[ RadSafe ] RE: Sandia test block movement

Jaro jaro-10kbq at sympatico.ca
Wed Mar 16 02:27:50 CET 2005


Hi Jay,

I should really be responding to your message off-list, since people seem to
be getting tired of the topic.
But I think this might be of interest to others interested in nuclear plant
containments. Apologies to all who I may be upsetting nonetheless.... my
hope is that the changed subject heading will allow you to discriminate.

You are right -- the concrete block was moved. In fact it was placed on an
air cushion, to approximate a frictionless surface. The technique is similar
to that of the well-known "ballistic pendulum," in that it allows you to
calculate precisely how much energy was used to move the block, versus how
much went into blast effects of destroying the impacting aircraft and in
blasting a hole (less that 2" deep) in the concrete.

This is a simple grade 12 high school physics problem. Conservation of
momentum and conservation of energy, with frictionless surfaces :

mass of jet = m

velocity of jet before impact = v1

mass of block = 25 * m     (i.e. block is 25 x the mass of the jet)

velocity of block before impact = 0

velocity of everything (jet + block) after impact = v2

Total energy before impact:

E1 = 1/2 * m * v1^2 + 1/2 * 25 * m * 0 = 1/2 * m * v1^2

Total momentum before impact:

m * v1 + 25 * m * 0 = m * v1

Total energy after impact:

E2 = 1/2 * m * v2^2 + 1/2 * 25 * m * v2^2 = 1/2 * 26 * m * v2^2

Total momentum after impact:

m * v2 + 25 * m * v2 = 26 * m * v2

Momentum before impact = Momentum after impact:

m * v1 = 26 * m * v2

Thus v1 = 26 * v2 (i.e. the final velocity of the jet + block was 1/26 of
the jet's velocity)

Energy before impact = Energy after impact + destructive energy

Thus DE = E1 - E2

or

DE/E1 = 1 - E2/E1

DE/E1 = 1 - (1/2 * 26 * m * v2^2 ) / (1/2 * m * v1^2 )

DE/E1 = 1 - (26 * v2^2 ) / ( v1^2 )

substituting for v1 = 26 * v2 from the momentum equation:

DE/E1 = 1 - (26 * v2^2 ) / ( 26^2 * v2^2 )

DE/E1 = 1 - 1 / 26 = 25 / 26

Thus the destructive energy was 96% of the total jet's energy --  96% of the
original jet's energy went into damaging the jet and the wall. The
*movement* of the wall was therefore essentially irrelevant to the
scenario - it absorbed only 4% of the jet's energy. The block immitated a
large immovable object (i.e. a containment) because of its huge mass.

Hope this helps.

Cheers,

Jaro
^^^^^^^^^^^^^^^^^^^^^^


-----Original Message-----
From: radsafe-bounces at radlab.nl [mailto:radsafe-bounces at radlab.nl]On Behalf
Of MacLellan, Jay
Sent: Tuesday, March 15, 2005 7:42 PM
To: radsafe at radlab.nl
Subject: Re: [ RadSafe ] Unidentified Helicopters Nearly Fired Upon
OverNuclearPower PlantBy


I get the digest, and I am afraid I grew tired of this thread long
before I finished all the messages.  However, there is one point about
the Sandia crash test that wasn't mentioned in what I read.  The
concrete block was moved about 3 feet by the impact.  I'm not an
engineer, but I think that makes a big difference.  It wouldn't seem to
accurately model a direct impact on a containment structure.
Jay MacLellan

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