Exercises — Bernoulli's equation — derivation from F = ma along streamline

• $P$ = static pressure — pressure of the fluid mein unit volume per stored energy.
• $\frac{1}{2}\rho v^2$ = dynamic pressure — unit volume per kinetic energy.
• $\rho gy$ = gravitational (hydrostatic) term — unit volume per potential energy.

Teeno pascal unit share karte hain, aur $1\text{Pa}=1{\text{J/m}}^3$ — energy per volume. Yahi equality isliye hai ki teeno ko add kiya ja sakta hai: ye same tarah ki quantity hain.

(a) Honey moti hai — inviscid fail hoti hai (zyada viscosity, internal friction se energy lost). Dekho Viscosity and Poiseuille Flow. (b) Velocity field time ke saath badal'ta hai — steady fail hoti hai ($\partial /\partial t\ne 0$). (c) Sound ki speed ke paas/upar air ki density bahut badal'ti hai — incompressible fail hoti hai ($\rho$ constant nahi).

Kam. Horizontal ⇒ $\rho gy$ term dono points par same hai, isliye $P+\frac{1}{2}\rho v^2$ constant hai. Bada $v$ chota $P$ force karta hai. (Yahi Venturi Meter principle hai.)

Free surface (1) se hole (2) tak streamline. Dono atmosphere ke liye open hain isliye $P_1=P_2$; wide surface muskil se move karti hai isliye $v_1\approx 0$; height mein drop $h$ hai. $\rho gh=\frac{1}{2}\rho v_2^2\Rightarrow v_2=\sqrt{2gh}=\sqrt{2(9.8)(5.0)}\approx 9.9\text{m/s}.$ Same hai jaise koi object $5\text{m}$ se drop kiya ho — dekho Torricelli's Law.

Stagnation muh par air rest mein aa jaati hai ($v=0$); horizontal, isliye height cancel. $P+\frac{1}{2}\rho v^2=P_0\Rightarrow v=\sqrt{\frac{2(P_0-P)}{\rho }}=\sqrt{\frac{2(3000)}{1.2}}\approx 70.7\text{m/s}.$

Horizontal, isliye height cancel: $P_2=P_1+\frac{1}{2}\rho (v_1^2-v_2^2)=150000+\frac{1}{2}(1000)(2^2-6^2).$ $P_2=150000+500(4-36)=150000-16000=134000\text{Pa}.$ Jahan flow speed up hoti hai wahan pressure $16\text{kPa}$ drop hota hai.

Step 1 — speeds ko continuity se link karo (dekho Equation of Continuity). Same volume per second dono areas se guzarta hai — figure mein, wahi red streamline wide part aur throat dono ko thread karti hai: $A_1{v}_1=A_2{v}_2\Rightarrow v_2=\frac{A_1}{A_2}{v}_1=\frac{8}{2}{v}_1=4v_1.$ Step 2 — Bernoulli, horizontal (streamline ek height par rehti hai, isliye $\rho gy$ points 1 aur 2 ke beech cancel ho jaata hai): $P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}\rho ((4v_1{)}^2-v_1^2)=\frac{1}{2}\rho (15v_1^2).$ Step 3 — solve: $6000=\frac{1}{2}(1000)(15)v_1^2=7500v_1^2\Rightarrow v_1=\sqrt{0.8}\approx 0.894\text{m/s}.$ (Tab $v_2=4v_1\approx 3.58\text{m/s}$ agar puchha jaaye — figure mein throat par "fast" label.)

Same width ⇒ same speed ⇒ dynamic terms cancel; sirf height term shift karta hai: $P_2=P_1-\rho g(y_2-y_1)=200000-(1000)(9.8)(4.0)=200000-39200=160800\text{Pa}.$ Rising fluid potential energy khareedne ke liye pressure "spend" karta hai — yahi Hydrostatic Pressure hai jo ek moving fluid ke andar reappear kar raha hai.

