Question

Water flows through a Venturi meter with a pipe diameter of 10.0 cm and a constriction diameter of 5.6 cm. The U-tube manometer is partially filled with mercury. Find the flow rate of the water in the pipe of 10.0 cm diameter if the difference in the mercury level in the U-tube is 3.0 cm.

Answer in L/s

Answer 1

Apply Bernoulli equation between the pipe and the throat ( constriction ) of theVenturi flow meter :

Vbar sub T = { [ - delta P/ rho ] [ 2 ] [ g sub C ] / [ 1 - B^4 ] }^1/2

B = D sub T / D sub P

B = 5.6 / 10.0 = 0.56

1 - B^4 = 0.9016

- delta P = ( R sub manometer ) ( rho sub Hg - rho sub H20 ) ( g / g sub C )

- delta p = ( 3 /100.0 ) (13550 - 1000) ( 9.807 / 1.000 )

- delta P = 3692.33 N. per sq. m. = 3692.33 pa.

- delta P / rho =3692.33 / 1000 = 3.692 N. - m. per kg


Vbar sub T = { [ 3.692 ] [ (2.000 ) ( 1.000 0)]/ [ 0.9016 ] }^1/2


Vbar sub T = 2.8617 m. per sec. <------------------------------

A sub FT = ( pi ) ( D sub T )^2 / ( 4 )

A sub FT = ( pi ) ( 0.05600 )^2 / 4 = 0.002463 sq. m.

Q sub F = [ Vbar sub T ] [ A sub FT ]

Q sub F = [ 2.8617 ] [ 0.002463 ] = 7.0486X10^-3 cu. m. per sec. =

Q sub F = 7.0486 L. per sec. <------------------------------------


mdot = [ Q sub F ] [ rho ]

mdot = [ 7.0486 L. / sec. ] [1.000 kg. / L. ] = 7.0486 kg. per sec.= 7.0486 L/s

Answer 2

According to Bernoulli's equation in case of incompressible flow (such as the flow of water), the theoretical pressure drop (p1- p2) at the constriction would be given by:
?p = p1 - p2 = ? (v2^2 - v1^2) / 2 .....(1)
where ? is the density of the fluid, v1 is the fluid velocity where the pipe is wider, v2 is the faster fluid velocity where the pipe is narrower.
When manometer is in state of balance, pressures in both columns of U-tube are equal (I took lower level of mercury as reference level):
p1+ ?gh = p2 + ?Hg gh
Difference in pressures is:
?p = p1 - p2 = gh (?Hg - ?)
where
?Hg is density of mercury, ?Hg= 13593 kg/m^3
? is density of water, ?=1000 kg/m^3
g is acceleration of gravity, g=9.81 m/s^2
h is difference in the mercury level in the U-tube, h=3.0 cm.
?p = 9.81 * 0.03 * (13593 - 1000) = 3706.1199 Pa

Volumetric flow rate in wider and narrower section of tube is the same:
v1 A1 = v2 A2 .....(2)
where A1 and A2 are areas of wider and narrower section of tube, respectively.
A1= 0.1^2 ? / 4 = 7.854 * 10^-3 m^2
A2= 0.056^2 ? / 4 = 2.4630 * 10^-3 m^2
From (2),
v2 = v1 A1/A2 .....(3)
Substitute (3) for v2 in (1)
?p = ? (v1^2 (A1/A2)^2 - v1^2) / 2
?p = ? v1^2 ((A1/A2)^2 - 1) / 2
v1^2 = 2 ?p / [? ((A1/A2)^2 - 1)]
because we can't use fractions here, denominator in this equation looks a bit messy, so I'll rewrite it as
v1^2 = 2 ?p A2^2 / [? (A1^2 - A2^2)]
and finally
v1 = A2 * sqrt { 2?p / [? (A1^2 - A2^2)] }
v1 = 2.4630 * 10^-3 * sqrt { 2 * 3211.97 / [1000 * (7.854 ^2 - 2.4630^2) * 10^-6] }
v1 = 0.83705m/s
Volumetric flow rate of the water:
Q= A1 v1
Q= 7.854 * 10^-3 * 0.83705= 6.574*10^-3 m^3/s
or
Q = 6.574liters per second
To get mass flow rate m_dot we simply multiply this by density of water to get
m_dot = 6.574 kg/s

Answer 3

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Answer 4

Answer 5

Answer 6

the pressure is P = p x g x h

where p is density of mercury,g = 9.8 m/s^2 and h = 3.0 cm = 3.0 x 10^-2 m

Also we know that

P = (1/2)p1 x v^2

where p1 is density of water(1000 kg/m^3) and v is speed of water

or v^2 = (2P/p1)

or v = (2P/p1)^1/2

the flow rate through the 10.0 cm diameter pipe is

A x v

where A = pi x (d^2/4),d = 10.0 cm = 10.0 x 10^-2 m