Total Acceleration of Fluid MCQ Quiz in தமிழ் - Objective Question with Answer for Total Acceleration of Fluid - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Acceleration of a Fluid Particle:

(i) Acceleration is defined as the rate of change of velocity with respect to time.

$Velocity(\overrightarrow{V})=u(x,y,z,t)\hat{i}+v(x,y,z,t)\hat{j}+w(x,y,z,t)\hat{k}$

Where, u, v and w are the components of velocity in x, y and z direction respectively.

Acceleration in x-direction:

$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}+\frac{\partial u}{\partial t}$

Acceleration in y-direction:

$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}+\frac{\partial v}{\partial t}$

Acceleration in z-direction:

$a_z=u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}+\frac{\partial w}{\partial t}$

Calculation:

Given,

Velocity component in x-direction; u = t2 + 3y

$\frac{\partial u}{\partial x}=0,\frac{\partial u}{\partial y}=3,\frac{\partial u}{\partial z}=0,\frac{\partial u}{\partial t}=2t$

Velocity component in y-direction; v = 4t + 5x

$\frac{\partial v}{\partial x}=5,\frac{\partial v}{\partial y}=0,\frac{\partial v}{\partial z}=0,\frac{\partial v}{\partial t}=4$

Acceleration in x-direction:

$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}+\frac{\partial u}{\partial t}$

a_x = (t² + 3y) × 0 + (4t + 5x) × 3 + 0 + 2t

a_x = 12t + 15x + 2t

acceleration at the point (5,3) at time t = 2 units

a_x = 12 × 2 + 15 × 5 + 2 × 2 = 103 units

Acceleration in y-direction:

$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}+\frac{\partial v}{\partial t}$

a_y =  (t2 + 3y) × 5 + 0 + 0 + 4

a_y = 5t² + 15y + 4

acceleration at the point (5,3) at time t = 2 units

a_y = 5 × 2² + 15 × 3 + 4 = 69 units

So, Resultant acceleration;

$a=\sqrt{a_x^2+a_y^2}$

$a=\sqrt{{103}^2+{69}^2}=123.97$

a ≈ 124 units

$a_x=u\frac{du}{dx}+v\frac{du}{dy}+w\frac{du}{dz}+\frac{du}{dt}$

= (-2x) × (2) = 4x

$a_y=u\frac{dv}{dx}+v\frac{dv}{dy}+w\frac{dv}{dz}+\frac{dv}{dt}$

= 0 + 2y × 2 + 0 + 0 = 4y

Hence, u = x + 2y, v = 4y + z, w = -x – 4z

$\left|\overrightarrow{a}\right|=\sqrt{{\left(\frac{du}{dt}\right)}^2+{\left(\frac{dv}{dt}\right)}^2+{\left(\frac{dw}{dt}\right)}^2}$

$\frac{du}{dt}=\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}$

= 0 + (x + 2y)(1) + (4y + z)(2) + 0

${\frac{du}{dt}|}_{\left(2,0,0\right)}=2$

$\frac{dv}{dt}=\frac{\partial v}{\partial t}+u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}$

= 0 +(x+2y) (0) + (4y + z) (4) + (-x – 4z) (1)

${\frac{dv}{dt}|}_{\left(2,0,0\right)}=-2$

$\frac{dw}{dz}=\frac{\partial w}{\partial t}+u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}$

= 0 + (x + 2y) (-1) + (4y+z) (0) + (-x – 4z) (-4)

${\frac{dw}{dz}|}_{\left(2,0,0\right)}=6$

$\left|a\right|=\sqrt{4+4+36}=6.633$

Concept:

For a velocity field, $\overrightarrow{V}=u\hat{i}+\nu \hat{j}+w\hat{k}$

Total acceleration is the vector sum of the convective acceleration vector (directional acceleration vector) and local acceleration vector.

$\overrightarrow{a}=\frac{D\overrightarrow{V}}{Dt}=u\frac{\partial \overrightarrow{V}}{\partial x}+\nu \frac{\partial \overrightarrow{V}}{\partial y}+w\frac{\partial \overrightarrow{V}}{\partial z}+\frac{\partial \overrightarrow{V}}{\partial t}$   ---(1)

Along x-direction, 

$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}+\frac{\partial u}{\partial t}$

Calculation:

Given:

u = 3 + 2xy + 4t2, v = xy2 + 3t.

$\frac{du}{dx}=2y,\frac{du}{dy}=2x,\frac{du}{dt}=8t$

$a_{conv}=u\cdot \frac{du}{dx}+v\cdot \frac{du}{dy}$

a_conv = (3 + 2xy + 4t²) (2y) + (xy² + 3t) (2x)

∴ a_conv = 2x²y² + 4xy² + 6y + 8t²y + 6xt

Concept:

The total accelration is given by,

${\mathrm{a}}_{\mathrm{n}}=\frac{\mathrm{d}{\mathrm{V}}_{\mathrm{n}}}{\mathrm{d}\mathrm{t}}=\left[\frac{\partial {\mathrm{V}}_{\mathrm{n}}}{\partial \mathrm{t}}\right]+\left[\mathrm{U}\frac{\partial {\mathrm{V}}_{\mathrm{n}}}{\partial \mathrm{x}}+\mathrm{V}\frac{\partial {\mathrm{V}}_{\mathrm{n}}}{\partial \mathrm{y}}+\mathrm{W}\frac{\partial {\mathrm{V}}_{\mathrm{n}}}{\partial \mathrm{z}}\right]$

Where,

$\left[\frac{\partial {\mathrm{V}}_{\mathrm{n}}}{\partial \mathrm{t}}\right]=$ Local accelration or temporal acceleration

$\left[\mathrm{U}\frac{\partial {\mathrm{V}}_{\mathrm{n}}}{\partial \mathrm{x}}+\mathrm{V}\frac{\partial {\mathrm{V}}_{\mathrm{n}}}{\partial \mathrm{y}}+\mathrm{W}\frac{\partial {\mathrm{V}}_{\mathrm{n}}}{\partial \mathrm{z}}\right]=$ Convective acceleration

Local acceleration:

Acceleration calculated due to variation of velocity with respect to time.

