Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline: Exercises

2.2.15 · D4 · Physics › Fluid Mechanics › Assumptions in Bernoulli — steady, inviscid, incompressible,

Intuition Is page ko kaise use karein

Neeche diye gaye har problem mein aapko dekhna hai ki Bernoulli ke chaar licenses (steady, inviscid, incompressible, along a streamline) mein se kaun sa hold kar raha hai ya kaun sa toot raha hai — aur phir uska consequence calculate karna hai. Solutions collapsible [!recall]- callouts ke andar chhupaaye gaye hain: pehle khud try karo, phir reveal karo. Problems "sirf assumption ka naam batao" (L1) se shuru hokar "teen ideas combine karo aur apna reasoning defend karo" (L5) tak jaate hain.

Agar koi symbol yahan unfamiliar lage, toh yeh signal hai ki parent topic dobara dekho: the parent topic. Hum sirf wahi use karte hain jo usne build kiya hai: four-term energy sum $P + \frac{1}{2} ρ v^{2} + ρ g h$ , material derivative $\frac{D v}{D t} = \frac{\partial v}{\partial t} + v \frac{\partial v}{\partial s}$ , aur continuity rule $A_{1} v_{1} = A_{2} v_{2}$ .

Definition Is page par use hone wale symbols (har ek earn kiya hua)

$P$ = pressure = force per unit area jo ek fluid push karta hai, pascals mein ( $Pa = N/m^{2}$ ).
$ρ$ (rho) = density = har cubic metre mein packed mass, $kg/m^{3}$ mein.
$v$ = apni streamline ke saath fluid ki speed, $m/s$ mein.
$h$ = ek chosen reference level se upar ki height, metres mein.
$g = 9.8 m/s^{2}$ = gravity ka pull, har kilogram par.
$A$ = pipe ka cross-sectional area (woh "window" jiske through fluid flow karta hai, uska size).
$s$ = streamline coordinate = distance jo streamline ke along measure ki jaati hai jis par parcel chal raha hai, metres mein. Ise ek water parcel ke odometer ki tarah socho jo apna khud ka path follow karta hai.
"Head" = ek energy term jise $ρ g$ se divide karke height ke roop mein likha jaata hai. Toh ek pressure $P$ , pressure-head $P / (ρ g)$ metres ban jaata hai. Hum ise use karte hain kyunki losses traditionally metres mein quote ki jaati hain.

Level 1 — Recognition

L1.1 — Failing assumption ka naam batao

Paani ek 50 m lambi rusty horizontal pipe of constant diameter se flow karta hai. Inlet gauge pressure $180 kPa$ hai, outlet $140 kPa$ hai. Speed aur height dono ends par identical hain. Plain Bernoulli equal pressures predict karta hai. Kaun sa assumption toot raha hai, aur bacha hua $40 kPa$ physically kya ban jaata hai?

Recall Solution L1.1

Constant area ⇒ continuity $A_{1} v_{1} = A_{2} v_{2}$ se, $A_{1} = A_{2}$ ke saath, $v_{1} = v_{2}$ milta hai. Horizontal ⇒ $h_{1} = h_{2}$ . Toh Bernoulli ke $\frac{1}{2} ρ v^{2}$ aur $ρ g h$ terms dono ends par equal hain aur yeh $P_{1} = P_{2}$ force karta hai. Reality: $P$ $40 kPa$ se drop hua. Toot a hua license inviscid hai. Wall friction (viscosity) mechanical energy ko irreversibly heat mein convert kar deta hai, isliye "constant" downstream decay ho jaata hai. Fix extended equation hai jisme head-loss term $h_{L}$ hota hai.

L1.2 — Steady hai ya unsteady?

Aap ek valve kholte ho; pehle aadhe second mein pipe mein water column har jagah ek saath speed up ho raha hai. Kaun sa single acceleration term ab non-zero hai, aur yeh kaun sa assumption khatam karta hai?

