Applications — Pitot tube, Venturi meter, orifice flow: Worked examples

2.2.16 · D3 · Physics › Fluid Mechanics › Applications — Pitot tube, Venturi meter, orifice flow

Yeh page ek drill hai. Hum parent note se teen master formulas lete hain aur har tarah ke cases par apply karte hain — normal numbers, zero inputs, degenerate geometry, limiting behaviour, ek real-world word problem, aur ek exam twist. Pehle hum poora territory map karte hain, phir har cell ko walk-through karte hain.

Sirf yahi tools hain, dobara clearly define karke:

The scenario matrix

Is topic ke har problem ka ek cell hota hai inhi mein se. Neeche ke examples mein har example ke saath uska cell label diya gaya hai.

Cell	Device	Kya cheez ise "ek case" banati hai
A	Pitot	Standard: diya $Δ P$ , dhoondo $v$
B	Pitot	Manometer height $Δ h$ se padha jaata hai (indirect $Δ P$ )
C	Venturi	Standard: diya $Δ P$ , dhoondo $Q$
D	Venturi	Degenerate geometry $A_{2} \to A_{1}$ (throat = pipe) — limiting behaviour
E	Orifice	Standard depth $H$ , dhoondo jet speed
F	Orifice	Zero input $H = 0$ (hole surface par hi)
G	Orifice	Real-world word problem: jet range floor par (Projectile Motion)
H	Orifice	Exam twist: kaun sa hole zyada door land karta hai? (max range)
I	Mixed	Sanity/limiting: $Δ P$ double karo → $Q$ $2$ badhta hai, $2$ nahi

A — Pitot, seedha case

Worked example Cell A: standard airspeed

Ek glider par lagaa pitot tube dynamic pressure $P_{s} - P_{0} = 1800 Pa$ read karta hai, air ki density $ρ = 1.2 kg/m^{3}$ hai. Airspeed $v$ dhoondo.

Forecast: bada $Δ P$ ya halka fluid → tez speed. Air bahut halki hai, toh "highway" jaisi speed expect karo, yaani tens of m/s.

Pitot formula likho. $v = \frac{2 ( P _{s} - P _{0} )}{ρ}$ . Yeh step kyun? Tube ka munh flow ko rok leta hai ( $v = 0$ wahan), saari kinetic energy extra pressure $P_{s} - P_{0}$ mein convert ho jaati hai; same height par Bernoulli exactly yahi deta hai.
Numbers daalo. $v = \frac{2 ( 1800 )}{1.2} = 3000$ . Yeh step kyun? Direct substitution — height term $ρ g h$ cancel ho jaata hai kyunki dono ports same level par hain.
Calculate karo. $3000 \approx 54.8 m/s$ .

Verify: Units: $Pa / (kg/m^{3}) = (N/m^{2}) (m^{3} / kg) = N\cdotm/kg = m^{2} / s^{2} = m/s$ . ✓ Aur $54.8$ m/s ≈ 197 km/h — ek believable glide speed hai.

B — Pitot manometer se padha jaaye

Worked example Cell B: dynamic pressure ek liquid height mein chhupa hua

Paani ( $ρ = 1000 kg/m^{3}$ ) mein ek pitot–static tube mercury manometer ( $ρ_{m} = 13600 kg/m^{3}$ ) se connected hai jo height difference $Δ h = 5 cm = 0.05 m$ dikhata hai. $g = 10$ lo. Paani ki speed $v$ dhoondo.

Forecast: mercury, paani se 13.6 guna dense hai, toh sirf $5$ cm ka column bhi ek badi pressure represent karta hai. Lekin paani bhi bhaari hai, toh speed bahut zyada nahi hogi — kuch m/s.

