## [answered] MAE 754 HW #2 Due date: Tuesday September 13, 2016 Chapter

i need help with question 1 (a,b). i have also attached a pdf which can help to solve the question.?

MAE 754 HW #2 Due date: Tuesday September 13, 2016

Chapter 3 Homework Solutions

1. Plot the lift and drag coefficients and L/D ratio for a flat plate, varying the angle of

attack from 1 to 20 degrees for ? = 1.4 and M? =15 using exact shock wave and

expansion theory, the approximate theory in the hypersonic limit, Newton?s theory, and

modified Newton?s theory. I want 3 charts, one for Cl, Cd, and L/D, with four curves on

each to compare the four methods. Comment on the accuracy of the methods, and

whether or not you had a good time doing it.

2. Giant mutant Mach 20 cicadas are on the attack! Assuming that they are 2D wedges with

a 10 m depth, their upper and lower surfaces are as shown in Figure 3. Note that the

lower two points are at -2.5 m. Use the shock expansion method to calculate the lift and

drag, assuming ?? =1 kg/m3 and T? is 273 K. Compare this with the tangent wedge

method, which keep in mind, can only be applied to the front upper and lower panels. 1 MAE 754 HW #2 Due date: Tuesday September 13, 2016

1. Plot the lift and drag coefficients and L/D ratio for a flat plate, varying the angle of

attack from 1 to 20 degrees for ? = 1.4 and M? =15 using exact shock wave and

expansion theory, the approximate theory in the hypersonic limit, Newton?s theory, and

modified Newton?s theory. I want 3 charts, one for Cl, Cd, and L/D, with four curves on

each to compare the four methods. Comment on the accuracy of the methods, and

whether or not you had a good time doing it.

First we summarize the equations needed in terms of the method used, and note that I use the

subscripts ?1? and ??? interchangeably. Also, I did not specify whether you can use the small

angle approximation or not. I?m going to assume that it is okay to do so for the deflection angle

? (angle of attack), but not necessarily the shock angle ?.

Exact Shock wave and Expansion theory

For the leeward surface, use the Prandtl-Meyer function ?M = ? +1 1 ? 1 2

M 1 tan 1 M 2 1

tan

? 1 ? +1 (1) Add to that value the angle of attack to obtain ?(M2). Solve Eq. numerically for M2., using

?(M2) for the LHS of the equation. Calculate the pressure drop with

? ? 1 2 ? 1

1+

M p2 (2)

2

= p ? 1 2 1+

M2

2 Calculate the pressure coefficient with Cp = 2

?M 2 P2 1 P (3) For the windward (lower) surface, obtain ? from M 2 sin 2 ? 1 tan? = 2cot? 2 M ? + cos2? + 2 where ? is the angle of attack. From oblique shock relations,

P2

2?

M 2 sin 2 ? 1

= 1+

? +1 P

Use Eq. 3 to calculate the pressure coefficient. Finally 2 (4) (5) MAE 754 HW #2 Due date: Tuesday September 13, 2016

C n = C p,l C p,u

Cl = C n cos?

C d = C n sin? (6) L / D = Cl / C d Hypersonic Shock wave and Expansion theory

For the leeward surface, use

M1

? 1

=1

M 1?

M2

2 (7) Then, p 2 M 12

=

p1 M 22 ? ? 1 (8) Use Eq.3 to calculate the pressure coefficient on the upper surface. For the windward (lower)

surface, obtain ? from

? +1

?=

?

(9)

2

where ? is the angle of attack. From oblique shock relations,

P2

2?

=

M 2 sin 2 ? P

? +1 (10) Use Eq. 3 to calculate the pressure coefficient, and then our final answer is obtained from Eqs. .

Newton?s method

The upper surface is Cpu = 0. The lower surface is calculated with Cpl = 2 sin2?. The final

answer is obtained from Eqs. .

