University Of Karachi

BINGHAM FLOW OVER A CIRCULAR TUBE

Syed Saad AhmedB-0933040B.E 3rd Year

Chemical Engineering

Submitted To:Sir Ahsan Ghani

BINGHAM FLOW OVER A CIRCULAR TUBE 2012

BINGHAM FLOW IN A CIRCULAR TUBE

A fluid that is nearly described by the Bingham model is flowing through a vertical tube as the result of a pressure gradient and/or gravitational acceleration. The radius and length of a tube are R and L, respectively. It is desired to obtain a relation between the volume of the flow Q and the combined pressure and gravity forces acting on the fluid.

Solution:

Consider a steady leminar flow of a fluid of constant density ρin a very long tube of length L and radius R. we select our system a cylindrical shell of thickness Δ rand the length L, then various combination to the momentum balance in the z direction are:

Rate of momentum in across cylindrical surface at ‘r’: 2πrL ( τ rz )|¿rRate of momentum out across cylindrical surface at ‘r+ Δr’:

2πrL ( τ rz )| ¿r+Δ r

Rate of momentum in across annular surface ‘z=0’:

2πr Δ r νz (ρν z )| ¿z=0

Rate of momentum out across annular surface ‘z=L’:

2πr Δ r νz (ρν z )| ¿z=L

Gravity force acting on cylinder: (2πr ΔrL ) ρg

Pressure force acting on annular surface at ‘z=0’: (2πr Δ r )Po

Pressure force acting on annular surface at ‘z=L’: −(2πr Δ r )PL

Putting all the values in the momentum balance equation,

rate of momentum∈−rate of momentumout+∑ of forces actingon system=0

{2πrL (τ rz )|¿r+2 πr Δ x νz (ρν z )| ¿z=0 }−{2 πrL (τ rz )| ¿

r+Δr+2πr Δ x ν z ( ρν z )| ¿

z=L }+ {(2πr ΔrL ) ρg+ (2πr Δr ) Po− (2πr Δr ) PL }=0

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BINGHAM FLOW OVER A CIRCULAR TUBE 2012

2πrL ( τ rz )|¿r−2 πrL ( τ rz )| ¿r+Δr

+(2πrΔr ) (ρgL+Po−PL)=0

2πrL ( τ rz )| ¿r+Δr

−2πrL ( τ rz )|¿r=(2 πrΔr ) (ρgL+Po−PL )

Dividing both sides by (2πLΔr):

r {( τ rz )| ¿r+Δr

−(τ rz )|¿r }Δr

=r {(Po−PLL )+ ρgLL }Taking limit Δr approaches to zero:

limΔr⟶0

{r {( τ rz )| ¿r+Δr

−( τ rz )|¿r }Δr }=r {(Po−PLL )+ ρgLL }dd

(r τ rz)=(℘o−℘L

L )rWhere, represents the combined effect of static pressure and gravitational force,℘ and ℘=P+ρgL

Upon integration:

∫ d (r τ rz )=(℘o−℘L

L )∫ rdr r τ rz=℘o−℘L

L×r2

2+C1

τ rz=r2 (℘o−℘L

L )+C1 …. equation (i)

Taking boundary condition 1:

At r=0, τrZ=0,

C1=0

Putting in equation (i),

τ rz=r2 (℘o−℘L

L )

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BINGHAM FLOW OVER A CIRCULAR TUBE 2012

For the non Newtonian fluid or bingham flow the velocity gradient is zero when the momentum flux is less than the value τ o. Hence once expect a plug flow region In the central part of the tube, outside the plug flow region the momentum flux ad the velocity gradient are related accordingly,

τ o−μod V Z

dr=(℘o−℘L

2 L )rdV Z

dr=τoμo

−(℘o−℘L

2μoL )r

d V Z=τoμodr−(℘o−℘L

2 μoL )rdrIntegraing,

∫ dV Z=∫τoμodr−(℘o−℘L

2μoL )∫r dr

d V Z=τoμor−(℘o−℘L

4 μoL )r2+C2 …equation (ii)

Taking boundary condition 2:

At Vz=0, r=R,

C2=(℘o−℘L

4 μoL )R2− τoμoR

Putting in equation (ii),

d V Z=τoμor−(℘o−℘L

4 μoL )r2+(℘o−℘L

4 μoL )R2− τoμoR

d V Z=(℘o−℘L

4 μoL )R2−(℘o−℘L

4 μoL )r2− τoμoR+

τoμor

After integration,

V Z≥(℘o−℘L

4 μoL )R2[1−( rR )2]− τo

μoR [1− r

R ] r>ro

Hence ro is the radius of the plug flow region, defined by,

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BINGHAM FLOW OVER A CIRCULAR TUBE 2012

