Answer to Question #107183 in Physical Chemistry for Bharti

Question

Show that the velocity profile is steady and laminar flow of Newtonian fluid in circular pipe is parabolic. Also prove that average velocity is half of maximum velocity of fluid. Determine kinetic energy correction factor and momentum co-reaction factor for laminar flow of a Newtonian fluid in pipeline.

Answer 1

1.Kinetic Energy Correction Factor

$(K.E.)_{actual}=\frac{1}2mv^2$

$(K.E.)_{actual}=\int_0^R\frac{1}2\rho (2πr)v^3 dr$

$(K.E.)_{actual}=\int_0^R\frac{1}2\rho (2πr)v^3 dr$

$(K.E.)_{actual}=-π\rho(\frac1{4\mu})^3(\frac{dp}{dx})^3\int_0^R(R^2-r^2)^3rdr$

$(K.E.)_{actual}=-π\rho(\frac1{4\mu})^3(\frac{dp}{dx})^3\int_0^R[rR^6-r^7-3R^4r^3+3R^2r^5]dr$

$(K.E.)_{actual}=-π\rho(\frac1{4\mu})^3(\frac{dp}{dx})^3[\frac{R^8}2-\frac{R^8}8-\frac{3R^8}{4}+\frac{3R^8}{6}]$

$(K.E.)_{actual}=-π\rho(\frac1{4\mu})^3(\frac{dp}{dx})^3[\frac{R^8}{8}]$

$(K.E.)_{avg}=\frac{1}2\rho A v^3$

$(K.E.)_{avg}=-\frac{1}{1024\mu^3}\rho πR^2(\frac{dp}{dx})^3R^6$

$\alpha=\frac{(K.E.)_{actual}}{(K.E.)_{avg}}=2$

2.Momentum Correction Factor

$P_{avg}=mv_{avg}=\rhoπR^2 (\frac{-1}{8\mu})^2(\frac{dp}{dx})^2R^4$

$P_{avg}=\rhoπ (\frac{1}{64(\mu)^2})^2(\frac{dp}{dx})^2R^6$

$P_{actual}=\rho dA v^2$

$P_{actual}=\rho\int_0^R (2πrdr) ((\frac{-1}{4\mu})^2(\frac{dp}{dx})^2(R^2-r^2)^2)$

$P_{actual}=2π\rho \frac{1}{16(\mu)^2} (\frac{dp}{dx})^2 \int_0^R( rR^4+r^5-2R^2r^3)dr$

$P_{actual}=2π\rho \frac{1}{16(\mu)^2} (\frac{dp}{dx})^2 [\frac{R^6}2+\frac{R^6}6-\frac{R^6}2]$

$P_{actual}=π\rho \frac{1}{48(\mu)^2} (\frac{dp}{dx})^2 R^6$

$\beta=\frac{P_{actual}}{P_{avg}}=\frac{4}{3}$