Chapter 12.1











12  cARGO
cALCULATION

12.1    CALCULATION OF MAXIMUM ALLOWED
LIQUID VOLUME

In this part, we will take a look at the
different methods in calculating cargo onboard. The quantities of cargo we will
load are specified in the charter party and this information is given directly
from the charter or from the operation in the ownerłs office. When we load and
transport liquefied gases there are some variables that we have to have in
mind, the setting of the safety valveÅ‚s “relief valve", the cargo temperature
when loading and at which temperature we should discharge the cargo. The type
of gas carrier and the equipment we have onboard is also important in the
flexibility of our transport.

 

12.4.1                  
Maximum filling limit

Maximum filling limit is the maximum volume
liquid we are allowed to load in the cargo tank. In chapter 15 of the IMO gas
code, we find that the maximum filling can be 98% of full tank volume. Filling
limit depends on the set point of relief valve and the density of the actual
cargo.

Formula for maximum volume liquid is as
follows:

Filling limit = rR
/ rL
* 98%




rR


Density of reference temperature
on the relief valve setting




rL


Density for actual cargo
temperature




 

This means that if the relief valve setting
is low, we can load more than if the setting is high. If there is a possibility
to take off one or more of the pilot valves, we can increase the liquid volume
loaded. We then have to calculate the difference between the pilot settings.
The time used for loading will also increase if we have a lower set point on
the cargo tankłs relief valves. What we always have to avoid is an uncontrolled
venting.

 

Uncontrolled venting is when we get such a
high pressure in the cargo tank that the relief valve opens.

 

If we look at some examples e.g. propane and
the first example relief valve setting is 4,5 bar and the other example relief
valve setting is 0,5 bar. Cargo temperature is
35oC.

 




Relief valve setting 4,5 bar


4,5 bar + 1 bar = 5,5 bar @
5oC


= 523,3 kg/m3




Relief valve setting 0,5 bar


0,5 bar + 1 bar = 1,5 bar @
-32oC


= 570,2 kg/m3




Cargo temperature
35oC


 


= 573,7 kg/m3




 

In our example with 1000 m3 tank,
we can see that the difference is about 45mt. With 4,5 bar setting we can load
513,259mt and with 0,5 bar setting we can load 558,900mt. If the freight rate
is 80 USD/mt we then miss USD 3651. If we are on a gas carrier on 10 000 m3
the loss of income will then be USD 36 510.

When we reduce the set point on cargo tank
relief valves, the time used for loading and discharging will increase. What we
have to avoid is letting the cargo tank fill 100% with liquid.

 

On semi-refrigerated gas carriers, normally
the lowest relief valve setting is 0,5 bar. There are two or more pilot valves
e.g. 3,6 bar and 5,2 bar. If we change the relief valve setting, we have to
mark that on the cargo tank and also note it in the deck-log book.

On fully refrigerated gas carriers the relief
valve setting is about 0,25 bar and there are often facilities for putting one
extra weight on the pilot, normally 0,2 to 0,3 bar. That means we have a relief valve setting of 0,45 bar. The extra
setter is allowed to be used only while loading or gas freeing.

 

In all cargo calculations in this compendium,
we use T0 = 273oC and atmospheric pressure to 1,013 bar
if nothing else is stated. In all calculations we have to use pressure in kilo
Pascal (kPA) that gives 1,013 bar Þ
101,3 kPa. For the cargo calculations, we use densities from thermodynamic
properties edited by Ocean Gas Transport.

 

When the vessel is at sea and we get a telex
that we are to set up to load propane at
30oC in Fawly. Our cargo
tank relief valve set point is 4,5 bar. To find out how mush we can load, we
then have to take a rough calculation. We can then use density for propane at

30oC and for 6oC, this only to get an overview of how
mush we can load.

