Show that the momentum correction factor and energy correction correction factor for laminar flow through circular pipe are 4343 and 2.0 respectively.

 Show that the momentum correction factor and energy correction correction factor for laminar flow through circular pipe are 

43${\displaystyle \frac{4}{3}}$ and 2.0 respectively.

Solution:-

(i) Momentum correction factor or β$\beta$

The velocity distribution through a circular pipe for laminar flow at any radius ‘r’ is given by,

u=14μ(−dPdx)(R2−r2)…………..(1)$u={\displaystyle \frac{1}{4\mu }}\left(-{\displaystyle \frac{dP}{dx}}\right)(R^2-r^2)\dots \dots \dots \dots ..(1)$

Consider an elementary area dA in the form of a ring at the radius ‘r’ and width ‘dr’ then

dA=2πrdr$dA=2\pi rdr$

Rate of fluid flowing through the ring,

=dQ = Velocity× area of the ring element

=u×2πrdr$=u\times 2\pi rdr$

Momentum of the fluid through ring per second

=mass × velocity

=ρ×dQ×u$=\rho \times dQ\times u$

=ρ×2πr×u×u$=\rho \times 2\pi r\times u\times u$

=2πρu2rdr$=2\pi \rho u^{2}rdr$

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Therefore the total actual momentum of the fluid per second across the section ∫R02πρu2rdr${\int }_0^{R}2\pi \rho u^{2}rdr$

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Substituting the value of ‘u’ from equation (1),

=2πρ∫R0[14μ(−dPdx)(R2−r2)]2rdr$=2\pi \rho {\int }_0^R{\left[{\displaystyle \frac{1}{4\mu }}\left(-{\displaystyle \frac{dP}{dx}}\right)(R^2-r^2)\right]}^{2}rdr$

=2πρ[14μ(−dPdx)]2∫R0(R2−r2)2rdr$=2\pi \rho {\left[{\displaystyle \frac{1}{4\mu }}\left(-{\displaystyle \frac{dP}{dx}}\right)\right]}^2{\int }_0^R(R^2-r^2{)}^{2}rdr$

=2πρ[116μ2(dPdx)2]∫R0(R4+r4−2R2r2)rdr$=2\pi \rho \left[{\displaystyle \frac{1}{16{\mu }^2}}{\left({\displaystyle \frac{dP}{dx}}\right)}^2\right]{\int }_0^R(R^4+r^4-2R^2{r}^2)rdr$

=πρ8μ2(dPdx)2∫R0(R4r+r5−2R2r3)dr$={\displaystyle \frac{\pi \rho }{8{\mu }^2}}{\left({\displaystyle \frac{dP}{dx}}\right)}^2{\int }_0^R(R^{4}r+r^5-2R^2{r}^3)dr$

=πρ8μ2(dPdx)2[R4r22+r66−2R2r44]R0$={\displaystyle \frac{\pi \rho }{8{\mu }^2}}{\left({\displaystyle \frac{dP}{dx}}\right)}^2{\left[{\displaystyle \frac{R^4{r}^2}{2}}+{\displaystyle \frac{r^6}{6}}-{\displaystyle \frac{2R^2{r}^4}{4}}\right]}_0^R$

=πρ8μ2(dPdx)2[R62+R66−2R64]$={\displaystyle \frac{\pi \rho }{8{\mu }^2}}{\left({\displaystyle \frac{dP}{dx}}\right)}^2\left[{\displaystyle \frac{R^6}{2}}+{\displaystyle \frac{R^6}{6}}-{\displaystyle \frac{2R^6}{4}}\right]$

=πρ8μ2(dPdx)2[6R6+2R6−6R612]$={\displaystyle \frac{\pi \rho }{8{\mu }^2}}{\left({\displaystyle \frac{dP}{dx}}\right)}^2\left[{\displaystyle \frac{6R^6+2R^6-6R^6}{12}}\right]$

=πρ8μ2(dPdx)2×R66$={\displaystyle \frac{\pi \rho }{8{\mu }^2}}{\left({\displaystyle \frac{dP}{dx}}\right)}^2\times {\displaystyle \frac{R^6}{6}}$

