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#1. energy, E, of the From declassified pictures of an atomic blast, Geoffrey Taylor was able to predict the atomic bomb solely from measurements of the radii of the blast, r, at a particular time, t, after the blast and knowing the ambient air density, a) Use the Buckingham Pi theorem with E, ρ, and t as core variables to ρ, and pressure, P. determine relevant dimensionless parameters and write the relation between them in functional form. b) Because the energy is very large, for small times after the explosi dimensionless parameter containing P is nearly zero and the functionalon the dependence on this dimensionless parameter can be considered constant. Knowing this and using your result from part a), you want to determine the either have to know or measure if you wanted to predict the energy for a different bomb. What parameters would you energy of this bomb? Solution a) M r 0 t 0 P 1 ρ 1 E 1 L T 10 01 - - 12 - 03 2 - 2 5 - 3 = 2 dimensionless groups
" 1 = rta # b^ E c^ = L [ T ] a^^ $ %& M L 3 ' ()^ b^ $ % &^ ML T 22 ' ( )^ c * a = + 2 / 5 , b = 1 / 5 , c = + 1 / 5 * " 1 = rt +^2 /^5 #^1 /^5 E +^1 /^5
" 2 = Ptd # e^ E f^ = LTM 2 [ T ] d^^ $ %& M L 3 ' ()^ e^ $ % &^ ML T 22 ' ( )^ f * d = 6 / 5 , e = + 3 / 5 , f = + 2 / 5 * " 2 = Pt^6 /^5 #+^3 /^5 E +^2 /^5
" 1 = rt +^2 /^5 #^1 /^5 E +^1 /^5 = f ( " 2 ) = f ( Pt^6 /^5 #+^3 /^5 E +^2 /^5 ) b) The functional form becomes: !
determine the energy/yield at a particular time after the explosion of a different bomb if^ "^1 =^ rt #^2 /^5 $^1 /^5 E #^1 /^5 =^ f^ (^ "^2 )^ =^ cons tan^ t , so you can you know the air density, time, and radius at that particular time.
#2 Solution: This is very similar to the Chevron dust release problem from lecture, and the gravity settling chamber section of Denn. a) The time for the particles to settle is simply the distance they must fall (5.2 m) divided by the terminal velocity of the spherical particle. The terminal velocity V be found iteratively (assume Re<1, use Stokes’ law to calculate Cp appears in both the drag coefficient CD and the Re, so VD, calculate Vp from Cp mayD, check Re with the V quantity that is independent of Vp calculated, etc.) or this canp. be done graphically, by finding a I calculated
!
CD Re^2 = 43^ gDp^3 (^ " $ p 2^ #^ ")^ " (which is independent of Vp) with the data given, this is:
!
CD Re^2 = 43^ gDp^3 (^ " $ p 2^ # ")^ "= 43 (9.81^ m^ s^2 )(^10 #^4 m (1.8)^3 (2000 x 10 # (^5) kg #1.25 / m %^ kg s )^ 2 m^3 )(1.25^ kg^ m^3 )= 101 so we could plot
!
go through a point where Re = 10 and C^ CD^ =^101 /Re^2 on our CD^ vs Re^ plot. It would have a slope ofD = 1.01. It would intersect the C^ - 2 on the logD- log plot, andvs Re plot at Re=3.43, CD =8.59. Rather than actually plotting on my would intersect, and realized it would be in the intermediate regime, where CD vs Re plot, I simply eye-balled where the curves
!
CD " 18 Re#0.6. Substituting into our equation !
CD = 101 /Re^2 , we find Re=3.43, CD =8.59. From Re=3.43, we can calculate Vp, since !
Vp = Re # D^ " p. The calculation yields Vp = 0.494 m/s The time to fall 5.2 m at this terminal velocity is simply 5.2 m/ V 10.5 s p = 5.2 m/(0.494m/s) = b) If the nearest town is 1600 m from the field, we want a wind speed such that the particles travel less than this distance in the time it takes them to fall to the ground, so Vwind < (1600 m)/(10.5 s) = 152 m/s
v 2 =^ Q A 22 =(! /^ Q 42 ) D 22 =(! 0 /. 410 )( 0 m. 123 /kg m) 2 = 8. 84 m/s
v 3 =^ Q A 33 =(! /^ Q 43 ) D 32 =(! 0 /. 415 )( 0 m. 173 /kg m) 2 = 6. 61 m/s
Now we decompose linear momentum equation into x and y components. Examine the x-component:
FFxx == " " vv 112 Q A 11 !! " " vv 22 Q^22 A cos 2 cos( # ( # 2 ) 2 !)! " v " 3 vQ 323 A cos 3 cos( è 3 ( è ) + 3 ) P + 1 AP 11 , Ax 1 !, xP! 3 AP 33 , Ax 3 , x
+! #! vv 21 QQ 21 cos=(( 1000 " 2 ) = ! m^ kg 3 ( 1000 ) ( 3. 54 m kg 3 )^ m s( 8 ).( 840. 25 m s)( 0 ms 3. 10 )= ms 8843 )cos. 2 (! N 45 o)=! 625. 1 N
!+ " Pv 1 3 AQ 1 , 3 x cos= P ( è 13 ( !) = /! 4 ( 1000 ) D 12 = m^ kg 3 () 7000 ( 6. 61 m sPa)( 0 ).( 15! m/s^34 ))cos ( 0 .( 3060 om)= ) 2! 495 = 494. 6 N. 8 N
! F xP 3 = A 8843 , x =. 2 !N P 3 !( " 625 / 4 .) 1 D N 32 cos! 495 ( # 3 .) 6 =N!+( 4000 494. 8 PaN)(! " 45 / 4 .) 4 ( 0 N. 17 =m 213 )^2 cosN ( 60 o)=! 45. 4 N
Examine the y Fy =! " v 2 2 A 2 - sincomponent: ( # 2 )! " v 32 A 3 sin( è 3 )! P 3 A 3 , y
Fy =! " v 2 Q 2 sin ( # 2 )! " v 3 Q 3 sin( è 3 )! P 3 A 3 , y
!! # " vv 32 QQ 32 sinsin( ( " è 32 )) ==!!(( 10001000 m m kg^ kg 33 ))(( 86 .. 8461 m^ m s s))(( 00 .. 1510 m^ mss 33 ))sinsin (( 60! 45 o)=o )!= 858625. 7. 1 NN
! F yP 3 = A 6253 , x =. 1 !N P 3 !( " 858 / 4 .) 7 D N 32 sin! 78 ( # 3. 6 ) =N!=( 4000! 312 PaN)( " / 4 )( 0. 17 m)^2 sin( 60 o)=! 78. 6 N
