Applications — Pitot tube, Venturi meter, orifice flow

2.2.16 · Physics › Fluid Mechanics

Teen classic devices jo Bernoulli's equation ko ek measuring tool mein badal dete hain. Ek idea master karo — pressure ko speed ke saath trade karo — aur teeno khud nikal aate hain.

0. Wo tools jinse hum sab derive karte hain

Hum sirf in dono ko combine karenge. Aur kuch zaroorat nahi hai.

1. Pitot tube — flow ki speed measure karna

Derivation (scratch se). Same height $h$ par ek streamline lo, door upstream wale point (1) se jahan speed $v$ aur static pressure $P_{0}$ hai, stagnation mouth (2) tak jahan $v_{2} = 0$ aur pressure $P_{s}$ hai: $P_{0} + \frac{1}{2} ρ v^{2} = P_{s} + 0$ Yeh step kyun? Heights equal hain, isliye $ρ g h$ cancel ho jaata hai; point 2, stagnation ki definition se rest par hai.

Toh dynamic pressure hai $P_{s} - P_{0} = \frac{1}{2} ρ v^{2}$ . Solve karo:

2. Venturi meter — pipe mein flow rate measure karna

Derivation. Horizontal pipe ( $h_{1} = h_{2}$ ), wide section (1) aur throat (2): $P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2} \Rightarrow P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$ Continuity deta hai $v_{1} = \frac{A _{2}}{A _{1}} v_{2}$ . Substitute karo: $P_{1} - P_{2} = \frac{1}{2} ρ v_{2}^{2} (1 - \frac{A _{2}^{2}}{A _{1}^{2}})$ Yeh step kyun? Humne $v_{1}$ ko replace kiya taaki sirf ek unknown $v_{2}$ bache. $v_{2}$ ke liye solve karo aur $A_{2}$ se multiply karo $Q$ paane ke liye:

Figure — Applications — Pitot tube, Venturi meter, orifice flow

3. Orifice flow (Torricelli) — fluid tank se drain hona

Derivation. Surface (1) upar, hole (2) depth $H$ par. Dono atmosphere ke liye open hain isliye $P_{1} = P_{2} = P_{a t m}$ . Agar hole area $≪$ tank area, toh $v_{1} \approx 0$ : $P_{a t m} + 0 + ρ g H = P_{a t m} + \frac{1}{2} ρ v^{2} + 0$ Yeh step kyun? Pressures cancel ho jaate hain; surface speed $\approx 0$ continuity ke dwara ( $A_{h o l e} ≪ A_{t ank}$ ); hole ko height reference set karo.

Forecast-then-Verify

Recall Compute karne se pehle predict karo

Agar venturi pressure drop double ho jaaye, kya $Q$ double hoga? Nahi — $Q \propto Δ P$ , isliye yeh $2 \approx 1.41 \times$ se badhega.
Do holes, ek depth $H$ par aur ek $4 H$ par. Speed ratio? $1 : 2$ (kyunki $v \propto H$ ).
Same $Δ P$ ke liye water vs air mein Pitot — kaunsa zyada $v$ dega? Air, kyunki $v \propto 1/ ρ$ aur air halki hai.

Common mistakes (steel-manned)

80/20 — sabse zyada yield karne wale takeaways

Har device = Bernoulli + Continuity, phir speed ke liye solve karo.
Pitot: $v = 2Δ P / ρ$ . Venturi: $Q = A_{1} A_{2} \frac{2Δ P}{ρ ( A _{1}^{2} - A _{2}^{2} )}$ . Orifice: $v = 2 g H$ .
Zyada tez fluid ⇒ kam pressure. Woh arrow yaad karo aur teeno re-derive kar sakte ho.

Recall Feynman: 12-saal ke bachche ko explain karo

Socho ek hallway mein daudo. Narrow part mein tumhe speed up karna padta hai aage badhne ke liye (woh continuity hai). Jab tum bhagte ho, side walls par kam push karte ho (low pressure) — jab tum khade hote ho toh wall par zyada lean karte ho (high pressure). Ek pitot tube waise hai jaise hawa mein seedha haath aage karo: tumhe ek hard push feel hota hai jo batata hai hawa kitni tez hai. Ek venturi woh narrow hallway hai: squeeze karne se paani tez bhaagta hai aur halka press karta hai, aur hum woh read karte hain. Ek orifice ek paani ki bucket mein hole hai: hole jitna gehra hoga, paani ko utna zyada "fall" milega, toh woh utni tezi se bahar niklega — bilkul waise jaise usi depth se ek pathar giraaya jaaye.

Connections

Bernoulli's Equation — parent principle.
Continuity Equation — $A_{1} v_{1} = A_{2} v_{2}$ ka link deta hai.
Torricelli's Law — orifice flow ka special case.
Projectile Motion — orifice jet range ke liye use hoti hai.
Manometers and Pressure Measurement — $Δ P$ ko $Δ h$ ke roop mein kaise padhte hain.
Dynamic vs Static vs Stagnation Pressure.

Flashcards

Stagnation point kya hota hai?

Ek aisa point jahan fluid ko rest par la diya jaata hai (

v = 0

), kinetic energy extra (stagnation) pressure mein convert ho jaati hai.

Pitot tube speed formula

v = 2 (P_{s} - P_{0}) / ρ

, jahan

P_{s} - P_{0}

dynamic pressure hai.

Venturi mein throat pressure kyun kam hota hai?

Continuity chhhote throat par zyada speed force karta hai; Bernoulli phir constant total energy ke liye kam pressure require karta hai.

Venturi volume flow rate

Q = A_{1} A_{2} \frac{2 ( P _{1} - P _{2} )}{ρ ( A _{1}^{2} - A _{2}^{2} )}

Torricelli's law batao

Depth

H

par hole se efflux speed

v = 2 g H

hai, bilkul

H

se free fall jaisi.

Kya orifice speed hole area par depend karti hai?

Nahi — speed

2 g H

hai; area sirf flow rate

Q = A 2 g H

badlata hai.

Orifice jet ki horizontal range (hole depth H, height y above floor)

x = 2 H y

Agar venturi

Δ P

double ho,

Q

kaise change hoga?

2 \approx 1.41 \times

se badhega, kyunki

Q \propto Δ P

Wo do equations jo teeno devices derive karte hain

Bernoulli (

P + \frac{1}{2} ρ v^{2} + ρ g h =

const) aur Continuity (

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

Δ P

ko manometer height mein convert karna

Δ P = (ρ_{m} - ρ) g Δ h

density

ρ_{m}

wale fluid column ke liye.