What is the difference between load line hydrometer and draught survey hydrometer ?
Load line and draft survey hydrometers are a pure nonsense, because none of them reads density - details are explained in Open Letter to AMSA, NCB and USCG.
Gdynia, 13th March 2017.
OPEN LETTER TO AMSA, NCB and USCG.
Re.: False load line and draft surveys in Australia and USA.
1. In your opinion FW where FWA applies has:
- SG15C/15C = 1
- density = specific weight in vacuum = specific Wv = 0.9991 T/m3;
- specific weight in air = specific Wa = 0.998 T/m3;
- air boyancy correction = ABC = 0.002
- SG = specific Wa + ABC = 1 T/m3.
Consequently for load line = technical purposes displacement = D = 1 T means allowed relative displacement = relative D = D/SG = 1 m3 of water of SG15C/15C = 1, while for draft survey = commercial purposes relative D = 1 m3 means Wa = 0.998 T.
Density in your opinion is of no use for stability purposes.
2. Assumption that for technical purposes water of SG15C/15C = 1 has SG = 1 T/m3 is however false, because in FPS and British Engineering system (Ref 1) used for technical purposes gravity force (Tons) = mass (Tons) (Ref 2).
Consequently water of SG15C/15C = 1 has SG = density= 0.9991T/m3.
In coherent SI system the same water has in standard location SG = 9.801171 kN/m3, while in the same location water of unit density has SG = 9.81 T/m3. Instead of absolute gravitational weight system in metric world relative system is used where water of SG15C/15C = 1 has at any location SG = 9.801171/9.81 = 0.9991, while water of unit density has SG = 1. Consequently in relative system water of SG = density = 1 T/m3 has SG = RD = 1.
Obviously in above systems D = 1T of vessel in water of SG 15C/15C = 1 and SG = 0.9991 means relative D = 1/density15C = 1.0009 m3 and Wa = relative D * specific Wa = 0.9989 T i.e. your Wa = 0.998 T for D = 1T is false.
3. If one assumes that weight = result of weighing of mass, then by OIML R33 one has to assume that 1 T = mass of water of SG20C/20C = SG62F/62F == SG15C/15C = = SGT1/T1 i.e. pure water:
- has for technical purposes weight in vacuo = Wv = mass = 1 T;
- has for commercial purposes: - weight in air = Wa = 0.9989 T (1 m3 = 0.99776 T, 1 UK gallon = 10 lb) in standard imperial conditions (Ref 3); - apparent mass = Ma = 1 * [(1 - 0.0012/0.998203)/(1-0.0012/8.3909)] = 0.998941 T in standard NIST conditions (Ref 4); - conventional value = Mc = 1 * [(1 - 0.0012/0.998203)/(1-0.0012/8)] = 0.998948 T in OIML standard conditions (Ref 4&5).
If imperial standard is international standard then:
- for technical purposes Wv = 1 T = mass of 1.000027 m3 of pure water at 4C = mass of 1.0009 m3 of pure water at 15C etc;
- for commercial purposes Wa = 0.9989 T = Wa of 1.000027 m3 of pure water at 4C = Wa of 1.0009 m3 of pure water at 15C etc,
Consequently water of SG4C/4C = SG15/15C = 1 has:
- specific Wv at 4C = 0.999973 T/m3, specific Wv at 15C = 0.9991 T/m3 etc for technical purposes;
- specific Wa at 4C = 0.998873 T/m3, specific Wa at 15C = 0.998 Tm3 etc for commercial purposes. Other way ABC = specific Wv at T1 - specific Wa at T1 = density at T1 - specific Wa at T1 = 0.0011 T/m3 and your assumption that ABC = 0.002 is false.
4. In view of the above a vessel of relative D = 1.002 m3 and with vacuum in cargo tanks has in water of unit density: a) displacement = D = mass = Wv = 1.002 m3 * 1 T/m3 = 1.002 T; b) Wa = 1.002 m3 * 0.9989 T/m3 = 1.0008978 T.
After filling tanks with 6 m3 of air of density = 0.0011 T/m3 the vessel has: a) initial D = 1.0086 T b) initial relative D = 1.0086 m3.
After loading 6m3 of water of unit density the vessel has: a) final D = 1.0086 T - 0.0066 T + 6T = 7.002 T; b) final relative D = 7.002 m3.
Other way loaded cargo changed relative D by 7.002 m3 - 1.0086 m3 = 5.9934 m3 and result of draft survey: a) 5.9934m3 * density = 5.9934 * 1 = 5.9934 T, if density is used for the survey; b) 5.9934 m3 * specific Wa = 5.9934 * 0.9989 = 5.9868 T, if specific Wa is used for the survey.
In fact the cargo of 6 m3 of water of unit density has: a) Wv = mass = volume of water * density = 6T b) Wa = volume of water * (density - ABC) = 5.9934 T. The conclusion is that result of draft survey = commercial weight = Wa of cargo is correct if density is used for the purpose, while it is false if specific Wa is used instead because in such case result is neither mass nor Wa.
