Applications — Pitot tube, Venturi meter, orifice flow: Exercises

2.2.16 · D4 · Physics › Fluid Mechanics › Applications — Pitot tube, Venturi meter, orifice flow

Symbols ka recap taaki kuch bhi unexplained na rahe. Height convention: saari heights ek chosen reference level se upar measure ki jaati hain jise datum kehte hain; hum alag letters sirf yeh signal karne ke liye use karte hain ki height kis cheez ki hai, lekin sab ka matlab hai "datum se metres upar" (ya kisi named surface se neeche depth jahan bataya gaya ho):

$P$ = pressure (force per area, units pascal $Pa$ ).
$v$ = fluid speed. $ρ$ = density.
$z$ = datum ke upar ek point ki elevation (tab use hota hai jab pipe/throat upar ya neeche ho).
$H$ = free surface se neeche depth (draining tanks ke liye use hota hai).
$y$ = floor ke upar hole ki height (sirf projectile range ke liye use hota hai).
$h$ = raw Bernoulli statement ke andar ek generic height.
$A$ = fluid jo cross-section area se flow karta hai. $Q = A v$ = volume per second flowing.
$Δ P = P_{1} - P_{2}$ , ek pressure difference; $Δ h$ = ek manometer height reading.

Level 1 — Recognition

L1.1

Ek tube ko is tarah se moda gaya hai ki uska open mouth oncoming river current mein seedha point kare. Mouth par ikattha hone wala fluid momentarily ruk jaata hai. Us point ka naam batao aur wahan fluid speed state karo.

Recall Solution

Mouth flow ki taraf face karta hai, isliye aane wala fluid decelerate hokar ruk jaata hai: yeh stagnation point hai, aur wahan speed $v = 0$ hai. Kyun: saari aane wali kinetic energy $\frac{1}{2} ρ v^{2}$ extra pressure mein convert ho gayi hai. Dekho Dynamic vs Static vs Stagnation Pressure. Answer: stagnation point, $v = 0$ .

L1.2

Ek horizontal pipe jo ek throat tak narrow hoti hai, usme narrow throat par pressure wide section ke comparison mein zyada hai ya kam? Ek sentence mein reasoning do.

Recall Solution

Kam. Continuity ke anusaar throat fluid ko speed up karne par majboor karta hai ( $A_{2} < A_{1} \Rightarrow v_{2} > v_{1}$ ); Bernoulli ke anusaar, same height par faster fluid lower pressure carry karta hai. Answer: throat par lower pressure.

L1.3

Paani ek tank se ek chhote hole ke through nikalta hai jo surface se $H$ depth neeche hai. Jet speed formula likho aur batao ki yeh kis cheez par depend nahi karta.

Recall Solution

Torricelli: $v = 2 g H$ (dekho Torricelli's Law). Yeh hole ki area ya shape par depend nahi karta — sirf depth $H$ par. Area rate $Q = A 2 g H$ ko affect karta hai, speed ko nahi.

Level 2 — Application

L2.1

Ek glider par pitot tube pressure difference $P_{s} - P_{0} = 1080 Pa$ read karta hai hawa mein ( $ρ_{a} = 1.2 kg/m^{3}$ ). Airspeed nikalo.

Recall Solution

KYA: pitot speed formula $v = 2 (P_{s} - P_{0}) / ρ$ apply karo. KYUN: mouth stream ki taraf face karta hai, isliye $P_{s} - P_{0} = \frac{1}{2} ρ v^{2}$ exactly dynamic pressure hai. $v$ ke liye solve karo. $v = \frac{2 ( 1080 )}{1.2} = 1800 \approx 42.4 m/s .$ Answer: $v \approx 42.4 m/s$ .

L2.2

Ek tank mein depth $H = 1.25 m$ par ek hole hai. Jet speed nikalo. Agar hole area $A = 4 cm^{2}$ hai, volume flow rate $Q$ nikalo.

