# Homework Help: Hydrostatic Blood Pressure

1. Nov 14, 2009

### robertsa

1. The problem statement, all variables and given/known data

a person is 1.8 m tall, standing upright. His heart is 1.4m above his feet. He is doing a head stand. Calculate the hydrostatic blood pressure in (a) his feet and (b) his head.

2. Relevant equations
P-Po= density x g x d

3. The attempt at a solution

When standing upright:

Brain
Pheart-Pbraind=1059.5 x 9.81 x 0.4
13500-Gauge P in brain= 9.3 KPa

Feet
Pfeet-Pheart=1059.5 x 9.81 x 1.4
feet-13500=1059.5 x 9.81 x1.4
P feet= 28 KPa

How do I know what distance to use when he is upside down? You would think these numbers would just be opposite?

2. Nov 14, 2009

### Delphi51

Interesting question! I'm wondering why the position of the heart matters? "Hydrostatic" suggests to me that it is the pressure ignoring the pumping, as if you have a column of blood 1.8 m tall.

3. Nov 14, 2009

### robertsa

We did the solution for the same height, when standing upright in class. We always to the distance to be from the heart..im just not sure how to approach when hes upside down.

4. Nov 14, 2009

### Delphi51

He is 1.8 m high and the heart is 1.4 m high, so it is 0.4 m from head to heart. The heart is 0.4 m above the head when he is "standing" on his head. Your calc is correct, if the calculation is to be done with reference to the heart.

Will the feet pressure then be negative because the feet are above the heart?

5. Nov 14, 2009

### robertsa

The calculation i did above was for when he was standing upright.. when he is upside down the pressure in the brain should be greater than that of the presure in the feet.

6. Nov 14, 2009

### Delphi51

Yes, that is why I suggested that the 28 kPa for the feet should be negative, since you are taking the pressure with respect to the heart level.

To my mind, it would make more sense to make the pressure zero at height 1.8 m, but of course you or your prof can take the reference pressure anywhere you like.