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 forcem / Inertial forcep = Pressure forcem / Pressure focep
= Viscous forcem / Viscous forcep → 1
Pressure force = (2fLρv2/D)D2 (i.e. = frictional pressure drop x area)→ 2
From 1,
NRem = NRep
Calculations:
NRem = NRep
1 x vm x 800 / 50 = 30 x vp x 1 / 1
vm = 1.875 vp
Pressure force ratio = (2fm x 800 x 1 x vm2 / 1) x 12 / ((2fp x 1 x 30 x (vm / 1.875)2 / 30) x 302)
= 1600fmvm2 / 512fpvm2
=3.125fm/fp
f = 0.079 (Dvρ/μ)-0.25
fm/fp = (1 x vm x 800 / 50)-0.25 / (30 x (vm/1.875) x 1 / 1)-0.25 = 16-0.25/16-0.25 = 1
pressure ratio = 3.125 x 302 / 12 = 2812.5 ( i.e. pressure = pressure force/area)
Pressure drop of the prototype = 2.25/2812.5 atm = 0.0008 atm = 81.06 N/m2
Last Modified on: 01-May-2024
Chemical Engineering Learning Resources - msubbu
e-mail: learn[AT]msubbu.academy
www.msubbu.in