Calculating Water Density at 10.92 km Depth

In summary, the pressure at the bottom of the ocean is 1.16*10^8 Pa. If the pressure is calculated ignoring the change of density, the pressure is 1.10*10^8 Pa. The water density at the bottom using the compressibility and the actual pressure is 1033 Pa but I should get 1080 Pa. Could someone please give me a hint to what I am doing wrong?
  • #1
Swatch
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At the bottom of the ocean at 10.92 km the pressure is 1.16*10^8 pa.
If the pressure is calculated ignoring the change of density the pressure is 1.10*10^8 pa
I have to calculate the water density at the bottom using the compressibility and the actual pressure.

Since k= - Delta V/Vo*DeltaP
DeltaP*k= - DeltaV/Vo and V=m/rho
DeltaP*k=-(rho1-rho2)/rho2 where rho1 and rho2 are the water density at the surface and at the bottom respectively.
I end up with:
rho2=-rho1/(DeltaP*k -1)

From this I get rho2 as 1033 pa but I should get 1080 pa

Could someone please give me a hint to what I am doing wrong?


Thanks.
 
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  • #2
I don't recognize your form of the definition of the bulk modulus. It should be in the form of:

[tex]k = -V \frac{\Delta P}{\Delta V}[/tex]

I am still working through the units. You should end up with kg/m^3 for density, not Pa. Pascals are units of pressure.
 
  • #3
OK. I finally went thru it. First, it will depend on what values you are using for the density of water at the surface and for the bulk modulus.

After substituting [tex]\rho = \frac{m}{V}[/tex] in the above definition and some algebra, I end up with:

[tex]k = - \frac{\Delta P}{\frac{\rho_1}{\rho_2} - 1}[/tex]

Like I said, it depends on what you use for k and [tex]\rho[/tex] at the surface. I used k = 2.2x10^9 and [tex]\rho_1[/tex] = 1000 kg/m^3 and ended up with a result of [tex]\rho_2[/tex] = 1055.6 kg/m^3 which is pretty close to what you say numerically you should get.
 
Last edited:
  • #4
I don't understand.
According to my textbook the bulk modulus is -DeltaP*V/DeltaV

The compressibility, k is the reciprocal of the bulk modulus and so is
k=-DeltaV/V*DeltaP
 
  • #5
Ahh. I see. I'm used to seeing "k" for bulk modulus and "c" for compressibility. It doesn't really matter though. It should all work out in the end. If you solve my equation for bulk modulus for [tex]\rho_2[/tex] and invert it, you will get the same result that you have in your original post.

Also, it might help if you were to present the equation as: k = -DeltaV/(V*DeltaP). I am admittedly horrible when having to do orders of operation, so I like to see parentheses.

What values did you use for the density at sea level and for compressibility? Did you see my comment about the units you ended up with?
 

1. How do you calculate water density at 10.92 km depth?

The formula for calculating water density at a specific depth is d = ρ0 * e^(-z/H), where d is the density, ρ0 is the density at the surface, z is the depth, and H is the scale height. In this case, z = 10.92 km and H = 8.5 km for seawater. Plug in the values and solve for d to get the water density at 10.92 km depth.

2. What is the density of water at 10.92 km depth?

Using the formula mentioned above, the density of water at 10.92 km depth can be calculated. However, it is important to note that the density of water can vary based on factors such as salinity, temperature, and pressure.

3. How does depth affect the density of water?

The deeper the depth, the higher the pressure, which can compress the water molecules and increase the density. Additionally, colder water tends to be denser than warmer water. Therefore, as depth increases, water density generally increases as well.

4. Why is it important to calculate water density at specific depths?

Calculating water density at specific depths is important for understanding ocean circulation patterns and the distribution of heat and nutrients in the ocean. It also helps in predicting and understanding the behavior of ocean currents and their impact on climate.

5. Can water density at 10.92 km depth change over time?

Yes, water density at a specific depth can change over time due to various factors such as changes in temperature, salinity, and pressure. These changes can be influenced by natural events like El Niño or human activities such as climate change. It is important to continuously monitor and study these changes to better understand their impact on the ocean and the planet as a whole.

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