TBR: artery pressure vs atmospheric pressure

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Aug 17, 2014
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I came across a set of questions TBR physics page 96 passage 3 that dealt with measuring the blood pressure by inserting cannula into the artery. The passage stated that average gauge pressure of an artery was 10^4 Pa. One question required the understanding of whether the atmosphere or the artery pressure was greater. I thought that the atmospheric pressure was a lot greater than the gauge pressure since atm pressure = 101.3 kPa. However the answer stated that "Since the average gauge pressure in the artery is greater than zero, the average pressure in the artery is larger than the atmospheric pressure". I know that P total = P gauge + P atm. In this case was I supposed to use the Ptotal (instead of just P gaguge) and compare it with atmospheric pressure? When I think about it conceptually it makes sense, when you get a cut, blood leaks out due to the difference in pressure.

Generally speaking when do I add P atm to P gague?

In that drawing that I attached, would I include P atm to calculate the total pressures of fluid in test tube A and test tube B, even though they're not open to the atmosphere directly?

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Well the gauge pressure is just what you read off of a pressure gauge. It sets the atmospheric pressure as the 0 point and only measures up from there. I think using the equation as P_gauge=P_actual-P_atm is a little more intuitive. I like to think of pressure as the force of particles hitting a specific area (P=F/A), I think it makes it easier to think about it from a microscopic point of view.

To answer your questions:
1. If you want to measure something in a perfect vacuum (a zero pressure environment) then you would add P_atm to P_gauge. Since most things are done at atmospheric level, almost never. If it helps you, think about pressure being a relative value like voltage.

2. Well the atmosphere is pushing down on the exposed fluid in the bucket which is what is bringing the fluid up into the test tube, but it depends on how you'd want to measure pressure. Usually for that setup you'd look at how high the water went (h) and use ρ_water*g*h=P (pressure head equation{or how mercury barometers work}), but you'd just find that P=P_atm for the test tube in the bucket.

Actually, I just saw that you drew a lid on the bucket, if that is the case, then your P in the ρ_water*g*h equation would be equal to 0 since the water wouldn't rise above the level of the water in the tub in the test tube (you need a force to push the water up, and if you take out the atmosphere there isn't anything that will push it up).
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