Exercises — Continuity equation — derivation (conservation of mass), ρAv = const

$\rho Av=\text{const}$ use karo (yeh mass flow rate hai). Kyun: fluid compressible hai — $\rho$ change hota hai — toh volume form $Av$ conserved nahi hai. Sirf mass kabhi create ya destroy nahi ho sakta, toh sirf $\rho Av$ hi guaranteed constant hai. Yahan $\rho$ drop karna silently ${\rho }_1={\rho }_2$ assume kar lega, jo steam ke liye galat hai.

Woh speed up hota hai. Rule: incompressible flow ke liye $v\propto 1/A$, toh chhota area $\Rightarrow$ bada speed. s01 figure dobara dekho — skinny slab zyada lamba hai, matlab zyada tez.

$Q=Av\Rightarrow v=\frac{Q}{A}=\frac{3\times 10^{-3}}{6\times 10^{-4}}=5\text{m/s}$. Units: $\frac{{\text{m}}^3/\text{s}}{{\text{m}}^2}=\text{m/s}$ ✓ — ek speed, jaise expected tha. (Areas already m² mein hain, toh koi conversion zarurat nahi.)

Incompressible, toh $A_1{v}_1=A_2{v}_2$ with $A=\pi r^2$. Yeh ek ratio problem hai — humein sirf $A_1/A_2$ chahiye, toh $r$ ke units (cm ya m) cancel ho jaate hain aur $\pi$ bhi cancel ho jaata hai: $v_2=v_1\frac{r_1^2}{r_2^2}=1.5\times \frac{6^2}{2^2}=1.5\times 9=13.5\text{m/s}.$ Kyun cm yahan safe hain: dono radii ratio $(r_1/r_2{)}^2$ mein hain, toh jab tak $r_1$ aur $r_2$ same unit use karte hain unit gayab ho jaati hai — ratios ke liye metres mein convert karna zaruri nahi. Kyun $r^2$: area radius squared ke saath badhta hai; radius factor 3 se ghata, toh area $3^2=9$ se ghata, toh speed ×9 badhi.

Density change hoti hai → mass form use karo: $v_2=\frac{{\rho }_1{A}_1{v}_1}{{\rho }_2{A}_2}=\frac{1.2\times 0.20\times 8}{0.80\times 0.20}=\frac{1.92}{0.16}=12\text{m/s}.$ Kyun speed up hoti hai: same mass per second, lekin ab har kilogram zyada volume leta hai (kam density), toh usi area ko clear karne ke liye zyada tez move karna padta hai.

Yeh real $Q$ produce karta hai, toh pehle area ko SI mein convert karo: $A=2{\text{cm}}^2=2\times 10^{-4}{\text{m}}^2$. $Q=Av=(2\times 10^{-4})(4)=8\times 10^{-4}{\text{m}}^3/\text{s}$. $t=\frac{V}{Q}=\frac{0.024}{8\times 10^{-4}}=30\text{s}.$ Kyun $t=V/Q$: continuity kehti hai ki wahi volume-per-second $Q$ jo tap se nikalti hai woh tank mein pahunchti hai (koi paani gayab nahi hota). Toh tap har second $Q$ cubic metres steadily deliver karta hai; total volume $V$ ko us per-second rate se divide karna exactly utne seconds batata hai — rate × time = volume, rearranged.

Junction par volume conservation: jo per second enter karta hai = jo per second nikalta hai. Yeh ek ratio-style balance hai — har area dono sides par cm² mein hai, toh cm² cancel ho jaate hain aur hume m² mein convert kiye bina clean speed m/s mein milti hai: $Q_0=2Q_b\Rightarrow A_0{v}_0=2A_b{v}_b.$ $v_b=\frac{A_0{v}_0}{2A_b}=\frac{10\times 3}{2\times 4}=\frac{30}{8}=3.75\text{m/s}.$ Kyun cm² safe hain: $A_0$ aur $A_b$ ratio $A_0/A_b$ ke roop mein appear hote hain, toh unka common unit gayab ho jaata hai — sirf numerical size ratio matter karta hai yahan. Kyun factor 2: continuity poore flow ke baare mein hai, toh dono branches milke puri incoming $Q$ carry karni chahiye. s02 dekho — incoming volume-per-second do streams mein fan out hota hai.

