澳洲代写｜CIVL3612｜Fluid Mechanics流体力学 悉尼大学
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课程介绍：
This unit of study aims to provide an understanding of the conservation of mass and momentum in differential forms for viscous fluid flows. It provides the foundation for advanced study of turbulence, flow around immersed bodies, open channel flow, pipe flow and pump design.
| Key Information | Details |
|---|---|
| Course Number | CIVL3612 |
| Pre-requisites | (Not mentioned in the text) |
| Academic Unit (Major) | Civil Engineering |
| Instructor | (Not mentioned in the text) |
| Credit Points | 6 |
Business Models商业模式问题集
A common observation in big rivers or other fast-flowing bodies of water (e.g. during floods) is shown in the figures and sketch below. A fast moving stream of water that is steadily flowing along suddenly decelerates and the position of the free surface ‘jumps’ upwards. After a lot of local turbulent motion, the flow settles down again but is now steadily moving at a significantly slower speed.
We will represent the free surface height as $h(x)$ and the velocity by the function $u(x)$. The fluid has constant density $\rho$ and we will treat the problem as one-dimensional. You can assume that viscous stresses along the control surfaces of the volume shown above are negligibly small, and neglect the density of air.
PART I:
a) consider a streamline drawn (line $\mathrm{AB}$ in the figure) just above the smooth flat lower surface of the channel. How is the static pressure in the fluid along this line related to the height of the river? How does the static pressure vary along the line DEA?
(a):
The pressure distribution on line $\mathrm{AB}$ follows the hydrostatic rule. It is true that the flow is not static but by picking an arbitrary control volume at any point on line AB (green dashed control volume in Figure 1) one can see that the balance of forces in the $y$-direction will tell us that the difference between the pressure at the bottom and the ambient pressure should balance the weight of the liquid inside the control volume. This simply implies that the static pressure on line $\mathrm{AB}$ should be equal
The pressure distribution on line DEA also follows the hydrostatic change merely due to the fact that there is no curvature in the streamlines as one integrates the Euler equation normal to them and thus the only change in pressure when one moves from $\mathrm{E}$ to A will be the hydrostatic part. Ignoring the density of air one can see that the pressure is constant from D to E and then start to grow linearly with height as we move from $\mathrm{E}$ to $\mathrm{A}$. The result is shown in Figure
b) Using the control volume shown in the sketch develop two expressions that relate the velocity and height of the stream at station 1 and the velocity and height of the stream at station 2. Developing a table of relevant quantities along each face of the control volume ABCDEA is highly recommended!
c) [2 points] Combine your expressions from (a) and (b) together to show that the speed of the river can be simply evaluated from simple measurements of the river height (e.g. using marked yardsticks attached to the channel floor):
$$
u_1=\sqrt{\frac{g h_2}{2 h_1}\left(h_1+h_2\right)}
$$
}(b) and (c): The selected control volume is shown in Figure 3 (dashed green line). One can subtract the ambient pressure from the entire problem and knowing that the net effect of uniform $P_a$ acting on the control volume is zero then there will be no change in the problem analysis if we only deal with gauge pressures $\left(P(x, y)-P_a\right)$Table 1 summarizes all the important parameters acting on different control surfaces for the selected control volume:
Now we can start by writing the conservation rules using the RTT. It is important to notice that due to the turbulent mixing happening in the region of the hydraulic jump, energy will not be conserved and thus either applying the conservation of energy or the Bernoulli equation will not be the right approach. If we write the conservation of mass for the selected control volume then we will have:
$$
\text { C.O.Mass: } 0=\frac{d}{d t} \int_{\text {c.v. }} \rho d V+\int_{\text {c.s. }} \rho\left(v-v_c\right) \cdot n d A
$$
Knowing that the problem is steady state and using the tabulated quantities, conservation of mass can be simplified to:
$\rho u_1 h_1=\rho u_2 h_2 \Rightarrow u_1 h_1=u_2 h_2$
The conservation of linear momentum in the $x$ direction can also be written in the RTT form:
$$
\text { C.O.Momentum: } \frac{1}{W} \sum F_x=\frac{d}{d t} \int_{c . v .} \rho v_x d V+\int_{\text {c.s. }} \rho v_x\left(v-v_c\right) \cdot n d A
$$
where $W$ is the width into the page.
The net of external forces acting in the $x$-direction on the control volume neglecting the wall shear effect is a result of pressure forces acting on the (AD) and (BC) control surfaces:
$$
\frac{1}{W} \sum F_x=\int_{A D}\left(P-P_a\right) d y-\int_{B C}\left(P-P_a\right) d y=\int_0^{h_1} \rho g y d y-\int_0^{h_2} \rho g y d y=\rho g\left(\frac{h_1^2}{2}-\frac{h_2^2}{2}\right)
$$
The right hand side of the RTT for the conservation of linear momentum can also be simplified to (knowing that the problem is steady and using the tabulated identities):
$$
\text { R.H.S. of RTT for C.O. Momentum }=\rho u_2^2 h_2-\rho u_1^2 h_1
$$
thus the conservation of linear momentum implies that:
$$
\rho g\left(\frac{h_1^2}{2}-\frac{h_2^2}{2}\right)=\rho u_2^2 h_2-\rho u_1^2 h_1 \Rightarrow \frac{g}{2}\left(h_1^2-h_2^2\right)=u_2^2 h_2-u_1^2 h_1
$$
using the result from conservation of mass (equation (1)) one can eliminate $u_2$ from equation (2) to give:
$$
\frac{g}{2}\left(h_1^2-h_2^2\right)=h_1 u_1^2\left(h_1 / h_2-1\right) \Rightarrow u_1=\sqrt{\frac{g h_2}{2 h_1}\left(h_1+h_2\right)}
$$
where we have used the identity $h_1^2-h_2^2=\left(h_1-h_2\right)\left(h_1+h_2\right)$.
A deeper question to answer is why is the water moving so fast locally to begin with. To answer this we must consider the topography of the river bed that is upstream of station 1 , as shown in the drawing below. We denote the height of the fluid stream above the river bed as $h(x)$ and the height of the riverbed by $b(x)$ :
d) [1 point] Consider a slice of river $d x$ and show that conservation of mass can be written in the form:
$$
u(x) \frac{d h(x)}{d x}+h(x) \frac{d u(x)}{d x}=0
$$
(d):
For the selected control volume (Figure 4 ) one can easily write the conservation of mass using Taylor series to obtain expressions for $u(x+\Delta x)$ and $h(x+\Delta x)$ :
$$
u(x) h(x)=u(x+\Delta x) h(x+\Delta x) \rightarrow u(x) h(x)=\left(u(x)+\frac{d u}{d x} \Delta x\right)\left(h(x)+\frac{d h}{d x} \Delta x\right)
$$
which after ignoring the second order terms such $\left(\Delta x^2\right)$ it can be rewritten as:
$$
\Delta x\left(u(x) \frac{d h}{d x}+h(x) \frac{d u}{d x}\right)=0 \Rightarrow u \frac{d h}{d x}+h \frac{d u}{d x}=0
$$
Another way to reach the same result is to say that since the flow is incompressible then the volumetric flow rate should remain unchanged thus $d(u h) / d x=0$ which will lead to the same result we just derived in equation (4).
Figure 4: An arbitrary control volume selected to derive the conservation of mass in the differential form.
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