Recall Solution L1.2

Missing term kahan se aata hai. Parent ne Bernoulli ko streamline coordinate $s$ ke saath ek parcel par Newton's 2nd law se derive kiya — "Euler equation" along the streamline: $ρ unsteady \frac{\partial v}{\partial t} + convective v \frac{\partial v}{\partial s} = - \frac{\partial P}{\partial s} - ρ g \frac{\partial h}{\partial s} .$ Ise $s$ ke along point 1 se point 2 tak integrate karne par ordinary Bernoulli plus ek extra piece milta hai unsteady term se: $unsteady / inertial term ρ \int_{1}^{2} \frac{\partial v}{\partial t} d s + (P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2}) - (P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1}) = 0.$ Yahan integral usi streamline ke along liya jaata hai (yahi " $d s$ " ka matlab hai — upar define kiya gaya streamline-odometer step), time par nahi. (Smooth irrotational flow mein yahi term aksar velocity potential $ϕ$ use karke $ρ \partial ϕ / \partial t$ likha jaata hai; dono equivalent forms hain.) Answer. Local (unsteady) term $\frac{\partial v}{\partial t} \neq = 0$ : velocity field har fixed point par time mein change ho rahi hai. Yeh steady assumption ko khatam karta hai, aur exactly yahi $ρ \int_{1}^{2} \partial v / \partial t d s$ (woh "water-hammer / inertial" term) hai jo plain Bernoulli drop kar deta hai. Jab flow settle ho jaata hai, $\partial v / \partial t \to 0$ aur Bernoulli phir valid ho jaata hai.

L1.3 — Gas hai ya nahi?

Air $30 m/s$ par flow kar rahi hai; local speed of sound $340 m/s$ hai. Mach number compute karo aur batao ki incompressible license hold karta hai ya nahi.

Recall Solution L1.3

Mach number $= \frac{v}{a} = \frac{30}{340} \approx 0.088$ . Kyunki $0.088 < 0.3$ hai, density change negligible hai: incompressible assumption hold karta hai, aur Bernoulli use kiya ja sakta hai.

Level 2 — Application

L2.1 — Venturi speed-up (valid case)

Paani ( $ρ = 1000 kg/m^{3}$ ) horizontally ek pipe se flow karta hai jo area $A_{1} = 8 cm^{2}$ se $A_{2} = 2 cm^{2}$ tak narrow hoti hai. Wide-section speed $v_{1} = 1.5 m/s$ hai. Yeh maante hue ki charo licenses hold karte hain, $v_{2}$ aur pressure drop $P_{1} - P_{2}$ nikalo.

Geometry neeche draw ki gayi hai — dashed plum streamline ko seedha throat se trace karo.

Figure — Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline

Figure ko aise padho: teal shading paani hai; jahan pipe small window $A_{2}$ (right) par pinch hoti hai, paani ko tezi se guzarna padta hai, isliye orange velocity arrow $v_{1}$ se $v_{2}$ tak bada hota hai. Do imagined pressure taps ( $P_{1}$ wide side par, $P_{2}$ throat par) wahi hain jo Venturi meter actually measure karta hai — faster throat ek lower pressure padhta hai.

Recall Solution L2.1

Step 1 — continuity (WHY: steady, incompressible flow mein paani ki same mass har second har window se cross karni chahiye, isliye chhoti window faster speed force karti hai — mass conservation). $A_{1} v_{1} = A_{2} v_{2} \Rightarrow v_{2} = \frac{A _{1}}{A _{2}} v_{1} = \frac{8}{2} (1.5) = 6 m/s .$ Step 2 — Bernoulli, horizontal isliye $h$ drop ho jaata hai (WHY: same streamline, sare licenses assume valid hain). $P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$ $P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} (1000) (6^{2} - 1. 5^{2}) = \frac{1}{2} (1000) (36 - 2.25) .$ $P_{1} - P_{2} = 16, 875 Pa \approx 16.9 kPa .$ Faster narrow section ⇒ lower pressure, bilkul waise jaisa throat gauge padhta.

