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A venturimeter equipped with a differential mercury manometer has an inlet diameter of 8cm and throat diameter is 4cm. Find rate of flow of water through the meter, if difference in mercury column is 15cm. Specific gravity of mercury is 13.6 and g=9.8m/s²

A venturimeter equipped with a differential mercury manometer has an inlet diameter of 8cm and throat diameter is 4cm. Find rate of flow of water through the meter, if difference in mercury column is 15cm. Specific gravity of mercury is 13.6 and g=9.8m/s²

Grade:11

3 Answers

Nirmal Singh.
askIITians Faculty 44 Points
7 years ago
Let the velocity at inlet is v1 and at outlet is v2
by the equation of continuity A1v1 = A2v2
so v1/v2 = A2 / A1 = pi*r1^2 / pi*r2^2 = r1^2 / r2^2 = (4x10^-2)^2 / ( 2x10^-2)^2 = 16/4 = 4
v1/v2 = 4
by Bernoulli equation
P1+1/2rho*v1^2 = P2+1/2rho*v2^2
P1 - P2 = 1/2rho'(v1^2 - v2^2) where rho' = specific gravity of water
P1 - P2 = rho g h, where rho = specific gravity of mercury
P1 - P2 = 13.6*9.8*15x10^-2 = 1.9992 N/m^2
Now 1.9992 = 1/2 *1000 [v1^2 - (v1/4)^2]
v1^2 = 0.00426
v1 = 0.0653 m/ s
Hence rate of flow of water = A1v1 = pi*(r1^2)*v1 = 3.14*(4x10^-2)^2*0.0653 = 3.28 m^3/s
Thanks & Regards,
Nirmal Singh
Askiitians Faculty
Pushkar Aditya
71 Points
7 years ago
Sir you did the wrong calculation and according to your method, The right answer is, Let the velocity at inlet is v1 and at outlet is v2, Given that d1=8cm-->r1=4cm d2=4cm-->r2=2cm by the equation of continuity A1v1 = A2v2 so p(r1)²v1=p(r2)²v2 (r1)²v1=(r2)²v2 16v1=4v2 v2=4v1. by Bernoulli equation, P1+1/2 rho×(v1)² = P2+1/2 rho`×(v2)² P1 - P2 = 1/2rho`(v2²-v1²)--->(1) where rho` = density of water P1 - P2 = rho g h---->(2), where rho = density of mercury Therefore from (1)and(2), rho gh=1/2rho`[(v2)²-(v1)²] rho/rho` gh=1/2[16v1²-v1²] We know that specific gravity of mercury =density of mercury(rho)/density of water(rho`) Therefore rho/rho`=13.6 Therefore 13.6×9.8×15×10^-2=1/2[15v1²] 13.6×9.8×2×10^-2=v1² Therefore v1²=2.6656 v1=v2.6656 v1=1.6326 m/s², v2=4v1=41×1.6326=6.5304 m/s². Therefore rate of flow of water = A1v1 =p(r1)²v1 =p×(4÷100)²×1.6326 =0.008209646 m^3/s =82.09×10^-4 m^3/s.
chaitnyakishore
26 Points
7 years ago
Let the velocity at inlet is v1 and at outlet is v2 inlet diameterd1=8cm-->r1=4cm=0.04m outlet diameterd2=4cm-->r2=2cm=0.02m by the equation of continuity A1v1 = A2v2 so pr1²v1=pr2²v2 v1=(r2²/r1²)v2-->v1=(4²/2²)v2 v1=v2/4 Therefore v2=4v1 by Bernoulli equation P1+1/2rho*v1^2 = P2+1/2rho*v2^2 P1 - P2 = 1/2rho((v2)²-(v1)²) where rho = density of water P1 - P2 = rho` g h, where rho` = density of mercury rho`gh=1/2 rho((v2)²-(v1)²) rho`/rho gh=1/2((v2)²-(v1)²) But rho`/rho =density of mercury/density of water = specific gravity of mercury=13.6 Therefore 13.6×9.8×15×10^-2=1/2 ((4v1)²-(v1)²) 13.6×9.8×15×10^-2=1/2 (16v1²-v1²) 13.6×9.8×15×10^-2=1/2 (15v1²) Therefore v1²=2.6656 v1=1.6326 m/s² Therefore v2=4v1=4×1.6326=6.5304 m/s² Therefore Rate of flow of water=A1v1 =pr1²v1 =(22/7)×(4/100)²×1.6326 =0.008209646 Therefore rate of flow of water=82.09×10^-4 m^3/s

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