=πρ48μ2(dPdx)2×R6……………(2)$={\displaystyle \frac{\pi \rho }{48{\mu }^2}}{\left({\displaystyle \frac{dP}{dx}}\right)}^2\times R^6\dots \dots \dots \dots \dots (2)$

Momentum of the fluid per second based on average velocity

=mass of fluidsecond×Average Velocity$={\displaystyle \frac{\text{mass of fluid}}{\text{second}}}\times \text{Average Velocity}$

=ρAu¯¯¯×u¯¯¯=ρAu¯¯¯2$=\rho A\overline{u}\times \overline{u}=\rho A{\overline{u}}^2$

Where A = area of cross-section = πR2$\pi R^2$

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Average velocity = u¯¯¯=Umax2$\overline{u}={\displaystyle \frac{U_{max}}{2}}$

=12×14μ(−dPdx)R2$={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{4\mu }}\left(-{\displaystyle \frac{dP}{dx}}\right)R^2$

=18μ(−dPdx)R2$={\displaystyle \frac{1}{8\mu }}\left(-{\displaystyle \frac{dP}{dx}}\right)R^2$

${\mathrm{<span\; style="color:\; \#351c75;\; font-family:\; georgia;\; font-size:\; medium;"><i><br></i></span>}}^{}$

Therefore Momentum per second based on average velocity,

=ρ×πR2[18μ(−dPdx)R2]2$=\rho \times \pi R^2{\left[{\displaystyle \frac{1}{8\mu }}\left(-{\displaystyle \frac{dP}{dx}}\right)R^2\right]}^2$

=ρ×πR2×164μ2(−dPdx)2R4$=\rho \times \pi R^2\times {\displaystyle \frac{1}{64{\mu }^2}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^2{R}^4$

=ρπ(−dPdx)2R664μ2.......................(3)$={\displaystyle \frac{\rho \pi {\left(-{\displaystyle \frac{dP}{dx}}\right)}^2{R}^6}{64{\mu }^2}}.......................(3)$

β=πρ48μ2(−dPdx)2×R6ρπ64μ2(−dPdx)2R6$\beta ={\displaystyle \frac{{\displaystyle \frac{\pi \rho }{48{\mu }^2}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^2\times R^6}{{\displaystyle \frac{\rho \pi }{64{\mu }^2}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^2{R}^6}}$

∴β=6448=43$\therefore \beta ={\displaystyle \frac{64}{48}}={\displaystyle \frac{4}{3}}$

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2) Energy correction factor ‘a’

Kinetic energy of the fluid flowing through the elementary ring of radius ‘r’ and of width ‘dr’ per second,

=12×mass×u2$={\displaystyle \frac{1}{2}}\times \text{mass}\times u^2$

=12×ρdQ×u2$={\displaystyle \frac{1}{2}}\times \rho dQ\times u^2$

=12×ρ×(u×2πrdr)×u2$={\displaystyle \frac{1}{2}}\times \rho \times (u\times 2\pi rdr)\times u^2$

=12ρ×2πru3dr$={\displaystyle \frac{1}{2}}\rho \times 2\pi r{u}^{3}dr$

=πρru3dr$=\pi \rho r{u}^{3}dr$

Therefore total actual kinetic energy of flow per second

=∫R0πρru3dr$={\int }_0^R\pi \rho r{u}^{3}dr$

=∫R0πρr[14μ(−dPdx)(R2−r2)]3dr$={\int }_0^R\pi \rho r{\left[{\displaystyle \frac{1}{4\mu }}\left(-{\displaystyle \frac{dP}{dx}}\right)(R^2-r^2)\right]}^{3}dr$

=πρ[14μ(−dPdx)]3∫R0(R2−r2)3rdr$=\pi \rho {\left[{\displaystyle \frac{1}{4\mu }}\left(-{\displaystyle \frac{dP}{dx}}\right)\right]}^3{\int }_0^R(R^2-r^2{)}^{3}rdr$