 

At 6oC r
is 522,0 kg/m3 and at
30oC r
is 567,9 kg/m3

 

Then we get 98% * rR/rL
Þ
98% * 522,0/567,9 = 90,07%

 

To calculate the accurate filling limit, we
have to know the actual cargo temperature and we must use density table.
However, as long as we do not know the exact cargo temperature, we use the
nearest values in the table. When we
know the exact temperature of the cargo, we can calculate more accurately. Relief valve setting is 4,5 bar and
atmospheric pressure 1,013 bar gives absolute pressure 5,513 bar.

 

In the thermodynamic table we find:

5,45 bar
Þ 5oC

 

5,61 bar
Þ 6oC

 

 

We have to interpolate between 5,45 bar and
5,61 bar to find the correct reference temperature and the correct density. The
reference temperature is 5,39oC and reference density is 522,79 kg/m3.

 





Then
we use -30oC and we find density to 567,9 kg/m3.

 





Filling limit = rR / rL * 98%

= 522,79 kg/m3 / 567,9 kg/m3 x 98% = 90,22%

 

In this example the filling limit will be
90,17% when we load propane at a temperature on -30oC. If the
loading temperature is colder than -30oC the filling limit will be
less than 90,17% and higher if the temperature is above -30oC.

 

12.4.2                  
Example 1

 




Cargo Propane


 


 


 


 




Temp in
oC


-30


oC density


567,9


kg/m3




Temp. reference rel. valve R


5,39


oC density


522,79


kg/m3




Tank #1, 100% Volume


1182,18


m3


 


 




Relieve valve set point


4,5


bar


 


 




Atmospheric pressure


1,013


bar


 


 




Absolute pressure relieve valve


5,513


bar


 


 




Filling limit =


rR


/ rL x


98 %


 




Filling limit


522,793


/ 567,900 x


98,00 %


90,22 %




When we have loaded propane on
30oC to the limit 90,17% we
are then sure that if the pressure in the cargo tank increases to 4,5 bar and
the temperature in the liquid increases to 5,39oC the liquid volume will
be 98%.

When we have
calculated the filling limit we can find the maximum volume of liquid that we
can load.

VL
= 0,98 * V * rR/ rL




VL


Volume liquid




V


100% Volume of the cargo tank




rR


Density of reference temperature on the relief valve setting




rL


Density for actual cargo temperature




 

 

When we have found the correct filling limit,
we can find the maximum volume to be loaded.
We have to find the cargo tank at 100% volume and multiply with the
actual filling limit.

If we have a cargo tank on 1182,18 m3 volume
at 100%, we find the maximum volume to be loaded by multiplying with 90,17%
filling limit.

 

 

 

 




Cargo tank 100% volume in m3


Filling limit in %


Volume to be loaded in m3




1182,18


90,17


1065,943




 

When we do this calculation we use the
formula: VL = 0,98 x V x rR / rL.

 

12.4.3                  
Example 2

 




Filling
limit


522,793


/
567,900


98 %


90,216
%




Filling
volume = Filling limit * Cargo tank 100% vol.


 


 


 




Filling
volume


90,22 %


x
1182,18


m3 =


1065,972
m3




 


 


 


 


 




Or
Filling Volume VL=


0,98 x 1182,18 x 522,793 / 567,9 =


1066,517 m3




 

After we have found the filling volume, we
find the ullage or sounding in the vessels ullage/sounding table.

 

Sounding is the level from tank bottom to the
liquid surface. Ullage is the
level from liquid surface to deck level.
In the following examples, we use sounding.

 

 










In this example, we
find the correct sounding to be 8,1662 meters. We have to do this calculation
on each cargo tank before we start loading.

 



In this example, the filling volume is
1065,943 m3 and that is in between 8,16 meters and 8,17 meters, so we have to
interpolate to find the correct sounding.

 

 

12.4.4                  
Example 3

 




Filling volume = Filling limit * Cargo tank
100% vol.