Pehle, gauge vs absolute. Gauge pressure atmosphere ke relative measure kiya jaata hai — yahi ek tyre gauge read karta hai, aur yeh negative ho sakta hai (atmospheric se neeche). Absolute pressure ek perfect vacuum se measure kiya jaata hai ($P_{\text{abs}}=P_{\text{gauge}}+P_{\text{atm}}$) aur physically zero se neeche nahi ja sakta, kyunki zero absolute pressure ka matlab pehle se hi "koi molecules push nahi kar rahe" hai. Isliye ek predicted absolute pressure of $-5000\text{Pa}$ impossible hai: iske liye fluid ko pull karna padega, aur liquids zyada tension sustain nahi kar sakte. Actually, jaise $P$ water ki vapour pressure (ek chota sa positive number) ki taraf girta hai, water boil/cavitate karta hai: vapour bubbles form hote hain aur smooth streamline toot jaati hai. Bernoulli ki assumptions (incompressible, single continuous streamline) tab fail ho jaati hain. Isliye ek negative absolute pressure ek warning flag hai ki flow us tak pahunchne se pehle cavitate karta hai — yeh koi real pressure nahi hai.

Top aur bottom ko (approximately) equal-height points on nearby streamlines treat karo, har ek par Bernoulli apply karo. Pressure difference: $P_{\text{bot}}-P_{\text{top}}=\frac{1}{2}\rho (v_{\text{top}}^2-v_{\text{bot}}^2)=\frac{1}{2}(1.2)(120^2-100^2).$ $=0.6(14400-10000)=0.6(4400)=2640\text{Pa}.$ Lift $=\Delta P\times \text{area}=2640\times 30=79200\text{N}\approx 7.9\times 10^4\text{N}$. (Dekho Aerodynamic Lift. Real lift ke liye circulation theory chahiye, lekin Bernoulli estimate sign aur rough size capture karta hai.)

Continuity: $A_1{v}_1=A_2{v}_2\Rightarrow v_1=\frac{A_2}{A_1}{v}_2=\frac{0.05}{0.5}{v}_2=0.1v_2$. Bernoulli (dono surfaces atmosphere par, drop $h$): $\rho gh=\frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}\rho (v_2^2-(0.1v_2{)}^2)=\frac{1}{2}\rho v_2^2(1-0.01).$ $gh=0.495v_2^2\Rightarrow v_2=\sqrt{\frac{9.8\times 1.8}{0.495}}=\sqrt{35.63}\approx 5.97\text{m/s}.$ Naive Torricelli se compare karo $\sqrt{2gh}=\sqrt{35.28}\approx 5.94\text{m/s}$ — correction tiny hai ($0.01$), isliye yeh justify karta hai ki kyun $v_1\approx 0$ usually kaam karta hai.

Uniform width ⇒ continuity se tube mein speed har jagah same hai; use $v$ kaho. (a) Outlet speed. Tank surface (1, $v\approx 0$, $P_{\text{atm}}$, $y=0$) se outlet (3, $P_{\text{atm}}$, $y=-D$) tak streamline — figure mein poora red path: $P_{\text{atm}}+0+0=P_{\text{atm}}+\frac{1}{2}\rho v^2+\rho g(-D).$ $\frac{1}{2}\rho v^2=\rho gD\Rightarrow v=\sqrt{2gD}=\sqrt{2(9.8)(2.5)}\approx 7.0\text{m/s}.$ Outlet speed sirf source ke neeche drop $D$ par depend karta hai (neeche wala double-arrow) — bend height speed ke liye cancel ho jaata hai. (b) Bend ke top par pressure (figure mein point 2, $y=+H$, speed $v$). Surface (1) se top (2) tak: $P_{\text{atm}}=P_2+\frac{1}{2}\rho v^2+\rho gH.$ $P_2=P_{\text{atm}}-\frac{1}{2}\rho v^2-\rho gH=101325-\frac{1}{2}(1000)(49)-(1000)(9.8)(1.0).$ $P_2=101325-24500-9800=67025\text{Pa}.$ Positive (vapour pressure se upar), isliye siphon kaam karta hai. Agar $P_2$ zero ki taraf jaata, water column toot jaata — siphon ki maximum lift height.

