MechanicalFluid MechanicsmediumFE ExamCoursework

Pressure Drop Across a Venturi

Water ( $\rho = 1000\ \text{kg/m}^3$ ) flows through a horizontal venturi meter.

The inlet diameter is $D_1 = 100\ \text{mm}$ and the throat diameter is $D_2 = 50\ \text{mm}$ . The inlet velocity is $V_1 = 2.0\ \text{m/s}$ .

Neglecting losses, find the pressure drop $P_1 - P_2$ between the inlet and the throat.

Given

Hint 1

Two relations are needed: one to find the throat velocity, one to convert velocities into pressures.

Hint 2

Continuity relates velocity to area, and area goes as diameter squared.

Hint 3

Halving the diameter quarters the area, so the velocity quadruples — to 8 m/s.

Worked solution — try the problem first

Step 1 — Continuity gives the throat velocity.

$A_1V_1 = A_2V_2 \quad \Longrightarrow \quad V_2 = V_1\frac{A_1}{A_2} = V_1\left(\frac{D_1}{D_2}\right)^2$

The area ratio goes as the square of the diameter ratio:

$V_2 = 2.0 \times \left(\frac{100}{50}\right)^2 = 2.0 \times 4 = 8.0\ \text{m/s}$

Step 2 — Bernoulli between inlet and throat. The meter is horizontal, so the elevation terms cancel:

$P_1 + \tfrac{1}{2}\rho V_1^2 = P_2 + \tfrac{1}{2}\rho V_2^2$

$P_1 - P_2 = \tfrac{1}{2}\rho\left(V_2^2 - V_1^2\right) = \tfrac{1}{2}(1000)\left(8.0^2 - 2.0^2\right)$

$= 500(64 - 4) = 500 \times 60 = 30{,}000\ \text{Pa}$

$\boxed{P_1 - P_2 = 30\ \text{kPa}}$

The pressure falls at the throat because the fluid speeds up — that is the whole operating principle of the meter.

Concepts:Bernoulli equationContinuityVenturi meter

Sign in to submit

Grading needs an account so your progress, attempts, and daily quota can be tracked. The problem statement and worked solution stay open to everyone.