Question

A pair of vertical, open-ended glass tubes inserted into a horizontal pipe are often used together to measure flow velocity in the pipe, a configuration called a Venturi meter. Consider such an arrangement with a horizontal pipe carrying fluid of density p. The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.) Express your answer in terms of quantities given in the problem introduction and g, the magnitude of the acceleration due to gravity.

Answer 1

Answer:

P1 = ρ g h1

v1 = √(2g(h2 - h1)/(1 - γ^2))

Explanation:

a)

P1 = ρ g h1

b)

P2 = ρ g h2

P2 - P1 = 0.5 ρ (v1^2 - v2^2)

ρ g h2 - ρ g h1 = 0.5 ρ (v1^2 - v2^2)

g h2 - g h1 = 0.5 (v1^2 - v2^2)

2 g (h2 - h1) = v1^2 - v2^2

v1 A1 = v2 A2 >>>>> v2=v1(A1/A2)

2 g (h2 - h1) = v1^2 - v1^2 (A1/A2)^2

2 g (h2 - h1) = v1^2 (1 - (A1/A2)^2)

2 g (h2 - h1) = v1^2 (1 - γ^2)

v1 = √(2g(h2 - h1)/(1 - γ^2))

Answer 2

The gauge pressure at the bottom of tube 1 is;

P1 = ρgh1

The velocity at the bottom of tube 1 is;

v1 = √(2g(h2 - h1))/(1 - γ²)

Formula for gauge pressure is;

P = ρgh

Where;

ρ is density

g is acceleration due to gravity

h is height

Thus;

P1 = ρgh1

P2 = ρgh2

Now, from bernoulli equation, we know that;

P2 - P1 = ½ρ((v_1)² - (v_2)²)

Thus, Plugging in the relevant values of P1 and P2 gives;

ρg(h2 - h1) = ½ρ((v_1)² - (v_2)²)

ρ will cancel out to give;

2g(h2 - h1) = ((v_1)² - (v_2)²)

We also know from bernoulli equation that;

V1A1 = V2A2.

Since we want to find v1, let us make V1 the subject of the formula;

v_2 = v1(A1/A2)

Thus;

2g(h2 - h1) = ((v1)² - v1²(A1/A2)²)

Factorizing gives;

2g(h2 - h1) = v1²(1 - (A1/A2)²)

A1/A2 can be expressed as γ. Thus;

2g(h2 - h1) = v1²(1 - γ²)

v1 = √(2g(h2 - h1))/(1 - γ²)

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