$A_1{v}_1=A_2{v}_2$ deta hai $\frac{A_1}{A_2}=\frac{v_2}{v_1}=3$. Upstream (slow) section ka area downstream (fast) section se teen guna hai — wide-slow, narrow-fast, speed ratio se exactly ulta. Kyun ratio flip hota hai: $A_1{v}_1=A_2{v}_2$ mein product $Av$ fixed hai, toh $A$ aur $v$ inversely tied hain — agar $v$ 3 se multiply hota hai, toh product constant rakhne ke liye $A$ ko 3 se divide hona padta hai. Isliye area ratio ($3$) speed change ka reciprocal hai, yaani yeh speed ratio $v_2/v_1$ doosri taraf se padha jaata hai. Koi units zaruri nahi kyunki sirf ratio poochha gaya hai.

Do cheezein change hoti hain, toh poori mass form rakho: $v_2=\frac{{\rho }_1{A}_1{v}_1}{{\rho }_2{A}_2}=\frac{1.0\times 0.30\times 20}{1.5\times 0.10}=\frac{6.0}{0.15}=40\text{m/s}.$ Kyun: area ×3 se ghata (speed upar push karta hai) jabki density ×1.5 se badhi (speed neeche pull karti hai). Net factor $=\frac{3}{1.5}=2$, toh $v$ 20 se 40 ho jaata hai.

General junction rule hai total in = sum of totals out: $A_0{v}_0=A_1{v}_1+A_2{v}_2={\sum }_i{A}_i{v}_i.$ Incoming: $Q_0=A_0{v}_0=10\times 3=30$ (cm²·m/s). Branch 1 leta hai $Q_1=A_1{v}_1=4\times 2=8$. Toh branch 2 ko baaki carry karna padega: $Q_2=Q_0-Q_1=30-8=22,v_2=\frac{Q_2}{A_2}=\frac{22}{3}\approx 7.33\text{m/s}.$ Kyun sum, factor 2 nahi: L3.1 mein "$2$" ek shortcut tha jo sirf tab valid tha jab dono branches identical theen. Generally har branch jo $Av$ carry karta hai woh carry karta hai, aur continuity bas demand karti hai ki woh sab add up karke incoming flow ke barabar hon. Equal-branch case woh special instance hai $A_1{v}_1=A_2{v}_2$. (Yahan saare $Q$ consistent cm²·m/s units mein hain, toh subtraction legitimate hai; sirf speed $v_2$ ko shared cm² cancel hone ki zarurat hai, jo ho jaata hai.)

Step 0 (kaun si form?): paani incompressible hai ($\rho$ constant), toh volume form $A_1{v}_1=A_2{v}_2$ legal hai — speeds ke liye humein $\rho Av$ ki zarurat nahi hai. Step 1 (continuity, ek ratio): areas $A_1/A_2$ ke roop mein appear hote hain, toh cm² cancel ho jaate hain: $v_2=v_1\frac{A_1}{A_2}=2\times \frac{8}{2}=8\text{m/s}.$ Step 2 (Bernoulli, ab SI mein): speeds m/s mein hain aur $\rho$ kg/m³ mein, toh pressure Pa mein aati hai: $p_1-p_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)=\frac{1}{2}(1000)(8^2-2^2)=500\times (64-4)=30000\text{Pa}.$ Kyun pehle continuity: Bernoulli ko dono speeds chahiye; continuity woh ek tool hai jo tumhe geometry se $v_2$ deta hai. Yeh chained logic exactly waise hi hai jaise ek Venturi meter kaam karta hai — narrow spot tez jaata hai, pressure drop hoti hai, aur tum drop ko ek gauge se read karte ho.

Free surface (section 1) aur hole (section 2) ke beech continuity, incompressible paani: $v_1=v_2\frac{A_2}{A_1}=6\times \frac{5\times 10^{-4}}{1.0}=3\times 10^{-3}\text{m/s}=3\text{mm/s}.$ Kyun itni tiny: surface hole se 2000× wide hai, toh yeh 2000× zyada dheere creep karti hai jitna tez jet shoot karta hai — yahi reason hai ki hum energy problems mein usually tank surface ko stationary treat karte hain.