Convective Acceleration:

Accleration calculated due to variation of velocity with respect to space is known as convective accelration.

Convective acceleration in x direction is given as

$\mathrm{u}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{v}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\mathrm{w}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}$

Calculation:

$\begin{array}{l}{V_{avg}}_1=1.5m/s\\ V_{av{g}_2}=15m/s\end{array}$

∂x = 0.375 m

Convective acceleration $=\mathrm{u}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}$

$\mathrm{u}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\frac{15\left(15-1.5\right)}{0.375}=540m/s^2$

Velocity field $\left(v\right)=2i+\left(x+y\right)j+\left(xyz\right)k$

$Acceleration\left(a_n\right)=\left(\frac{\delta v}{\delta t}\right)+\left(u\frac{\delta v}{\delta x}+v\frac{\delta v}{\delta y}+w\frac{\left(\delta v\right)}{\delta z}\right)$

Where $\left(\frac{\delta v}{\delta t}\right)\Rightarrow Localaccelaration$

$\left(u\frac{\delta v}{\delta x}+v\frac{\delta v}{\delta y}+w\frac{\left(\delta v\right)}{\delta z}\right)\Rightarrow Convectiveaccelaration$

Here, $u=2,v=x+y,w=xyz,t=0,$ n is direction

$a_x=0+2\times \frac{\partial }{\partial x}\left(2\right)+\left(x+y\right)\times \frac{\partial \left(2\right)}{\partial y}+\left(xyz\right)\times \frac{\partial \left(2\right)}{\partial z}$ 

a_x = 0

$a_y=0+2\times \frac{\partial \left(x+y\right)}{\partial x}+\left(x+y\right)\times \frac{\partial \left(x+y\right)}{\partial y}+\left(xyz\right)\times \frac{\partial \left(x+y\right)}{\partial z}$ 

a_y = 2 + x + y

$a_z=0+2\times \frac{\partial \left(xyz\right)}{\partial x}+\left(x+y\right)\times \frac{\partial \left(xyz\right)}{\partial y}+xyz\times \frac{\partial \left(xyz\right)}{\partial z}$ 

= 0 + 2yz + (x + y) × (xz) + xyz × xy

a_z = 2yz + (x + y) × (xz) + x²y²z

Acceleration = a_xi + a_yj + a_zk

$a=0+\left(2+x+y\right)j+\left(2yz+\left(x+y\right)xz+x^2{y}^{2}z\right)k$

$a_{\left(1,1,2\right)}=0+\left(2+1+1\right)j+\left(2\times 1\times 2+\left(1+1\right)\times 1\times 2+1^2\times 1^2\times 2\right)k$

Concept:

Convective acceleration is given by,

$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}$

$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}$

$a_z=u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}$

Calculation:

Given:

u = 2x2 + y2 and v = -4xy

To get, $a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}$

At point (1, 2)

$u\frac{\partial u}{\partial x}=\left(2x^2+y^2\right)\times \frac{\partial \left(2x^2+y^2\right)}{\partial x}=\left(2x^2+y^2\right)\times \left(4x\right)$

Now putting the coordinates of x and y in the above equation (x = 1, y = 2)

$u\frac{\partial u}{\partial x}=\left(2x^2+y^2\right)\times \left(4x\right)=\left(2\times 1^2+2^2\right)\times \left(4\times 1\right)=24\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$

Same

$v\frac{\partial u}{\partial y}=-4xy\times \frac{\partial \left(2x^2+y^2\right)}{\partial y}=-8x{y}^2$

Now putting the coordinates of x and y in the above equation (x = 1, y = 2)

$v\frac{\partial u}{\partial y}=-8x{y}^2=-8\times 1\times 2^2=-32$

$w\frac{\partial u}{\partial z}=0\mathrm{S}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e},\mathrm{u}\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{p}\mathrm{e}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{f}\mathrm{z}$

Therefore,

Concept:

When a tank partially filled with liquid is given a constant horizontal acceleration, the horizontal free surface of liquid will become inclined as shown in the below figure,

The dotted line is original surface and the inclined line is new surface.

As there is just no spilling of water, the new surface will touch the left side top of the tank.

The tan of the inclination is given by 

$tan\theta =\frac{a}{g}$

Where a - acceleration imparted;

Calculation:

From the given figure,

$tan\theta =\frac{a}{g}=\frac{0.8}{1}$

⇒ a = 0.8g = 8 m/s²

Concept:

The convective acceleration in y-direction is given by:

$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}$

Where,

u, v, w are the velocity in x, y and z direction respectively.

Calculation:

u = 10 x + 5 y

at, (1, 0, -3)

u = 10 units

v = 15 xy²z

at, (1, 0, -3)

v = 15 × 1 × 0 = 0 units

w = 13 x + 20 y

at, (1, 0, -3)

w = 13 × 1 + 0 = 13 units

$\frac{\partial v}{\partial x}=15y^{2}z=0$

$\frac{\partial v}{\partial y}=30xyz=0$

$\frac{\partial v}{\partial z}=15x{y}^2=0$

a_y = 10 × 0 + 0 + 13 × 0