above $M=v/a$, $a_0=\sqrt{\gamma R{T}_0}$, ${\rho }_0=p_0/(R{T}_0)$. The isentropic relations give $\rho ={\rho }_0{\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{-\frac{1}{\gamma -1}},v=Ma=M{a}_0{\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{-\frac{1}{2}}.$ Their product, normalised by ${\rho }_0{a}_0$, is $G(M)\equiv \frac{\rho v}{{\rho }_0{a}_0}=M{\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{-\frac{\gamma +1}{2(\gamma -1)}}.$ Differentiate. Write $u=1+\frac{\gamma -1}{2}{M}^2$ and $n=\frac{\gamma +1}{2(\gamma -1)}$, so $G=M{u}^{-n}$ and $u^{\prime }=(\gamma -1)M$. By the product rule: $G^{\prime }(M)=u^{-n}+M\cdot (-n)u^{-n-1}{u}^{\prime }=u^{-n}[1-n(\gamma -1)\frac{M^2}{u}].$ Set $G^{\prime }(M)=0$. Since $u^{-n}>0$, the bracket must vanish: $1=n(\gamma -1)\frac{M^2}{u},n(\gamma -1)=\frac{\gamma +1}{2}.$ So $u=\frac{\gamma +1}{2}{M}^2$, i.e. $1+\frac{\gamma -1}{2}{M}^2=\frac{\gamma +1}{2}{M}^2$. Rearranging: $1=(\frac{\gamma +1}{2}-\frac{\gamma -1}{2})M^2=M^2\Rightarrow \boxed{M=1.}$ That is the proof: the only stationary point is $M=1$, and since $G\to 0$ as $M\to 0$ and $G$ decreases for large $M$, it is a maximum. This is why the throat chokes at $M=1$. Numerical confirmation ($\gamma =1.4$, exponent $-3.0$, $\frac{\gamma -1}{2}=0.2$):

• $M=0.8:G=0.8(1.128{)}^{-3}=0.5573$
• $M=1.0:G=1.0(1.2{)}^{-3}=0.5787$
• $M=1.2:G=1.2(1.288{)}^{-3}=0.5616$ The middle value ($M=1$) is largest, matching the derivative result. The figure plots the full dome.

WHAT: apply continuity ${\rho }_1{A}_1{v}_1={\rho }_2{A}_2{v}_2$ (same $\dot{m}$ everywhere). WHY: mass is conserved — nothing is added between stations. $\dot{m}=3.0\times 0.01\times 700=21\text{kg/s}$ $A_2=\frac{\dot{m}}{{\rho }_2{v}_2}=\frac{21}{0.6\times 2100}=\frac{21}{1260}=0.016\overline{6}{\text{m}}^2$ So $A_2\approx 0.0167{\text{m}}^2>A_1$. Comment: the area grew even though the flow sped up — the hallmark of supersonic flow, where a diverging nozzle accelerates gas. This is exactly the geometry behind Nozzle Area Ratio and Expansion.

WHAT: $\dot{m}\propto p_0$ (linear). A $+10\%$ error in $p_0$ gives a $+10\%$ error in $\dot{m}$. WHY linear? $p_0$ appears to the first power in the choked formula. Result: computed $\dot{m}$ is $10\%$ too high. Contrast: a $+10\%$ error in $T_0$ would give only a $\dot{m}\propto 1/\sqrt{T_0}$ change: $1/\sqrt{1.10}=0.9535$, i.e. $\dot{m}$ reads about $4.65\%$ too low. Temperature errors are gentler than pressure errors — worth knowing when you trust a sensor.

Choked $\dot{m}$ scales like $A^{\ast }{p}_0/\sqrt{T_0}$ and is blind to back-pressure (provided $p_b/p_0\le r_{\text{crit}}$). If you can re-derive that sentence from $\dot{m}=\rho Av$, you own this topic. ::: $\dot{m}=\rho Av$ + set $M=1$ at the throat + write everything in chamber conditions $\Rightarrow \dot{m}=\frac{A^{\ast }{p}_0}{\sqrt{R{T}_0}}\Gamma (\gamma )$.