Question

A venturi tube is measuring the flow of water; it has a main diameter of 4.5 cm tapering down to a throat diameter of 1.1 cm . If the pressure difference is measured to be 80 mm-Hg, what is the velocity of the water?

Answer 1

Assuming incompressible and frictionless flow Bernoulli's equation applies to this problem:

p/ρ + g∙h + (1/2)∙v² = constant
=>
p₁/ρ + g∙h₁ + (1/2)∙v₁² = p₂/ρ + g∙h₂ + (1/2)∙v₂²

(Here subscript 1 refers to inlet flow, subscript 2 to the throat)

Because both sections have the same average elevation, i.e. h₁= h₂ this simplifies to
p₁/ρ + (1/2)∙v₁² = p₂/ρ + (1/2)∙v₂²
<=> 2∙(p₁ - p₂)/ρ = v₂² - v₁²
<=> 2∙∆p/ρ = v₂² - v₁²

Due to conservation of mass and incompressibility of fluid the volumetric flow rate through each section is the same. Volumetric flow rate equals velocity time cross section area
Q = v∙π∙d²/4
hence:
v₁∙π∙d₁²/4 = v₂∙π∙d₂²/4

So the speeds are related as
v₂ = v₁∙(d₁/d₂)²

Substitute this expression for v₂ to the equation above and solve for v₁:
2∙∆p/ρ = v₁²∙(d₁/d₂)⁴ - v₁²
=> v₁ = √[ 2∙∆p / (ρ∙((d₁/d₂)⁴-1) ]

For proper computation you need to convert pressure difference to Pascals (Pa = kg/(m∙s²))
Since 1atm = 760mmHg = 101325Pa
∆p = 80∙101325 / 760 Pa = 10666 Pa

Hence:
v₁ = √[ 2 ∙ 10666Pa / (1000kg/m³ ∙ ((4.5 /1.1)⁴-1) ]
   = 0.276 m/s