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Asked: 2020-05-13T16:59:32+05:30 In:

GATE MT 2020 Q45. Figure shows schematic of a venturi meter. The cross sectional area is 100 mm^2 at A and is 50 mm^2 at B. If air is flowing through the venturi meter at a flow rate of 10^-3 m^3. S^-1, the height H in the air over-water manometer is _____________ mm (round off to the nearest integer). A. Incompressible flow with no friction losses.B. Density of air is 1 kg m^-3. C. Density water is 1000 kg m^-3. D. Acceleration due to gravity is 9.8 m s^-2.

GATE MT 2020 Q45. Figure shows schematic of a venturi meter. The cross sectional area is 100 mm^2 at A and is 50 mm^2 at B. If air is flowing through the venturi meter at a flow rate of 10^-3 m^3. S^-1, the height H in the air over-water manometer is _____________ mm (round off to the nearest integer). A. Incompressible flow with no friction losses.B. Density of air is 1 kg m^-3. C. Density water is 1000 kg m^-3. D. Acceleration due to gravity is 9.8 m s^-2.
GATE MT 2020 Q45. Figure shows schematic of a venturi meter. The cross sectional area is 100 mm^2 at A and is 50 mm^2 at B. If air is flowing through the venturi meter at a flow rate of 10^-3 m^3. S^-1, the height H in the air over-water manometer is _____________ mm (round off to the nearest integer). A. Incompressible flow with no friction losses.B. Density of air is 1 kg m^-3. C. Density water is 1000 kg m^-3. D. Acceleration due to gravity is 9.8 m s^-2.
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    5 Answers

    1. According to Bernoulli's Principle at A and B points, P(A)/ρ*g + v(1)^2/2*g + Z(1) = P(B)/ρ*g + v(2)^2/2*g + Z(2) ----> Eq(1) Z(1) = Z(2) = 0 (since.. datum line is axial line of the pipe) ρ = 1 kg/m^3 Given Q(volumetric flowrate) = 10^(-3) m^3/s. v(1) = Q/A(1) = 10^(-3)/(100*(10^-3)^2) = 10 m/sRead more

      According to Bernoulli's Principle at A and B points,

P(A)/ρ*g + v(1)^2/2*g + Z(1) = P(B)/ρ*g + v(2)^2/2*g + Z(2) ----> Eq(1)

Z(1) = Z(2) = 0 (since.. datum line is axial line of the pipe)
ρ = 1 kg/m^3

Given Q(volumetric flowrate) = 10^(-3) m^3/s.
v(1) = Q/A(1) = 10^(-3)/(100*(10^-3)^2) = 10 m/s
v(2) = Q/A(2) = 10^(-3)/(50*(10^-3)^2) = 20 m/s

Substituting above values in Eq - 1;

P(A) - P(B) = 150 ---> Eq(2)

Air-over-Water Manometer;

P(A) + ρ(air)*g*(H+h) = ρ(water)*g*H + ρ(water)*g*h + P(B) ---> Eq(3)

cancelling like terms and Equating Eq(2) & Eq(3)

150 = ρ(water)*g*H
H = 150 / (1000 * 9.8) = 0.0153m = 15.3mm

       

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    2. Best Answer
      Platinum Gate aspirant.

      The problems in fluid are basically composed of threemthree principles: Continuity equation and Bernoulli theorem, pressure principles. In this problem we see a horizontal venturi meter connected with a Manometer. The difference in height of manometer liquid is required. This can be easily calculateRead more

      The problems in fluid are basically composed of threemthree principles: Continuity equation and Bernoulli theorem, pressure principles.

      In this problem we see a horizontal venturi meter connected with a Manometer. The difference in height of manometer liquid is required.

      This can be easily calculated by applying Bernoulli theorem and getting the Height(H) relationship with velocity of air flowing in venturi meter.

      By given flow rate i.e. continuity equation and area at A and B(Note: to carefully change the units of area)  we can find velocity at A(v1) and Velocity at B(v2). Now solving we get the required height as shown:

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    3. Platinum GATE 2021 QUALIFIED
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    4. Platinum Metallurgical and Materials Engineering student

      15.3 mm

      15.3 mm

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    5. Gold

      15mm

      15mm

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