Answers to Sample Problems From Pitot Tube Project

1. Pitot Tube consists of two concentric pipes and has a small diameter overall diameter (about 5 mm). The inner pipe is open and is pointed in the direction of flow – this is known as the stagnation point as fluid is brought to rest at this pipe and velocity is zero. The outer pipe is closed at the front but has holes drilled in the side of it. Fluid flows past these holes like wind blowing across the face of a chimney pipe. The velocity at the face of these holes is equal to the mainstream velocity in the pipe. This difference in velocities creates a difference in pressures. The difference in pressures is measured by means of a manometer in the lab or differential pressure transmitter in industry. The pressure difference is converted to the velocity of fluid.

Here are two sketches of the Pitot Tube connected to a manometer:

2. Take Bernoulli’s equation and assume

1. V1 = 0 at stagnation point

2. h1 = h2, both points at same elevation

Rewrite Bernoulli to get:

Rearrange as follows:

3. This is a square root relationship as v increases with the square root of deltaP. Have a think about this yourself and you’ll see it’s not a straight line but a curve. It might be easier to square both sides to get v² = deltaP and then think about how that would look as it’s the same thing.

4. 6 mbar = 6 hectoPa = 600 Pa

Velocity = (2(600)/1000)^1/2 = 1.095 m/s

Diameter = 200 mm, c.s.a. = πd²/4 = 3.14(0.20)²/4 = 0.031m²

Flowrate, Q = va = 1.095 x 0.031 = 0.034 m³/sec

Mass flowrate = ρQ = 1000 x 0.034 = 34 kg/sec

(I kept everything to 3 decimal places).