Draft alterations depend on water density

Good day

It happened that I forgot the formula for the calculation of vessel's draft at different water density, though I searched the Internet and found nothing in English. So I decided to translate the article, which I found in Russian.

When the vessel passes from one water basin to another, the salinity (seawater density) changes.  It's very useful to know the way vessel's draft will change. And for the determination of vessel's draft delta we will turn to physics.
When steaming in the water density ρ and ρ' vessel's displacement accordingly will be:

D = ρ×V     and     D = ρ'×V',

where V - is vessel's volume before transition of water density;

V' - is vessel's volume after transition of water density.
When we equate the right parts of the formula, we receive:

ρ×V = ρ'×V'     or     V/V' = ρ'/ρ

The vessel's volume can be represented by the main dimensions L, B, T and coefficient δ:

V = δ×L×B×T     and     V' = δ'×L'×B'×T'

It may be assumed that L, B and δ do not change their amounts when water salinity changes, due to the fact that vessel's volume does not change considerably. In this case the volume changes due to the change of vessel's draft in following way:

ρ×T = ρ'×T'     or     T/T'= ρ'/ρ

So, when vessel transits from a sea water to fresh water, or vice versa, the draft will change in inverse ratio to the according water density. The change of volume is determined in following way:

ΔV = V' - V = D/ρ' - D/ρ = D(ρ - ρ')/(ρ×ρ')     or     ΔV = V×(ρ - ρ')/ρ'

It can also be determined as the volume with area of the current waterline (which does not change considerably) and height, which equals to difference of drafts ΔT, i.e. ΔV = S×ΔT. Therefore:

S×ΔТ = V×(ρ - ρ')/ρ'

and so

ΔТ = V/S × (ρ - ρ')/ρ'     or     ΔТ = D/(S×ρ) × (ρ - ρ')/ρ'

Example:

When the vessel transits from fresh water (ρ = 1,000 mt/m³) into sea water (ρ = 1,025 mt/m³):

ΔT = D/(S×1,000) × (1,000 - 1,025)/1,025

As the figure will be negative, accordingly vessel's draft will become less and the vessel will float. And when the vessel transits vice versa:

ΔT = D/(S×1,025) × (1,025 - 1,000)/1,000

In this case the result will be positive, the vessel will submerse and vessel's draft will increase.

References:

this article in Russian at moryak.biz

P.S. some time has passed and I learned that the formula is not suitable for the usual operator, as it contains the "area of waterline", which is still pending to be determined and is not usually known to an operator. So I decided to amend the formula for it to be suitable.

In order to determine S we should use the waterplane coefficient: 

ɑ = S/(L×B), therefore S = ɑ×L×B.

The waterplane coefficient for dry seagoing bulk vessels should be around 0,82 - 0,85; so we may assume it to be fixed as 0,84. As such the final formula will be amended as follows:

ΔТ = V/(0,84×L×B) × (ρ - ρ')/ρ'

or

ΔТ = D/(0,84×L×B×ρ) × (ρ - ρ')/ρ'

where

D - vessel's displacement

L - vessel's length

B - vessel's breadth

ρ - water density before draft alteration

ρ' - water density after draft alteration

Hope this should clarify a bit. In case you have any further questions, please ask them below.