L2.2 — Torricelli drain

Ek badi open tank mein paani $H = 1.8 m$ gehrayi tak ek small side hole ke upar hai. Hole se nikalne wali jet speed $v$ nikalo. (Tank surface speed $\approx 0$ kyunki tank wide hai.)

Recall Solution L2.2

Free surface (point 1) se hole (point 2) tak ek streamline choose karo. Dono atmosphere ke samne exposed hain, isliye $P_{1} = P_{2} = P_{atm}$ — pressure terms cancel ho jaate hain. $\approx 0 \frac{1}{2} ρ v_{1}^{2} + ρ g H = \frac{1}{2} ρ v_{2}^{2} + 0.$ $v_{2} = 2 g H = 2 (9.8) (1.8) = 35.28 \approx 5.94 m/s .$ Yeh Torricelli's law hai — height $H$ se free-fall speed, kyunki pressure energy yahan koi net role nahi nibhati.

Level 3 — Analysis

L3.1 — Real head loss wali rough pipe

Paani ek horizontal constant-area pipe mein flow karta hai. Measured pressure drop $Δ P = 40 kPa$ hai, $ρ = 1000 kg/m^{3}$ . Is loss ko head loss $h_{L}$ metres mein express karo, aur batao yeh kaun se extended balance mein aata hai.

Recall Solution L3.1

Streamline ke along extended (viscous) Bernoulli: $P_{1} + \frac{1}{2} ρ v^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ v^{2} + ρ g h_{2} + friction ρ g h_{L} .$ Constant area + horizontal ⇒ $v$ aur $h$ terms cancel ho jaate hain, $P_{1} - P_{2} = ρ g h_{L}$ bacha rehta hai: $h_{L} = \frac{Δ P}{ρ g} = \frac{40 , 000}{1000 \times 9.8} = 4.08 m .$ Toh viscous friction $4.08 m$ head "kha" leta hai — mechanical energy heat mein gayi. Yahi woh term hai jo plain Bernoulli inviscid flow assume karke chhod deta hai.

L3.2 — Wing par inviscid kahan break down karta hai?

Air $30 m/s$ (Mach $0.09$ ) par steady cruise mein ek wing ke upar flow karti hai. Bernoulli faster top surface se lift sahi predict karta hai — sirf ek thin region ko chhodkar. Woh region ka naam batao, woh assumption ka naam batao jo wahan fail hoti hai, aur explain karo ki baaki flow kyun abhi bhi fair game hai.

Recall Solution L3.2

Fail hone wala region boundary layer hai — surface ke right against woh paper-thin film jahan velocity free-stream value se wall par zero tak drop karti hai. Wahan inviscid fail hoti hai: shear (viscosity) dominate karta hai. Is layer ke bahar bulk flow effectively frictionless, steady, aur low-Mach (incompressible) hai, isliye har outer streamline ke saath Bernoulli legitimate hai aur faster-top ⇒ lower- $P$ ⇒ lift argument valid rehta hai. Dono regions simply alag streamlines hain alag physics ke saath.

Level 4 — Synthesis

L4.1 — Venturi meter do failures apart padhna

Ek horizontal Venturi ( $ρ = 1000$ ) $A_{1} = 10 cm^{2}$ se $A_{2} = 4 cm^{2}$ tak narrow hoti hai. Ideal Bernoulli pressure drop continuity se neeche calculate kiya gaya hai, lekin real manometer $8.0 kPa$ padhta hai. $v_{1} = 2 m/s$ diya hai: (a) ideal $Δ P$ compute karo, (b) friction loss $Δ P_{loss}$ nikalo, (c) har woh assumption ka naam batao jo valid rahi aur jo fail hui.

