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

WHAT: hum count karte hain ki har second line cross karne wale kilograms kitne hain. WHY: yeh raw definition hai $\dot{m}=\rho Av$ — koi choking nahi, koi thermodynamics nahi, bas "stuff sweeping past." $\dot{m}=0.8\times 0.05\times 400=16\text{kg/s}$ Units check: $\frac{\text{kg}}{{\text{m}}^3}\cdot {\text{m}}^2\cdot \frac{\text{m}}{\text{s}}=\frac{\text{kg}}{\text{s}}$ ✔.

Depend karta hai: throat area $A^{\ast }$, chamber pressure $p_0$, chamber temperature $T_0$, aur gas type through $\gamma$ aur $R$. Depend NAHI karta: downstream (exit / back) pressure par, ek baar throat choke ho jaaye toh. WHY: $M=1$ par pressure signals sonic throat ke upstream nahi ja sakti, isliye chamber kabhi lower back-pressure "sunta" nahi. Dekho Choked Flow and Sonic Conditions.

Step A — coefficient (sirf gas). Exponent hai $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.3}{0.6}=3.8333$: $\Gamma =\sqrt{1.3}{\left(\frac{2}{2.3}\right)}^{3.8333}=1.14018\times (0.86957{)}^{3.8333}=1.14018\times 0.59454=0.67793$ Step B — $\sqrt{R{T}_0}$ 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 ise split kiya? Gas-only part ($\Gamma$) ko engine part ($A^{\ast }{p}_0/\sqrt{R{T}_0}$) se alag karne se errors visible hoti hain aur reuse aasaan ho jaata hai.

WHAT: choked formula ko $A^{\ast }$ ke liye invert karo. WHY: $\dot{m}$ linear hai $A^{\ast }$ mein, isliye yeh ek clean division hai. $A^{\ast }=\frac{\dot{m}\sqrt{R{T}_0}}{p_0\Gamma }$ $\Gamma$: $\sqrt{1.2}{\left(\frac{2}{2.2}\right)}^{\frac{2.2}{0.4}}=1.09545\times (0.90909{)}^{5.5}=1.09545\times 0.58471=0.64051$. $\sqrt{R{T}_0}:$ $\sqrt{340\times 3200}=\sqrt{1088000}=1043.07$. $A^{\ast }=\frac{250\times 1043.07}{7\times 10^6\times 0.64051}=\frac{260767}{4483570}=0.05816{\text{m}}^2$ Toh roughly $A^{\ast }\approx 0.0582{\text{m}}^2$ (throat radius of $\sqrt{A^{\ast }/\pi }\approx 0.136\text{m}$).

WHAT: poore numbers ki jagah proportionalities use karo. $\dot{m}\propto \frac{A^{\ast }{p}_0}{\sqrt{T_0}}$

• $p_0$: $\times 3$
• $A^{\ast }$: $\times \frac{1}{2}$
• $T_0$: $\times 4\Rightarrow \sqrt{T_0}\times 2\Rightarrow$ contributes $\times \frac{1}{2}$ $\text{factor}=3\times \frac{1}{2}\times \frac{1}{2}=\frac{3}{4}=0.75$ Result: $\dot{m}$ apni original value ka 75% ho jaata hai. WHY $\sqrt{T_0}$ denominator mein? Fixed pressure par hotter gas less dense hoti hai; tum jo mass shove kar sakte ho woh kam ho jaata hai chahe gas tezi se move kare. Dekho Isentropic Flow Relations.

Sirf gas-dependent group badlega: $\dot{m}\propto \frac{\Gamma (\gamma )}{\sqrt{R}}$. Gas P: ${\Gamma }_P=\sqrt{1.4}{\left(\frac{2}{2.4}\right)}^{3.0}=1.18322\times (0.83333{)}^3=1.18322\times 0.57870=0.68475$. Phir ${\Gamma }_P/\sqrt{R_P}=0.68475/\sqrt{287}=0.68475/16.9411=0.040419$. Gas Q: ${\Gamma }_Q=\sqrt{1.2}{\left(\frac{2}{2.2}\right)}^{5.5}=1.09545\times 0.58471=0.64051$. Phir ${\Gamma }_Q/\sqrt{R_Q}=0.64051/\sqrt{320}=0.64051/17.8885=0.035805$. Ratio P:Q $=0.040419/0.035805=1.1289$. Result: Gas P identical chamber aur throat conditions mein roughly 12.9% zyada mass flow karta hai. WHY $1/\sqrt{R}$? Bada $R$ (halke molecules) matlab same $p,T$ par lower density, isliye zyada mass per second nahi — chahe halki gas exhaust speed aur Specific Impulse ke liye achi ho.

WHAT: mass flow ko exhaust velocity ke saath combine karo. WHY: pressure-matched exit ke saath, thrust simply $F=\dot{m}c$ hai — dekho 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 $F=\dot{m}c$ se: $\dot{m}=\frac{F}{c}=\frac{500000}{3000}=166.67\text{kg/s}$ Step 2 — 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 — $A^{\ast }$ ke liye invert karo: $A^{\ast }=\frac{\dot{m}\sqrt{R{T}_0}}{p_0\Gamma }=\frac{166.67\times 1039.23}{8\times 10^6\times 0.65792}=\frac{173205}{5263360}=0.03291{\text{m}}^2$ Toh $A^{\ast }\approx 0.0329{\text{m}}^2$ (throat radius $\approx 0.102\text{m}$). WHY yeh order? Thrust $\dot{m}$ fix karta hai; phir choked formula (inverted) geometry fix karta hai. Do clean steps, do alag physics laws chained.