Recall Solution

Speed: $v = 2 g H = 2 (10) (1.25) = 25 = 5 m/s$ . Rate: area convert karo $A = 4 cm^{2} = 4 \times 1 0^{- 4} m^{2}$ , phir $Q = A v = (4 \times 1 0^{- 4}) (5) = 2 \times 1 0^{- 3} m^{3} / s .$ Answer: $v = 5 m/s$ , $Q = 2 \times 1 0^{- 3} m^{3} / s$ .

L2.3

Ek pitot ka pressure difference ek mercury manometer par read kiya jaata hai jiske dono legs same flowing water mein tapped hain dono sides par, isliye paani mercury surfaces tak dono legs mein tubes ko bharta hai. Mercury height difference $Δ h = 5 cm$ , mercury $ρ_{m} = 13600 kg/m^{3}$ , flowing water $ρ = 1000 kg/m^{3}$ . Water speed nikalo. (Dekho Manometers and Pressure Measurement.)

Recall Solution

KYA: manometer height ko pressure mein convert karo. Kyunki paani mercury ke upar dono legs mein baithta hai, paani ke columns ek doosre ko partly balance karte hain aur net reading mercury ke paani ke upar excess weight se drive hoti hai. Isliye correct relation hai: $Δ P = (ρ_{m} - ρ) g Δ h .$ $(ρ_{m} - ρ)$ kyun? Height $Δ h$ par jahan ek leg mein mercury hai aur doosre mein paani, sirf density difference unbalanced hai; agar tumne paani bhool diya aur $ρ_{m} g Δ h$ use kiya to $ρ g Δ h$ se over-count karte. (Jab mercury ke upar fluid hawa jaisi gas ho, $ρ$ negligible hai aur simpler $ρ_{m} g Δ h$ theek hai — yeh woh geometry hai jo shortcut ko justify karta hai.) $Δ P = (13600 - 1000) (10) (0.05) = 12600 \times 0.5 = 6300 Pa .$ Phir pitot speed water density ke saath (woh fluid jo actually flow kar raha hai): $v = \frac{2 ( 6300 )}{1000} = 12.6 \approx 3.55 m/s .$ Answer: $v \approx 3.55 m/s$ .

Level 3 — Analysis

L3.1

Ek horizontal venturi paani carry karta hai. Wide area $A_{1} = 8 cm^{2}$ , throat $A_{2} = 2 cm^{2}$ , measured pressure drop $P_{1} - P_{2} = 3000 Pa$ . Flow rate $Q$ nikalo.

Neeche figure (s01): ek lavender-filled pipe jo left par wide draw ki gayi hai, middle mein ek narrow throat tak pinch hoti hai, phir phir wide ho jaati hai. Ek slow coral inlet arrow $v_{1}$ throat par ek longer, faster arrow $v_{2}$ ban jaata hai; wide section ko " $P_{1}$ high" tag kiya gaya hai aur throat ko " $P_{2}$ low," saath mein dono se pressure taps uthte hue. Yeh visually dikhata hai kyun pressure exactly wahan dip karta hai jahan fluid speed up hoti hai.

