u k 5 , (7.91) the effect on the resistance is negligible; in this case
we speak of hydraulically smooth surfaces. If the protrusion height is
considerably greater than the thickness of the buffer region we speak of
a dynamically completely rough surface,
(8.2) Here h is a mean distance between the upper and lower walls,
which in bearings is typically of the order h L . (8.3) For 1 we
ignore the first term in brackets on the right-hand side and for the ratio
of the convective terms to the remaining frictio
can also extract this directly from the thermal equation of state = (p, T )
if we consider that the change in state of a material particle is isobaric if
the change in density as a result of pressure change is ignored. Using
(7.27) the energy equation (4.
viscosity ( = 1/3lv) of dilute gases. In carrying these ideas over to
turbulent exchange motion it is assumed that turbulent fluid parcels,
i.e. fluid masses which move more or less as a whole, behave like
molecules, thus moving over the distance l unaffe
This arises from the flow channel of a simple shearing flow if the upper
wall is inclined to the x-axis at an angle . Since the fluid adheres to the
wall it is pulled into the narrowing gap so that a pressure builds up in
the gap; this pressure is quite s
z = 6 (h U) x + (h W) z . (8.18) If the plates are rigid bodies the
derivatives U/x and W/z on the righthand side vanish. Further the
plate velocity W in the z direction is often zero. 8.2 Statically Loaded
Bearing 8.2.1 Infinitely Long Journal Bearing To
du/dy d2u/dy2 2 du dy 2 . (7.66) Comparing this with the mixing length
formula (7.57), we deduce a formula for the mixing length from
dimensional analysis l = du/dy d2u/dy2 . (7.67) We make no
further use of this here. Using t = w = u2 in (7.66) we obtain
unidirectional flow with a vanishing pressure gradient along a smooth
flat wall. In laminar flow with a vanishing pressure gradient and with the
basic assumptions of unidirectional flow (u1 = f(x2), u2 = u3 = 0) the
NavierStokes equations simplify to 0 =
Reynolds stress and the mean velocity. It is clear that we should find
this relation using some turbulence model. One such model is the
Boussinesq formulation of the Reynolds stresses u v = A u y , (7.56)
where A is the turbulent transport coefficient, or
known as the velocity defect law. By subtracting (7.79) from the law of
the wall, we acquire the expression Umax u = 1 lnu R + B , (7.81)
which shows explicitly how the maximum velocity depends on the
Reynolds number uR/. For given Umax and R, (7.81) is a
return to the law of the wall. Now if we take the limit uR/ for
fixed R it also means that y would no longer appear in the relation. In
order not to lose relevant information in these limits, we form the
entirely equivalent form u u = F u R , y R . (7.74)
stresses arise from molecular exchange of momentum. Molecular
momentum exchange occurs when a molecule at position y with a
velocity u in the x direction moves to position y l under thermal
motion where its velocity is u du. Therefore the molecule carries
turbulent flow was not realized in the experiments, or that the shear
stresses were not constant because of very large pressure gradients
(see (7.41). In the region 30 y u 1000 (7.72) we find reasonably
good agreement for 0.4 and B 5. (The values given in
2 0 p sin R d . (8.32) Since cos is an even function, so too are h()
and all powers of h(). From (8.28), p/ is then also an even
function, and the pressure itself must be an odd function of . The X
component of the force then vanishes. The Y component bal
Prandtls mixing length formula. Since viscosity has no effect in the
region in which we are interested, the fluid properties are only
described by the density . In the relation between the constant shear
stress and the velocity distribution u = f(y), asid
1. Since the magnitude of the velocity, in contrast to the velocity
distribution, is directly dependent on the Reynolds number we require
that the derivatives of the velocity distributions agree in the overlap
region y
1, y/R 1: du dy = u R dF d = u2 df d
= e h (8.23) is the relative eccentricity. Since is very small the fact that
the lubrication gap or the fluid film is curved is not important; let us
consider the fluid film to be unwrapped (Fig. 8.3) and set dx = Rd.
With the notation we have introduced,
obtain the limit curve l = f(Re) (dashed line in Fig. 7.4). 8 Hydrodynamic
Lubrication 8.1 Reynolds Equation of Lubrication Theory The geometric
characteristics of the unidirectional flows discussed in Chap. 6 are their
infinite extension in the flow dire
ME 152 B Homework 3 Solutions
Boundary layers and flow about immersed bodies
Problem 1: Modeling airfoil performance using Xfoil
Part (a): Running Xfoil for an angle of attack of 5 degrees.
There are small differences in the two sets of curves for the pre
which, although the fluctuating motion itself is not zero, the distribution
of the average velocity is mainly influenced by the viscous shear
stresses. Thus the name viscous sublayer is justified. For dimensional
reasons the thickness of this layer must b
the Reynolds stress u v , is constant and therefore independent of y.
We have already identified the constant of integration as the shear
stress w at the wall, since for y = 0 the Reynolds stress vanishes as a
consequence of the no slip condition. Because
dy du dy , (7.58) and the eddy viscosity t = l 2 du dy , (7.59)
which, from Prandtls mixing length model, therefore depends on the
velocity gradient. 7.3 Turbulent Shear Flow Near a Wall 219 At first sight
the unknown eddy viscosity in (7.59) has only bee
= 2(T + T ) xixi . (7.30) 7.3 Turbulent Shear Flow Near a Wall 213
Noting the rules (7.10), averaging leads us to the equation for the mean
temperature c ui T xi = c u i T xi + 2T xixi , (7.31) which,
because of (7.12), can also be expressed in the form c
integration which appears when we solve (7.68) is fixed so that du/dy
tends to infinity when y tends to zero. Now (7.62) does not hold in this
region (as just stated), but taking the limit y 0 in (7.62) corresponds
to a very thin viscous sublayer where du
to p x = 2u y2 . (8.7) Using (8.1) and since v U, the
component of the Navier-Stokes equations in the y direction leads to
the order of magnitude equation 2U2 h + 2U2 h p y + 3 U
h 2 + U h 2 , (8.8) from which we infer the equation 0 = p y . (8.9)
However
find the equation corresponding to (7.84) is U u = 1 ln R k + 8.5
3.75 . (7.96) Using (7.87) we obtain the resistance law of the completely
rough pipe as = 8 2.5 ln R k + 4.752 , (7.97) or using the logarithm to
the base ten as before = 2 lg R k + 1.742
neglected; the simple Eq. (7.37) is here applicable. This does not only
hold for the channel and pipe flows already mentioned, but also for
turbulent boundary layer flows. In all these flows, a layer close to the
wall exists where the outer boundaries of
that p/x is only a function of x since the Reynolds stress v2 by
assumption only depends on y. Since the second term in (7.38) does not
depend on x, it also follows that p/x is a constant. The entire shear
stress, which we now abbreviate to = du dy u v (7
closed form expressions which describe the entire wall region. We shall
not go any further into these, because the resistance laws which we will
discuss shortly do not require the exact distribution of the mean
velocity in the buffer layer. The exact dist