homework 6 solution manual

7/31/2019 Homework 6 Solution Manual

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HOMEWORK 6 SOLUTION MANUAL

Marat K

Problem 1 (5 points):

The Bernoulli equation is:

Eq1Since the velocity at the stagnation point is zero and maximum pressure is found at locations with

minimum velocity we should rewrite the Bernoulli equation as:

Eq2

( ) Problem 2 (5 points):

Problem 2 Schematic

Start with the Bernoulli equation (Eq1) again and rewrite it as:

Eq3

And realize that in an incompressible flow


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= QEq4

Combining Eq3 with Eq4 will yield Eq5

Eq5

Solving for Q in Eq5 will yield the following:

( )( )

Eq6

Eq6 is a little messy so keep track of all the units and make sure they simplify to units of volumetric flow

rate. This is the easiest way to verify that your final solution makes sense. An alternative way to obtain Q

is presented below. Combine Eq3 and 4 to yield Eq7:

() * + ( )( ) * ( ) +

Eq7

Then realize that:

( ) Eq8


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Problem 3 (10 points):

Please read pages 227 to 237 in the textbook since they discuss this problem in detail. Also, refer to the

diagram in Figure 3.22 for this problem. You can solve this problem using potential flow or stream

function but we will use the latter in this derivation. First derive the stream function for the entire flow

by summing the stream function for uniform flow and a source. If do that correctly you should end up

with Eqn 3.75, which is Eq10 here.

Eq9

Eq10

Remember that stream lines are always parallel to the velocity vectors so fluid particles can never pass

through the stream lines. Because of this we can think of a region that is completely enclosed by a single

stream function as a solid body. To find the enclosing stream function you should first compute

velocities from the global stream function as:

Eq11

Eq12

At the stagnation points Vr and V are zero. Solving Eq11 and Eq12 simultaneously will yield:

Eq13

Eq14


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Eq 13 is the answer for part c. Next, plug Eq13 and 14 into Eq 10 to find the stream function that passes

through the stagnation points:

Eq15

Since the stream function that passes through the stagnation points is also the stream function that lies

on the body you can use Eq10 and 15 to find r as a function of on the body .

Eq16

Rewrite the above equations in terms or x and y that will be easier to plot.

( )Eq17

Now to find the height of the body above the centerline, take a limit of the previous equation at x

( )

Eq18

The width of the body is twice the height above the centerline.

Eq19

Eq19 is the answer to part a. You already found the radial and tangential velocity fields in Equations 11

and 12. The total velocity is simply: Eq20

Equation 20 is a function of r and but the two are related on the body. This relation was presented in

Eq16.and rewritten as Eq21.


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Eq21

If you plug Eq11 and 21 into Eq 20 you will find the Velocities on the body as a function of . To find themaximum velocity simply take the derivative of the Velocity equation and set it equal to zero

Eq22

You can avoid long derivations in this problem by plotting V and finding the max V in MATLAB. The

MATLAB code and plot are presented in Fig1 and 2. Note that I assigned Vinf = 1 in the beginning of the

MATLAB script but this will not change the location of max theta. This will change the value of Vmax and

after playing with Vinf you should realize the following: You can try to solve this part analytically using MAPLE as is shown in figure 3

MaxThet = 1.1000

>> Vnet(k) = 1.2596

Figure 1 V (theta)


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%Arbitrary ConditionsQ = 1;V = 1;

i = 0;

for theta = 0:.01:2*pi

i = i+1;thet(i) = theta;r(i) = (Q / 2) * ( 1 - (theta / pi)) / ( V * sin (theta));

Vr = V * cos (theta) + Q / ( 2 * pi * r(i));Vt = V * sin (theta) ;Vnet(i) = sqrt ( Vr ^ 2 + Vt ^2);

x(i) = r (i) * cos (theta);y(i) = r (i) * sin (theta);

endplot (thet, Vnet)xlabel ('theta')ylabel ('V')

%Find where max velocity occurs[j,k] = max (Vnet)

MaxThet = thet(k)

Figure 2 Matlab code for 2c

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Figure 3. Problem 3 solved using MAPLE

If you never used MAPLE before then note that text in black is what you type in and text in blue is the

output. On line 1 I defined Psi and let MAPLE solve for r as a function of theta on line 2. I then used that

r to solve for V. According to MAPLE, there is no analytical solution for MaxTheta in this problem but this

approach is not completely fruitless. The derivative plot crosses the theta intercept at 1.1 and states

that the maximum velocity must occur at a theta value of 1.1. This matches the MATLAB solution.


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Problem 4(10 points):

Please refer at Figure 3.23 in your book for this problem. Just as before, we first need to sum the stream

function of the uniform flow, source and sink to get the stream function for the entire flow (as discussed

on page 234-235 in your book.) This stream function is presented again in Eq23

Eq23

Next, we need to find the stream equation for the body that contains the stagnation point. You can do

this as we did in problem 3 or you can simply realize that there is a stagnation point is at the nose ( =

1 = 2= ) . Plug these theta values into Eq23 to get

= 0 Eq23Therefore, along the body surface:

Eq24

Since we were given Q/Uinf parameter, we need to normalize Eq24 as is shown in Eq25.

