Find the manometer head to two decimal places.

Question

Air is flowing through a circular tube with a changing cross-section as shown above. A water manometer is connected between sections 1 and 2. The rate of air flow is 170 l/s and the diameters are d_1= 20 cm and d_2=21 cm. Assuming the density of air to be 1.2 kg/m3, calculate the manometer head h (in m to two decimal places). Assume that the flow has no losses.

Answer 1

First, the volumetric flow rate, which is a constant 170 liters per second, equals the cross sectional area times the velocity:

dV/dt = A v = (pi d² / 4) v = 170,000 cm³/s

v = (4 / pi d²) (dV/dt)

Note that one liter equals a thousand mL and, equivalently, a thousand cubic cm. Therefore,

v₁ = 541 cm/s = 5.41 m/s

v₂ = 491 cm/s = 4.91 m/s

So, when the pipe widens, the flow slows. According to Bernoulli's law, representing conservation of energy along a streamline within the flow,

½ p v² + P = constant

Recalling that the density of air is stated to remain constant,

½ p₁ v₁² + P₁ = ½ p₂ v₂² + P₂

½ p_air (v₁² - v₂²) = P₂ - P₁ = ∆P

Now, that pressure differential pushes water down according to hydrostatic equilibrium:

∆P = p_water g h

Therefore,

h = (p_air (v₁² - v₂²)) / (2 p_water g)

Substituting the appropriate values (using g=9.81 m/s²):

h = 0.0316 cm ≈ 0.00032 m

The slight change in the speed of the lighter air generates a barely perceptible imbalance in the heavier water of the manometer of about a third of a mm.