Filling limit


90,17 %


1182,180


m3 =


1065,943 m3




 


 


 


 


 




sounding in
m


volume in m3


 


 


 




8,16


1065,25


m3


 


 




8,17


1066,36


m3


 


 




8,1662


1065,94


m3


 


 




 

When we have found the correct sounding/ullage
we have to find which corrections we must use to get the actual
sounding/ullage. The corrections can be found in the sounding/ullage table for
each vessel.

 

12.4.5                  
Corrections

There are normally four corrections to be
used: the correction on the float, correction on the sounding tape, list and
trim correction. The float correction depends on the liquid density; with a
higher density the float becomes lighter in the liquid. The tape correction
depends on the temperature in the vapour phase. List and trim correction depend
on how the vessel is in the water. We have to study the corrections carefully
so we use the correct sign character. Spherical floats have the highest
corrections on float. All corrections we do, we find in the sounding/ullage
table for each cargo tank. On the next page, we found an example of a spherical
float.

 

Example of a spherical float





The table for float correction is calculated
against different densities and when we have a cargo with density in between
the table values, we have to interpolate to find the correct correction.

Out of the table above we can see that
lighter liquid will give a higher correction.

 

The
Float correction table


Specific
gravity (kg/dm3) Corrections in 
 meter




 





If we have cargo density 0,55 kg/dm3,
we have to interpolate between 0,50 and 0,60 and

the correction will then be 0,160 meter.
Small floats will give the lowest corrections. A tank equipped with spherical
float will have higher corrections than tanks equipped with a flat float.

A correction on the sounding tape depends on
the temperature in the vapour phase in the tank. High temperature and a small
vapour volume give a small correction, low temperature and big vapour volume
gives a higher correction.

 

Correction on trim is either a correction to
be added or multiplied to the measured sounding/ullage or the volume table is
calculated with the trim directly. The trim correction is higher on long tanks
than on short tanks. This means that small transverse tanks have a trim
correction near to zero and long tank has higher corrections.

 

Correction on list is either correction to be
added or multiplied to the measured sounding/ullage or the volume table is
calculated with the list directly. The list corrections are highest on wide
transverse tanks and small on narrow longitudinal tanks.

How we should use the corrections are explained
in each sounding/ullage table. Earlier in this chapter, we found the corrected
sounding to be 8,1662 meter. We will now continue using this example to find
the sounding that we will read on the sounding tape. Normally the corrections
are used directly on the sounding measurement, but when we calculate the other
way we have to use the correctionłs signs the opposite way.



 

12.4.6                  
Example 4

 




Corrected sounding


8,1662




Trim correction from table


-0,021




List correction from table


0




Sounding w. 20oC


8,1872




Correction for vapour temperature


-0,001




Float correction from table


0,1564




Read sounding


8,0318




 

To find the correct corrections we have to
know the density of the cargo, in this case, propane at
30oC and
density 567,9 kg/m3 = 0,5679kg/dm3, aft trim on 0,5 meter
zero list and
15oC in the vapour phase. When we are completely loaded on this tank, we will have a
sounding of 8,0318 meter.

The 98% maximum filling is to prevent liquid
getting in the relief valve, if the tank pressure reaches the relief valve
setting.

On vessels with relief valve setting of 0,5
bar we do not have any possibilities to heat the cargo at sea. On
semi-refrigerated or fully pressurised vessels, we have opportunity to heat the
cargo while the vessel is at sea. When we are heating the cargo, we have to
follow the tank pressure carefully to avoid uncontrolled venting. Vessels with
a low relief valve setting can have a higher filling limit than vessels with a
high relieve valve setting.





The sketch below shows how the filling limit
changes with the cargo temperature, as long as the relief valvełs set point is
the same.






12.1   
Calculation of cargo with use of ASTM-IP tables

In this chapter we will look at the tables
and corrections we use when calculating weight of cargo onboard gas carriers.
We then start to look at how we calculate weight in air at 15oC by
using the correct tables.

