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Grade: 11
        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
6 years ago

Answers : (3)

Nirmal Singh.
askIITians Faculty
44 Points
							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
6 years ago
Pushkar Aditya
71 Points
							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.
						
6 years ago
chaitnyakishore
26 Points
							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
						
6 years ago
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