$Q=A_1{v}_1=12\times 1=12$ (cm²·m/s mein — har section ke liye same kyunki incompressible; cm² cancel ho jaate hain jab hum har speed ke liye wapas divide karte hain). $v_2=\frac{Q}{A_2}=\frac{12}{4}=3\text{m/s},v_3=\frac{Q}{A_3}=\frac{12}{6}=2\text{m/s}.$ Sabse tez jahan sabse narrow → section 2 ($3\text{m/s}$). Notice karo $v_3<v_2$: 4 se 6 cm² tak widening flow ko phir slow kar deta hai. Continuity local hai — har section ki speed uske apne area se set hoti hai.

(a) Incompressible: $v_2=v_1\frac{A_1}{A_2}=30\times \frac{0.40}{0.20}=60\text{m/s}.$ (b) Full mass form: $v_2=\frac{{\rho }_1{A}_1{v}_1}{{\rho }_2{A}_2}=\frac{1.2\times 0.40\times 30}{1.6\times 0.20}=\frac{14.4}{0.32}=45\text{m/s}.$ (c) Incompressible guess (60) over-estimate karta hai. Jab gas zyada dense compress hoti hai, har kilogram ab kam volume mein pack hota hai, toh usi mass ko move karne ke liye kam speed chahiye — sahi speed (45) kam hai. Area ratio alone (×2) density rise (×1.33) se partly "absorb" ho jaata hai: net speed factor $=\frac{2}{1.33}=1.5$, deta hai $30\times 1.5=45$. Dekho Incompressible vs compressible flow.

Saari quantities already SI mein hain (m², m³, m³/s), toh koi conversion zaruri nahi. (a) Empty hone ka time. Har second wahi $Q$ nikalti hai, toh total volume ÷ rate time deta hai: $t=\frac{V}{Q}=\frac{0.90}{1.5\times 10^{-3}}=600\text{s}=10\text{min}.$ Kyun $t=V/Q$: continuity guarantee karti hai ki per second nikalne wala volume exactly $Q$ hai (koi paani gayab nahi hota), toh $0.90{\text{m}}^3$ ko $1.5\times 10^{-3}{\text{m}}^3$ har second ki dar par empty karne mein $V/Q$ seconds lagte hain — rate × time = volume, rearranged. (b) Exit pipe area. Wahi $Q$ chote pipe se squeeze hoti hai, toh $Q=A_{\text{exit}}v$: $A_{\text{exit}}=\frac{Q}{v}=\frac{1.5\times 10^{-3}}{3}=5\times 10^{-4}{\text{m}}^2=5{\text{cm}}^2.$ (c) Surface drop speed. Surface (area $A_1$) bhi wahi $Q$ pass karti hai: $Q=A_1{v}_{\text{surface}}$, toh $v_{\text{surface}}=\frac{Q}{A_1}=\frac{1.5\times 10^{-3}}{0.50}=3\times 10^{-3}\text{m/s}=3\text{mm/s}.$ Kyun $Q$ teeno parts ko link karta hai: wahi volume-per-second tank empty karta hai, jet ko feed karta hai, aur surface ko neeche laata hai — continuity ek shared number $Q$ ke through badi slow surface ko chote fast jet se tie karta hai.

Mass flow rate $\rho Av$ — kabhi bhi volume form $Av$ nahi.

Total in = sum of totals out: $A_0{v}_0={\sum }_i{A}_i{v}_i$. Equal branches mein har ek ko aadha milta hai; unequal branches mein har ek apna $A_i{v}_i$ carry karta hai.

Bernoulli ko dono speeds chahiye; continuity woh hai jo area geometry ko doosri speed mein convert karti hai.

Ratio-only problems (area ratio se speed) mein cm² cancel ho jaate hain; koi bhi problem jo real $Q$, time, ya pressure yield kare usse SI (m²) chahiye.

$v$ ka kya matlab hai? Chosen positive direction ke relative reversed flow — fluid us section ko peeche ki taraf cross kar raha hai.