 

The tables we are using are the ASTM-IP-API
tables for light hydrocarbons. Density is mass divided by volume. The mass has
either kilo (kg) or metric ton (mt) as unit. Volume has either cubic meter (m3)
or litre (lt) as unit. Unit for density is either kg/m3 Þ
tonn/m3 or kg/dm3 Þ
kg/lt. Density and specific gravity is often given in vacuum, then we need tables
or calculations to convert to weight in air at 15oC.

 

12.4.7                  
Liquid calculation

We start calculation of the liquid in air and
then we look at the vapour calculation. For LPG cargoes and some chemical
cargoes it is normally accepted to calculate the weight in air at 15oC,
as we do in the crude oil trade. We then get either specific gravity 60/60oF
or density at 15oC from shore and we have to use the ASTM-IP-API
tables.

In table ASTM-IP no. 21, we find density at
15oC when the gravity 60/60oF is given. In table ASTM-IP
no. 54, we find the reduction factor to the actual cargo temperature compared
with density at 15oC. In table ASTM-IP no 56, we find the factor to
be used to find weight in vacuum from weight in air. If we take an example with
propane, liquid temperature is -25oC and specific gravity 0,5075, we
will calculate the weight in air at 15oC.

 

We then start with table ASTM-IP-API no. 21
to find density at 15oC from specific gravity 60/60oF
0,5075.

 




We look in
the column for Specific gravity 60/60oF 0,507 and find density at 15oC to


0,5073 kg/lt




We then look
in the column for specific gravity 60/60oF 0,508 an find density at 15oC
to


0,5083 kg/lt




The density
has now increased with


0,0010 kg/lt




Our Specific
gravity is 0,5075, we then have to interpolate as follows 0,5073 + (0,0010 /
0,001 x 0,0005) that give


0,5078 kg/lt






 

12.4.8                  
Example on table ASTM IP-API 21

 




Specific


API


 


 


 


 


 




Gravity


Gravity


Density


 


 


 


 




60/60oF


60oF


15oC


 


 


 


 




0,506


-


0,5063


 


 


 


 




0,507


-


0,5073


 


 


 


 




0,508


-


0,5083


 


 


 


 




 


 


 


 


 


 


 




Specific
gravity 60/60oF


 


0,5075


that gives


0,5078


kg/lt




 

We have now find the density at 15oC
to 0,5078 kg/lt which is equal to 507,8 kg/m3, which we use in table ASTM IP
no.54 to find the reduction factor to
25oC. In table ASTM IP no.54,
we look in the column for actual liquid temperature
25oC. The table
is divided in three columns and we have to interpolate between the 0,505 and
0,510 columns.

 

12.4.9                  
Example from table 54

 




Table 54C


 


 


 


 


 




Observed


Density 15 oC





temperature,


0,500


 


0,505


 


0,510




oC


Factor to reduce
volume to 15 oC




 

When we do the interpolation, we find the
reduction factor to 1,10432. When we
have different temperatures on the different cargo tanks, we have to do this
calculation on each tank. Below, we have an example on table ASTM IP no. 54




12.4.10             
Example on table ASTM IP-API 54

 




Table 54C


 


 


 


 


 




Observed


Density 15 oC





temperature,



0,500


 


0,505


 


0,510




oC


Factor to
reduce volume to 15 oC




-26


1,111


3


1,108


3


1,105




-25,5


1,109


2


1,107


3


1,104




-25


1,108


2


1,106


3


1,103




-24,5


1,107


2


1,105


3


1,102




-24


1,106


3


1,103


2


1,101




 


 


 


 


 


 




-25


0,5078


 


1,10432


 


 




 

The next correction is the shrinkage factor,
which is a thermal factor on the tank steel. Shrinkage factor is normally 1 at
20oC and is less than one when the steel is colder than 20oC.

The shrinkage factor is the correction for
the thermal expansion on the cargo tank steel. It is the correction between 20oC
and the actual steel temperature. With different steel, we have different
shrinkage factors, but on one vessel the shrinkage factor is similar on all
cargo tanks if they are made of equal steel. Aluminium and invar steel have a shrinkage
factor near 0 and mild steel has higher factor. Shrink factor for a vessel
depends on the material of the cargo tank. There is a shrinkage table on each
vessel. Only vessels with equal quality of steel and tank thickness have equal
shrinkage factors. When we calculate cargo, we use shrinkage factor both on the
liquid and the vapour.