Modified Newton?s method C

sin2, where

The upper surface is Cpu = 0. The lower surface is calculated with C

p

p

m

ax Cp max = 2

?M 2 pO2 1 p and 3 (11) MAE 754 HW #2 Due date: Tuesday September 13, 2016 ? ? + 1 M ? 1 1 ? + 2?M 2 2

= p 4?M 2 2? 1

? +1 pO 2 2 (12) The final answer is obtained from Eqs. .

Results

Figure 1 shows a plot of the results as a function of the angle of attack. For inviscid flows, the

exact shock wave and expansion wave theory is taken to be the correct answer. Qualitatively, all

methods correctly predict an increase in Cl and Cd with angle of attack, and the lift to drag ratio

is correct for all methods. At high angles of attack, the hypersonic approximation is within 10%

of the exact method for Cl and Cd and is much better than either Newton or modified Newton,

Figure 2. All methods give very poor predictions of the coefficients for low angles of attack.

The hypersonic approximation also gives negative lift and drag coefficients for ?&lt;2o. In

summary, none of the approximate methods are satisfactory for flat plates at low angles of

attack. Above ?=14o, the hypersonic approximation gives reasonable results within 10% of the

exact method, while the other methods were within 20 to 25%, with Newton?s result in better

agreement than the modified method. I had a great time working this problem. (a) (b) (c) Figure 1. Plots of lift coefficient (a), drag coefficient (b), and L/D (c) vs angle of attack ? using exact oblique

shock wave and expansion theory, hypersonic approximations, Newton?s theory, and Modified Newton?s

theory. Figure 2. Comparison of errors for each method, with respect to exact oblique shock and expansion wave

theory. 4 MAE 754 HW #2 Due date: Tuesday September 13, 2016

2. Giant mutant Mach 20 cicadas are on the attack! Assuming that they are 2D wedges with

a 10 m depth, their upper and lower surfaces are as shown in Figure 3. Note that the

lower two points are at -2.5 m. Use the shock expansion method to calculate the lift and

drag, assuming ?? =1 kg/m3 and T? is 273 K. Compare this with the tangent wedge

method, which keep in mind, can only be applied to the front upper and lower panels. Figure 3. Cicada upper and lower surfaces. The scale is in units of meters. See the attached code in Appendix B starting on p. 9. The lift and drag force were similar for

both methods, since the forces are dominated by the pressure behind the oblique shock wave on

the upper panel. For the tangent wedge method, only the first upper and lower panels are

considered, and give -7.56108 N and 6.24108 N, respectively for lift and drag. The shock

expansion method gives -7.97108 N and 6.21108 N. In level flight, this would require over

7.5108 N of lift force generated by the wings and 6.2108 N of thrust. The total force vector

would be ~100 MN, which is about twenty times the thrust generated by the 5 Saturn V F1

engines at full throttle. Strong bugs1! 1 True bugs, like cicadas and aphids, possess mouthparts where the mandibles and maxillae have evolved

into a proboscis, sheathed within a modified labium to form a &quot;beak&quot; or &quot;rostrum&quot; which is capable of piercing

tissues (usually plant tissues) and sucking out the liquids ? typically sap. Cicadathras, we will discover, have been

genetically engineered to lose the rostrum to be more hydrodynamically efficient. 5 MAE 754 HW #2 Due date: Tuesday September 13, 2016 Appendix A Matlab for #1

function hw301

%calculation of Cl, Cd, and L/D for exact shockwave and expansion theory,

%hypersonic and small angle approximations, newton's theory, and modified

%newton's theory

%constants

g = 1.4;

M1 = 15;

t = 1:.1:20;

%--------------------------------------------exact theory

%==upper surface==

nu1 = hyp('prandtl','gamma',g,'M1',M1);

nu2 = nu1 + t;

%allocate memory

M2 = t*0;

%solve for M2 using inverse prandtl meyer function

for tt = 1:length(t(t&lt;=15))

M2(tt) = hyp('aprandtl','gamma',g,'nu',nu2(tt),'M1',M1);

end

%for t&gt;15, M2 is really big, and the pressure ratio p2/p1 ~ 0, so just set

%M2 values for t&gt;15 to something really big.

