Tanker Cargo calculations
TANKER CARGO CALCULATIONS
The volume of cargo on board can be determined by means of soundings or ullage measurements and
calibration tables (tank tables).
The purpose of the cargo calculations is to convert the observed volume into weight.
Calculation of trim, stability, freeboard, shear forces, bending moments is based on weights.
On the B/L the quantity of cargo is stated as a weight (Metric Tons , Long Tons, Short Tons, Pounds, etc. )
When making the stowage plan a lot of information has to be gathered, a lot of factors have to be taken into
account
Cargo calculations are important because they are the link between the available space and the weight to be
loaded
The relationship between the volume and the mass or weight can be expressed by the density (specific gravity,
litre weight, API, relative density etc.)
Density and volume change is function of the temperature
The weight of a cargo is of course independent of the temperature but the weight in air (apparent weight) ≠ the
weight in vacuum (true weight)
Density
Fundamentally
Density: Mass per unit volume [kg/m3 or kg/litre]
When calculating cargo
True density: Weight per unit of volume in vacuum
Apparant density: Weight per unit of volume in air
Mass is a measure of the quantity of material in a body and is constant regardless of geographical location,
altitude or atmospheric conditions
Weight is the force with which a body is attracted to the earth and varies from place to place with « g », the
acceleration of gravity
Standard “g” = 9,81m/sec2
Weight = Mass x 9,81m/sec2
All commodities are sold by weight and this means weight in air.
The term «weight» in general practice has been accepted as being the value secured when an object is weighed
in air
RSR Page 1
This weight or «weight in air» is often converted to «weight in vacuum» by the application of an air buoyancy
correction (vacuum factor)
Volume and density corrections
Weight = volume x density
Only true if volume and density are known at the same temperature
3 solutions
1)Correct the volume to the temperature of the density
VCF = Volume correction factor
Volume correction to 60°F or 15°C
2)Correct density to the temperature of the volume
DCC = Density correction per degree centigrade
3)Correct both to the same reference temperature, 60°F or 15 °C
Use of ASTM Tables
Normally the density or API is provided by the terminal or surveyor in the load ports and what is used will be
dependent on the region / port of loading.
For example in USA / Canada, Persian Gulf, API usage is prevalent, while entire of Europe and Asia uses Density
at 15C.
PROCEDURE OF CALCULATIONS
Working with Density at 15oC :
1) Observed Ullage ‐ apply corrections ‐ get Corrected Ullage
2) Observed Interface ‐ apply corrections ‐ get Corrected Interface
3) From Corrected Ullage, find Total Observed Volume TOV (in M³)
4) From Corrected Interface, find Volume of Water (in M³ )
5) TOV ‐ Water = Gross Observed Volume (GOV) of Cargo (in M³)
6) Use Density at 15C and Observed Temperature (oC) and find Volume Correction Factor (VCF) from Table 54
7) Gross Standard Volume (GSV) = GOV x VCF (cubic metres)
8) Weight Correction Factor (WCF) = Density at 15C in vacuum ‐ 0.0011 (or the Density at 15C in air)
9) Weight in Air (Metric Ton) = GSV x WCF(Density at 15C in air)
10) Weight in Vaccum (Metric Ton) = GSV x Density at 15C in vacuum
RSR Page 2
TANKER CARGO CALCULATIONS
On board a Tanker ullage is measured either by an ullage tape, Ullage Temperature Interface, (UTI) / Sonic tape,
Whessoe gauge, Radar gauge etc.
This Ullage is to be corrected for Trim, & list
Correction depends the location of ullage port, its height , how far it is ford of aft bhd and by what distance it is
displaced from the C/L of tank.
Use of the Whessoe Tank Gauge
The function of the gauge is to register the ullage of the tank at any given time, in particular when the liquid level
in the tank is changing during the loading and discharge periods.
The gauge is designed to record the readings not only at the top deck level of the tank but also remotely at a
central cargo control room. A transmitter is fitted on the head of the gauge for just this purpose.
The unit is totally enclosed and various models manufactured are suitable for use aboard not only oil tankers,
but chemical and gas carriers as well .
Tank Radar System
This is a totally enclosed measuring system which can only be employed if the tank is fully inerted. Systems are
generally fitted with oxygen sensor and temperature sensor switches, so if the atmosphere in the tank is hot or
flammable the radar will not function.
RSR Page 3
The main unit of the system is fitted on the deck with an inserted cable tube into the tank holding a transducer.
Cable then carries the signal to a control unit in the cargo control room where the signal is converted to give a
digital read‐out for each tank monitored .
The transducer would be fitted as close to the centre of the tank area as was possible. Such siting tends to
eliminate errors due to trim and list.
Q 1 A crude oil tanker of LBP 228m has a box shaped tank of Dimensions 32 x 20 x 20 m. trimmed 1 m by stern.
