ME3332 HOMEWORK 2 1. The diagram shows water flowing by KathleenKress

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									                                   ME3332 HOMEWORK 2




1. The diagram shows water flowing in a round pipe. In the portion of the pipe that is situated just
   downstream of the inlet, the flow tends to undergo some complex processes which die away with
   increasing distance from the inlet. When these processes have disappeared, a special type of flow
   regime is encountered. It is the regime of fully developed flow. Fully developed flow can be
   recognized by means of two characteristics: (a) the pressure decreases linearly so that dp/dx is a
   constant and (b) the velocity profiles at all locations are identical. These features are shown in the
   diagram.
   For fully developed flow, it is a standard design approach to predict the pressure drop by making
   use of a dimensionless pressure gradient called the friction factor, which is defined as:




    The friction factor depends on another dimensionless quantity called the Reynolds number. By
    any standard, the Reynolds number is the most commonly used parameter in all of fluid
    mechanics and heat transfer. It is defined as:


    In these equations, , , and (=            ) are fluid properties,   is the mean velocity,     is the
    diameter of the pipe, and          is the axial pressure gradient. When flow in the pipe is laminar,
    the equation that connects the friction factor and the Reynolds number is         .

    Problem statement: Air flows in a round pipe at a mean velocity of                . The temperature
    of the air is     , and the pipe diameter is          .
         (a) Find the Reynolds number of the flow. (air properties are in Table 2.1)
         (b) What is the magnitude of the friction factor?
         (c) What is the pressure drop for a     length of pipe in the fully developed region?

Solution:

(a) From Table 2.1,                    and                              , so
(b) For laminar flows,


(c) From the definition of friction factor,




     Since   is constant in the fully developed region,




2. A uniquely shaped U-tube manometer is used to measure the
   pressure drop between locations 1 and 2 for air flowing in a
   round pipe. The inner diameter of leg 1 is          while that of
   leg 2 is          . To test the accuracy of this manometer, first
   we have measured the pressure drop using a “normal” U-tube
   which has inner diameters of both legs equal to         , and we
   have found that the pressure drop is            .
       (a) What was the difference between the water levels in
            two legs for the “normal” U-tube? (the experiment was
            conducted at        )
       (b) What is the expected height difference between two
            legs for our newly designed U-tube.

Solution:

Before answering the problem, let’s look at two important force balances.

Force Balance 1: Consider a cylinder of fluid of
small diameter whose axis is horizontal,as shown
in the diagram to the right. As the cross section
of the cylinder is very small (area       ) we can
assume the pressure to be uniform on the two end faces, namely and , as shown in figure. As the
fluid is stationary, this cylindrical mass of fluid is also stationary, so the forces acting on it should
cancel out. If we focus only on the horizontal forces, the following force balance can be written.



Or
This result shows that the pressures at the outboard ends of an stationary cylinder of fluid with its axis
horizontal must be the same.

Force Balance 2: Consider the fluid mass shown in this figure, a
vertical cylindrical mass of fluid. The cross-sectional area at the top and
bottom faces is , and the pressures on those faces are uniform
(because those faces are horizontal) and are equal to               and
respectfully, as shown in figure. As the fluid is stationary, this
cylindrical chunk of fluid is also stationary, so the forces acting on it
should cancel out. If we focus only on the vertical forces, the following
force balance can be written.



where     is the weight of fluid inside the cylinder. So,



where is the volume of the cylinder which is equal to                                                    .
Rewriting the equation yields,



Notice that the cross-sectional area is cancelled out from the final equation, and the pressure
difference between two horizontal planes only depends on their vertical distance from each other.

Now we can go back to answer the question, but let’s solve the
problem for a very general case where the two legs can have any
arbitrary cross section (one example is shown in this figure). Using
the two force-balances mentioned above, we can write the following
pressure equations.

                    from 2nd force balance

                    from 2nd force balance

                     from 1st force balance

Or by substituting the first two equations in the third one:



Since this result is true for any manometer, we can be sure that for a given pressure drop the height
difference in the two legs will be the same regardless of the legs’ cross sectional areas. So,

(a) and (b) From Table 2.1, at            ,                     , therefore,
            As we said this result is correct for both the “normal” and “special” manometers.

3. A fluid flows from left to right
   through the annular space defined
   by               . Imagine that the
   velocity is measured at 10
   locations across the width of the
   annulus. One of these locations is
   at        , and another is at       . At both of these locations, the velocity is zero. At the other
   eight locations, numerical values of the velocity are available. The radial spacing between the
   points of velocity measurement are equal all across the width of the annulus.

    Write a formula for the mean velocity in terms of the known velocities at the ten locations and
    in terms of the geometry used to set up the calculation.

    Solution:

    I will try to solve this problem using four different methods, and then to see how they compare in
    a case for which we know the exact mean velocity. Before that, let’s see what the mean velocity
    is and devise a strategy to solve our problem.