WHAT: normalised flux $G(M)$ banao, differentiate karo, $G^{\prime }(M)=0$ solve karo. WHY: peak ka location exactly physical choke point hai — ise prove karne ke liye calculus chahiye, teen sample points nahi. Upar ki definitions se yaad karo $M=v/a$, $a_0=\sqrt{\gamma R{T}_0}$, ${\rho }_0=p_0/(R{T}_0)$. Isentropic relations dete hain $\rho ={\rho }_0{\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{-\frac{1}{\gamma -1}},v=Ma=M{a}_0{\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{-\frac{1}{2}}.$ Unka product, ${\rho }_0{a}_0$ se normalise kiya hua, hai $G(M)\equiv \frac{\rho v}{{\rho }_0{a}_0}=M{\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{-\frac{\gamma +1}{2(\gamma -1)}}.$ Differentiate karo. $u=1+\frac{\gamma -1}{2}{M}^2$ aur $n=\frac{\gamma +1}{2(\gamma -1)}$ likho, toh $G=M{u}^{-n}$ aur $u^{\prime }=(\gamma -1)M$. Product rule se: $G^{\prime }(M)=u^{-n}+M\cdot (-n)u^{-n-1}{u}^{\prime }=u^{-n}[1-n(\gamma -1)\frac{M^2}{u}].$ $G^{\prime }(M)=0$ set karo. Kyunki $u^{-n}>0$ hai, bracket vanish hona chahiye: $1=n(\gamma -1)\frac{M^2}{u},n(\gamma -1)=\frac{\gamma +1}{2}.$ Toh $u=\frac{\gamma +1}{2}{M}^2$, yaani $1+\frac{\gamma -1}{2}{M}^2=\frac{\gamma +1}{2}{M}^2$. Rearrange karne par: $1=(\frac{\gamma +1}{2}-\frac{\gamma -1}{2})M^2=M^2\Rightarrow \boxed{M=1.}$ Yahi proof hai: eklauta stationary point $M=1$ hai, aur kyunki $G\to 0$ jab $M\to 0$ aur $G$ large $M$ ke liye decrease karta hai, yeh ek maximum hai. Yahi wajah hai ki throat $M=1$ par choke karta hai. Numerical confirmation ($\gamma =1.4$, exponent $-3.0$, $\frac{\gamma -1}{2}=0.2$):

• $M=0.8:G=0.8(1.128{)}^{-3}=0.5573$
• $M=1.0:G=1.0(1.2{)}^{-3}=0.5787$
• $M=1.2:G=1.2(1.288{)}^{-3}=0.5616$ Middle value ($M=1$) sabse bada hai, jo derivative result se match karta hai. Figure poora dome plot karta hai.

WHAT: continuity apply karo ${\rho }_1{A}_1{v}_1={\rho }_2{A}_2{v}_2$ (same $\dot{m}$ har jagah). WHY: mass conserved hota hai — dono stations ke beech kuch add nahi hota. $\dot{m}=3.0\times 0.01\times 700=21\text{kg/s}$ $A_2=\frac{\dot{m}}{{\rho }_2{v}_2}=\frac{21}{0.6\times 2100}=\frac{21}{1260}=0.016\overline{6}{\text{m}}^2$ Toh $A_2\approx 0.0167{\text{m}}^2>A_1$. Comment: area bada hua chahe flow speed up hua — yeh supersonic flow ki hallmark hai, jahan diverging nozzle gas ko accelerate karta hai. Yahi geometry hai Nozzle Area Ratio and Expansion ke peeche.

WHAT: $\dot{m}\propto p_0$ (linear). $p_0$ mein $+10\%$ error se $\dot{m}$ mein $+10\%$ error aata hai. WHY linear? $p_0$ choked formula mein first power par appear karta hai. Result: computed $\dot{m}$ $10\%$ zyada hai. Contrast: $T_0$ mein $+10\%$ error se $\dot{m}\propto 1/\sqrt{T_0}$ change aata: $1/\sqrt{1.10}=0.9535$, yaani $\dot{m}$ roughly $4.65\%$ kam read karta hai. Temperature errors pressure errors se gentler hote hain — yeh jaanna helpful hai jab tum kisi sensor par trust karte ho.

Choked $\dot{m}$ scale karta hai $A^{\ast }{p}_0/\sqrt{T_0}$ ki tarah aur back-pressure ke baare mein andha hai (provided $p_b/p_0\le r_{\text{crit}}$). Agar tum us sentence ko $\dot{m}=\rho Av$ se re-derive kar sako, toh yeh topic tumhara hai. ::: $\dot{m}=\rho Av$ + $M=1$ set karo throat par + sab kuch chamber conditions mein likho $\Rightarrow \dot{m}=\frac{A^{\ast }{p}_0}{\sqrt{R{T}_0}}\Gamma (\gamma )$.