Velocity of fluid in venturi tube

[Dennydont]

Homework Statement

A venturi tube is a means of measuring fluid velocity. One particular device consists of a level, circular pipe of radius 5.7 cm which at one point narrows to a radius of 3.7 cm. An ideal, incompressible liquid flows through the tube and the pressure difference between the wide and narrow parts is found to be 3.5 kN/m^2. If the density of the liquid is 1120 kg/m3 what is the velocity of the fluid in the wide part of the pipe?
ρ = 1120 kg/m^3
∆P = 3.5kN/m^2
narrow radius = 3.7cm
wide radius = 5.7cm

Homework Equations

Continuity: ρA₁v₁ = ρA₂v₂
Bernoulli: P + 1/2ρv^2 + ρgh

The Attempt at a Solution

I tried finding v₁ by using the rearranging Bernoulli's equation v = √((2∆P)/ρ) = 2.5m/s
Then using continuity equation, tried to find v₂ which happened to equal 1.05m/s. Incorrect.

 

[ehild]

Homework Statement

A venturi tube is a means of measuring fluid velocity. One particular device consists of a level, circular pipe of radius 5.7 cm which at one point narrows to a radius of 3.7 cm. An ideal, incompressible liquid flows through the tube and the pressure difference between the wide and narrow parts is found to be 3.5 kN/m^2. If the density of the liquid is 1120 kg/m3 what is the velocity of the fluid in the wide part of the pipe?
ρ = 1120 kg/m^3
∆P = 3.5kN/m^2
narrow radius = 3.7cm
wide radius = 5.7cm

Homework Equations

Continuity: ρA₁v₁ = ρA₂v₂
Bernoulli: P + 1/2ρv^2 + ρgh

You need the Bernoulli equation. What you wrote it was not an equation.