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fluidmech

# A dr 2 x r where a is a constant of integration

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Unformatted text preview: plifying and ignoring the product of two small quantities we have the following result. δp τ dτ du τ =µ for Newtonian fluids. =+ δx r dr dy du If y is measured from the inside of the pipe then r = -y and dy = - dr so τ = − µ dr 2 δp µ du du =− −µ 2 δx r dr dr 2 1 du d u 1 δp + 2 =− µ δx r dr dr du rd 2 u r δp + =− 2 µ δx dr dr ⎛ du ⎞ d⎜ r ⎟ du rd 2 u dr ⎠ Using partial differentiation to differentiate ⎝ yields the result + dr dr dr 2 ⎛ du ⎞ d⎜ r ⎟ r δp dr ⎠ hence ⎝ =− µ δx dr r 2 δp du Integrating we get r =− +A dr 2 µ δx du r δp A =− + ...........(A) dr 2 µ δx r where A is a constant of integration. © D.J.DUNN 18 Integrating again we get r 2 δp + A ln r + B................( B) 4 µ δx where B is another constant of integration. Equations (A) and (B) may be used to derive Poiseuille's equation or it may be used to solve flow through an annular passage. u=− 2.10.1 PIPE At the middle r=0 so from equation (A) it follows that A=0 At the wall, u=0 and r=R. Putting this into equation B yields R 2 δp 0=− + A ln R + B where A = 0 4 µ δx B= R 2 δp 4 µ δx u=− r 2 δp R 2 δp 1 δp 2 R − r 2 and this isPoiseuille' s equation again. + = 4 µ δx 4 µ δx 4 µ δx { } 2.10.2 ANNULUS Fig.2.11 r 2 δp + A ln r + B 4 µ δx The boundary conditions are u = 0 at r = R i and r = R o . u=− Ro2 δp + A ln Ro + B...........................(C ) 0=− 4 µ δx Ri2 δp + A ln Ri + B .........................(D) 0=− 4 µ δx subtract D from C 1 δp − Ro2 + Ri2 + A{ln Ro − ln Ri } 0= 4 µ δx { 0= } ⎧R ⎫ 1 δp 2 Ri − R02 + A ln ⎨ o ⎬ 4 µ δx ⎩ Ri ⎭ { } © D.J.DUNN 19 A= { 1 δp Ro2 − Ri2 4 µ δx ⎧R ⎫ ln ⎨ o ⎬ ⎩ Ri ⎭ } This may be substituted back into equation D. The same result will be obtained from C. { } Ri2 δp 1 δp Ro2 − Ri2 + 0=− ln Ri + B 4 µ δx 4 µ δ x ⎧ Ro ⎫ ln ⎨ ⎬ ⎩ Ri ⎭ ⎡ ⎧ ⎢ ⎪ 1 δp ⎢ 2 ⎪ Ro2 − Ri2 B= Ri − ⎨ 4 µ δx ⎢ ⎪ ln ⎧ Ro ⎫ ⎢ ⎪ ⎨ Ri ⎬ ⎢ ⎩⎩⎭ ⎣ { ⎤ ⎫ ⎥ ⎪ ⎪ ⎥ ⎬ ln Ri ⎥ ⎪ ⎥ ⎪ ⎥ ⎭ ⎦ } This is put into equation B ⎡ ⎧ ⎢ ⎪ - r 2 δp 1 δp R − R 1 δp ⎢ 2 ⎪ Ro2 − Ri2 + ln r + Ri − ⎨ u= 4 µ δx 4 µ δx 4 µ δx ⎢ ⎧ Ro ⎫ ⎪ ln ⎧ Ro ⎫ ⎢ ln ⎨ ⎬ ⎪ ⎨ Ri ⎬ ⎢ ⎩ Ri ⎭ ⎩⎩⎭ ⎣ ⎤ ⎡ ⎥ ⎢ 2 2 2 2 Ro − Ri 1 δp ⎢ 2 Ro − Ri 2 −r + u= ln r + Ri − ln Ri ⎥ ⎥ ⎢ 4 µ δx ⎧R ⎫ ⎧R ⎫ ⎥ ⎢ ln ⎨ o ⎬ ln ⎨ o ⎬ ⎥ ⎢ ⎩ Ri ⎭ ⎩ Ri ⎭ ⎦ ⎣ { { 2 o 2 i } } { { ⎤ ⎫ ⎥ ⎪ ⎪ ⎥ ⎬ ln Ri ⎥ ⎪ ⎥ ⎪ ⎥ ⎭ } } ⎤ ⎡ ⎥ ⎢2 r 1 δp ⎢ Ro − Ri2 2 2⎥ u= ln + Ri − r ⎥ Ri 4 µ δ x ⎢ ⎧ Ro ⎫ ⎥ ⎢ ln ⎨ ⎬ ⎥ ⎢ ⎩ Ri ⎭ ⎦ ⎣ { } For given values the velocity distribution is similar to this. Fig. 2.12 © D.J.DUNN 20 ASSIGNMENT 2 1. Oil flows in a pipe 80 mm bore diameter with a mean velocity of 0.4 m/s. The density is 890 kg/m3 and the viscosity is 0.075...
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