Height ko pressure mein badlo. $P_{s} - P_{0} = ρ_{m} g Δ h$ . Yeh step kyun? Ek manometer unknown pressure difference ko known liquid column ke saath balance karta hai; column ka weight per area $ρ_{m} g Δ h$ hota hai. (Dekho Manometers and Pressure Measurement.)
Calculate karo. $P_{s} - P_{0} = 13600 \times 10 \times 0.05 = 6800 Pa$ . Yeh step kyun? Yeh woh dynamic pressure hai jo moving paani carry karta hai.
Pitot mein daalo. $v = \frac{2 ( 6800 )}{1000} = 13.6 \approx 3.69 m/s$ . Yeh step kyun? Same formula jaise Cell A, lekin $Δ P$ ab manometer se aaya hai, direct gauge se nahi.

Verify: $v^{2} = 13.6$ ; wapas multiply karo: $\frac{1}{2} ρ v^{2} = \frac{1}{2} (1000) (13.6) = 6800$ Pa = wahi $Δ P$ jahan se shuru kiya tha. ✓ Loop close ho gaya.

C — Venturi, seedha case

Worked example Cell C: constriction se flow rate

Paani ( $ρ = 1000$ ) ek venturi se flow karta hai jisme wide area $A_{1} = 20 cm^{2} = 2 \times 1 0^{- 3} m^{2}$ aur throat $A_{2} = 8 cm^{2} = 8 \times 1 0^{- 4} m^{2}$ hai. Throat mein pressure drop $P_{1} - P_{2} = 1500 Pa$ hai. $Q$ dhoondo.

Forecast: ek real pipe mein litres per second ke order mein kuch hoga, toh $Q$ around $1 0^{- 3} m^{3} / s$ hoga.

$Q$ likho. $Q = A_{1} A_{2} \frac{2 ( P _{1} - P _{2} )}{ρ ( A _{1}^{2} - A _{2}^{2} )}$ . Yeh step kyun? Continuity ( $A_{1} v_{1} = A_{2} v_{2}$ ) ek unknown hata deta hai; Bernoulli pressure ka link deta hai. Dono milake $Q$ sirf measurable areas aur $Δ P$ ke terms mein deta hai.
Areas. $A_{1} A_{2} = (2 \times 1 0^{- 3}) (8 \times 1 0^{- 4}) = 1.6 \times 1 0^{- 6}$ . Aur $A_{1}^{2} - A_{2}^{2} = 4 \times 1 0^{- 6} - 6.4 \times 1 0^{- 7} = 3.36 \times 1 0^{- 6} m^{4}$ . Yeh step kyun? Yeh formula ke liye do geometry pieces hain.
Root ke andar. $\frac{2 ( 1500 )}{1000 \times 3.36 \times 1 0 ^{- 6}} = \frac{3000}{3.36 \times 1 0 ^{- 3}} = 8.93 \times 1 0^{5}$ .
Complete karo. $Q = 1.6 \times 1 0^{- 6} 8.93 \times 1 0^{5} = 1.6 \times 1 0^{- 6} \times 944.9 \approx 1.51 \times 1 0^{- 3} m^{3} / s$ . Yeh step kyun? $8.93 \times 1 0^{5} \approx 945$ ; area product se multiply karo.

Verify: $1.51 \times 1 0^{- 3} m^{3} / s = 1.51$ L/s — garden-hose-sized pipe ke liye realistic hai. Root ke andar units: $Pa / (kg/m^{3} \cdot m^{4}) = s^{- 2} m^{- 2}$ , root se $m^{- 1} s^{- 1}$ milta hai, $m^{4}$ se multiply karo toh $m^{3} / s$ . ✓

D — Venturi bina constriction ke (degenerate limit)

Worked example Cell D: throat = pipe,

$A_{2} \to A_{1}$ Venturi formula kya predict karta hai $Q$ ke liye agar "throat" bilkul narrow nahi hua, yaani $A_{2} = A_{1}$ , lekin phir bhi koi pressure drop $P_{1} - P_{2} > 0$ read ho raha hai? Physically interpret karo.

Forecast: agar pipe kabhi narrow nahi hoti, toh continuity kehta hai $v_{1} = v_{2}$ , toh Bernoulli force karta hai $P_{1} = P_{2}$ . Ek uniform pipe mein nonzero $Δ P$ impossible hona chahiye — formula ko blow up karna chahiye ya ise forbid karna chahiye.