=πρ[164μ3(−dPdx)3]∫R0(R6−r6−3R4r2+3R2r4)rdr$=\pi \rho \left[{\displaystyle \frac{1}{64{\mu }^3}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^3\right]{\int }_0^R(R^6-r^6-3R^4{r}^2+3R^2{r}^4)rdr$

=πρ[164μ3(−dPdx)3]∫R0(rR6−r7−3R4r3+3R2r5)dr$=\pi \rho \left[{\displaystyle \frac{1}{64{\mu }^3}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^3\right]{\int }_0^R(r{R}^6-r^7-3R^4{r}^3+3R^2{r}^5)dr$

=πρ[164μ3(−dPdx)3][R6r22−r88−3R4r44+3R2r66]R0$=\pi \rho \left[{\displaystyle \frac{1}{64{\mu }^3}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^3\right]{\left[{\displaystyle \frac{R^6{r}^2}{2}}-{\displaystyle \frac{r^8}{8}}-{\displaystyle \frac{3R^4{r}^4}{4}}+{\displaystyle \frac{3R^2{r}^6}{6}}\right]}_0^R$

=πρ64μ3(−dPdx)3[R82−R88−3R84+3R86]$={\displaystyle \frac{\pi \rho }{64{\mu }^3}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^3\left[{\displaystyle \frac{R^8}{2}}-{\displaystyle \frac{R^8}{8}}-{\displaystyle \frac{3R^8}{4}}+{\displaystyle \frac{3R^8}{6}}\right]$

=πρ64μ3(−dPdx)3[R8(12−3−18+1224)]$={\displaystyle \frac{\pi \rho }{64{\mu }^3}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^3\left[R_8\left({\displaystyle \frac{12-3-18+12}{24}}\right)\right]$

=πρ64μ3(−dPdx)3(R88)………….(4)$={\displaystyle \frac{\pi \rho }{64{\mu }^3}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^3\left({\displaystyle \frac{R^8}{8}}\right)\dots \dots \dots \dots .(4)$

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Kinetic energy of floor based on average velocity

=12×mass×u¯¯¯2$={\displaystyle \frac{1}{2}}\times \text{mass}\times {\overline{u}}^2$

=12×ρAu¯¯¯×u¯¯¯2$={\displaystyle \frac{1}{2}}\times \rho A\overline{u}\times {\overline{u}}^2$

=12×ρ×A×u¯¯¯3$={\displaystyle \frac{1}{2}}\times \rho \times A\times {\overline{u}}^3$

Substituting the value of A=πR2$A=\pi R^2$

And u¯¯¯=18μ(−dPdx)R2$\overline{u}={\displaystyle \frac{1}{8\mu }}\left(-{\displaystyle \frac{dP}{dx}}\right)R^2$

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Kinetic energy of the flow per second

=12×ρ×πR2×[18μ(−dPdx)R2]3$={\displaystyle \frac{1}{2}}\times \rho \times \pi R^2\times {\left[{\displaystyle \frac{1}{8\mu }}\left(-{\displaystyle \frac{dP}{dx}}\right)R^2\right]}^3$

=12×ρ×πR2×164×8μ3(−dPdx)3R6$={\displaystyle \frac{1}{2}}\times \rho \times \pi R^2\times {\displaystyle \frac{1}{64\times 8{\mu }^3}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^3{R}^6$

=ρ×π128×8μ3(−dPdx)3R8……………(5)$={\displaystyle \frac{\rho \times \pi }{128\times 8{\mu }^3}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^3{R}^8\dots \dots \dots \dots \dots (5)$

∴a=equation(4)equation(5)$\therefore a={\displaystyle \frac{equation(4)}{equation(5)}}$

a=πρ64μ3(−dPdx)3(R88)ρ×π128×8μ3(−dPdx)3R8$a={\displaystyle \frac{{\displaystyle \frac{\pi \rho }{64{\mu }^3}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^3\left({\displaystyle \frac{R^8}{8}}\right)}{{\displaystyle \frac{\rho \times \pi }{128\times 8{\mu }^3}}{\left(-{\displaystyle \frac{dP}{dx}}\right)}^3{R}^8}}$

=128×864×8$={\displaystyle \frac{128\times 8}{64\times 8}}$

a=2.0