M2(t&gt;15) = max(M2);

%calculate the pressure drop

p2p1 = hyp('p2p1wave','gamma',g,'M1',M1,'M2',M2);

%Cpu

cpu = cpfun(g,M1,p2p1);

%==lower surface==

%allocate memory

beta = t*0;

%find beta from theta-beta-M relations

for tt = 1:length(t)

beta(tt) = hyp('betafun','gamma',g,'M',M1,'theta',t(tt));

end

%calculation the pressure rise

p2p1 = hyp('p2p1shock','gamma',g,'M',M1,'b',beta);

%Cpl

cpl = cpfun(g,M1,p2p1);

%==Cl, Cd, L/D==

[Cl1, Cd1, LD1] = aerofun(cpl,cpu,t);

%keyboard 6 MAE 754 HW #2 Due date: Tuesday September 13, 2016

%-------------------------------------------hypersonic approximations

%==upper surface==

%calculate the pressure drop

p2p1 = ((1-(g-1)/2 .*M1.*(t/180*pi)).^2).^(g./(g-1));

%Cpu

cpu = cpfun(g,M1,p2p1);

%==lower surface==

beta = (g+1)/2 .*t;

p2p1 = 2*g/(g+1).*M1.^2 .*sind(beta).^2;

%Cpl

cpl = cpfun(g,M1,p2p1);

%==Cl, Cd, L/D==

[Cl2, Cd2, LD2] = aerofun(cpl,cpu,t);

%-------------------------------------------Newton's method

%==upper surface==

cpu = 0;

%==lower surface==

cpl = 2*sind(t).^2;

%==Cl, Cd, L/D==

[Cl3, Cd3, LD3] = aerofun(cpl,cpu,t);

%----------------------------------modified Newton's method

%==upper surface==

cpu = 0;

%==lower surface==

p20p1 = ((g+1).^2 .*M1.^2 ./(4*g.*M1.^2-2.*(g-1))).^(g./(g-1)) .*((1g+2*g.*M1.^2)./(g+1));

Cpmax = 2./(g.*M1.^2) .*(p20p1-1);

cpl = Cpmax*sind(t).^2;

%==Cl, Cd, L/D==

[Cl4, Cd4, LD4] = aerofun(cpl,cpu,t);

%-----------------------------------plot results

figure(1), plot(t,Cl1,t,Cl2,t,Cl3,t,Cl4)

legend('Exact','Hypersonic','Newton','Modified','location','northwest');

legend boxoff

xlabel('\theta (degrees)'), ylabel('Cl')

figure(2), plot(t,Cd1,t,Cd2,t,Cd3,t,Cd4)

legend('Exact','Hypersonic','Newton','Modified','location','northwest');

legend boxoff

xlabel('\theta (degrees)'), ylabel('Cd')

figure(3), plot(t,LD1,t,LD2,t,LD3,t,LD4)

legend('Exact','Hypersonic','Newton','Modified','location','northeast');

legend boxoff

xlabel('\theta (degrees)'), ylabel('L/D')

e1 = (Cl2-Cl1)./Cl1*100;

e2 = (Cl3-Cl1)./Cl1*100;

e3 = (Cl4-Cl1)./Cl1*100;

figure(4), plot(t,e1,t,e2,t,e3),xlabel('\theta (degrees)'), ylabel('error') 7 MAE 754 HW #2 Due date: Tuesday September 13, 2016

legend('Hypersonic','Newton','Modified','location','southeast'); legend

boxoff

function cp = cpfun(g,M,p2p1)

cp = 2/g./M.^2 .*(p2p1-1);

function [cl, cd, LD] = aerofun(cpl,cpu,t)

cn = cpl-cpu;

cl = cn.*cosd(t);

cd = cn.*sind(t);

LD = cl./cd; 8 MAE 754 HW #2 Due date: Tuesday September 13, 2016 Appendix B. Matlab Code for #2

function [L,D] = shockexpansion(xu,yu,xl,yl,freestream)

%function [L,D] = shockexpansion(x,y,side,freestream)

%2D version of shock expansion method. Takes a vector of xu, yu, xl, and yl

points

%which describe the 2D surface, top and bottom.