Recall Solution L4.1

(a) Ideal drop. Pehle continuity (WHY: steady incompressible flow mein mass conservation — same volume per second chhote window se guzarta hai, isliye woh tezi se move karta hai): $v_{2} = \frac{A _{1}}{A _{2}} v_{1} = \frac{10}{4} (2) = 5 m/s .$ $Δ P_{ideal} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} (1000) (25 - 4) = 10, 500 Pa = 10.5 kPa .$ (b) Loss. Manometer ek chhota drop padhta hai ( $8.0 kPa$ ) ideal ( $10.5 kPa$ ) se — friction pressure recovery kharab karta hai, isliye tappings ke beech measured mechanical-energy drop ideal se kam hai: $Δ P_{loss} = Δ P_{ideal} - Δ P_{measured} = 10.5 - 8.0 = 2.5 kPa .$ Yeh $2.5 kPa$ viscous cost hai. (c) Jo hold kiye: steady (fixed flow), incompressible (water), along a streamline (centre-line). Jo fail hua: inviscid — woh $2.5 kPa$ heat mein friction hai.

L4.2 — Line mein pump

Paani ko ek lower reservoir (surface at rest, atmospheric) se ek upper outlet tak pump kiya jaata hai jo $12 m$ upar hai, atmosphere mein $3 m/s$ par exit karta hai. Friction ignore karte hue, pump head $h_{P}$ (metres of water jo pump add karna chahiye) nikalo. $g = 9.8$ .

Recall Solution L4.2

Energy balance pump term ke saath (plain Bernoulli illegal hai — pump "constant" ko jump karata hai): $lower surface P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} + ρ g h_{P} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2} .$ WHY $P_{1} = P_{2}$ cancel hota hai: reservoir surface aur outlet dono same atmosphere ke samne open hain, isliye $P_{1} = P_{2} = P_{atm}$ . Equal numbers equation ke opposite sides par subtract hokar zero ho jaate hain — atmosphere dono ends par equally push karta hai aur koi net work nahi karta, isliye woh simply drop out ho jaata hai. $v_{1} \approx 0$ (wide reservoir), $h_{1} = 0, h_{2} = 12$ ke saath: $ρ g h_{P} = \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2} \Rightarrow h_{P} = \frac{v _{2}^{2}}{2 g} + h_{2} = \frac{3 ^{2}}{2 ( 9.8 )} + 12.$ $h_{P} = 0.459 + 12 = 12.46 m .$ Velocity head ( $0.459 m$ ) plus lift ( $12 m$ ) exactly wahi hai jo pump supply karta hai.

Level 5 — Mastery

L5.1 — Free vortex: do streamlines, ek galat constant

Ek free vortex (water draining) mein speed radius ke saath $v = \frac{k}{r}$ vary karti hai (constant $k$ ke saath). Inner streamline $r_{1} = 0.1 m$ par aur outer $r_{2} = 0.4 m$ par lo, $k = 0.6 m^{2} / s$ ke saath, sab same height par, $ρ = 1000$ . Ek student do radii ke across ek Bernoulli constant equate karta hai $Δ P$ nikalne ke liye. (a) Naive $Δ P = \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2})$ compute karo. (b) Precisely explain karo yeh yahan sirf kyun legitimate hai, aur woh extra property batao jo ise rescue karti hai.

Recall Solution L5.1

(a) $v_{1} = \frac{0.6}{0.1} = 6 m/s$ , $v_{2} = \frac{0.6}{0.4} = 1.5 m/s$ . $Δ P = \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2}) = \frac{1}{2} (1000) (36 - 2.25) = 16, 875 Pa \approx 16.9 kPa .$ (b) Points $r_{1}$ aur $r_{2}$ alag streamlines (concentric circles) par hain. Bernoulli ka constant sirf ek streamline ke along guaranteed equal hota hai, isliye unhe cross karte hue equate karna normally illegal hai. Yeh sirf isliye rescue hota hai kyunki ek free vortex ( $v \propto 1/ r$ ) irrotational hai — irrotational flow ke liye Bernoulli constant global hai (har streamline par same). Isliye along a streamline license upgrade hokar "anywhere" ban jaata hai, aur number valid rehta hai. Ek forced vortex ( $v \propto r$ , rotational) is equation ko nonsense bana deta.

