Applications — Pitot tube, Venturi meter, orifice flow: Question bank

2.2.16 · D5 · Physics › Fluid Mechanics › Applications — Pitot tube, Venturi meter, orifice flow

Definition Symbol glossary (shuru karne se pehle ek nazar daalo)

$P_{s}$ = stagnation pressure, wahan padha jaata hai jahan pitot mouth par fluid ko rest mein laya jaata hai.
$P_{0}$ = undisturbed stream ka static pressure (side-on port).
$P_{atm}$ = atmospheric pressure — khuli hawa ka pressure. Kisi jet ya tank surface ke liye jo hawa ke samne khuli ho, yahi local static pressure hota hai, isliye $P_{atm}$ kisi bhi open-to-air point par $P_{0}$ ka role play karta hai.
$Δ P$ = woh pressure difference jo ek device measure karta hai (pitot ke liye $P_{s} - P_{0}$ , venturi ke liye $P_{1} - P_{2}$ ).
$ρ$ = flowing fluid ki density (air, water, …).
$v$ = fluid ki local flow speed; $v_{1}, v_{2}$ = ek device ke do labelled points par speeds (venturi mein inlet 1 vs throat 2; draining tank mein surface 1 vs hole 2).
$g$ = gravitational acceleration ( $\approx 9.8 m/s^{2}$ ), gravity ki strength jo har $ρ g h$ aur $2 g H$ term mein aati hai.
$A_{1}, A_{2}$ = venturi ke wide inlet aur narrow throat ke cross-section areas.
$A_{hole}, A_{tank}$ = drain hole aur tank surface ka area.
$H$ = free surface ke neeche drain hole ki depth (Torricelli speed drive karta hai).
$h$ = Bernoulli ke $ρ g h$ term mein ek point ki height (streamline par elevation).
$Δ h$ = manometer liquid column height jo $Δ P$ read off karne ke liye use hoti hai.
$ρ_{m}$ = manometer liquid ki density; $ρ$ (phir se) = $P_{1} - P_{2} = (ρ_{m} - ρ) g Δ h$ mein flowing fluid ki density.
$x$ = floor par jet ki horizontal range; $y$ = hole se floor tak jet ki vertical fall height (projectile motion mein use hoti hai).

True or false — justify karo

Venturi meter ke narrow throat mein zyada pressure hota hai kyunki fluid squeeze hota hai.

False. Fluid kisi solid ki tarah compress nahi hota — continuity se woh throat mein speed up karta hai, aur Bernoulli se faster flow ka matlab lower pressure hai. Throat ek pressure dip hai.

Torricelli ki speed

v = 2 g H

drain hole ki size par depend karti hai.

False. Formula mein koi area hai hi nahi;

v

purely depth

H

se set hoti hai. Hole size flow rate

Q = A_{hole} v

badalta hai, har escaping particle ki speed nahi.

Pitot mouth par stagnation pressure atmospheric pressure ke barabar hoti hai.

False. Mouth stream ki taraf face karta hai aur fluid ko rest mein laata hai, isliye woh static plus dynamic term read karta hai:

P_{s} = P_{0} + \frac{1}{2} ρ v^{2} > P_{0}

. Woh extra bit exactly speed encode karta hai.

Venturi pressure drop

Δ P

ko double karne se measured flow rate

Q

double ho jaata hai.

False. Bernoulli deta hai

Δ P = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})

, isliye

Δ P

speed ke square ke saath scale karta hai; continuity (

Q = A_{2} v_{2}

) phir

Q \propto v_{2} \propto Δ P

banata hai.

Δ P

double karne se

Q

sirf

2 \approx 1.41

se multiply hota hai.

Same measured

Δ P

ke liye, air mein ek pitot tube water mein se zyada speed report karta hai.

True.

v = 2Δ P / ρ

, isliye

v \propto 1/ ρ

; air paani se kaafi halki hai, isliye same

Δ P

bahut tez flow se correspond karta hai.

Orifice derivation mein hum surface speed

v_{1}

ko hole size se independent hokar ignore kar sakte hain.

False. Hum

v_{1}

tab drop karte hain jab

A_{hole} ≪ A_{tank}

force karta hai

v_{1} = (A_{hole} / A_{tank}) v_{2} \approx 0

. Agar hole tank cross-section ke comparable hai, toh

v_{1}

rakhna padega.

Bernoulli's equation mein absolute pressures use karne zaroori hain, gauge pressures kabhi nahi.

False. Teeno devices sirf difference

P_{1} - P_{2}

use karte hain, isliye koi bhi consistent reference cancel ho jaata hai. Rule yeh hai: ek convention chunno aur problem ke beech mein Pa aur atm mix mat karo.

Bernoulli mein

ρ g h

height term air mein ek pitot tube ke liye matter karta hai.

True in principle, practice mein negligible. Ek halki gas mein tube ke tiny vertical span par

ρ g h

dynamic term ke comparison mein bahut chhota hai, isliye hum use cancel kar dete hain — lekin yeh ek size argument hai, exact zero nahi.

