- #1
Spartanlol
- 19
- 0
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?
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?