Project Session Lift and Weight Group E11B
Solutions to the problems:
Task 1: Please refer to spreadsheet, all angles and extra calculations and plottings are ther. ther. Task 2: 2.1 Why do we plot ! p "negati#e$ instead of ! p "positi#e$ upward%
&n the formula of the pressur p ressure e coe'cient:
C p =
p − p∞ 0.5 ρ V ∞
2
( you can see that
when p is smaller than p )( !p will be negati#e* So the plot will also be negati#e* +ence( they use ! p so that the cur#e will be a positi#e cur#e* ,his way you can compare it better with the downward ! p cur#e*
-*-$ a$ the mach number decreases from from .*./0 to .*.0* 2ccording 2ccording to Bernoulli3s law 1
2
1
2
P1+ ρV ρ V 1= P 2+ ρ V 2 2
2
P1 doubles when 4 - 5 416-* b$ 2t 05.* 05.*7( 7( the the 8ow 8ow bec become omes s comp compre ress ssib ible le and and the the follo followi wing ng e9ua e9uati tion on holds: −γ /( γ − 1)
P γ −1 2 =(1 + M ) Pt 2
Which means P is in#ersely proportional to 0-( so the pressure drops 9uadratically with the speed*
2.3) es( es( there is a relation relation between the thic;ness and the pressure pressure distribution of the the airfoil*
2.4: What is the highest value of p that is to !e found in the test and "h#$ %o" do "e call this point$ =ormules used: !p 5 "P>Peff )/(0,5*ρ∞*V∞2) Peff =p∞+(85/135)*(Patm-p∞) Peff is the corrected ip!t press!re "eca!se the ip!t press!re meas!red ma# ot "e acc!rate eo!$h% & tas' 1 #o! ca see that the maim!m p is at a a$e of attac' of 12%5 de$rees% (.(?/@7A??-$ ,he point at maimum !p is called the stagnation point* ,his is when !p 5 1* &n this case it is about 1-*A degrees* 2t a higher angle of attac;( the !p drops again* ,he reason for this is because the pressure under the airfoil is higher than the pressure on the airfoil( because of the diCrence between those pressures( lift is generated* 2t the maimum !p( the wind is in a 8ow that is called attached 8ow* When the angle of attac; is higher dan it would be on the maimum !p it creates a saperate 8ow* Because of this seperate 8ow the pressure under the airfoil drops* ,herefore there is no more lift generated
T&'( 3 3.1: please refer to the spreadsheet attached 3.2 Plot the pressure distribution from the pro#ided data in ecel* Dpper and lower surface ha#e to be plotted in the same graph*
Pressure istri!ution 1*1 .*7 .*/ p
.*@
Dpper Surface
Lower Surface
.*. . >.*-
.*-
.*@
.*/
.*7
1
1*-
>.*@ >.*/ *+
Dse the ,rapeoid Fule to nd the lift coe'cient of the 0+@@ aerofoil* ,he three column titles are gi#en to point you in the right direction*
pper surface x+c . .(..A.A .(.1@@? .(.-7.(.@/1H
p .(??.? .(A7H11 .(-..H? >.(./7? >.(--?/H
'tep si-e .(..A.A .(..?@@ .(.1H1 .(.1H?H .(.--1
&verage height .(H7?..(??A .(./7@A >.(1@/H7 >.(-H?/@
&rea covered .(..?7@AA1 .(..H17777 .(...?7@A >.(..-/H/H >.(../177@