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

Recall Solution

Pehle formula derive karo (KYUN, sirf plug mat karo). Horizontal pipe isliye $z_{1} = z_{2}$ aur $ρ g z$ terms cancel ho jaate hain. Wide (1) aur throat (2) ke beech Bernoulli: $P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2} \Rightarrow P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) .$ KYUN continuity ab: hamare paas ek equation hai lekin do unknown speeds hain. Continuity $A_{1} v_{1} = A_{2} v_{2}$ humein $v_{1} = \frac{A _{2}}{A _{1}} v_{2}$ likhne deta hai, ek unknown ko khatam karke: $P_{1} - P_{2} = \frac{1}{2} ρ v_{2}^{2} (1 - \frac{A _{2}^{2}}{A _{1}^{2}}) .$ $v_{2}$ ke liye solve karo, phir volume rate $Q = A_{2} v_{2}$ paane ke liye $A_{2}$ se multiply karo. Algebra karne par (root ke andar $A_{1}^{2}$ se top aur bottom dono multiply karo) compact form milta hai: $Q = A_{1} A_{2} \frac{2 ( P _{1} - P _{2} )}{ρ ( A _{1}^{2} - A _{2}^{2} )} .$ Ab plug in karo. Areas convert karo: $A_{1} = 8 \times 1 0^{- 4} m^{2}$ , $A_{2} = 2 \times 1 0^{- 4} m^{2}$ . $A_{1}^{2} - A_{2}^{2} = 64 \times 1 0^{- 8} - 4 \times 1 0^{- 8} = 60 \times 1 0^{- 8} = 6 \times 1 0^{- 7} m^{4} .$ $Q = (8 \times 1 0^{- 4}) (2 \times 1 0^{- 4}) \frac{2 ( 3000 )}{1000 ( 6 \times 1 0 ^{- 7} )} .$ Root ke andar: $\frac{6000}{6 \times 1 0 ^{- 4}} = 1 0^{7}$ , isliye $1 0^{7} \approx 3162$ . $Q = (1.6 \times 1 0^{- 7}) (3162) \approx 5.06 \times 1 0^{- 4} m^{3} / s .$ Answer: $Q \approx 5.06 \times 1 0^{- 4} m^{3} / s$ (lagbhag $0.5 L/s$ ).

L3.2

L3.1 ke venturi ke liye, throat speed $v_{2}$ aur wide-section speed $v_{1}$ nikalo, aur confirm karo ki woh continuity follow karte hain.

Recall Solution

$v_{2} = Q / A_{2} = \frac{5.06 \times 1 0 ^{- 4}}{2 \times 1 0 ^{- 4}} \approx 2.53 m/s$ . $v_{1} = Q / A_{1} = \frac{5.06 \times 1 0 ^{- 4}}{8 \times 1 0 ^{- 4}} \approx 0.632 m/s$ . Continuity check: $A_{1} v_{1} = (8 \times 1 0^{- 4}) (0.632) = 5.06 \times 1 0^{- 4} = Q$ , aur $v_{2} / v_{1} = A_{1} / A_{2} = 4$ . ✓ Throat $4 \times$ faster chalta hai, exactly jaisa areas demand karte hain.

L3.3

Ek tank itna bhara hai ki free surface uski side wall mein ek hole se $H = 0.80 m$ depth par hai. Hole floor se $y = 1.25 m$ upar hai. Floor par horizontal jet kitni door land karta hai?

Neeche figure (s02): ek lavender water tank jisme surface marked hai; ek coral dot hole mark karta hai. Ek mint double-arrow $H$ depth ko surface se hole tak label karta hai, ek butter double-arrow $y$ height ko hole se floor tak label karta hai, aur ek coral parabola jet ko neeche arching karte hue floor par ek landing mark $x$ tak trace karti hai. Yeh visually do ideas alag karta hai: $H$ speed set karta hai, $y$ fall time set karta hai.

Recall Solution

KYA: do alag physics steps — exit speed nikalo, phir jet ko ek projectile ki tarah treat karo. KYUN split karo? Ek baar fluid hole se nikalta hai, sirf gravity kaam karta hai; horizontally koi force nahi, isliye yeh ordinary Projectile Motion hai. Exit speed (Torricelli): $v = 2 g H = 2 (10) (0.80) = 16 = 4 m/s$ . Height $y$ se fall time: $y = \frac{1}{2} g t^{2} \Rightarrow t = 2 y / g = 2 (1.25) /10 = 0.25 = 0.5 s$ . Horizontal range: $x = v t = 4 \times 0.5 = 2 m$ . (Compact form: $x = 2 H y = 2 0.80 \times 1.25 = 21 = 2 m$ .) ✓ Answer: $x = 2 m$ .