    By definition, the mean velocity is a velocity which, when multiplied by the density and the
    cross-sectional area gives the same mass flow rate as the real non-uniform velocity profile. In
    mathematical terms, the mean velocity is the area-weighted average (if density is constant) of the
    velocity profile, or:




    so,




    As we do not have the algebraic equation for the velocity profile and only know the velocity
    magnitude at certain given locations, we cannot integrate the equation given above. Therefore, we
    have to approximate the integral in the numerator by a summation over some finite areas.
    (Integrating is summing over infinite number of infinitesimal areas, and unlike finite areas, the
    velocity is constant at infinitesimal areas because of their being very small. Thus, after
substitution of infinitesimal areas by finite areas and replacing integration with summation, some
error is introduced into our calculation.)




where is the number of small finite areas. Within each such element, the velocity is assumed to
be uniform.

The following four methods use the above-mentioned summation to calculate the mean velocity,
and they only differ in the way we choose the finite areas ( ’s) for which we are going to apply
the summation and/or the magnitude of the mean velocity in those areas.

Method I:

In this method we divide the cross sectional area as it has been
shown in figure. This is very similar to what we saw in class
where we applied this method to flow inside a circular pipe. The
measurement points in this method lie halfway between the two
lines that define each    . Exceptions to this designation of the
velocity must be made for the first and last points at which
velocity is measured. The first point lies on the inner surface of
its element, and the last point lies on the outer boundary of its
element.

For each of the rings:



For the first ring:




For the second to the ninth rings:




And finally for the tenth ring:
Also,



As               , so




Method II:

This method, and the next two methods, uses another choice of the
finite areas. The areas of the elements used in these three methods
are defined by circles passing through the measurement points.
The only difference between the methods is the choice of constant
velocity over these areas. The first method among these three
methods uses the velocity at the inner boundary of every finite
area as the constant velocity in that area.

For this method and the next two methods,




So,




Method III:

In this method, we use the velocity at the outer edge of each element, so that.




Method IV:

For method IV, we will use the average of the velocities at the inner and the outer boundaries of
each element. The resulting equation for follows as,
   Comparison:

   In order to compare the three methods introduced above, we are going to apply them to a situation
   where we know the exact mean velocity. We have measured the local velocities at the 10 points
   mentioned above for a laminar flow in an annular space (outer pipe’s inner radius being      and
   inner pipe’s outer radius being        ). The true mean velocity for this flow is          . The
   following table lists the measured velocities:

      Measurement point      Velocity (      )   Measurement point           Velocity (       )
                1                    0                       6                     1.455
                2                  0.654                      7                    1.279
                3                  1.107                      8                    0.974
                4                  1.380                      9                    0.546
                5                  1.492                     10                      0


   Now if we apply the four methods, we will get the following results:

               Method    Calculated mean velocity (      )        Relative error (%)
                  I                       0.98                              2
                  II                      1.02                              2
                 III                      0.94                              6
                 IV                       0.98                              2


4. A dam creates a lake that is
   upstream of it. The diagram
   shows the situation in which the
   lake water height is below the
   top of the dam, while a
   receiving body of water below
   the dam is at a substantially
   lower level. The respective
   levels are at   and . The water is at a temperature that is approximately              .
   Find the net horizontal force on the dam.

   Solution:

   Again we are going to try two different methods in solving this problem. Before we present the
   solution, it is worth noting that the first method is more general and in addition to the horizontal
   force, it can be used to calculate the vertical force too, but the second method can be used for
   calculating the horizontal force only.

   Method I:
In this method, we are going to sum all the horizontal
forces on the dam. For the purposes of this problem, the
only force acting on the dam is the pressure forces. But, as
we know, pressure acts normal to the surface, so for the
curved face of the dam we need to calculate the horizontal
component of the pressure force exerted on the dam. If we
enlarge a portion of dam and look at the forces acting on a
very small part of dam lying at a depth equal to , we will
see a picture very similar to one shown here. From this
picture:



where    is the width of the dam. Also, from the figure,                 , so that



Note that this result can be applied to any surface, no matter if the surface is curved or straight (it
only has to be an extrusion of a two-dimensional curve). Therefore, if we had neglected the
curvature of the surface and considered it as a vertical line, we would have arrived at the same
result.

The force exerted by the upstream lake and ambient air is:




and that exerted by the downstream lake and the ambient air is:




Therefore the net horizontal force can be calculated as:
Note that the ambient pressure has cancelled out, and we could have got to the same result if we
had neglected the air pressure (or in other words, if we had used the gauge pressure instead of
absolute pressure)

Performing the integration yields,




Method II:

In this method instead of summing the forces
acting on the dam, we will consider the illustrated
control volume which encompasses the dam and
cuts its base, so that the external horizontal force
acting on the assumed control volume is equal to
the net horizontal force acting on the dam.

Using this method we do not need to project the
pressure force on the horizontal plane, as the
faces of the control volume are vertical.

The force balance is exactly the same as above:

								
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