Denominator mein $A_{2} = A_{1} = A$ rakho. $A_{1}^{2} - A_{2}^{2} = A^{2} - A^{2} = 0$ . Yeh step kyun? Denominator $ρ (A_{1}^{2} - A_{2}^{2})$ exactly woh "kitna fluid speed up hota hai" factor hai. Narrowing nahi → yeh zero ho jaata hai.
Formula padho. $Q = A^{2} \frac{2 ( P _{1} - P _{2} )}{ρ \cdot 0} \to \infty$ kisi bhi $Δ P > 0$ ke liye. Yeh step kyun? Zero se divide → unbounded. Mathematically model yeh input reject kar deta hai.
Physical reading. Bernoulli akele ek uniform pipe mein $P_{1} \neq = P_{2}$ sustain nahi kar sakta (heights equal, speeds equal). Toh $Δ P = 0$ hi ek consistent value hai; jo $\frac{0}{0}$ result hota hai woh model ka yeh kehna hai ki "unconstrcited pipe se $Q$ ke baare mein koi information nahi milti." Yeh step kyun? Degenerate geometry ka matlab hai measurement principle gayab ho gayi — tumhe squeeze chahiye.

Verify: consistency check — agar $Δ P = 0$ aur $A_{2} = A_{1}$ ho, toh numerator bhi $0$ hai, giving $Q = A^{2} 0/0$ , the honest "indeterminate": meter koi reading nahi deta, exactly sahi. ✓

E — Orifice, plain depth

F — Orifice with zero depth (zero input)

Worked example Cell F: hole bilkul surface par,

$H = 0$ Hole exactly water line par hua hai, $H = 0$ . Jet speed kya hogi?

Forecast: koi depth nahi matlab jet ko power karne ke liye koi "fall" nahi — yeh zero speed par ooze out hona chahiye.

$H = 0$ substitute karo. $v = 2 g (0) = 0$ . Yeh step kyun? Zero head hone par motion mein convert karne ke liye koi potential energy nahi hai.
Physical reading. Surface par fluid ke paas kuch nahi jo use horizontally bahar push kare — woh bas spill karta hai. Jaise jaise hole neeche jaata hai, $v$ $H$ ki tarah badhta hai: pehle steep, phir flatten hota hai. Yeh step kyun? Yeh boundary/zero case hai jo poore $H$ curve ko anchor karta hai.

Verify: limit continuous hai — $lim_{H \to 0} 2 g H = 0$ , direct substitution se match karta hai. ✓ Koi sign issue nahi: $H < 0$ (surface ke upar) physically meaningless hai, toh domain $H = 0$ se shuru hoti hai.

G — Orifice + projectile: floor par range (word problem)

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

Jet speed (Torricelli). $v = 2 g H = 2 (10) (0.20) = 4 = 2 m/s$ . Yeh step kyun? Yeh horizontal launch speed hai — figure mein red jet dekho.

Fall time (free fall from $y$ ). Launch par jet ki koi vertical speed nahi hoti, toh $y = \frac{1}{2} g t^{2} \Rightarrow t = 2 y / g = 2 (0.45) /10 = 0.09 = 0.30 s$ . Yeh step kyun? Horizontal aur vertical motions independent hain (Projectile Motion); vertical pure free fall hai.

Range. $x = v t = 2 \times 0.30 = 0.60 m$ . Yeh step kyun? Koi horizontal force nahi → constant horizontal speed, toh distance = speed × time.

Verify: shortcut $x = 2 H y = 2 0.20 \times 0.45 = 2 0.09 = 2 (0.30) = 0.60 m$ agree karta hai. ✓

H — Exam twist: kaun sa hole sabse door land karta hai?

Range function likho. $x (y) = 2 (D - y) y = 2 D y - y^{2}$ . Yeh step kyun? Torricelli speed aur free-fall time combine karke range purely hole height $y$ ka function ban jaati hai.

