Hydrostatic Equation/Finding depth

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One atmosphere of pressure is equal to 101,325 Pa. If the density of water is 998 kg/m3, what is the necessary depth to reach 2 atm of pressure relative to the surface



Hydrostatic equation: p=-wh where p is change in density, w is specific weight (density*gravity), and h is change in altitude.




Now the hydrostatic equation is p=-wh where p is change in density, w is specific weight, and h is change in altitude. Now, to get w, it is simply w=(998)(9.81)=9790.38. So we now have p=-(9790.38)h. Now it is (101325*2)=-(9790.38)h. Multiply and divide and we get -20.69889014=h.


My question is whether I did this right or not. If I was successful, should I put it as positive 20.698 meters?
 
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I do not think your equation applies. For this problem, the density of water can be considered constant (incompressible fluid assumption).
 
Well, he gave us a hint of: "Hint: what equation relates altitude (or depth) with pressure?"

This equation would make sense. I'm not sure what you mean by the density being a constant. It already is. However, when calculating for pressure you need the specific weight of water which requires gravity times density. Also, from what I was told, 1 atm is 10.3 meters, so this answer should make sense.
 
The answer does make sense. I am just questioning how you got there.
Hydrostatic equation: p=-wh where p is change in density...
So, what was the change in density... of a constant-density fluid?
 
My apologies, I meant p is change in pressure.
 
Well, now that I think about it, wouldn't this answer be wrong? If it is the change in pressure, and we are going down to 2atm, wouldn't it still come out at positive 1atm, since the starting atm is 1? Or do we just count the starting pressure as 0?!
 
what is the necessary depth to reach 2 atm of pressure relative to the surface
They are looking for relative pressure. Surface pressure is your reference.
Prelative = Pabsolute@20.6 - Pabsolute@0