Venturimeter

VENTURIMETER

Venturimeter is a device which is used to measure discharge of liquid flowing through the pipeline.

Construction

Converging Cone

It is tapered in a section whose diameter reduces from d1 to d2. The angle for converging cone is 21͒ to 23͒.

2. Throat

The portion after converging cone of diameter d₂ is known as the throat. This portion is used to calculate pressure difference with help of U tube manometer.

3. Diverging Cone

It's diameter increases from d2 to d1 and angle for diverging cone is 6͒ to 8͒.

The expression for Rate of Flow through Venturimeter

Consider a venturimeter fitted in a horizontal pipe through which fluid is flowing is as shown in fig.

Let d1 = diameter of inlet or at section 1

p1 = pressure at section 1

v1 = velocity of fluid at section 1

a = area at section 1 = π(d1xd1)/4

and d2, p2, v2, a2 are corresponding values at section 2

a1v1 = a2v2

According to Bernoulli's theorem

P₁/ρ + ½ x ( v₁)² + gh₁ = P₂/ρ + ½ x ( v₂)² + gh₂

As the tube is horizontal, therefore
h1 = h2

Substituting in above equation we get

P₁/ρ + ½ x ( v₁)² = P₂/ρ + ½ x ( v₂)²

By solving the above equations we get the rate of flow through venturimeter is

$V = a_{1}\times v_{1} = a_{1}a_{2} \sqrt{2gh} / \sqrt{a_{1}^{2}- a_{2}^{2}}$

Problems

1) A horizontal venturimetrer with inlet and throat diameter 30 cm and 15 cm respectively is used to measure the flow of water. The reading of differential manometer connected to the inlet and the throat is 20 cm of mercury. Determine the rate of flow. Take Cd = 0.98

Solution
Given d1 = 30 cm

$a_{1} = \frac{\Pi }{4} d_{1}^{2} = \frac{\Pi }{4} 30^{2} = 706.85 cm^{2}$

d2 = 15 cm

$a_{2} = \frac{\Pi }{4} d_{2}^{2} = \frac{\Pi }{4} 15^{2} = 176.7 cm^{2}$

Reading for differential manometer = x = 20 cm of mercury

Hence difference of pressure head of mercury is given by

$h = x \frac{s_{h}}{s_{o}} - x$

where sh is specific gravity of mercury = 13.6, so = specific gravity of water = 1

h = 20 x 12.6 = 252 cm of water

$Q = c_{d}a_{1}a_{2} \times \sqrt{2gh}/ \sqrt{a_{1}^{2}}- \sqrt{a_{2}^{2}}$

$Q = 0.98 \times \frac{706.85\times 176.7}{\sqrt{706.85^{2}-176.7^{2}}}\times \sqrt{2\times 9.81\times 252}$

Q = 125756 c.cm/s = 125.756 lit/s

2) A 20 cm x 10 cm venturimeter is inserted in vertical pipe carrying oil of sp. gr. 0.8, the flow of oil is in upward direction. The difference of levels between the throat and inlet section is 50 cm. The oil mercury differential manometer gives a reading of 30 cm of mercury. Find the disharge of oil. Neglect losses.

Solution

Given d1 = 20 cm

$a_{1} = \frac{\Pi }{4}(20)^{2} = 314.16 cm^{2}$