The other factor we are failing to consider is intraabdominal pressure.
The damage to internal organs (or likely hood of their removal) is going to be a function of the total pressure differential. So the shop-vac is a few PSI less than atmospheric, what is the pressure in the abdomen? The organs in question are below the diaphragm so it should be pretty steady, but isn't there some abdominal fluid in there with a little pressure on it to hold everything? Skin, when cut, sure acts like it's higher pressure inside than out.
Millcreek, or one of the med folks, can you chime in? We'll also need the average uterine circumference so we can go from PSI to actual expelling force.
I dealt with internal pressure on another matter. "Normal" systolic blood pressure is supposedly around 120mmHg, which is
(120/760) x 14.7 = 2.3 PSI
so take it from there.
But that's what brought up the question in my original post in this thread although unstated: What's the pressure differential between a person's internal pressure and the "vacuum" cleaner's ultimate pressure.
Calling it 11 PSI as I concluded above, that brings us to
(14.7PSI - 11PSI) + 2.3 PSI = 6 PSI total pressure differential between the person and the vacuum cleaner hose..
(14.7PSI - 11PSI) is the "vacuum" in the vacuum cleaner hose.
Ick.
The area of application is significant, but I guess you can figure about 1.75 in
2 for a typical hose diameter.
(1.75/2)
2 X pi= 2.40 in
2 area .
So 6 PSI X 2.40 in
2 = 14.4 pounds of force on that area.
Seems like enough to do substantial damage to a
sensitive area. Lots more than a Prom Date hickey. I note you can hold a vacuum cleaner hose against the palm of your hand without leaving much more than a reddish area for a couple of minutes.
Beyond that, we get into force distribution over a diaphragm (the skin), and I don't want to go any further.
My original question has been answered: Pressure in a closed-off vacuum cleaner hose is approximately 11PSI.
Terry, 230RN