 

12.4.11             
Example on shrinkage factor at
different temperatures

 






Temp.


Sh.fact.


 


Temp.


Sh.fact.


 


Temp.


Sh.fact.




20


1


 


-21


0,99879


 


-62


0,99759




19


0,99997


 


-22


0,99876


 


-63


0,99756




18


0,99994


 


-23


0,99873


 


-64


0,99753




17


0,99991


 


-24


0,99870


 


-65


0,99750




16


0,99988


 


-25


0,99868


 


-66


0,99747






 

The last table ASTM-IP no 56 is used to find
mass of liquid and vapour in air from mass in vacuum or vice versa. We have to
use the liquid density at 15oC, which in this example is 0,5078
kg/ltr, and find the factor for propane to 0,99775. We have to multiply this
factor with the mass in vacuum to get mass in air. If we have the mass in air
we must divide with the factor. When the cargo calculations are completed, on
the bill of lading and the other cargo papers we have to note if the loaded
mass is in vacuum or air. We must always use liquid density at 15oC
on the actual cargo to find the correct factor.



 

12.4.12             
Example on table ASTM IP-API 56

 






Table
56


 


 


 




Density at 15oC
kg/ltr


Factor for mass in vacuum to mass in air




 


 


 


 




0,5000


to


0,5191


0,99775




0,5192


to


0,5421


0,99785




0,5422


to


0,5673


0,99795




0,5674


to


0,5950


0,99805




Factor
is


 


0,99775






 

We can look at one example where we have
loaded 1089,556m3 propane with specific gravity 60/60F 0,5075 and
liquid temperature is
25oC.

From table ASTM-IP-API no. 21 we find the
cargo density at 15oC to 0,5078 kg/ltr. Þ 507,8 kg/m3

From table ASTM IP no. 54, we find reduction
factor from 15oC to
25oC to 1,10432.

From table ASTM IP no. 56, we find factor
from mass in vacuum to mass in air to 0,99775.

From the cargo tank shrinkage table, we find
shrinkage factor to 0,99868 at
25oC.

The calculation gives us 610 994 kg in vacuum
at 15oC that gives us 609 619 kg in air. We have to note on all
cargo documents that the mass is in air and also note the specific gravity
60/60F.

12.4.13             
Calculation of the liquidłs mass






Volume
loaded


1089,556


m3




Shrinkage
factor for -25oC


0,99868


 




Corrected
volume at -25oC


1088,118


m3




Reduction
factor to 15oC


1,10432


 




Volume at 15oC


1201,630


m3




Density at
15oC


507,8


kg/m3




Mass in
vacuum at 15oC


610188


kg




Factor from
table 56


0,99775


 




Mass in air
at 15oC


608815


kg






 

12.4.14             
Calculation of vapour

We will now calculate mass of the vapour in
air at 15oC. We always have to calculate the density of the vapour
as the density change with the pressure. When we are calculating the mass in
air on the vapour we need the following values, 288 K which is equal to 15oC,
101,325 kPa which is equal to 1,013 bar. Molar volume of ideal gas at 288 K is
23,6382 m3/kmol. We also need molar weight of the actual cargo and
for propane it is 44,1 kg/kmol. Then we use the actual cargo temperature and
pressure. We can take an example with Propane with vapour temperature at
18oC
and cargo tank pressure at 1,5 bar.

 

·     
Ts is standard temperature 288
K

·     
Tv is average temperature on
vapour in K

·     
Pv is absolute pressure of
vapour in kPa

·     
Ps is standard pressure
101.325 kPa Þ
1,013 bar

·     
Mm is molecular mass of the
product in kg/kmol

·     
I is molar gas volume at 288 K and
standard pressure 1,013bar Þ 23,6382 m3/kmol

rv
= (Ts x Pv x Mm) / (Tv x Ps x
I) kg/m3

When we insert the values in the formula we
find the following vapour density.