Initial ullage of 0.60m is measured by Radar beam level gauge fitted 3 m ford of aft bhd on centerline of the tank.
Given the density of oil at 15°C = 0.810t/m³ and observed temp is 24.5°C. On completion of unloading v/l
trimmed 3.8 m by stern & ullage of tank was 19.88 m Calculate the Qty of oil discharged.
In Δ PTE
PT / TE = tan ϴ also tanϴ = t/LBP
PT = 13 x 1/228 = 0.0570 m
UT = UP + PT = 0.60 + 0.0570 = 0.6570m
Sounding of (mean level) = 20 – 0.657 = 19.343
GOV = 32 x 20 x 19.343 = 12379.52 m³
Now after cargo is discharged
To establish if wedge exits
RSR Page 4
To establish if DL ˂ 32 m (Wedge forms)
TM / ML = tan ϴ
TM = UM – UT & tan ϴ = trim / LBP tan
Or ML = 7.2m
DL = DM + ML = 3.0 + 7.2 = 10.2m
DL ˂ 32 m hence Wedge forms
In Δ KDL
KD/DL = tan ϴ & tan ϴ = 3.8/228
KD =10.2 x 3.8/228 = 0.17m
Vol of wedge = ½ x 10.2 x 0.17 x 20 = 17.34 m³
Vol discharged = 12379.52‐ 17.34 = 12362.18m³
Qty discharged = 12362.18 x Vcf x Wcf = 9915.78 MT
Q.2 A box shaped tank 28x18x16 m is to be loaded with crude oil at a temp of 24°C . The δat 15°C = 0.8250, if
2% of the vol of tank is to be left for expansion calculate Final obs ullage by the measuring tape at the ullage port
located 2m ford of aft bhd , 1m above and 3m to port of CL.Given LBP 240m, Trim 3m by stern, list ½°(s).
First correct for Trim then for list
Since 2% of the vol of tank to be left for exp
Actual mean height of oil = 16 x 98/100=15.68
UT = 16 +1 ‐15.68 = 1.32m
In Δ PTE
PT / TE = tan ϴ also tanϴ = t/LBP
RSR Page 5
PT = 12 x 3/240
PT = 0.15 m
UP = UT ‐ PT = 1.32 – 0.15 = 1.17m
TAPE CORRECTION :‐
In Δ UOP
UO / UP = Cos ϴ
UO = UP Cos ϴ
Also Tan ϴ = trim/LBP or ϴ = tanˉ¹ (trim/LBP)
Or ϴ = tanˉ¹ (3/240) = 0.86°
UO = 1.1699m
Now tackling List
In Δ OCE OC/OE = tanϴ
OC = OE tanϴ = 3 x tan 0.5° = 0.02618
Therefore UC = UO + OC
= 1.1699 + 0.02618 = 1.19608
In Δ UDC UD/UC = cosϴ
UD = UC cosϴ
= 1.196 X COS 0.5° = 1.196034 which will be the observed ullage
GOV = 28 x 18 x 15.68 = 7902.72 m³
VCF = 0.9924
GSV = 7902.2 x 0.9924 = 7842.659 m³
RSR Page 6
Weight in air = 7842.659 x (0.8250 – 0.0011)
= 6461.5923 mt
********************************************************************************************
Q.3 A box shaped tank 35m x 24.3m x 15m had 83m³ of ballast on arrival at load port on a ship of LBP 219m .
This tank was then loaded with a) 5000m³ of oil at 27°C (δ15°c =0.8193) b) 5000 mt of oil (δ15°c =0.8867)
Calculate the final ullage by UTI gauge (temp 12°C, Trim zero, List 2.5°P) . Ullage port located 6m ford of aft
bulkhead & 4m to port of centre line and 88 cms above the tank top.
Parcel A
Obs Vol at 27°C = 5000m³, temp = 27°C, δ15°c =0.8193 Vcf at 27°C = 0.9897,
Nov at 15°C = 5000 x 0.9897 = 4948.5 m³ Vcf at 12°C = 1.0026
Obs Vol at 12°C = Nov at 15°C / Vcf at 12°C = 4948.5 / 1.0026 = 4935.7 m³
Parcel B
Qty of oil = 5000 mt , temp = 27 deg, δ15°c =0.8867, WRF = 0.8867 – 0.0011 = 0.8856
Nov at 15°c = Mass / WRF = 5000 / 0.8856 = 5645.9 m³
Vcf at 12°c = 1.0022
Obs Vol at 12°c = Nov at 15°c / Vcf at 12°c = 5645.9 /1.0022 = 5633.5 m³
Total Obs Vol of oil & water
= 4935.7 + 5633.5 + 83 = 10652.2 m³
Depth of oil & water = 10652.2 / 35 x 24.3 = 12.525 m
Therefore Corrected ullage at 12°c = (15 + 0.88) – 12.525 = 3.355 m
Therefore Obs ullage (corr’d for list only) = 3.355 – 0.175 ( list correction = b x Tan ϴ) = 3.180 m
************************************************************************************
Q.4 A box shaped bunker tank 28 x 15 x 12 m was loaded at 40°C to 98% capacity. δ15°c (in vac) 0.8825. If this
fuel is now heated to 50°C and 20% of bunker is used, Calculate the ullage from tank radar gauge. Tank radar
gauge is fitted 5m ford of the aft bulkhead, 3m port of C/L, 1.5m above the deck, trim 4.19m, list 2 deg (s), LBP
200m. Also calculate Qty remaining now in metric tonnes.