L5.2 — Compressibility kab bite karta hai? (design threshold)

Air ( $a = 340 m/s$ , $ρ = 1.2 kg/m^{3}$ ) ek body ke upar speed up hoti hai. Kis free-stream speed $v$ par flow pehli baar Mach $0.3$ compressibility threshold cross karta hai, aur rest se us speed tak accelerate karte hue corresponding ideal Bernoulli pressure drop kya hai?

Recall Solution L5.2

Threshold speed: $Mach = v / a = 0.3 \Rightarrow v = 0.3 (340) = 102 m/s .$ Rest se $v$ tak accelerate karte hue ideal pressure drop (incompressible Bernoulli, edge par abhi just valid): $Δ P = \frac{1}{2} ρ v^{2} = \frac{1}{2} (1.2) (102)^{2} = \frac{1}{2} (1.2) (10404) = 6242.4 Pa \approx 6.24 kPa .$ $102 m/s$ ke upar incompressible license fail ho jaata hai: $\int d P / ρ$ ab $Δ P / ρ$ ke equal nahi rehta aur ek compressible correction chahiye.

L5.3 — Material derivative se steady-check

Ek pipe flow mein velocity apne streamline coordinate ke saath $v (s, t) = (2 + 0.5 s) (1 + 0.1 t) m/s$ di gayi hai, jahan $s$ streamline ke along distance (metres) hai aur $t$ time (seconds) hai. Sab kuch $s = 4 m$ , $t = 0$ par evaluate karo: (a) kya flow steady hai? (b) full material acceleration $\frac{D v}{D t}$ compute karo, aur (c) batao plain Bernoulli kaunsa piece galti se ignore karta.

Recall Solution L5.3

(a) $\frac{\partial v}{\partial t} = (2 + 0.5 s) (0.1)$ . $s = 4$ par: $= (2 + 2) (0.1) = 0.4 \neq = 0$ . Steady nahi — Bernoulli licensed nahi hai. (b) Convective part: $\frac{\partial v}{\partial s} = 0.5 (1 + 0.1 t)$ ; $t = 0$ par, $= 0.5$ . Speed at $(s = 4, t = 0)$ : $v = (2 + 2) (1) = 4 m/s$ . $v \frac{\partial v}{\partial s} = 4 (0.5) = 2 m/s^{2}, \frac{\partial v}{\partial t} = 0.4 m/s^{2} .$ $\frac{D v}{D t} = \frac{\partial v}{\partial t} + v \frac{\partial v}{\partial s} = 0.4 + 2.0 = 2.4 m/s^{2} .$ (c) Plain Bernoulli quietly $\partial v / \partial t = 0$ set karta hai, isliye woh $0.4 m/s^{2}$ unsteady piece ignore kar deta aur exactly us contribution se pressure field galat predict karta.

Recall Self-test wrap-up

Har broken license ke liye one-line diagnosis: Constant-area pipe, pressure phir bhi drop ::: inviscid fail (friction → heat) Valve abhi khula, sab kuch accelerate ho raha ::: steady fail ( $\partial v / \partial t \neq = 0$ ) Mach $> 0.3$ ::: incompressible fail ( $\int d P / ρ \neq = Δ P / ρ$ ) Alag streamlines par points compare karna ::: along-a-streamline fail (unless irrotational) Do points ke beech ek pump ya fan hai ::: energy inject hui, "constant" jump karta hai

Parent: Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline · Core law: Bernoulli's Equation