Homework
#10 GEOL 4880 Humphrey Fall 2015
These are not due until AFTER
thanksgiving break
1)
I talked in class about energy in river flow,
and showed that the Froude number can be interpreted as the ratio of kinetic
energy to potential energy in the flow. Many questions in river flow can be addressed
by examining the total energy of the flow.
Energy, per unit width and unit length, (in other words a square meter
column of water) of the flow can be expressed as a height of water (similar to
the concept of hydraulic head in Darcy’s groundwater flow). To be precise the energy in a column is equal
to the potential energy above the bed (rgh)
plus the kinetic energy (rv2/2),
‘h’ is the depth, and ‘v’ is the velocity of the water. Dividing this by density*gravity turns this
into the energy head of the flow: E = h + v2/(2g)
In
most river flows, by far the largest part of this energy head is just the depth
of flow, with the kinetic energy only a few extra centimeters of head. Because of this we often ignore the kinetic
energy.
To
get a sense for the amount of energy in river flow, we calculate several
energies (expressed as water head). Use the Laramie river in flood (use these
values for all questions on the Laramie in this homework), with a depth of 1m,
flow velocity of 1m/s, slope of 2x10-4. Calculate:
a.
the potential energy per width, per meter
length of the Laramie river (as a head, relative to local elevation).
b.
the
kinetic energy of the flow (as a head)
c.
the
potential energy lost by a column of water per meter of flow down river (as a
head change)
d.
the
Froude number for the Laramie river
e.
what is the super elevation expected on a
bend in the Laramie river (assume the bend has a radius of 20m and the width is
about 4 meters).
2)
a. Estimate the Hydraulic Radius of the Laramie
River at flood (1m deep and 4m wide, assume a rectangular channel shape)
b. Search the web for an appropriate value of Mannings ‘n’ for the Laramie river. Calculate the Manning velocity of the Laramie river
in flood
c. We talked
about the logarithmic velocity profile that develops above a rough bed. For rivers such as the
d.
High
Froude number flow (super critical or shooting flow) is relatively rare and
usually only found in steep bedrock rivers.
Calculate the velocity that would be needed in the Laramie river to achieve super critical flow.
e.
(hard,
mini puzzle) Assuming the discharge and the roughness stay the same, how steep
would the Laramie river have to be to reach a Fr of 1?
(use Mannings)
3)
While
we are talking about energy: energy is expended by river water to move its load
of sediment. This sort of calculation is
very difficult to do correctly, but we can approximate the energy by saying
that the water flow has to counteract the settling velocity. We will try this for the Laramie River (flow
parameters in previous question). We
will make several assumptions: the only sediment in transport is sand (0.2mm),
and the amount in transport (mass qsed) is
0.1kilograms/(m3*s). Remember, energy is Force * distance. Force is easy; it is the weight of the
sediment, distance is less obvious, but think of the settling velocity of the
sand.
Compare
the energy to move the sediment with the energy of the flow from ‘1c’
above. (You will need to calculate the
energies in the same units!)
4)
In
class I quickly sketched the energy argument for the difference in behavior of
super-critical and sub-critical flow.
The development in class was fairly fast, but I want you to understand
this important concept. I would like you
to follow the logic outlined here and find solutions to illustrate this
behavior. Consider a rectangular channel
of constant width and steady discharge, which has a small bump of height ‘h’ which the flow must flow over. Assume the slope of the flow is so low that
we can ignore it and set the elevation of the bed of the incoming flow to be 0,
while the bed of the flow over the bump is just h
The
incoming depth is d1 and
the incoming velocity is v1,
since the width is constant, we just look at a 1m wide section of this
flow. The flow over the bump will have
depth d2 and velocity v2, and the water surface
height over the bump is d2 +
h. The mass of water per time
incoming is r*v1*d1 and the mass over the bump is the same
(or equivalently r*v2*d2).
The
kinetic energy of the water is the mass times the square of the velocity
divided by 2, while the potential energy is the mass times gravity times the
average height (d/2). We assume no
energy is lost between points 1 and 2, and so we can write the energy balance
for K.E. and P.E. at the two places as [equation
A]:
r*v13*d1/2
+ r*g*v1*d1
*d1/2 = r*v23*d2/2
+ r*g*v2*d2
*(d2/2 + h) , where we have been careful to include the
extra height of the bump (h) in the
potential energy at 2. This simplifies
considerably since r*v1*d1 is equal to r*v2*d2
. Further simplification occurs if we write v2 = v1*d1/d2 and multiply the whole equation by d22, to get [equation B]:
0
= [g*v1*d1] *d23
+ [2*g*d1*v1*h
- v13*d1 - g*v1*d12] *d22
+ [v13*d13 ], where all terms in ‘[ ]’ brackets are
known if we know v1*d1
. This is a cubic equation in d2 .
a. Write equation A, and produce equation B by the indicated steps.
b. To see what the Laramie river might do in encountering a step in the
bed, try putting
an input velocity of 1m/s, a depth of 1m, and a bump height of 0.1m. How much does the water surface drop over the
bump? (don’t
forget to add the height of the bump, and to help you, try a d2 of
0.6m to start, improve on that if you can)
c. Finally, just to see the opposite behavior for high Fr
number flow, try this with a flow velocity of 4m/s, depth of 1m and bump of
0.1m. (try d2
of 1.2m and see if you can improve).
Just so you can see the difference, calc the
Fr for these two cases (b and c).