.(./7 .(.?@@7 .(1-@A@ .(1A7-H .(1?A@ .(-A/1 .(-H7A/ .(-?.(H1-7 .(@-.- .(@H. .(A-.?H .(AH1/7 .(/-1?/ .(/H1-H .(H1?..(H/@H? .(7.7@ .(7@?1@ .(77/@ .(?1?- .(?@H1? .(?/?H .(?7/-? .(??/A 1
>.(-?/1 >.(7/./ >.(@1-/ >.(@-1./ >.(@1H?>.(@.AA7 >.(7H.1 >.(A?1 >.(-7/? >.(-?AAH >.(-/.@7 >.(--/?A >.(1?-1? >.(1A?A >.(11H?@ >.(.7.@H >.(.-7/7 .(.--? .(.7-./ .(1-1./ .(1A/.@ .(17A@ .(-1.-H .(--@A .(1//. .(??1A-
.(.-/17 .(.../ .(.H .(.H1 .(
[email protected] .(.@-?A .(.@A/ .(.@H/ .(.@7?A .(.A..H .(.A./H .(.A.H1 .(.A.-7 .(.@?1 .(.@HHA .(.@AHH .(.@AA .(
[email protected] .(.H.(.-7? .(.-H?/ .(.--A1 .(.1/A? .(.1.-@ .(..@H >1
>.(AH7A >.(?? >.(@1/7 >.(@1?@? >.(@11HA >.(?/-?A >.(H.AA >.(@7?A >.(1-1 >.(-H7.-A >.(-@H1A >.(-.?AH >.(1H.H >.(1A?@A >.(.??-.A >.(.A@AHA .(..17.A .(.AH1HA .(1.1A/ .(17AA .(1H.HA .(1?H7A .(-1H1 .(1?A1? .(AH7HHA .(@?AH/
>.(..?/71>.(.1-..7/ >.(
[email protected]?/H/ >.(.1AAHA//@ >.(.1/AA/@/7 >.(.1H.-.7H >.(.1/?-1HHA >.(.1/-7/7/H >.(.1A-H7H/@ >.(.1?-.H1>.(.1-@?.? >.(.1./-H-?A >.(..7H.1?/ >.(../H.@@7 >.(..@HH.? >.(..-@?H7?7 H(7/.HHE>.A .(..--H@ .(..HH7..(..@AA/?1 .(..@HHHA1 .(..@@A/.@ .(../.A1H .(..1??7H@/ .(..-..7@? >.(@?AH/
o"er surface x+c . .(...-.(../A .(.17 .(.@/@ .(.AA??
p .(??.? .(-7@/? >.(1--H >.(--1?7 >.(-HH- >.(11-/
'tep si-e .(...-.(../1 .(.11/A .(.1//@ .(.-1A .(.-A7
&verage height .(/H71 .(.7.H1 >.(1H-/-A >.(-@?/.A >.(-?@-@A >.(1?7
&rea covered .(...1@.17 .(...@?@HA>.(..-.11.71 >.(..@1A@-H >.(../-7-11 >.(..7-A.7@
.(.71H? .(1117 .(1@/..(17@11 .(--A7A .(-H.7A .(17/.(/7/ .(@-.A .(@H-1 .(A-/AH .(AH?7.(/--? .(/7A .(H-A .(HH7H.(7-17H .(7/1-? .(7?/A .(?-H./ .(?A-A? .(?H-HH .(?7H@ .(??/H 1
>.(-7@ >.(-7>.(-???A >.(-/?/7 >.(-//1 >.(-.AA>.(1H7/H >.(1A-A>.(1-?H@ >.(1.7.? >.(.7717 >.(.H-A >.(.AHAA >.(.@/.7 >.(.@H >.(.-/H>.(.1H1A >.(.11. >.(..-H.(...-.(..A1 >.(.. .(.1?? .(.AH/1 .(??1A-
.(...@ .(.@1? .(.7.? .(.@1H@ .(.@A .(.@HHH .(.A..1 .(.A1H.(.A-7/ .(.A/ .(.A-A .(.A-@H .(.A1./ .(.@? .(.@/H .(.@1A .(.?@.(.A-1 .(..A/ .(.-AA .(.-.17 .(.1@// .(..?-H .(.. >1
>.(-/.7 >.(1177A >.(-7@71A >.(-A1@A >.(--1./A >.(1?-.?A >.(1/AA?A >.(1@11 >.(117?1A >.(.?71A >.(.7.-/A >.(./@?A >.(.A171A >.(.@
[email protected] >.(..H-A >.(.-1?A >.(.1@.? >.(../7HA >.(..1-A .(..17/A .(...-AA .(..7@/A .(.7HH .(A-@A/A .(@?AH/
>.(..?H?A@@ >.(.1.//@7 >.(.1.7@7/. >.(.1.A//-H>.(..??@H?-A >.(..?1H/H7 >.(..7-71@./ >.(..H-??-@@ >.(../-7A7@H >.(..A-/@7@ >.(..@-H@111 >.(..@.H?-H >.(..-/@A/H@ >.(..1?H?7@A >.(..1@-@H17 >.(...?@/@?A >.(...AAA@-7 >.(...-@-./? >.(....7@(H/1AE>.A A(1@A?E>./ .(...1-@.?H .(...A??7 .(..1H1./@ >.(@?AH/
Step sie is the diCerence between the 6cn and 6cnI1
2#erage height 5
C pn+ C pn +1 2
2rea co#ered 5 step sie J area co#ered* Where !L is the lift coefficient( K is the angle of attac; and ! a is the aial force coefficient* &n 2nderson "-..7$ it is stated that for small angles of attac; "K AM$ you can assume that ! L ≈ !n*