Step 1 — manometer se pressure difference. U-tube mein, do arms do pipe sections se connect hain. Neeche wali mercury surface ke level par, dono sides se pressure match karni chahiye (same fluid, same height ⇒ same pressure — Hydrostatic Pressure). Arm 1 ke neeche jaao: $P_1$ plus ek water column; arm 2 ke neeche jaao: $P_2$ plus ek chota water column plus height $\Delta h$ ka ek mercury column. Inhe balance karo aur common water heights cancel karo to mercury column pressure difference ki jagah khada rehta hai, lekin humein woh paani subtract karna hai jo mercury displace karta hai doosri side mein: $P_1-P_2=({\rho }_{\text{Hg}}-{\rho }_{\text{water}})g\Delta h.$ $-{\rho }_{\text{water}}$ isliye appear hota hai kyunki ek side par $\Delta h$ mercury doosri side par $\Delta h$ paani ke against balanced hai; sirf density difference net pressure produce karta hai. Numerically: $P_1-P_2=(13600-1000)(9.8)(0.15)=12600\times 9.8\times 0.15=18522\text{Pa}.$ Step 2 — continuity: $v_2=\frac{A_1}{A_2}{v}_1=4v_1$. Step 3 — Bernoulli, horizontal: $P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1000)(16v_1^2-v_1^2)=7500v_1^2.$ Step 4 — solve: $v_1^2=\frac{18522}{7500}=2.4696\Rightarrow v_1\approx 1.571\text{m/s}.$ Step 5 — flow rate ($A_1=40{\text{cm}}^2=40\times 10^{-4}{\text{m}}^2$): $Q=A_1{v}_1=40\times 10^{-4}\times 1.571\approx 6.29\times 10^{-3}{\text{m}}^3/\text{s}.$ Yeh lagbhag $6.3$ litres per second hai.

Bend par, surface (1) se bend top (2) tak pressure with $v\approx 0$: $P_2=P_{\text{atm}}-\rho gH.$ Failure point ke liye $P_2=P_{\text{vap}}$ set karo: $H_{\max }=\frac{P_{\text{atm}}-P_{\text{vap}}}{\rho g}=\frac{101325-2300}{(1000)(9.8)}=\frac{99025}{9800}\approx 10.1\text{m}.$ Yahi $\sim 10\text{m}$ ceiling hai jo limit karti hai kitna high suction akela paani raise kar sakta hai — Hydrostatic Pressure ke saath ek khoobsurat cross-check.

Horizontal ⇒ $\rho gy$ term drop karo. Har line par Bernoulli sum ("constant") evaluate karo: $C_A=P_A+\frac{1}{2}\rho v_A^2=120000+\frac{1}{2}(1000)(5^2)=120000+12500=132500\text{Pa}.$ $C_B=P_B+\frac{1}{2}\rho v_B^2=120000+\frac{1}{2}(1000)(1^2)=120000+500=120500\text{Pa}.$ Kyunki $C_A=132500\ne 120500=C_B$, Bernoulli constant generally alag alag streamlines par alag hota hai — bilkul wahi parent-note warning. Tum line A se koi value line B par nahi le ja sakte. Equality ke liye extra condition: agar flow additionally irrotational ho (kahin bhi koi swirl nahi), to wahi constant fluid mein har jagah hold karta hai, sirf har streamline ke saath nahi. Inhi do lines ke alag-alag constants hain, isliye yeh flow rotational hai.

Continuity speeds ko tie karta hai; Bernoulli ek streamline ke saath pressure ↔ speed ↔ height trade karta hai; signs watch karo (negative pressure = cavitation) aur kabhi streamlines hop mat karo jab tak flow irrotational na ho.