Incompressible Bernoulli ka form kisi bhi airspeed par pitot tube ke liye safe hai.

False. Bernoulli yahan constant density

ρ

assume karta hai. Roughly Mach 0.3 (air mein lagbhag

100 m/s

) se upar gases ke liye compressibility matter karti hai aur

v = 2Δ P / ρ

under-read karta hai; ek compressible correction zaroori hai. Edge-case section dekho.

Error pakdo

"Venturi mein fluid throat mein slow down ho jaata hai kyunki wahan kam jagah hai."

Error continuity ko invert karta hai. Kam area matlab zyada speed, kam nahi:

A_{1} v_{1} = A_{2} v_{2}

jahan

A_{2} < A_{1}

force karta hai

v_{2} > v_{1}

. Same volume per second ko ek chhote darwaze se squeeze through karna padta hai.

"Ek pitot tube pressure measure karta hai, isliye uska output ek pressure hai, speed nahi."

Woh ek pressure difference measure karta hai, lekin physics use convert karta hai:

v = 2 (P_{s} - P_{0}) / ρ

. Device ko speed output karne ke liye calibrate kiya jaata hai kyunki dynamic pressure speed ka ek known function hai.

"Torricelli kehta hai jet gehre hole se tez niklati hai kyunki upar ka paani zyada weigh karta hai."

Reasoning sahi answer par galat wajah se pahunche hai. Yeh weight pushing nahi hai balki falling surface height

H

ki potential energy ka kinetic energy mein conversion hai —

v = 2 g H

exactly

H

ke through ek free-fall speed hai.

"Horizontal jet ke liye, sabse gehra hole hamesha floor par paani sabse door throw karta hai."

Galat — range hai

x = 2 H y

jahan

y

hole ke neeche fall height hai. Ek gehra hole (bada

H

) ka remaining fall

y

chhota hota hai, isliye range mid-height par peak karti hai, bottom par nahi.

"Kyunki venturi ke dono ports horizontal pipe par hain, manometer fluid matter nahi karta."

Manometer fluid density

ρ_{m}

conversion set karta hai

P_{1} - P_{2} = (ρ_{m} - ρ) g Δ h

, jahan

ρ

flowing fluid hai. Ek alag manometer liquid same

Δ P

ke liye alag

Δ h

deta hai, isliye yeh bahut matter karta hai.

"Pitot-static tube ka static port bhi hawa ki taraf face karna chahiye."

Nahi — static port side-on (flow ke parallel) hona chahiye taaki woh undisturbed static pressure

P_{0}

read kare. Use hawa ki taraf face karne se woh bhi stagnation pressure read karega, aur difference vanish ho jaayega.

Why questions

Pitot tube ko doosre (static) port ki zaroorat hi kyun hai?

Kyunki speed difference

P_{s} - P_{0}

par depend karti hai. Stagnation port akela

P_{s}

deta hai, lekin static reference

P_{0}

ke bina tum dynamic term

\frac{1}{2} ρ v^{2}

isolate nahi kar sakte.

Derivation mein hum dono pitot points ko same height par kyun set karte hain?

Taaki

ρ g h

terms equal hon aur cancel ho jaayein, ek clean balance chhod ke dynamic pressure aur jis speed ko hum dhundh rahe hain ke beech. Equal height gravity ko bookkeeping se hata deta hai.

Orifice speed same height se freely falling body se kyun match karta hai?

Bernoulli surface(1)→hole(2) likho: dono points air ke samne khule hain, isliye

P_{atm} + \frac{1}{2} ρ v_{1}^{2} + ρ g H = P_{atm} + \frac{1}{2} ρ v_{2}^{2}

. Dono taraf identical

P_{atm}

cancel hota hai;

v_{1} \approx 0

ke saath yeh rehta hai

ρ g H = \frac{1}{2} ρ v_{2}^{2}

, yaani

v = 2 g H

— exactly

H

se giraa hua ek pathar. Dekho Torricelli's Law.

Venturi solve karne se pehle hume

v_{1} = (A_{2} / A_{1}) v_{2}

substitute kyun karna padta hai?

Bernoulli akele do unknown speeds

v_{1}

aur

v_{2}

rakhta hai. Continuity doosra relation deta hai, jisse hum

v_{1}

eliminate kar sakte hain aur ek single unknown

v_{2}

solve karne ke liye chodh sakte hain.

Actual venturi/orifice flow ideal formula se thoda kam kyun hota hai?

Real fluids viscosity ki wajah se energy lose karte hain aur jet contract hoti hai (vena contracta), aisi effects jo ideal Bernoulli+continuity model ignore karta hai. Engineers ise 1 se kam discharge coefficient se patch karte hain.

"Faster ⇒ lower pressure" energy conservation violate kyun nahi karta?

