### Fluid Mechanics

# Similarity - Pressure Drop

A geometrically similar model of an air duct is built 1:30 scale and tested with water which is 50 times more viscous and 800 times more dense than air. When tested under dynamically similar conditions, the pressure drop is 2.25 atm in the model. Find the corresponding pressure drop in the full-scale prototype.

Data:

Dimensions of the model = 1/30 times the dimensions of prototype

Viscosity of fluid in the model = 50 times the viscosity of fluid in prototype

Density of fluid in the model = 800 times the density of the fluid in prototype

Formulae:

For dynamic similarity, ratio of forces to be equal

Inertial force_{m} / Inertial force_{p} = Pressure force_{m} / Pressure foce_{p}

= Viscous force_{m} / Viscous force_{p} à 1

Pressure force = (2frLv^{2 }/ D)D^{2} (i.e. = frictional pressure drop x area) à 2

From 1,

NRe_{m} = NRe_{p}

**Calculations:**

NRe_{m} = NRe_{p}

1 x v_{m} x 800 / 50 = 30 x v_{p} x 1 / 1

v_{m} = 1.875 v_{p}

Pressure force ratio = (2f_{m} x 800 x 1 x v_{m}^{2} / 1) x 1^{2 }/ ((2f_{p} x 1 x 30 x (v_{m} / 1.875)^{2} / 30) x 30^{2})

= 1600f_{m}v_{m}^{2} / 512f_{p}v_{m}^{2}

=3.125f_{m}/f_{p}

f = 0.079 (Dvr/m)^{-0.25}

f_{m}/f_{p} = (1 x v_{m} x 800 / 50)^{-0.25} / (30 x (v_{m}/1.875) x 1 / 1)^{-0.25} = 16^{-0.25}/16^{-0.25} = 1

pressure ratio = 3.125 x 30^{2} / 1^{2} = 2812.5 ( i.e. pressure = pressure force/area)

Pressure drop of the prototype = 2.25/2812.5 atm = **0.0008 atm = 81.06 N/m ^{2}**