τ o=(℘o−℘L)2L

ro

Putting In above,

V Z≤(℘o−℘L

4 μoL )R2[1−( roR )2]− (℘o−℘L )

2 LμoroR [1− roR ]

V Z≤(℘o−℘L

4 μoL )R2[(1−( roR )2)−2 roR (1− roR )]

V Z≤(℘o−℘L

4 μoL )R2[1−( roR )2

−2 roR

+2( roR )2]

V Z≤(℘o−℘L

4 μoL )R2[1−2 roR +( roR )2]

V Z≤(℘o−℘L

4 μoL )R2[1− roR ]2

The volume rate of the flow can be calculated from,

Q=∫0

2 π

∫0

R

Vz r dr dθ

Q=2π∫0

R

Vzr dr

Integrating by parts,

∫uv dx=u∫ vdx−¿∫ dudx.∫ v dx dx¿

Q=Vz<∫r dr−∫ dVz> ¿dr.∫ r dr dr¿

Q=2π [Vz∫0

R

r dx−∫0

RdVzdr.∫ r dr dr ]

Q=2π [ [Vz r22 dx]R0−∫0RdVzdr.∫r dr dr ]

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BINGHAM FLOW OVER A CIRCULAR TUBE 2012

Q=2π [ [Vz r22 dx]R0−∫0RdVzdr.r2

2dr ]

Q=2π [ 12 [Vzr2dx ]R0−12∫0

RdVzdr

r2dr ]The quantity Vzr 2 is zero (0) at both limit and the integral the lower limit may be

replaced by ro because dVzdr

=0 for r<ro

Q=2π [−12 ∫0

RdVzdr

r2dr ]Q=π∫

0

R

−dVz> ¿drr2dr ¿

Since,

−dV Z

dr=(℘o−℘L

2μoL )r− τoμo

Putting in above,

Q=π∫0

R [(℘o−℘L

2 μoL )r− τoμo ]r2dr

Q=π∫0

R

(℘o−℘L

2 μoL )r3− τoμor2dr

Integrating,

Q=π (℘o−℘L

2 μoL )[ r 44 ] Rro− τoμo [ r33 ] Rro

Q=π (℘o−℘L

8 μoL ) [R4−ro4 ]− τ o3 μo

[R3−ro3 ]

Q=π (℘o−℘L

8 μoL )R4−(℘o−℘L

8μoL )r4− τo R3

3 μo+τ oro

3

3 μo

Putting ro=

2 Lτo℘o−℘L

,

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BINGHAM FLOW OVER A CIRCULAR TUBE 2012

Q=π [(℘o−℘L

8 μoL )R4−(℘o−℘L

8 μoL ) 16 L4 τo

4

(℘o−℘L)4,−τoR

3

3 μo+τo3

3μo ( 8 L3 τ o3

(℘o−℘L )3 )]Q=π [(℘o−℘L

8 μoL )R4− 2 L3τ o4

μo (℘o−℘L )3,−τoR

3

3μo+

8 L3 τo4

3μo (℘o−℘L )3 ]Q=π [(℘o−℘L

8 μoL )R4− τoR3

3 μo−

2L3 τ o4

μo (℘o−℘L )3,+

8 L3 τo4

3μo (℘o−℘L )3 ]Q=π [(℘o−℘L

8 μoL )R4− τoR3

3 μo−

2L3 τ o4

μo (℘o−℘L )3,+

8 L3 τo4

3μo (℘o−℘L )3 ]Q=π [(℘o−℘L

8 μoL )R4− τoR3

3 μo+

2 L3 τo4

3μo (℘o−℘L)3 ]Q=π (℘o−℘L

8 μoL )R4[1− 8 τ oL

3 (℘o−℘L) R+

16L4 τo4

3 (℘o−℘L )4 R4 ]Since the τ ris the rate of momentum flux at the wall,

τ r=(℘o−℘L) R

2 L

Putting in above,

Q=π (℘o−℘L

8 μoL )R4[1−43 ( τoτ r )+ 13 ( τo4

τ r4 )]

Q=π (℘o−℘L

8 μoL )R4[1−43 ( τoτ r )+ 13 ( τoτ r )4]

This equation is known as Bingham Reiner Equation, when τ ois zero than this model reduces to the Newtonian model.

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