5. Water of SG15C/15C = 1.0009 and water of SG4C/4C = 1.000027 are for technical = stability purposes the same, because both have the same actual SG = density = 1 T/m3, what means that 1 m3 of such waters provides the vessels with the same boyancy force = BF = 1 T. If the vessel of allowed relative D = 7.002 m3 and D = 7.002 T in water of SG15C/15C= 1.0009 (SG = 1) uses FWA, then it keeps the same allowed relative D, D and FWA in water of SG4C/4C = 1.000027 (unit SG) i.e. in water which for drinking purposes is fresh, while for stability purposes is water of unit density = unit SG = unit RD. Consequently one has to assume that a vessel in both waters above complies with Convention on Load Lines, because the Convention allows the vessels to keep FWA in "fresh water of unit density" i.e. in water of SG = density = 1 T/m3 and/or of SG = RD = 1. If water of "density other than unity" referred to by the Convention is dock water, then: a) pure water of SG15C/15C = 1 is dock water where vessel above is allowed to use DWA = FWA * (1.025-0.9991)/0.025 = 1.0236 * FWA to keep D = 7.002T; b) pure water in Gatun Locks of Panama Canal is dock water where vessel above is allowed to use DWA = FWA * (1.025 - 0.9954)/0.025 = 1.184 * FWA to keep D = 7.002T. Other way for load line purposes actual density is true, while SGT1/T1 is false.
6. The fact is that in oil industry (Ref 6) a volume = 6.302595113 barrels = 1.002033m3 of water of API60F = 10 and SG60F = 1 (by definition of API60F) has Wa = 1 T and mass = 1.0011 T, because such water has: a) specific Wa = 0.997971 T/m3; b) density = Specific Wa + 0.0011 = 0.99907 T/m3. Water of SG60F = 1.000030028 has density = 0.9991 T/m3 and specific Wa = density - 0.011 = 0.998 T/m3 i.e. a volume = 1.002 m3 of such water has Wa = 1T and mass = 1.0011 T.
The same density and specific Wa has water of SG15C/15C = 1 i.e. pure water at 15C. Consequently: a) Wa = 1 T means for commercial purposes weight of 1.002 m3 of water of SG15C/15C; b) D = mass = 1.0011 T means for load line purposes relative D = D/actual density and SG15C/15C = 1 is useless for stability purposes. Simple like umbrella if one knows what 1 T means.
7. Obviously you do not know what 1 T means. That's why you can't understand what the Convention means when it says 1 T and unit density. That's why you instruct the armies of your surveyors to use: a) SG15C/15C = 1 = reading of so called load line hydrometer for load line purposes; b) Specific Wa at 15C = reading of so called draft survey hydrometer for draft survey purposes. In fact however for both purposes i.e. for stability purposes density is true and SGT1/T1 and specific Wa are false as explained above. Conseqently the use of readings of both above mentioned hydrometers i.e. so called marine hydrometers is a reason why every load line survey and draft survey of your surveyors is false. Remark. For the sake of order USCG is not involved in draft surveys, while it's subsidiary = NCB is formally not involved in load line surveys. Kind regards, Capt A. Bratek email@example.com
PS. For more details see http://www.sites.google.com/site/marinehydrometers/
Well as the name suggests,
Draft survey hydrometer is used to know the density of the sea water that will be used during draft survey for calculating the weight of cargo loaded.
And Loadline hydrometer is used to know the density of sea water that will be used for calculating the compliance with the loadline convention.
But isn't the density of water same ? So if these two hydrometers are supposed to give different densities for the same water, is one of these incorrect ?
Absolutely not !!!
Draft survey measures the actual density of the water. Whereas Loadline hydrometer measures the Apparent density (also called specific gravity) of the water.
Let us see what is the difference !!!
Specific gravity is the density in relation to something. When we measure the specific gravity of water, it is in relation to the fresh water at 15 C.
So if specific gravity of sea water as measured with loadline hydrometer is 1.025, it means that the density of water in relation to the fresh water at 15 C is 1.025.
Specific gravity is a ratio, a Number and does not have any unit.
Now what is the actual density of water if the specific gravity is 1.025.
The fresh water density at 15 C is 0.999 Kg/L in vacuum. So the actual density of water with specific gravity 1.025 will be
1.025 x 0.9991 = 1.024 Kg/L in vacuum
The correction for converting this density from vaccum to the air is 0.0011. So actual density of sea water (with specific gravity of 1.025) in air will be
1.024-0.0011 = 1.023 Kg/L in air
Now If we measure the density of the same sea water with draft survey hydrometer, it will give the reading of 1.023 Kg/L in air.
When to use which Hydrometer ?
If you want the dock water density to know how much you can submerge the load line and still comply with the load line convention, use loadline hydrometer. This will give you the specific gravity (apparent density) of the sea water.
But when calculating the weight of cargo loaded by draft survey, we want to know the actual sea water density (and not specific gravity). In this case we must use draft survey hydrometer.
How to distinguish between Loadline and Draft survey hydrometer ?
Both the hydrometers can be distinguished by the marking on the hydrometer.
As loadline hydrometer measures specific gravity or Relative density, it will have the marking of notation RD or Sp. Gr.
Draft survey hydrometers have the unit of the density and temperature marked on these. For example it may have the marking as Kg/L at 15 C.
Ship staff must use the correct hydrometer for the intended use. If another hydrometer is used, proper corrections must be applied to it.