Bengham Fluid Rate Equation
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Transcript of Bengham Fluid Rate Equation
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Derivation offlow rate equation for
Bingham-fluids
Bingham-fluids are characterized by a yield stress. Contrary to Newtonian fluids they cantransmit a shear stress also without a velocity gradient. But in order to make the Bingham-fluid
flowing, the driving shear stress has to be larger than the yield stress. Below this yield stress thefluid will behave almost like a solid body and above as a liquid. Examples of Bingham-fluids are
tooth paste or paint.
Before determining the velocity profile, the following consideration has to be made:
The velocity profile of channel flows of Newtonian fluids features a velocity gradient whichdecreases towards the centre of the channel. Hence the shear stress transmitted by a fluid layer
also decreases toward the channel centre. For the reason that Bingham-fluids become solid when
the applied shear stress falls below the yield stress it will be clear that the Bingham-fluid willbecome solid in the centre layers of the channel. There a solid 'plug' will be moving within the
flow. In the process of deriving the velocity profile also the height of this solid area has to be
determined.
The following figures show the coordinates used and give an idea of the developing velocityprofile:
Fig.1: Used coordinate system
Fig 2: fully developed velocity profile
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Forces acting on a infinite small fluid element:
Fig. 3: Forces acting on the fluid element in x-direction
In putting up the forces acting on the fluid element the follwing conditions have already been
considered:
mass flow is constant along channal length (coservation of mass) (=> no acceleration forces inx-direction)
stationary case
From necessity for the forces to be balanced it results fromFig. 3:
(1)The shear stess for Bingham-fluids is:
(2)
Setting eq.(2)equal to eq.(1)leads to:
(3)
Equation(3)can be rearranged:
Using the usual assumption:Integrating twice results in:
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(4)
(5)
In order to determine the constants C1 and C2 knowledge of the boundary condition is necessary:
At y=0: No slip condition, e.g. u(y=0)=0.
Thus follows with eq.(5):
The friction force at the walls has to be balanced by the driving pressure force.
Thereby the friction force is given by the product of the wall surface area with shear
stress at the wall. The pressure force results from multiplying the crossectional area of thechannel with the pressure difference between channel entrance an channel end. The
pressure difference is here also labeled with p..
The wall surface area is: W=2bl.
The crossectional area is: S=bh.
Leading to:
Rearranging results in: (6)Because the pressure loss is constant and because the pressure decreases with increasing x it
follows: (7)
Setting eq.(7)equal to eq.(6)yields: (8)Introducing this relation into eq.(4)gives the constant C1:
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Using the constants C1 andC2 as well as eq.(5)results in u(y):
(9)
and:
Determination of y-grenz
After deriving the velocity profile the magnitude of y-grenz has to be determined.
Above y-grenz the Bingham-fluid is like a solid body, resulting in:
Introducing this condition into eq.(4)using C1 leads to:
From this equation, using the pressure loss(eq.7), y-grenz can be determined:
(10)
Derivation of u-max
For the maximum velocity u-max it is valid:
This means simply eq.(10)has to be introduced into eq.(9):
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Rearranging results in the expression for u-max:
(11)
Volume flow rate
The total volume flow rate within the channel can be written as the sum of two partial flow rates.One part comes the moving plug in the centre of the channel and the other part results from the
areas h y (h - y-grenz ) and y-grenzy where the velocity profile eq.(9)has to be applied.In this range the Bingham-fluid moves as a liquide.
Using the symmetry between upper and lower half channel the total volume flow rate can bewritten as:
(12)The solid part is to determined easily from eq.(12)because u-max and y-grenz are already known. Itis:
(13)
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For the liquid part in eq.(12)the following integral has to be solved (y-grenz is known):
(14)
Summing of the solid part eq.(13) and liquid part eq.(14)yields the total volume flow rate:
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