Problem 1 1- To verify whether the flow satisfies conservation of mass, we consider the continuity

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1 Problem 1 1- To verif whether the flow satisfies conservation of mass, we consider the continuit u v equation for D incompressible flow: 0 x u x x x 1 x 1 x v x 1 x 1 Therefore: u v 1 x x1 0, x which demonstrates that conservation of mass is satisfied. - Matlab code: %plot velocit vector field clear all; %create the matrices containing the x and coordinates of all points in the domain [X,Y]=meshgrid(-10:1:10, -10:1:10); r=size(x,1); c=size(x,); %at each point, calculate the x- and -components of the velocit for i=1:r for j=1:c U(i,j)=Y(i,j)^-X(i,j)*(1+X(i,j)); V(i,j)=Y(i,j)*(*X(i,j)+1); %plot the vector field over the domain quiver(x,y,u,v) axis square xlabel('x'); label('y');

2 Plot: 3- The components of the acceleration vector can be obtained b taking the material derivative of the velocit components: Similarl: Du u u ax V u u v u Dt t t x Dv v v a V v u v v Dt t t x u u u t x t x a x u v x x u x x v x x 0 1 x a 1 x x1 x x 1 x 1 x1 x x a x 1 u x 1 v x 1 t x 0 x1

3 a x 1 x x 1 Therefore, the acceleration field is defined as: In particular, at point 1,1 : 1 1 ax x x x a x 1 x x 1 ax 1,1 9 a a 1,1 7 1, m/s Problem 1- One velocit component, the other is unknown. The continuit equation can be used to determine the unknown velocit component. For an incompressible flow: u v 0 x u Since u x, x and substituting this in the continuit equation ields: v Integrating both sides with respect to : v( x, ) f ( x) In order to determine the unknown function f( x ), we consider the boundar condition provided in the problem (i.e., along the x-axis, the -component of the velocit is zero). Translated mathematicall: Using the expression derived for v( x, ) : v 0 at f( x) f( x) 0 Therefore, the unknown velocit component is: v( x, )

4 - Matlab code: %plot velocit vector field clear all; %create the matrices containing the x and coordinates of all points in the domain [X,Y]=meshgrid(0:1:10, 0:1:10); r=size(x,1); c=size(x,); %at each point, calculate the x- and -components of the velocit for i=1:r for j=1:c U(i,j)=*X(i,j); V(i,j)=-*Y(i,j); %plot the vector field over the domain quiver(x,y,u,v) axis square xlabel('x'); label('y'); Plot: S3- The volumetric flow rate (per unit depth of the page) q across the line AB is related to the average velocit V of the fluid across that line b the following relationship:

5 where L is the length of the line AB. q VL, q is obtained b taking the difference in the value of the streamfunction between point A and point B (see streamfunction properties): q (1,1) (0,0). Therefore: (1,1) (0,0) V L Let s determine the streamfunction expression: u x v x Integrating the 1 st relationship with respect to : ( x, ) x f ( x) Substituting this result in the second relationship: ( x f ( x)) x x f ( x) f( x) 0 f ( x) C We obtained the final form of the streamfunction: ( x, ) x C The constant C to zero: does not serve an purpose in this expression and can therefore be set equal Using this expression: Substituting in the velocit expression: ( x, ) x (0, 0) 0 (1,1) Quantitativel: V 1.41 m/s V L

6 Problem 3 1- A stagnation point is a point where the velocit is zero. It can be found b solving the sstem of equations: u 0 v x x There is one single stagnation point. - Matlab code: %plot velocit vector field clear all; %create the matrices containing the x and coordinates of all points in the domain [X,Y]=meshgrid(-:0.:, 0:0.5:5); r=size(x,1); c=size(x,); %at each point, calculate the x- and -components of the velocit for i=1:r for j=1:c U(i,j)= *X(i,j); V(i,j)= *Y(i,j); %plot the vector field over the domain quiver(x,y,u,v) axis square xlabel('x'); label('y'); hold on; %plot streamline passing through (1,) X=-:0.05:; Y=zeros(1,length(X)); for i=1:length(x) Y(1,i)=(1.5/0.8)-(1.3*0.7)/(0.8*( *X(i))); plot(x,y); axis([- 0 5]); Plot:

7 3- Acceleration field: ˆ ˆ u v u v u ˆ v ˆ DV V a V V i j i j Dt t t x ˆ ˆ a x x x ˆ ˆ t i j x i j a x x ˆ ˆ x ˆ x ˆ i j i j x0.8 ˆ i ˆj a xˆi a ax a 4- Streamline equation The first step is to establish the streamfunction expression. We start with: ˆj

8 u x Integrating on both sides with respect to : x f ( x) Using this intermediate result, the other relationship v x can be written: x Substituting in the intermediate result ields: x f ( x) f ( x) f( x) 1.5 f ( x) 1.5x C x 1.5x C Since the constant C to zero: does not pla an role in the streamfunction, it can be set arbitraril equal x 1.5x The streamline equation is obtained b setting the streamfunction equal to a constant (i.e., A): x 1.5x A A1.5x x