The Venturi Meter

A **Venturi meter**, named after the Italian physicist **Giovanni Battista Venturi** (1746 – 1822), is a differential pressure flow meter that generates a flow measurement by measuring the pressure difference at two different locations in a pipe. The device demostrates the Venturi effect, which states that the velocity of the flowing fluid increases due to a pressure drop created by constricting the diameter of the pipe.

## **Problem**

Consider an ideal fluid of density

ρ

{\displaystyle \rho}

 flowing through a horizontal pipe of variable cross-sectional area (Figure 1). The fluid at the portion of the pipe with cross-sectional area

A

1

{\displaystyle {A}\_{1}}

 is moving at

v

1

{\displaystyle {v}\_{1}}

 and has a measured pressure of

P

1

{\displaystyle {P}\_{1}}

. The fluid at the portion of the pipe with cross-sectional area

A

2

{\displaystyle {A}\_{2}}

 is moving at

v

2

{\displaystyle {v}\_{2}}

 and has a measured pressure of

P

2

{\displaystyle {P}\_{2}}

.

**Part 1:** Determine

v

1

{\displaystyle {v}\_{1}}

 in terms of

ρ
,

A
1

,

A
2

,

P
1

{\displaystyle \rho, A\_1, A\_2, P\_1}

 and

P
2

{\displaystyle P\_2}

.

**Part 2:** Two different fluids of density

ρ

1

{\displaystyle {\rho}\_{1} }

 (fluid in the horizontal pipe) and

ρ

2

{\displaystyle {\rho}\_{2} }

 (fluid in the u-shaped manometer), where

ρ

1

<

ρ

2

{\displaystyle {\rho}\_{1} < {\rho}\_{2} }

. The height difference of the fluid in the manometer is

h

{\displaystyle h}

. It helps to let

# h

h
1

−

h
2

{\displaystyle h = h\_1 - h\_2 }

. Determine

v

1

{\displaystyle {v}\_{1}}

 in terms of

ρ

1

,

ρ

2

,

A
1

,

A
2

{\displaystyle {\rho}\_{1}, {\rho}\_{2}, A\_1, A\_2 }

 and

h

{\displaystyle h}

.

## **Hints**

There are two important equations for this problem:

(1) The continuity equation

ρ

A
1

v
1

=
ρ

A
2

v
2

{\displaystyle \rho A\_1 v\_1 = \rho A\_2 v\_2 }

(2) Bernoulli’s equation

P
1

ρ
g

h
1

1
2

ρ

v
1
2

=

P
2

ρ
g

h
2

1
2

ρ

v
2
2

{\displaystyle P\_1 + \rho gh\_1 + \frac{1}{2} \rho v\_1^2 = P\_2 + \rho gh\_2+\frac{1}{2} \rho v\_2^2 }

## **Solution**

### **Part 1**

From the first equation we find

v
2

=

A
1

v
1

A
2

{\displaystyle v\_2 = \frac{A\_1 v\_1}{A\_2} }

Since there is no change in the altitude of the pipe, we can deduce that

P
1

1
2

ρ

v
1
2

=

P
2

1
2

ρ

(

A
1

v
1

A
2

)

2

{\displaystyle P\_1 + \frac{1}{2} \rho v\_1^2 = P\_2 + \frac{1}{2} \rho \left(\frac{A\_1 v\_1}{A\_2} \right)^2 }

Isolate for

v
1

{\displaystyle v\_1}

P
1

−

P
2

=

1
2

ρ

v
1
2

[

(

A
1

A
2

)

2

−
1

]

{\displaystyle P\_1 - P\_2 = \frac{1}{2} \rho v\_1^2 \left[\left(\frac{A\_1}{A\_2} \right)^2-1\right] }

v
1

=

2
(

P
1

−

P
2

)

ρ

[

(

A
1

A
2

)

2

−
1

]

{\displaystyle v\_1 = \sqrt{\frac{2(P\_1-P\_2)}{\rho \left[\left(\frac{A\_1}{A\_2}\right)^2-1 \right]}} }

### **Part 2**

Treat the manometer as two static column problems.

P
1

ρ

1

g

h
1

=

P
2

ρ

1

g

h
2

ρ

2

g
(

h
1

−

h
2

)

P
1

−

P
2

=
−

ρ

1

g

h
1

ρ

1

g

h
2

ρ

2

g
(

h
1

−

h
2

)

P
1

−

P
2

=
(

ρ

2

−

ρ

1

)
g
(

h
1

−

h
2

)

P
1

−

P
2

=
(

ρ

2

−

ρ

1

)
g
h

{\displaystyle \begin{align}
P\_1 + {\rho}\_{1} g h\_1 &= P\_2 + {\rho}\_{1} g h\_2 + {\rho}\_{2} g (h\_1 - h\_2) \\[1.5 pt]

P\_1 - P\_2 &= -{\rho}\_{1} g h\_1 + {\rho}\_{1} g h\_2 + {\rho}\_{2} g (h\_1 - h\_2) \\[1.5 pt]

P\_1 - P\_2 &= ({\rho}\_{2} - {\rho}\_{1}) g (h\_1 - h\_2) \\[1.5 pt]

P\_1 - P\_2 &= ({\rho}\_{2} - {\rho}\_{1}) gh
\end{align}
}

Substitute the difference of pressure into the result from part 1 to get

v
1

=

2
(

ρ

2

−

ρ

1

)
g
h

ρ

1

[

(

A
1

A
2

)

2

−
1

]

{\displaystyle v\_1 = \sqrt{\frac{2({\rho}\_{2} - {\rho}\_{1})gh}{{\rho}\_{1} \left[{\left(\frac{A\_1} {A\_2}\right)}^{2} - 1 \right]}} }