·     
Ts = 288 K

·     
Tv = 273 + (-18) = 255 K

·      Pv = (Ps + PT
) x 100 = (1,013 + 1,5) x 100 = 251.3 kPa

·      Ps = 101.3 kPa

·      Mm = 44,1 kg/kmol for
propane

·     
I = 23,6382 m3/kmol

 

 

12.4.15             
Density calculation of vapour






Ts


288


 


288


K


 


 


 


 




Ps


1,013


 


101,3


kPa


 


 


 


 




Pv


1,013


1,5


251,3


kPa


 


 


 


 




Mm


44,1


 


44,1


kg/kmol


 


 


 


 




I


23,6382


 


23,6382


m3/kmol


 


 


 


 




 


 


 


 


 


 


 


 


 




 


288


x


251,3


x


44,1


 


 


 




rv =


255


 


101,3


 


23,6382


=


5,227


kg/m3






 

When we have calculated the vapour density,
we have to calculate the mass of the vapour. We continue with the calculation
of propane loading. The cargo tank 100% volume is 1182,18m3 and we
have loaded 1089,556m3 liquid.
The vapour volume is then 100% cargo tank volume minus liquid
volume. That gives us 1182,18m3

1089,556m3 = 92,624m3.

We have a vapour temperature on
18oC,
which gives us a shrinkage factor (cargo tank expansion factor) on 0,99888
taken from the vesselłs shrinkage table. The vapour density is in kg/m3 and
the mass will then be in kilos.

When we calculate the mass of liquid in
kilos, we also calculate the mass of vapour in kilos. If we use mass of liquid
in metric ton, we have to calculate the vapour in metric ton also. In this
example, the vapour density is 5,227 kg/m3, which is equal to
0,005227 mt/m3. In this example, the mass of vapour is 484 kilos.

 

12.4.16             
Calculation of vapour mass at 15oC
in kilo

 






Cargo tank
100% volume


1182,180


m3




Liquid volume


1089,556


m3




Gas volume


92,624


m3




Shrinkage
for - 18oC


0,99888


 




Corrected
Gas volume


92,520


m3




Density of
gas at 15oC


5,227


kg/m3




Mass of gas
in vacuum at 15oC


484


kg






 

To find the total mass of liquid and vapour
in the cargo tank, we have to add mass of liquid 610 994 kg + 484 kg = 611 478
kg. Then we use ASTM-IP table 56 and find the conversion factor to mass in air.
Cargo density at 15oC is 507,8 kg/m3 with a factor of
0,99775. Then we multiply total mass in vacuum 611 478 kg with 0,99775 which
gives us 610 102 kg in air.

12.4.17             
Calculation of total mass in air at
15oC

 






Mass of
liquid in vacuum at 15oC


610 994


kg




Mass of gas
in vacuum at 15oC


484


kg




Total mass
in vacuum at 15oC


611 478


kg




Factor from
ASTM-IP 56 table


0,99775


 




Total mass
in air at 15oC


610 102


kg






 

We will take an example on a full calculation
and find the total mass in air, the cargo is propane and we have the following
information:

 






Molecular
mass


44,1


kg/kmol




Liquid
temperature


-24


oC




Vapour
temperature


-20


oC




Atmospheric
pressure


1,017


bar




Relief valve
setting


4,5


bar




Cargo tank
pressure


1,550


bar




Spes.Grav.60/60F


0,5072


 




Liquid density
at -24oC


560,6


kg/m3




Density at
Relief valve setting


522,756


kg/m3




Trim by
stern


1


meter




Sounding


8,152


meter




100 % Volume
of cargo tank


1468,180


m3




ROB before
loading


3114


kg






 

With a set point on the relief valve at 4,5
bar we can load maximum 91,38% with liquid temperature
24oC.

