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Prove that for viscous flow through a circular pipe the kinetic energy correction factor is equal to 2 while momentum correction factor is equal to 4/3.

Prove that for viscous flow through a circular pipe the kinetic energy correction factor is equal to 2 while momentum correction factor is equal to 4/3.

(10 Marks) May-2018

Subject Fluid Mechanics 2

Topic Laminar Flow

Difficulty Medium

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A) Momentum correction factor or B.

The velocity distribution through a circular pipe for laminar flow at any radius 'r' is

$u = \frac{1}{4 \mu} (\frac{-\delta p}{\delta x}) (R^2 - r^2)$ ------- (1)

Now consider an elementary area dA in the form of a ring at a radius 'r' and of width 'dr' then,

$dA = 2 \pi r d r$

Rate of fluid flowing through ring,

= dQ = Velocity x area of ring element

= $\mu \times 2\pi rdr$

Momentum of the fluid through ring per second.

= mass x velocity

= $\rho \times dQ \times \mu = \rho \times 2\pi rdr \times u \times u$

$= 2\pi \rho u^2 rdr$

$\therefore$ total actual momentum of fluid per second across the section,

= $\int^R_o 2\pi \rho u^2 rdr$

substituting value of u from (1),

$= 2\pi \rho \int^R_o [\frac{1}{4 \mu} (\frac{- \delta p}{\delta \pi })(R^2 - r^2]^2$ rdr

$= 2\pi \delta [\frac{1}{(16 \mu)^2}{(\delta p}{\delta x})^2 \int^R_o (R^4 + r^4 - r R^2 r^2) rdr$

(taken out all the constants and used $(a - b)^2 a^2 - 2ab + b^2)$

$= \frac{\pi \rho }{8 \mu^2} (\frac{\delta p}{\delta x})^2 \int^R_o (R^4r + r^5 - 2R^2 r^3) dr$

$= \frac{\pi \rho }{8 \mu^2} (\frac{\delta p}{\delta x})^2 [\frac{R^4 r^2}{2} + \frac{r^6}{6} - \frac{ 2 R^2 r^4}{4} ]^R$

$= \frac{\pi \rho }{8 \mu^2} (\frac{\delta p}{\delta x})^2 [ \frac{R^6}{2} + \frac{R^6}{6} = \frac{2R^6}{4}]$

$= \frac{\pi \rho }{48 \mu^2} (\frac{\delta p}{\delta x})^2 R^6$ --- (2) (on solving above equation)

Now, Momentum of the fluid per second based on average velocity,

$= \frac{ \text{mass of fluid} }{sec} \times \text{avg velocity}$

$= \rho A ū \times ū = \rho A ū^2$

where, A = Area of cross section = $\pi R^2$

ū = avg velocity = $\frac{u_{max}}{2}$

$= \frac{1}{2} \times \frac{1}{4 \mu} (\frac{- \delta p}{\delta x}) R^2$

($\therefore = \frac{1}{4 \mu} (\frac{- \delta p}{\delta x}) R^2)$

$= \frac{1}{8 \mu} (\frac{- \delta p}{\delta x}) R^2$

$\therefore$ momentum/sec based on avg velocity,

$= \rho \times \pi R^2 \times (\frac{1}{8 \mu} (\frac{-\delta p}{\delta x}) R^2 ) ^2$

= $\rho \pi (\frac{\delta p}{\delta x}) R^6$ -------- (3)

$64 \mu^2$

$\therefore \beta = \frac{ \text{momentum/sec based on actual velocity (2)} }{ \text{momentum/sec based on avg velocity (3) }}$

$= \frac{\pi \rho }{48 \mu^2} (\frac{\delta p}{\delta x})^2 R^6$

$\frac{\pi}{64 \mu^2} (\frac{-\delta p}{\delta x})^2 R^6$

$= \frac{64}{48} = 4/3$

B) Energy correction factor, $\alpha$

kinetic energy of the fluid flowing through the elementary ring radius 'r' and width 'dr' per sec.

$= \frac{1}{2} \times mass \times u^2 = \frac{1}{2} \times sdQ \times u^2$

$= \frac{1}{2} \times \rho \times (u \times 2 \pi rdr) \times u^2 = \frac{1}{2} \rho \times 2 \pi r u^3 dr = \pi \rho r u^3 dr$

$\therefore$ total actual kinetic energy of flow per second

$= \int^R_o \pi \rho r u^3 dr = \int^R_o \pi \rho r (\frac{1}{4 \mu} (\frac{- \delta p}{\delta x}) (R^2 - r^2))^3 dr$

$= \int^R_o \pi \rho r u^3 dr = \int^R_o \pi \rho r (\frac{1}{4 \mu} (\frac{- \delta p}{\delta x}) (R^2 - r^2))^3 dr$

$= \pi \rho \times (\frac{1}{4 \mu} (\frac{-\delta p}{\delta x}))^3 \int^R_o (R^2 - r^2)^3 rdr$

$= \frac{ \pi \rho }{64 \mu^3} (\frac{- \delta p}{ \delta x})^3 \int^R_o (R^6r - r - 3 R^4 r^3 + 3 R^2 r^5) dr$

$= \frac{ \pi \rho }{64 \mu^3} (\frac{- \delta p}{ \delta x})^3 [ \frac{R^6 r^2}{2} - \frac{-r^3}{8} - \frac{3 R^4 r^4 }{4} + \frac{3 R^2 r^6}{6}]^R$

$= \frac{ \pi \rho }{64 \mu^3} (\frac{- \delta p}{ \delta x})^3 \frac{R^8}{8}$ --- (4)

(on solving above equation).

Kinetic energy of flow based on average velocity.

$= \frac{1}{2} \times mass \times ū^2$

$= \frac{1}{2} \times \rho A ū \times ū^2 = \frac{!}{2} \times \rho A ū^3$

$A = \pi R^2$ and ū = $\frac{1}{8 \mu} (\frac{- \delta p}{ \delta x})^R$ $^2$

Now, kinetic energy of the flow/sec.

$= \frac{1}{2} \times \rho \times \pi R^2 (\frac{1}{8 \mu} (\frac{ -\delta p}{\delta x}) R^2 ) ^3$

$= \frac{1}{2} \times \rho \times \pi R^2 \times \frac{1}{64 \times 8 \mu^3} (\frac{-\delta p}{\delta x})^3 \times R^6$

$= \frac{\rho \pi}{128 \times 8 \mu^3} (\frac{- \delta p}{ \delta x}) R^8$ ---- (5)

$\therefore \alpha = \frac{ \text{K.E/ sec based on actual velocity(4) }}{ \text{K.E/ sec based on avg velocity (5) }}$

$= \frac{\pi \rho }{64 \mu^3} (\frac{- \delta p}{\delta x})^3 \frac{R^8}{8}$

$\frac{\rho }{128 \times 8 \mu^3} \times (\frac{-\delta p}{\delta x})^3 \times R^8$

$= \frac{128 \times 8}{64 \times 8} = 2$

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