Eq25

Next, realize that all the angles in Eq25 are related in Figure 3.23. The relations are reproduced in Figure

4


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Figure 4. Theta1, theta2, theta3 relations

Eq26

( )Eq27

( )Eq28

Rewrite Eq25 using Eq26 through 28 to get Eq30 and realize that that you should be able to use Eq30 to

solve for r as a function of theta. This is easier to do numerically by plugging in theta values and using a

root solving scheme to find the corresponding r values.

(( ) ( ) ) Eq30


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To find Cp we need to first find the velocities around the oval. Do this first rewriting Eq23 as the

following:

Eq31And remember that

Eq32

Eq33 u and v were solved in MAPLE and presented in Fig 5

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Figure 6

You can also solve for u and v by using the potential flow Eq34

Eq34

Then realize that: Eq35

Eq36And end up with:

Eq37

( ) ( )Eq38

( ) ( )Eq39

Coefficient of pressure is was derived in your book as Eq3.38 and repeated again here as Eq40

( )Eq40

Note the following:


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( ) ( ) ( ) Eq41

After deriving Eq30 through 41 you can finally write a MATLAB code to solve this problem. The codes

and their output are presented bellow.

function F=funny(r,th)% return PSI values for each r and theta values

Qby2piU=.2/(2*pi); %Givex=r.*cos(th);y=r.*sin(th);th1=atan2(y,x+1);th2=atan2(y,x-1);F=1 + Qby2piU*(th1-th2)./y; %PSI

Figure 7. Psi Function

% FIND ZEROES OF A funk FUNCTION USING THE SIMPLE BISECTION METHOD% Written by MARAT KULAKHMETOV% AERO Grad Student/AAE333 TA.% Email comments to [email protected]

function [G3] = bisect (th)

%Compute Zero using Bisection

G1 = 0; %Lower BoundG2 = 30; %Upper Boundtol = .001; %Zero ToleranceFF3 = 1; %Initialize Bisection Scheme

while abs(FF3)> tol

G3 =( G1 + G2 )/2;

FF1 = funk (G1,th); % funk is the name of the function you areFF2 = funk (G2,th); %solving for. Change this to your function name.FF3 = funk (G3,th);

if FF1 *FF3 < 0G2 = G3;

elseif FF2 *FF3 < 0G1 = G3;

else %Error Checking'Error: Bisection does not converge'break

end


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end

Figure 8. Bisection Function

% AAE333 HW7 Problem 4% Written by MARAT KULAKHMETOV% AERO Grad Student/AAE333 TA.% Email comments to [email protected]

Qby2piU=.2/(2*pi); %Givenj = 0;

for (th = 0:.01:2*pi)j = j+1;theta(j) = th;rad(j) = bisect(th); %Use Bisection Scheme to find Roots of Eq. 30

x = rad(j) * cos(th);y = rad(j) * sin(th);

%u and v obtained in MAPLE using the stream functionu(j) = 1 + ( Qby2piU) * (( 1 / ( (x+1) *( 1 + (y^2)/(x+1)^2))) - 1 / ( (x-

1) *( 1 + (y^2)/(x-1)^2)));v(j) = ( Qby2piU)* (( y / (((x+1)^2)*(1+ (y^2)/((x+1)^2))))- (y / (((x-

1)^2)*(1+ (y^2)/((x-1)^2)))));

V(j) = sqrt (u(j) ^2 + v(j) ^2); %Total Velocity/Free Stream VCp(j) = 1 - (V(j))^2; %Coefficient of Pressure

xx(j) = x;yy(j) = y;

endsubplot(2,2,1)plot (xx,yy)axis([-1.2 1.2 -1.2 1.2])xlabel ('x')ylabel ('y')title ('shape of the body')

subplot(2,2,2)plot (xx,V)axis([-1.2 1.2 0 1.5])xlabel ('x')ylabel ('Velocity / Free Stream Velocity')title ('Velocity Profile')

subplot(2,2,3)plot (xx,Cp)


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axis([-1.2 1.2 -1.2 1.2])xlabel ('x')ylabel ('Cp')title ('Cp')Figure 9. MATLAB script for HW7 Prob4

Figure 10. Solution for Problem 4

Professor Williams has also solved this problem and his solution is presented bellow. His script used

vectors instead of loops to sweep across the oval and his u and v values were computed using the

potential function. He also used the MATLABs fsolve command to find for r values but this command is

sensitive to the initial r0 command. Despite all these differences the two methods provide identical

solutions.


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function hw6_4% Uniform flow + Source at x=-1 + Sink at x=1% Plot body shape and pressure distribution

Qby2piU=.2/(2*pi);

th=linspace(.001,(2*pi-.001),500);r0=th;

% PSI=Uinf*y + Q/2pi(TH1-TH2)% =0 on dividing streamline% F==PSI/(Uinf*y) is also 0 on dividing streamline% (divide by y to avoid 0=0 at stagnation points)

r=fsolve('funk',r0,[],th);x=r.*cos(th);y=r.*sin(th);subplot(2,1,1)plot(x,y);axis([-1.2 1.2 -1.2 1.2])xlabel('x');ylabel('y')axis equal

r12=(x+1).^2+y.^2;r22=(x-1).^2+y.^2;

u=1+Qby2piU*((x+1)./r12 -(x-1)./r22);v=Qby2piU*(y./r12-y./r22);

cp=1-(u.^2+v.^2);subplot(2,1,2)plot(x,cp)axis([-1.2 1.2 -1.2 1.2])xlabel('x');ylabel('cp')axis equal

Dr. Williams Solution


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Dr. Williams Plots

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