Vol of tank = 5040 m³
Gov = 5040 – 100.8 (2%) = 4939.2 m³
Mass of oil = Gov x Vcf (40°C) x Wcf = 4939.2 x 0.9812 x 0.8814 = 4271.564 t
20% of bunker consumed = 854.31 t, ROB = 4271.564 ‐ 854.31 = 3417.25 t
Gsv = Mass of oil / Wcf = 3417.25 / 0.8814 = 3877.07 m³
RSR Page 7
Gov = Gsv / Vcf (50°C) = 3877.07 / 0.9738 = 3981.38 m³
Height of oil = Gov / Area = 3981 / 28x15 = 9.479 m
True ullage = 13.5 – 9.479 = 4.021 m
Now Trim Corr’n = (L/2 – a) x trim / LBP = (28/2 – 5) x 4.19 / 200 = 0.1885 m
List Corr’n = Dist off ullage port x Tan (list) = 3 x tan 2° = 0.1047 m
True ullage = Obs ullage ± Trim corr’n ± List corr’n
4.021 = Obs ullage + 0.1885 – 0.1047
Observed ullage = 3.9372 m
***********************************************************************************************
Q.5 A rectangular slop tank 10 x 10 x 20 m upon arrival was found to have 1m of water at 20°c and 1m of Arabian
light at 20°c. Density of Arabian light δ15°c (in vac) 0.8750. Arabian heavy crude of δ15°c (in vac) 0.8950 has to be
loaded on top such that ullage of slop tank will not fall below 20 cm at a max voyage temp of 38 °c. Calculate the
ullage at load port if temp of both the oils is 12°c. Find the weight of Arabian heavy oil assuming (1) there is no
comingling (2) negligible change in vol of water with change in temp (3) No trim change
Vol of tank = 10 x 10 x 20 = 2000 m³
Available vol at Max voy temp = 10x10x19.8 = 1980 m³
Gov of water = 10 x10 x1 = 100 m³
Gov of Arab light @20°c = 10 x 10 x 1 = 100 m³
Gsv of A/L = 100 x Vcf(0.9962) = 99.62 m³
Now Gov of A/L @38°c = Gsv / Vcf (0.9825) = 101.394
Gov of water @38°c = 100 m³
Total Gov of tank = Gov (A/L) + Gov (water) + Gov (A/H)
1980 = 101.394 + 100 + Gov (A/H)
Therefore Gov of A/H = 1778.605 m³ (@38°c)
Gsv of A/H = Gov x Vcf(38°c) = 1778.605 x 0.9830 = 1748.369 m³
Mass of A/H = Gsv x Wcf (0.8950 – 0.0011)
= 1748.369 x 0.8939 = 1562.867 t
Now Gov of A/H @12°c = Gsv / Vcf(12°c) = 1748.369/1.0022 = 1744.53 m³
Gov of A/L @12°c = Gsv / Vcf(12°c) = 99.62 / 1.0023 = 99.391 m³
Gov of water @12°c = 100 m³
RSR Page 8
Total Gov of oil = Gov (A/L) + Gov (A/H) + Gov (water)
= 99.391 + 1744.53 + 100 = 1943.922 m³
Gov = Area x Height of tank
1943.922 = 10 x 10 x Height of oil
Height of oil = 19.439 m
Therefore ullage = 20 – 19.439 = 0.5607 m
=============================================================================================
Recapitulate
When ullage is taken by Ullage tape / Sonic tape / UTI
Ullage Corrected for Tape (UP) = UO/Cos ϴ
Ullage Corrected for Tape & Trim (UT)
= UO/Cos ϴ +/‐ [{ (L/2) – a} x{ trim / LBP} ]
Ullage Corrected for Tape, Trim & List
= UO/Cos ϴ +/‐ [{ (L/2) – a} x{ trim / LBP} ]+/‐ b x tanΦ
Trim correction:‐ +ve if ullage pipe towards trimmed end
‐ve if ullage pipe is opposite to the trimmed end
List correction:‐ +ve if ullage pipe towards List side
‐ve if ullage pipe is opposite to the Listed side
‘L’ length of tank
‘a’ location of ullage port ford of bhd
‘b’ location of ullage port away from CL
‘ϴ’ angle of Trim
‘Φ’ angle of List
RSR Page 9
