ind blowing past a flag causes it to flutter in the breeze. The frequency of thi

Question

ind blowing past a flag causes it to flutter in the breeze. The frequency of this fluttering, , is assumed to be a function of the wind speed V , the air density , the accel- eration due to gravity g, the length of the flag l, and the area density A (with dimensions of mass per unit area) of the flag material. It is desired to predict the flutter frequency of alarge12mflagina10m/swind. Todothisamodelflagwithl=1.2mistobetested in a wind tunnel.

(a) Express the non-dimensional frequency as a function of the other non-dimensional groups.
(b) Determine the required area density of the model flag material if the large flag has A =1kg/m2.

(c) What wind tunnel velocity is required for testing the model?
(d) If the model flag flutters at 6Hz, predict the frequency for the large flag.

Explanation / Answer

>> GIVEN:

>> Flag Length, L" = 12 m

>> Model flag length, Lm = 1.2 m

>> Velocity, V = 10 m/s

>> Model flag frequency, wm=6 Hz

>> Area density, A = 1 kg/m2

As, By Dimesional Analysis

Frequency, w=f(p,g,l,pA)

=> w=T^-1

=> v=LT^-1

=> p=ML^-3

=> g=LT^-2

=> L" = L

=> A=ML^-2

>> Now,as

>> No of reference dimensions=3

>> No of variables=6

=> No of pi-terms=6-3=3

=> w(l/g)1/2 = [V/(g*L)1/2 , A/L)

Part a) By SImilarity

=> (A)m/mLm=(A/L)

As, m =

=> (A)m/Lm=(A/L)

=> Model Area density, (A)m = (A/L)*Lm=(1/12)*(1.2) = 0.1 Kg/m2

(A)m = 0.1 Kg/m2 ....ANSWER.........

Part (b). Again, By Similarity

=> Vm/(gmLm)1/2 =V/(gL)1/2

and, gm = g

=> Model velocity, Vm = (Lm/L)1/2 *(V) = (1.2/12)1/2 *(10) = 3.162 m/s

Vm = 3.162 m/s ........ANSWER.........

Part (c) .. Again, By Similarity,

=> wm*(Lm/gm)1/2 =w*(L/g)1/2

As , gm = g

=> w = wm*(Lm/L)1/2

=> w = 6*(1.2/12)1/2 = 1.897 Hz = 1.9 Hz

=> w = 1.9 Hz ........ANSWER................

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