Yahi toh energy conservation hai: total energy per volume

P + \frac{1}{2} ρ v^{2} + ρ g h

fixed hai, isliye kinetic term mein rise pressure term mein drop se pay honi chahiye. Dekho the pressure trinity.

Jet range

x = 2 H y

g

kyun vanish ho jaata hai jabki gravity clearly flow drive kar rahi hai?

Gravity do baar opposite effect ke saath appear hoti hai: woh exit speed up karti hai (

v = 2 g H

g

numerator mein) aur fall time short karti hai (

t = 2 y / g

g

denominator mein). Product

x = v t

mein

g

ke dono factors exactly cancel ho jaate hain.

Edge cases

Pitot formula kya deta hai agar stream rest mein ho (

v = 0

Tab

P_{s} = P_{0}

, isliye

Δ P = 0

aur

v = 0 = 0

— self-consistent. Koi flow nahi, koi dynamic pressure nahi, koi reading nahi.

Venturi

Q

ka kya hoga agar throat area

A_{2}

pipe area

A_{1}

ke paas pahunche?

Q = A_{1} A_{2} 2Δ P / [ρ (A_{1}^{2} - A_{2}^{2})]

mein denominator

A_{1}^{2} - A_{2}^{2} \to 0

, isliye formula blow up karta hai — lekin physically

Δ P \to 0

bhi hota hai, kyunki koi constriction nahi hai. Device kaam karna band kar deta hai: koi squeeze nahi, koi signal nahi.

Exactly water surface par rakhe ek hole (

H = 0

) par drain speed kya hai?

v = 2 g \cdot 0 = 0

. Surface par paani ke paas convert karne ke liye potential energy ki koi depth nahi hai, isliye woh essentially zero speed se dribbkle out hota hai.

Horizontal jet range

x = 2 H y

ke liye, jab hole floor ke paas aata hai (

y \to 0

) toh kya hota hai?

Range

\to 0

: jet ka almost koi vertical fall

y

nahi hai, isliye woh chahe kitni bhi tez nikle, tank ke base ke paas hi land karta hai.

Kis airspeed par incompressible pitot formula ek gas ke liye fail hona shuru hota hai?

Roughly Mach 0.3 (sea-level air mein lagbhag

100 m/s

) se upar, density changes ab negligible nahi rahtein, isliye

v = 2Δ P / ρ

mein

ρ

ko constant treat karna error introduce karta hai; real airspeed indicators ek compressible-flow correction apply karte hain. Yahi warning high-speed gas venturi par bhi apply hoti hai.

Kya venturi throat pressure itna low drop ho sakta hai ki formula unreliable ho jaaye?

Haan — agar throat speed itni zyada ho, toh

P_{2}

liquid ke vapor pressure tak gir sakta hai aur paani vapor bubbles mein flash ho jaata hai (cavitation). Bernoulli ek single incompressible phase assume karta hai, isliye us limit ke baad measurement (aur pipe surface) degrade ho jaata hai.

Kya yahi low-pressure danger orifice flow mein bhi aata hai?

Yeh ek submerged ya venturi-nozzle orifice mein ho sakta hai jahan jet ki high speed ek local pressure vapor pressure se neeche create karti hai; ek simple open-air tank hole atmospheric ke paas rehta hai aur safe hota hai. Cavitation real-world floor hai ki pressure ideal model fail hone se pehle kitna low drop ho sakta hai.

Bernoulli kya predict karta hai agar ek venturi tilt ho jaaye taaki throat inlet se upar ho?

ρ g h

terms ab cancel nahi hote; pressure drop phir speed increase aur height gain dono ko reflect karta hai, isliye horizontal-pipe formula ko

ρ g (h_{2} - h_{1})

term ke saath correct karna padega.

Recall Har trap ki ek-line summary

Speed energy se aati hai, rate geometry se. Pitot $v = 2Δ P / ρ$ ne "air vs water" aur "at rest → zero" items power kiye; Torricelli $v = 2 g H$ ne "hole size irrelevant", " $H = 0$ ", aur "range peaks mid-height" items power kiye; aur $Q = A_{1} A_{2} 2Δ P / [ρ (A_{1}^{2} - A_{2}^{2})]$ (rate = area × speed) ne " $Δ P$ " aur " $A_{2} \to A_{1}$ " traps power kiye. Hamesha pehle continuity, phir Bernoulli trace karo, pressure differences use karo (references cancel ho jaate hain), yaad rakho incompressible model gases mein Mach 0.3 se upar break karta hai, aur cavitation low pressure ki physical floor hai.

Connections

Bernoulli's Equation — har trap ke peeche energy balance.
Continuity Equation — venturi "squeeze" confusion resolve karta hai.
Torricelli's Law — free-fall analogy.
Projectile Motion — jet-range edge cases.
Manometers and Pressure Measurement — manometer fluid kyun matter karta hai.
Dynamic vs Static vs Stagnation Pressure — static-vs-stagnation port distinction.