%Freestream conditions are specified by the structured array 'freestream',

%the 5th argument, which contains the following children:

%

%rho

density, kg/m^3

%gamma

specific heat ratio

%M

Mach number

%T

temperature, K

%MW

molecular weight

%

%if freestream is not specified the defaults are

%rho = 1 kg/m^3

%gamma = 1.4

%M = 10

%T = 273;

%set defaults if freestream not specified

if(nargin&lt;5)

freestream.rho = 1;

freestream.gamma = 1.4;

freestream.M = 20;

freestream.T = 273;

freestream.MW= 28.97;

end

rho = freestream.rho;

gamma = freestream.gamma;

g = gamma;

M = freestream.M;

M1 = M;

T = freestream.T;

R = 8314.5/freestream.MW;

p = rho*R*T;

%calculate angles of the surfaces

theta_u = atand(diff(yu)./diff(xu));

theta_l = atand(diff(yl)./diff(xl));

%calculate the deflection on downstream surfaces

dtheta_u = diff(theta_u);

dtheta_l = diff(theta_l);

%use oblique shock theory to obtain pressure at the nose

%find beta from theta-beta-M relations

beta_u = hyp('betafun','gamma',gamma,'M',M,'theta',abs(theta_u(1)));

beta_l = hyp('betafun','gamma',gamma,'M',M,'theta',abs(theta_l(1))); 9 MAE 754 HW #2 Due date: Tuesday September 13, 2016

%calculation the pressure rise on the nose section

pnp1_u = hyp('p2p1shock','gamma',g,'M',M1,'b',beta_u);

pnp1_l = hyp('p2p1shock','gamma',g,'M',M1,'b',beta_l);

%calculation the Mach number on the nose section

Mn_u = hyp('machshock','gamma',g,'M',M1,'b',beta_u,'t',abs(theta_u(1)));

Mn_l = hyp('machshock','gamma',g,'M',M1,'b',beta_l,'t',abs(theta_l(1)));

%allocate memory for pressure

p_u = 0*theta_u;

p_l = 0*theta_l;

%calculate the pressure at the first section

p_u(1) = pnp1_u*p;

p_l(1) = pnp1_l*p;

%------------------Shock expansion method---------------------------------%pressure distribution

M2u = Mn_u;

M2l = Mn_l;

for ii = 1:length(dtheta_u)

%set upstream mach number

M1u = M2u;

M1l = M2l;

nu1u = hyp('prandtl','gamma',g,'M1',M1u);

nu1l = hyp('prandtl','gamma',g,'M1',M1l);

nu2u = nu1u + abs(dtheta_u(ii));

nu2l = nu1l + abs(dtheta_l(ii));

%solve for M2 using inverse prandtl meyer function. Set to something

%huge if the angle is so large that the inverse function blows up

try

M2u = hyp('aprandtl','gamma',g,'nu',nu2u,'M1',M1u);

catch me

M2u = 1000;

end

try

M2l = hyp('aprandtl','gamma',g,'nu',nu2l,'M1',M1l);

catch me

M2l = 1000;

end

%calculate the pressure drop

p2p1u = hyp('p2p1wave','gamma',g,'M1',M1u,'M2',M2u);

p2p1l = hyp('p2p1wave','gamma',g,'M1',M1l,'M2',M2l);

p_u(ii+1) = p_u(ii)*p2p1u;

p_l(ii+1) = p_l(ii)*p2p1l;

end 10 MAE 754 HW #2 Due date: Tuesday September 13, 2016

%calculate the panel areas

Au = sqrt(diff(yu).^2 + diff(xu).^2);