Level 4 — Synthesis

L4.1

Ek tall tank mein jis ki total water depth $D = 1.8 m$ hai, do holes drill kiye gaye hain: hole A surface se $H_{A} = 0.2 m$ neeche, hole B surface se $H_{B} = 1.6 m$ neeche. Dono ek hi vertical wall par hain. Dikhao ki woh floor par same spot par land karte hain, aur woh distance nikalo. (Floor tank ke base par hai.)

Neeche figure (s03): wahi tank ek baar draw kiya gaya hai, do holes ke saath — ek coral wala upar (depth $H_{A}$ ) aur ek mint wala neeche (depth $H_{B}$ ). Do jets, coral aur mint, bahar arch karte hain aur floor par same landing mark par milte dikhaye gaye hain, jisse "mirror-depth" symmetry ek nazar mein visible ho jaati hai.

Recall Solution

Setup: depth $H$ par surface ke neeche ek hole, floor se $y = D - H$ height par hota hai. Uski exit speed $v = 2 g H$ hai aur fall time $t = 2 y / g$ , isliye range $x = v t = 2 g H \cdot \frac{2 ( D - H )}{g} = 2 H (D - H) .$ Symmetry insight: $H (D - H)$ tab unchanged rehta hai jab $H \to D - H$ . Hole A mein $H_{A} = 0.2$ hai; mirror depth hai $D - H_{A} = 1.6 = H_{B}$ . Isliye A aur B mirror partners hain ⇒ equal range. Numerically: $x_{A} = 2 0.2 (1.8 - 0.2) = 2 0.2 \times 1.6 = 2 0.32 \approx 1.13 m$ . $x_{B} = 2 1.6 (1.8 - 1.6) = 2 1.6 \times 0.2 = 2 0.32 \approx 1.13 m$ . ✓ Same. Answer: dono wall se $\approx 1.13 m$ door land karte hain.

L4.2

Usi tank ke liye, kaunsi single depth $H_{⋆}$ par ek hole maximum possible range dega, aur woh range kya hogi?

Recall Solution

KYA: $x = 2 H (D - H)$ ko $H$ ke upar maximise karo. KYUN calculus/symmetry? Product $H (D - H)$ $H$ mein ek downward parabola hai; uski peak exactly midpoint $H_{⋆} = D /2$ par hai. $H_{⋆} = 1.8/2 = 0.9 m$ . Max range: $x_{m a x} = 2 0.9 (1.8 - 0.9) = 2 0.9 \times 0.9 = 2 (0.9) = 1.8 m$ . Neat tarike se, $x_{m a x} = 2 (D /2) (D /2) = D$ — maximum horizontal range water depth ke barabar hoti hai. Answer: mid-depth par hole $H_{⋆} = 0.9 m$ , range $= 1.8 m$ .

Level 5 — Mastery

L5.1

Ek venturi ek pipe mein vertically mounted hai jo paani upar ki taraf carry karta hai. Wide inlet (1) neeche hai, area $A_{1} = 12 cm^{2}$ ; throat (2) $Δ z = 0.30 m$ upar hai, area $A_{2} = 4 cm^{2}$ . Ek gauge true pressure difference $P_{1} - P_{2} = 4200 Pa$ read karta hai. Flow rate $Q$ nikalo, height difference ko correctly account karte hue. (Yahan $z_{1}$ aur $z_{2}$ points 1 aur 2 ki elevations ek common datum ke upar hain; datum inlet par lo isliye $z_{1} = 0$ aur $z_{2} = Δ z = 0.30 m$ .)

