The motion of a fluid particle is specified by the equations x = e\' and v = e-1

Question

The motion of a fluid particle is specified by the equations x = e' and v = e-1 in the Lagrangian description. Determine the equation for the path of the fluid particle. Determine the velocity of the fluid particle. Determine the acceleration of the fluid particle. Determine the velocity of the flow field. State under what conditions such a flow field can be defined. Determine whether the flow field is steady or not. Determine the equation of the streamline on which the fluid panicle is located. Determine the equation of die streakline on which the fluid particle is located. Determine whether the flow field is compressible or not. Determine whether the flow field is rotational or not. Determine the equation of the streamline passing through point Q(2. 2). Is it possible to define a stream function for this flow field? If yes. state why? and determine the stream function such that psi = 0 at the origin. Sketch the streamlines and show their direction. Give a physical interpretation of the flow field. If the pressure at the origin 0(0. 0) is 100 kPa. determine the pressure at point P(2. 2). Assume that the density of the fluid is 1000 kg/m3. Determine the local acceleration of the flow field. Determine the convective acceleration of the flow field, Determine the acceleration of the flow field. (Ans. xy = 1 e'i - e-1j ; e'i + c-1j; xi - yi incompressible; steady; xy= 1; xy = 1; incompressible; irrotational; xy = 4; yes, two-dimensional, steady and incompressible, xy; flow about a 90 degree corner located at the origin; 96 kPa; 0; xi + yj; xi + yj )

Explanation / Answer

a) Since we know that x = et and y = e-t. We just need to eliminate t from x and y to get the path line equation. It is easy to see that

xy = et e-t = et-t = e0 = 1. So the equation is

xy=1

b) v = dx/dt i +dy/dt j= eti - e-tj.

c) a = dv/dt =d( eti - e-tj)/dt = d( eti )/dt –d( e-tj)/dt = eti + e-tj.

d) Rewrite the velocity of the particle with x and y, we have

v = eti - e-tj = x i - y j

e) From the velocity field, vx = x, vy = -y. Then

vx/x+ vy/y = 1+(-1) = 0. It is incompressible flow.

f) Steady, since the velocity field is not a function of t.

g) Streamlines satisfy: dx/vx= dy/vy, i.e.,

dx/x = -dy/y

Integrate both side to have

lnx = -lny +C (C is a constant)    or

lnx+lny = C or lnxy = C or xy = eC.

The constant can be determined from the particle velocity. Use x = et and y = e-t in xy = eC. we have C = 0, and the streamline becomes

xy=1

h)For steady flow, the pathline, streamline and streakline are identical, so, the streakline is

xy =1

i) Since vx/x+ vy/y = 1+(-1) = 0. It is incompressible flow.

j) (i /x+ j /y) x v = (i /x+ j /y) x (x i - y j) = 0. irrotational.

k) From part g). The equation is xy = eC = D (D is another constant). Use x =2, y=2, we have

2(2)=D. or D = 4. Hence the equation is

Xy =4

l)Yes, it is possible, since it is a steady, incompressible and irrotational flow.

/x = vy     and /y = vx. That is

/x = y     and /y = x

d =/x dx+ /y dy= ydx +xdy

Integrate to give = xy.

m) equations available. It should be easy.

n) It’s a flow about 90 corner located at the origin.

o) From Navier-Stokes equation, From part r) below, we have the

-p/x = ax =x    -p/y = ay =y  

dp = p/x dx + p/y dy= -x - y . Integrate

p= -x2/2 –x2/2 + C (C: constant)

Use p(0,0) = 100 kPa = 100,000 Pa, we have C = 100,000. Hence

p= -x2/2 –y2/2 + 100,000.

At (2,2)

p =-1000(22/2 + 22/2) + 100,000 = 96,000 Pa = 96 kPa

p) alocal = =v/t =(x i - y j)/ t = 0

q) aconv = =vxv/x+ vyv/y=x (x i - y j)/ x- y (x i - y j)/ y = x i + y j

r) a = alocal + aconv = x i + y j

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