# Homework Help: Problem with pressure

1. Jul 14, 2010

### astenroo

1. The problem statement, all variables and given/known data
Recently I stumbled on an exercise in a physics book. My problem is, that I am confident there is a fault somewhere in the textbook, as my numbers don't add up with the ones in the "Answers" segment. The assignment:

A ship that sank off the coast of Aland during the Crimean war lies on a 40 m depth. Researchers want to investigate the the ships cargo from a small submarine. In the sub there is a "pressure equalization chamber" (airlock?) from which entrance to the water is possible from 1.00 m² hatch. a) calculate the hydrostatic pressure at this depth b) What is the pressure in the pressure chamber if the hatch is influenced by a 50.0 kN force? $$\rho$$=1.03*10³ kg/m³ and p0=101 kPa

2. Relevant equations

The hydrostatic pressure p=$$\rho$$gh
Total pressure at depth h p=p0 + $$\rho$$gh
Pressure p=F/A
3. The attempt at a solution

a) Hydrostatic pressure at 40 m : 101*10³ + 1.03*10³*9,81*40 = 505 kPa
The answer provided by the textbook is 392 kPa

b) I assume that the force acting on the hatch is directed outwards (towards the sea) creating a a slight overpressure in the chamber to keep it from flooding as soon as the hatch is opened.

Then: p=F/A, 50000 N/ 1.00 m² = 50 kPa. The total pressure in the chamber should then be 50 kPa + 505 kPa resulting in a total pressure of 555 kPa in the chamber. The pressure in the chamber according to the textbook is 455 kPa.

I know the hydrostatic pressure at 40 m can in no way be less than 5 atm (505 kPa). Am I right or am I thinking wrong somewhere?

2. Jul 14, 2010

Total pressure = atmospheric pressure+hydrostatic pressure.Part a of the question asked you to calculate hydrostatic pressure only i.e. P=h*rho*g.

3. Jul 14, 2010

### astenroo

quote: "Total pressure = atmospheric pressure+hydrostatic pressure.Part a of the question asked you to calculate hydrostatic pressure only i.e. P=h*rho*g. "

Yes, the total pressure would be atmospheric+hydrostatic. I am, however, quite sure that you have to consider the atmospheric pressure in order to have an accurate answer. Why else would they provide you with p0=101 kPa? Anyway, omitting the atmospheric pressure will not give a hydrostatic pressure of 392 kPa with the density of water being 1.03x10³ kg/m³. It would, however, give a pressure of approximately 455 kPa in the chamber (assuming it is of airlock type and the 50 kN force acting on it is generated by the internal pressure and not external).

4. Jul 14, 2010

P=40 *1000*9.8=392kPa
From this I'm assuming that the book told you to take the density of water to be
1000kg/m^3.Have a look.Of course the density of sea water is higher and your answer would be closer to the true value.
To calculate total pressure you do need to add atmospheric pressure but part (a) asked you to calculate hydrostatic pressure only.

5. Jul 14, 2010

### astenroo

Ah of course :) ! I am now wondering whether the numbers for the density for water and the air pressure was meant for another assignment in the book. Now that it's settled, then part (b). If the force, F, acting on the hatch (1.00 m²) is 50 kN, the pressure exerted on it would be 50 kPa, right? Then, assuming F is projected on the hatch from the inside, it means the chamber is overpressurized. Then, 392 kPa + 50 kPa = 442 kPa. Answer in the book is 455 kPa. If the values for density (1.03*10³ kg/m³) and air pressure (101 kPa) was meant for another assignment, then you have helped me solve the puzzle. The only other values given by the book was depth (40 m), area of the hatch (1.00 m²) and the force (50 kN) acting on the hatch.

Thank you!

6. Jul 14, 2010

### astenroo

I did some thinking. It seems that the only way 455 kPa in the chamber is possible, is if the 50 kN force acts on it from the outside. The book did not provide any information on the direction of the 50 kN force. Hence I assumed it was directed outward and not inward.

Thank you!

7. Jul 14, 2010