∫ C
=∑ Area coveredlower=−0,12155
∫ C
=∑ Area covered upper=−0.66495
p ,l
p ,u
C l ≈ C n=∫ C p , l −∫ C p ,u = 0,543401
3.3 !alculate the lift coe'cient*
( ) ( )) ( ( ) ( )) ( ) ( ))−(¿¿)=[( )− ( ) ] −([( )− ( ) ] +[− ( )+ ( ) ] ( − ) ( ) =∫ ¿ 0.13
∫ 0
(
x x d 1− 0.84 c c
(
2
x 1−200 c
x c
x 0.42 c
1
x d c
2
+ ∫ −2.47 +2.63 x d x
1
x c
0
C p ,l
c
0,13
C p , u
200 x 3
x d c
3
c
0.13
0
c
x 2.47 c
x 1.315 c
2
1
1
0
1
∫¿
C l=C n=
0
,2SN @ O2!2 ..-. "Wind tunnel$
C l =0
.
α
DC l / dα C l
@*@HA/ 2bout 1* at 1H
max
C l =0
•
,he point where
•
,his is logical as it is a symmetric airfoil* DC l / dα is the gradient of the graph( which was calculated to be about
•
α
is when the graph crosses the ais* ,his is .*
@*@HA/ "calculations shown in the spreadsheet$ C l is the point where the graph stops increasing "its maimum$* We do max
not ha#e enough angles to determinate this accurately* +owe#er( we got while in the lab that the stall angle was about 1H>17 degrees* =rom this information( we linearied the
C l
max
to be around 1* at 1H degres*
+owe#er this is not accurate as the gradient actually starts decreasing before reaching the maimum point of the cur#e*
,2SN A
)
0.13
=1.453
A*1 O2!2 ..-. has a lift cur#e that is going through the origin( because this aerofoil is symmetrical( and symmetrical aerofoils "non>cambered$ ha#e ero lift at an angle of attac; of ero( whereas cambered aerofoils ma;e for a lower pressure at the top of the wing and thus creating a pressure diCerence* ,his results in a lift coe'cient at an angle of attac; of ero* A*When the angle of attac; gets too high( attached 8ow turns into separated 8ow on the upper surface of the aerofoil( which causes the pressure diCerence to drop signicantly and !L drops precipitously* 2t this stage( the airplane is in a stall*
A* ,he formula for !l( when the aircraft has a stationary( le#el 8ight( is m∗g 1 2
2 ρ V S * Gi#en is that S( 4 and m constants are when comparing 0ars to
Earth* ,he difference is thus g6( therefore the ratio is "gmars6mars$6 "gearth6earth$( which gi#es a ratio of *1 A*@ 2ccording to the ,hin 2erofoil ,heory the slope of the Lift cur#e !L( K 5 -Q*
/* 2t steady horiontal 8ight: Weight5Lift and Rrag5,hrust
05-*A ;g G5*H-m6s5.*1--AH? ;g6m 45Am6s S5.*HHm!L5 0JG6"*A4-JS$ !L51*@?
,2SN / Build the ecel sheet capable of calculating the total lift force in Oewton* See ecel sheet !alculate E using E9uation 1-* 1
A ellipse = a! 2
A 1 2
= ellipse
1 4
a!
A rectangular =a
"=
A rectangular A 1 2
ellipse
=
1 1 4
= 4
!
!
=or tas; H( 7 and ? see the ecel sheet* Teroen U Turian Bart and Louis Tin Vian U 0a Tonathas la'ta Loedy
-*1( *-( *( ,as; H( 7( ? -*-( ,as; A U / -*-( tas; A( tas; / ,as;1( ,as;@ -*@( and helped in all( assisting with data calculation and more
