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1. A 2D inviscid flow field is represented by the velocity potential function: ° = Ax + Bx2 – By2. Where A = 1m/s, B = 15-7, and the coordinates are measured in meters. The flow density is p = 1.2 kg/m3. (a) (2 points) Calculate the velocity field. (b) (2 points) Verify that the flow is irrotational. (c) (2 points) Verify that the flow is incompressible. (d) (2 points) Obtain the expression of stream function. (e) (2 points) Calculate the pressure difference between point (0,0) and point (1, 2), assume x - y plan is horizontal. Solution:

Answer 1

GIVEN:

$\tiny \phi$ = Ax+Bx²-By²

A= 1 m/s, B = 1 S^-1

$\tiny \rho$ = 1.2 kg/m³

TO FIND :

a) VELOCITY FIELD

b) FLOW IS IRROTATIONAL

c) FLOW IS INCOMPRESSIBLE

d) EXPRESSION OF STREAM FUNCTION

e)PRESSURE DIFFERENCE BETWEEN THE POINTS(0,0) AND (1,2)

SOLUTION:

a) VELOCITY POTENTIAL

Function of space such that negative derivative of this function with respect to a given direction will give the component of velocity in that direction.

T=
_;
9 V= ay
;  
W = z

therefore

T=

(Ax + Bx2 - By2) U= ax

u = -(A+2Bx) .............(1)

substituting A and B IN .....(1)

U = (1+2(1)X)

U = -(1+2X)

subs points(0,0) and (1,2)

U_(0,0) = -(1+ 2(0))

U_(0,0) = -1 m/s

similarly,

U_(1,2) = - (1+ 2(1))

U_(1,2) = -(1+2 )= -3 m/s

U_(1,2)= -3 m/s

_and

9 V= ay

V= O(Ax + Bx2 - By2) ay

v = -(-2y)

V= 2By

V =2y

V_(0,0) = 2(0)

V_(0,0) = 0 m/s

V_{(1,2) = 2(2)}

V_(1,2) = 4 m/s

the velocity equation is V = ui + vj + wk

since it is 2D (w=0)

V_(0,0) = -i at origin is along negative x axis

at (0,0) the velocity is V_{1 =$\small \left ( \sqrt{-1} \right )^{2}$ = 1 m/s}

V_(1,2) = -3i +4j   

at (1,2) the velocity is V_{2 = $\small \sqrt{\left ( -3^{2} + 4^{2}\right )}$ =
(9+16)}

V_{2 =$\small \sqrt{\left ( 25}\right )}$}  

V₂ = 5 m/s

b) FLOW IS IRROTATIONAL

condition for flow is  irrotational (
_{ند 0}
)

= 1 (ди 2\дх ди ду,

u = -(A+2Bx) ; v = 2y

av ax 0
;
ди ду о

(0-0) II

ند 0

therefore, the flow is irrotational

c) FLOW IS INCOMPRESSIBLE

condition for incompressible flow  .........................is that continuty equation should be satisfied

2) 2) + = 0 дX2 ду?

=   
22(Ax + Bx? – By?) ax2 + (Ax + Bx? – By?) ay2

=
д(A + 2Bх) дх + д-2By) ду

=
2B-2B

= 0

since continuity equation is satisfied , it is incompressible flow.

d) EXPRESSION OF STREAM FUNCTION

for 2D flow

=V ax
;   
дар = — и ду
;  

..................................................................(eqn A)

д дх 2y

др || 2удх

..............................................................(eqn B)

V = 2yx + c

similarly

дар = — и ду

...............................................(eqn c)

(A + 2Bx) ду

the constant of integration c is a function of y in .....eqn B

differentiating with respect to y

V = 2yx + c

...........................................................(eqn E)

д ay 2x + дс ay

subsituting eqn C in eqn E

(A + 2Bх) – 2x = дС ду

substituting A and B values

_{1 + 2x – 2x = да ду}

_{де 1 ду}

_{1ду - Јә =}

= с

subs c= y in eqn B

U = 2yx + y

therefore the stream function is

U = 2yx + y

e)PRESSURE DIFFERENCE BETWEEN THE POINTS(0,0) AND (1,2)

$\Delta p = \frac{1}{2}\rho \left ( V_{2}^{2}-V_{1}^{2} \right )$

$\Delta p = \frac{1}{2}*1.2* \left (5^{2}-1^{2} \right )$

$\Delta p = \frac{1}{2}*1.2* \left (24 \right )$

$\Delta p = 14.4 N/m^{2}$

RESULT :

  

a) VELOCITY FIELD is V_(0,0) = -i at origin is along negative x axis and V_(1,2) = -3i +4j   

b) FLOW IS IRROTATIONAL , since   
ند 0

c) FLOW IS INCOMPRESSIBLE , since it satisfies bernoulli equation  

2) 2) + = 0 дX2 ду?

d) EXPRESSION OF STREAM FUNCTION is

U = 2yx + y

e)PRESSURE DIFFERENCE BETWEEN THE POINTS(0,0) AND (1,2) is $\Delta p = 14.4 N/m^{2}$ .