Question

5 solved examples for (differential equation in fluid dynamics )

*the exampls (proplems) should be have differential equation in Operative of the question (It is preferable to be for the highest order)

and The answer should be a solution to these differential equations

Answer 1

Question 1-

Consider a pollutant that degrades by participating in chemical reactions. In such case, the ‘source’ term s is properly speaking a ‘sink’ and has a negative value. Under first-order chemical kinetics, the rate of removal of the chemical is proportional to its own concentration, because the more chemical is present, the more reactions take place. Thus, a realistic model is s = −Kc, where K is a
constant of decay, with dimension equal to the inverse of time (example: K =0.2/day).
Consider now a one-dimensional system with uniform and constant velocity u.
Equation becomes-
∂c/∂t + u∂c/∂x = − Kc. (1.1)
To solve this equation, we need to know the initial concentration of the pollutant.
Let us denote it as c(x, t = 0) = c_{​​​​​​0}(x).
If there were no chemical degradation (K = 0), this patch would simply move
downstream without change, and the solution would be c(x, t) = c₀(x − ut), which
corresponds to a mere translation down the x–axis by the traveled distance ut. On
the other hand, if there were no movement (u = 0) and only chemical degradation,
the solution would be c(x, t) = exp(−Kt) c_{​​​​​​0}(x), which is the initial distribution
attenuated over time. Combining these two limiting solutions, it is not difficult to
show that the distribution of concentration corresponding to the combined case of
advection and degradation at any later time t is given by
c(x, t) = exp(−Kt )c_{​​​​​​0}(x − ut). (1.2)
As an application, let us consider an accidental release of benzene (C6H6,carcinogenic substance) in a river moving at the uniform speed of 15 cm/s. From
water, benzene volatilizes into the air, at a rate that corresponds to a decay con-
stant of 0.20/day (a function of water depth). If the initial concentration reaches
a maximum of 25 ppb (= 0.025 mg/L) at the center of the spill, and the drinking
standard is 5 ppb (= 0.005 mg/L), which downstream length of the river will be
subjected at some time or other to a concentration exceeding the drinking standard
and how long will the episode last?
To answer these questions, we track the maximum of the concentration over
time. Equation 1.2 tells that the maximum at time t is the initial maximum
times the factor exp(−Kt). The contamination episode ends when the maximum
concentration falls to the drinking standard and thus at time t such that (5 ppb)
= exp(−Kt)× (25 ppb). The solution is exp(−Kt) = 5/25 = 0.20, giving Kt =1.609 and t = 8.047 days = 6.953 × 10^5 s
. Over this time, the river water has
traveled a distance ut = (0.15 m/s)(6.953 ×10^5
s) = 104.3 km. Thus, more than
100 kilometers of river are being contaminated, and the episode will last slightly more than 8 days.

Question 2-

In water there is a flat horizontal bottom, a wave propagates in the x-direction
with the associated horizontal flow field:
u = U sin(kx − ωt) , v = 0,
where U is the velocity amplitude, 2π/k the wavelength and 2π/ω the period. The
question is: What is the vertical velocity?
Since water is incompressible, conservation of mass reduces to the equation of
continuity , which for v = 0 becomes
∂w/∂z = −∂u/∂x

=− kU cos(kx − ωt).

For a flat and impermeable bottom, at z = 0, this equation can be integrated,
providing:
w = − kUz cos(kx − ωt).
Thus, at some level H above the bottom, the vertical velocity exhibits oscillations
of amplitude kUH, which is not equal to that of the horizontal velocity. (Particle
trajectories are ellipses, not circles.)

Question 3-

A steady wind from a plain blows toward a ridge and enters a wind gap. By so doing, it accelerates from 12 to 40 m/s over a distance of 2 km. How does this horizontal acceleration compare to the vertical gravitational acceleration? And, which term in the corresponding momentum equation balances the acceleration term?
If we take x as the direction of the wind, then the wind speed is u(x). Steadiness
removes any time dependence, and we can assume that the cross-wind velocity
components, v and w are negligible. Friction is not expected to play a major role
in a rapidly adjusting situation and is neglected, too. What is left of the set of
momentum equations are the following two terms of the momentum equation given as

u∂u/∂x = −1/ρ_{​​​​​​0}(∂p/∂x)
The acceleration term is the one caused by the change in speed, thus u∂u/∂x. For
the given values, it is equal to
u∂u/∂x =∂/∂x (u_{​​​​​​^{​​​​​​2}}​ /2)
=(u^​​​2 _{downstream -} u^{​​​​​​2} _{upstream)/(2*distance)}
=
(40^2 − 12^2)/(2*2000)
= 0.364 m/s^{​​​​​​2}
which is very modest (about 4%) compared to the gravitational acceleration g =9.81 m/s^2.
According to the differential equation, the force accompanying (driving) the acceleration is the pressure force. The pressure gradient is negative:
∂p/∂x = − ρ₀u∂u/∂x
telling us that the pressure drops as the air accelerates. Physically, the air accel-
erates by being pushed from the rear. The pressure drop occurring over the 2 km
distance is given by an integration over that distance:

Such pressure difference is small compared to the standard ground atmospheric pressure of 101300 pa.

Question 4 - Sea Breeze

Because water is relatively transparent but land opaque, the solar radiation penetrates to greater depth in the sea than in the ground by the shore. In addition,the heat capacity of water is greater than that of soil, implying that for equal heat
input, the temperature of the sea increases less than that of land. During daytime by the seashore, these differences generally translate into a warmer land next to a colder sea. Warm air rises over the land, drawing cooler air from the sea .

Consider a vertical rectangular loop consisting in two horizontal segments span-
ning the coastline and in two vertical segments, one over the sea and the other over
the land, as drawn in .Since the horizontal segments (where dz = 0) do not contribute to the loop integral, only the vertical components do, and the Bjerknes circulation theorem given as

Question 5

where u is the velocity and p is the pressure of fluid and is dynamic viscosity.

In the case of an unbounded planar domain with two-dimensional — incompressible and stationary — flow in polar coordinates (r,φ), the velocity components (ur,uφ) and pressure p are:

where A and B are arbitrary constants. This solution is valid in the domain r ≥ 1 and for A < −2ν.