Recall Solution

KYA/KYUN: ab do points alag elevations par hain, isliye Bernoulli mein $ρ g z$ terms cancel nahi hote. (1) aur (2) ke beech full Bernoulli: $P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g z_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g z_{2} .$ Rearrange karo, $z_{2} - z_{1} = Δ z = 0.30 m$ ke saath: $effective drop Δ P_{eff} (P_{1} - P_{2}) - ρ g Δ z = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) .$ KYUN yeh grouping: measured pressure difference ka kuch hissa sirf fluid ko height $Δ z$ tak uthane mein kharch hota hai; sirf bacha hua speed change drive karta hai. Lift term compute karo: $ρ g Δ z = (1000) (10) (0.30) = 3000 Pa$ , isliye $Δ P_{eff} = 4200 - 3000 = 1200 Pa .$ Yahan se yeh L3.1 jaisi derivation hai (continuity $v_{1}$ ko substitute kar deta hai), sirf $P_{1} - P_{2}$ ki jagah $Δ P_{eff}$ ke saath: $Q = A_{1} A_{2} \frac{2 Δ P _{eff}}{ρ ( A _{1}^{2} - A _{2}^{2} )} .$ Areas: $A_{1} = 12 \times 1 0^{- 4}$ , $A_{2} = 4 \times 1 0^{- 4} m^{2}$ ; $A_{1}^{2} - A_{2}^{2} = (144 - 16) \times 1 0^{- 8} = 128 \times 1 0^{- 8} = 1.28 \times 1 0^{- 6} m^{4}$ . Root ke andar: $\frac{2 ( 1200 )}{1000 ( 1.28 \times 1 0 ^{- 6} )} = \frac{2400}{1.28 \times 1 0 ^{- 3}} = 1.875 \times 1 0^{6}$ ; $\approx 1369$ . Prefactor: $A_{1} A_{2} = (12 \times 1 0^{- 4}) (4 \times 1 0^{- 4}) = 4.8 \times 1 0^{- 7}$ . $Q = (4.8 \times 1 0^{- 7}) (1369) \approx 6.57 \times 1 0^{- 4} m^{3} / s .$ Answer: $Q \approx 6.57 \times 1 0^{- 4} m^{3} / s$ . ( $Δ z$ ignore karne se drop $3000 Pa$ se overstated hota aur $Q$ inflate ho jaata.)

L5.2

Ek wind tunnel mein pitot–static tube $P_{s} - P_{0} = 800 Pa$ deta hai hawa mein ( $ρ_{a} = 1.2$ ). Wahi tube ek water tunnel mein same reading $800 Pa$ deta hai. Dono speeds aur ratio nikalo, aur physically explain karo ki equal $Δ P$ ke liye kaunsa fluid faster move karta hai.

Recall Solution

Air: $v_{a} = 2 (800) /1.2 = 1333.3 \approx 36.5 m/s$ . Water: $v_{w} = 2 (800) /1000 = 1.6 \approx 1.265 m/s$ . Ratio: $\frac{v _{a}}{v _{w}} = \frac{ρ _{w}}{ρ _{a}} = \frac{1000}{1.2} = 833.3 \approx 28.9$ . Physical kyun: dynamic pressure $\frac{1}{2} ρ v^{2}$ hota hai. Same pressure tak pahunchne ke liye, halka fluid (hawa) apne chote $ρ$ ki wajah se bahut faster move karna padta hai — speed $1/ ρ$ ke anusaar scale hoti hai. Answer: $v_{a} \approx 36.5 m/s$ , $v_{w} \approx 1.27 m/s$ , ratio $\approx 28.9$ .

Recall Ek-line self-test

Kaunsa single equation, Bernoulli ke saath combine hokar, ek measured area ratio ko ek speed ratio mein convert karta hai? ::: Continuity equation $A_{1} v_{1} = A_{2} v_{2}$ .

Connections

2.2.16 Applications — Pitot tube, Venturi meter, orifice flow (Hinglish) · Bernoulli's Equation · Continuity Equation
Torricelli's Law · Projectile Motion · Manometers and Pressure Measurement · Dynamic vs Static vs Stagnation Pressure