Al = sqrt(diff(yl).^2 + diff(xl).^2);

%calculate the lift and drag forces

Lu = -cosd(theta_u).*p_u.*Au;

Ll = cosd(theta_l).*p_l.*Al;

L = (-cosd(theta_u).*p_u.*Au + cosd(theta_l).*p_l.*Al)

D = (sind(theta_u).*p_u.*Au - sind(theta_l).*p_l.*Al)

L = sum(L);

D = sum(D); function a = hyp(funstr,varargin)

%this is your one stop shop for supersonic and hypersonic relations

g = ;

M1 = ;

M2 = ;

b = ;

theta = ;

nu = ;

%parse arguments

for ii = 1:2:length(varargin)

if

(strcmpi(varargin{ii},'gamma'))

g = varargin{ii+1};

elseif(strcmpi(varargin{ii},'M1'))

M1 = varargin{ii+1};

elseif(strcmpi(varargin{ii},'M2'))

M2 = varargin{ii+1};

elseif(strcmpi(varargin{ii},'M'))

M = varargin{ii+1};

elseif(strcmpi(varargin{ii},'b'))

b = varargin{ii+1};

beta = b;

elseif(strcmpi(varargin{ii},'beta'))

b = varargin{ii+1};

beta = b;

elseif(strcmpi(varargin{ii},'theta'))

theta = varargin{ii+1};

elseif(strcmpi(varargin{ii},'t'))

theta = varargin{ii+1};

elseif(strcmpi(varargin{ii},'nu'))

nu = varargin{ii+1};

end

end

if(strcmpi(funstr,'prandtl'))

if(isempty(M1))

a = prandtl(g,M);

else

a = prandtl(g,M1); 11 MAE 754 HW #2 Due date: Tuesday September 13, 2016

end

elseif(strcmpi(funstr,'arcprandtl')||strcmpi(funstr,'aprandtl'))

a = aprandtl(g,nu);

elseif(strcmpi(funstr,'p2p1wave'))

a = p2p1wave(g,M1,M2);

elseif(strcmpi(funstr,'p2p1shock'))

a = p2p1shock(g,M,b);

elseif(strcmpi(funstr,'betafun'))

a = betafun(g,M,theta);

elseif(strcmpi(funstr,'machshock'))

a = machshockfun(g,M,theta,beta);

else

display('invalid function specified');

end

function nu = prandtl(g,M)

%prandtl-meyer function

nu = sqrt((g+1)./(g-1)).*(atand(sqrt((g-1)./(g+1).*(M.^2-1))))atand(sqrt(M.^2-1));

function M = aprandtl(g,nu)

%inverse prandtl meyer function, solving for M

%g, nu, keyboard

M = fzero(@(M) sqrt((g+1)./(g-1)).*(atand(sqrt((g-1)./(g+1).*(M.^2-1))))atand(sqrt(M.^2-1)) - nu,[1.000001,1e6]);

function a = p2p1wave(g,M1,M2)

a = ((1+(g-1)./2 .*M1.^2)./(1+(g-1)./2 .*M2.^2)).^(g./(g-1));

function a = p2p1shock(g,M,b)

a = 1+2.*g./(g+1).*(M.^2 .*sind(b).^2 - 1);

function M2 = machshockfun(g,M,t,b)

%mach number behind oblique shock

M2 = sqrt(((g-1).*M.^2 .*sind(b).^2 + 2)./ ...

(2*g.*M.^2 .*sind(b).^2 - (g-1))./(sind(b-t).^2));

function a = betafun(g,M,theta)

a = fzero(@(b) 2*cotd(b).*((M.^2 .*sind(b).^2 - 1)./(M.^2 .*(g+cosd(2*b)) +

2))-tand(theta),[theta, 60]); 12

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