Andar maximize karo. Parabola $f (y) = D y - y^{2}$ wahan peak karta hai jahan $f^{'} (y) = D - 2 y = 0 \Rightarrow y = D /2$ . Yeh step kyun? $x$ , $f$ ke saath badhta hai, toh $f$ maximize karna $x$ maximize karta hai; ek downward parabola apne vertex par peak karta hai.

Best height aur range. $y = D /2 = 0.5 m$ (mid-height). Phir $H = D - y = 0.5 m$ aur $x_{m a x} = 2 0.5 \times 0.5 = 2 (0.5) = 1.0 m$ . Yeh step kyun? $y = D /2$ par depth height ke barabar hoti hai — wahi balance point jo forecast ne predict kiya tha.

Verify: vertex par $x_{m a x} = D = 1.0 m$ (ek known result: max range fill height ke barabar hoti hai). Ek neighbour check karo: $y = 0.6 \Rightarrow x = 2 0.4 \times 0.6 = 2 0.24 = 0.98 m < 1.0$ . ✓ Aur symmetric bhi hai: $y = 0.4$ wahi $0.98$ m deta hai jaise $y = 0.6$ — mid-height ke baare mein symmetric holes equally door land karte hain.

I — Sanity limit: pressure drop double karna

Worked example Cell I: kya

$Δ P$ double karne se $Q$ double ho jaata hai? Cell C ke venturi mein pressure drop $1500$ Pa se double hokar $3000$ Pa ho jaata hai, baaki sab same hai. $Q$ kis factor se change hoga?

Forecast: $Q$ , $Δ P$ ke square root ke neeche hai, toh yeh $2$ se kam badhna chahiye — yaani $2 \approx 1.41$ .

Dependence isolate karo. $Q \propto P_{1} - P_{2}$ (baaki sab constant hai). Yeh step kyun? Sirf $Δ P$ change ho raha hai, toh ratios mein saare constant factors cancel ho jaate hain.
Ratio lo. $\frac{Q _{new}}{Q _{old}} = \frac{3000}{1500} = 2 \approx 1.414$ .
Naya value. $Q_{new} = 1.51 \times 1 0^{- 3} \times 1.414 \approx 2.14 \times 1 0^{- 3} m^{3} / s$ . Yeh step kyun? Cell C ke answer ko $2$ se scale karo.

Verify: scratch se recompute karo — root ke andar ab $\frac{2 ( 3000 )}{1000 \times 3.36 \times 1 0 ^{- 6}} = 1.786 \times 1 0^{6}$ , $= 1336$ , $Q = 1.6 \times 1 0^{- 6} \times 1336 = 2.14 \times 1 0^{- 3}$ . ✓ $2$ shortcut se match karta hai.

Recall Har cell ke liye ek-line recall

Pitot direct ::: $v = 2Δ P / ρ$ Pitot via manometer ::: $Δ P = ρ_{m} g Δ h$ , phir Pitot Venturi ::: $Q = A_{1} A_{2} 2Δ P / [ρ (A_{1}^{2} - A_{2}^{2})]$ Venturi with $A_{2} = A_{1}$ ::: $Δ P \neq = 0$ forbid hai ( $Q \to \infty$ / indeterminate) Orifice speed ::: $v = 2 g H$ , koi area nahi, koi density nahi Orifice at $H = 0$ ::: $v = 0$ Jet range ::: $x = 2 H y$ Max range hole height ::: $y = D /2$ , giving $x_{m a x} = D$ Double $Δ P$ ::: $Q$ , $2$ se badhta hai

Connections

Bernoulli's Equation — har cell yahan se shuru hota hai.
Continuity Equation — venturi cells mein area link deta hai.
Torricelli's Law — cells E, F, G, H.
Projectile Motion — range cells G aur H.
Manometers and Pressure Measurement — cell B ka $Δ h \to Δ P$ .
Dynamic vs Static vs Stagnation Pressure — Pitot aur Venturi ke peeche pressure vocabulary.