FULLY DEVELOPED LAMINAR FLOW IN A PIPE Let us consider fully developed laminar flow in a pipe. Here the flow is axisymmetric. Consequently, it is convenient to work in cylindrical coordinates . The control volume will be chosen as a differential annulus.
Find:
a) Velocity distribution b) Shear stress distribution c) Volume flow rate d) Average velocity e) Point of maximum velocity
Assumptions:
1. Fully developed flow (du/dx=0 ) 2. Steady flow 3. Laminar flow 4. Incompressible flow 5. There is no property change in - direction. 6. Radial velocity component is zero. 6. Neglect body forces
MM302 4
1
Boundary conditions
at r = 0 the velocity must be finite (from physical consideration) at r = R u = 0 (no slip condition) a) Velocity profile: Velocity distribution can be found by using the integral or differential form of the momentum equation. We will find the velocity distribution by using both methods. Method I:Application of integral momentum equation The control volume will be chosen as a differential annulus.
If we apply the x - component of momentum equation for the control volume shown in the figure. F S x F B x u d u V d A t CV CS
0
0
0
For fully developed flow, the net momentum flux through the control surface is zero.
F S x 0
The normal (pressure) force and the tangential (shear) forces act to the control volume. The surface forces acting on the differential fluid element in x -direction are
MM302 4
2
MM302 4
3
II. Method: By using the differential form of momentum equation in x direction. 1 u z 1 2u z 2u z u z u z u u z u z p ur u z 2 r 2 2 t r r z z r r r z r 0 0 0 0 0 0
Note: By replacing x z and uz u, and simplifying the above differential equation
or
1 u p r 0 x r r r
u r p r r r x
By integrating twice, r 2 p C 1 u ln r C 2 4 x
This equation is the same as the equation found by using integral momentum equation. b) Shear Stress Distribution:
c) Volume Flow Rate:
MM302 4
4
d) Average Velocity:
e) Point of Maximum Velocity:
FULLY DEVELOPED TURBULENT FLOW In turbulent flows, there is no universally acceptable relation between shear stress and velocity gradients. Therefore, the analytical solutions of turbulent flow problems are impossible, we must rely on semi-empirical data and numerical solutions.
MM302 4
5
INCOMPRESSIBLE INVISCID FLOW MOMENTUM EQUATION FOR FRICTIONLESS FLOW: EULER’S
EQUATIONS The equations of motion for frictionless flow are called Euler’s equations. These equations can be obtained from Navier-Stokes equations (by setting = 0 ).
u p u u u u v w x x y z t v p v v v u v w g y y x y z t g x
w p w w w u v w g z z x y z t We can also write the above equations as a single vector equation V V V V u v w g p x y z t
or
DV g p Dt
If the z coordinate is directed vertically upward, then since, z k , ˆ
g g k g z
ˆ
MM302 4
6
In cylindrical coordinates, Euler equations in the component form, with gravity is the only body force, are V r V r V V r V r V 2 g r ar V r V z r t r r z r 1 p
V V V V V V V V r V z r t r z r r r V z V z V V z V z 1 p g z a z V r V z t r r z z g
1 p
a
If the z-axis is directed vertically upward, then gr = g = 0 and gz = -g .
MM302 4
7
EULER’S EQUATION IN STREAMLINE COORDINATES
In this section, the Euler’s equation will be first derived in the streamline coordinates, and then integrated along a streamline. For this reason, consider an infinitesimal fluid element, which is moving along an instantaneous streamline, as shown in the figure. For simplicity, must be tangent to consider the flow in yz plane. Since velocity vector the streamline, the velocity field is given by V V ( s, t ) .
Figure. Fluid particle moving along a streamline.
If we apply Newton’s second law of motion in the streamwise (the s-) direction to the fluid element of volume dsdndx , then neglecting viscous forces we obtain
p ds p ds p dndx p dndx g sin dsdndx a s dsdndx s 2 s 2 where “as” is the acceleration of the fluid particle along the streamline. Simplifying the equation, p g sin a s s since
sin
z , we can write s
1 p
z g a s s s MM302 4
8
Along any streamline V=V(s,t), then the total acceleration in s-direction a s
DV Dt
V V V t s
Then, the Euler’s equation becomes
z V V 1 p g V s s t s
For steady flow, and neglecting body forces, it reduces to 1 p
V V s s which indicates that a decrease in velocity is accompanied by an increase in pressure and conversely.
If we apply Newton’s second law in the n-direction to the fluid element. Neglecting viscous forces, we obtain
p dn p dn p dsdx p dsdx g cos dsdndx an dndsdx n 2 n 2 Simplifying the equation, we get
p g cos an n z Since cos , we can write n
1 p
n
g
z an n
The centripetal acceleration, an, for steady flow can be written an where R is the radius of the curvature of the streamline. Then, equation normal to the streamline is written for steady flow as
V 2 R
Euler’s
1 p
V 2 z g R n n MM302 4
9
For steady flow in a horizontal plane, streamline becomes 2
1 p
n
Euler’s
equation normal to
V
R
It indicates that pressure increases in the direction outward from the center of curvature of the streamlines. Example: An ideal fluid (zero viscosity and constant density) flowing through a planar converging nozzle that lies in a horizontal plane, shown in the figure. Compare the pressures at points 1 and 2, at 3 and 4, and at 5 and 6.
MM302 4
10
BERNOULLI EQUATION: INTEGRATION OF EULER’S EQUATION ALONG A STREAMLINE FOR STEADY FLOW Consider the streamwise Euler equation in a streamline coordinates for steady flow. The equation is V
V 1 p z g 0 s s s
Multiplying by ds we get V z 1 p V ds ds g ds 0 s s s In general, the total differential of any parameter of the flow field (say pressure p) is given by
p p dp ds dn s n because p is function of both s and n. If we restrict ourselves to remain on the same streamline, (d s dxı dy j dz k ) d s dsı s dnın
dn 0 and dp
p ds s
Similar relations hold for other properties. With restriction of staying on the same streamline , Euler equation becomes
VdV Integrating V 2 2
dp
dp
gdz 0
gz C
(a constant)
If the density is constant, we obtain the Bernoulli equation
p
V 2 2
gz constant
It is subject to restrictions : 1. Steady flow 2. Incompressible flow 3. Frictionless flow 4. Flow along a streamline
MM302 4
11
Example: A hole is pierced at the bottom of a large reservoir, which is initially filled with an incompressible fluid of density to a depth of h, as shown in the figure. As a first approximation, fluid may be considered as inviscid, and the reservoir is large enough so that the change in its level may be neglected. Determine the velocity of the fluid leaving the hole , which is pierced at the bottom of the reservoir.
z Datum
MM302 4
12
Example: A hole pierced at the bottom of a large reservoir, which is initially filled with an incompressible fluid of density to a depth of ho. The area of the tank and the hole are At and Ah, respectively. For the quasi-steady flow of the fluid, develop an expression for the height of the fluid, h, at any later time, t .
ho
h
z
MM302 4
13