 

 

 

 

 

 

Maximum filling volume is, as follows:

 

Maximum filling volume = 0,98 x VT
x rR / rL

 

Maximum filling volume = 0,98 x 1468,18 x
522,7 / 560,6 = 1341,69 m3



We always have to start with the calculation
of maximum filling volume. This
calculation is based on figures we got before we start loading. If the
temperature and pressure changes, while we are loading, we have to recalculate
the maximum filling volume. Warmer cargo gives a higher filling volume; colder
cargo gives a lower filling volume. When the loading is completed, we do the
final calculation. We have to find the maximum filling limit on all tanks.

 

12.4.18             
Example on a full calculation on mass
at 15oC

 






 


PROPAN Tank #


2


 




1


100 %Volume
cargo tank


1468,180


m3




2


Liquid
temperature


-24,0


oC




3


Sounding


8,152


m




4


Float
correction


0,158


m




5


Correction for vapour temperature


-0,001


m




6


List
correction


0,000


m




7


Trim
correction


-0,059


m




8


Sounding at
20oC


8,250


m




9


Liquid
volume at 20oC


1341,373


m3




10


Shrinkage
factor tank steel at
24oC


0,99871


 




11


Corrected
liquid volume


1339,643


m3




12


Reductions
factor from table 54C


1,102


 




13


Liquid
volume at 15oC


1476,287


m3




14


Liquid
density at 15oC table 21


507,5


kg/m3




15


Mass of
liquid in vacuum at 15oC


749 215


kg




16


Uncorrected
vapour volume


126,807


m3




17


Shrinkage
factor vapour phase
20oC


0,99882


 




18


Corrected
vapour volume


126,657


m3




19


Tank
pressure


1,550


bar




20


Atmospheric
pressure


1,017


bar




21


Molecular mass
Propane


44,1


kg/mol




22


Vapour
temperature


-20,0


oC




23


Vapour
density at 15oC


5,382


kg/m3




24


Mass of
vapour in vacuum at 15oC


682


kg




25


Total mass
of cargo in the tank in vacuum


749 897


kg




 


Mass in air 749
897 kg x 0,99775


748 210


kg




 


ROB in air


3 114


kg




 


Total loaded
in air at 15oC


745 096


kg






 

 

After we complete the cargo calculation, we
have a ships figure which is the one the chief officer must calculate and one
shore figures, which is the one that the surveyor has calculated. Those two
figures will be nearly equal or equal. The one we use in the Bill of Lading is
the surveyorłs figure. In our example, we have loaded 745 096 kg in air at 15oC
in the actual cargo tank. It must be specified on the Bill of Lading that the
mass is in air at 15oC.

When we discharge the cargo, we will have
311,4 kg vapour left in the tank. At a
minimum, we are allowed to discharge is 99,5% of Bill of Lading, in this example
741 370 kg. It is important for the
vessel to calculate which temperature and cargo tank pressures will remain when
we finish discharging. In this example,
we must have maximum 0,16 bar pressure and vapour temperature
27oC.

When we load on an atmosphere from a previous
cargo, we call that ROB (Remaining on Board) or heel. That means when we have calculated the total mass of cargo in a
tank we have to subtract the ROB. When
the discharging is completed, that means we are finished pumping liquid. We have blown hot vapour to shore and tank
pressure, and vapour temperature is equal to what we estimated before loading.

It is important to remember that the tank
pressure has a big influence on the vapour density. If we transport an ambient cargo, we have to remove the tank
pressure before we commence the calculation of the ROB. Tank pressure is removed with the vesselłs
compressors and the condensate is sent directly to the discharge line.

We can look at two examples on density
calculation of a cargo with equal temperature but different tank
pressures. We use propylene as example
and vapour temperature is
25oC molecular mass 42,08 kg/kmol.

The first example tank pressure is 0,3 bar
and the other example tank pressure 1,5 bar.
The atmospheric pressure is 1,020 bar vessels total volume is 12000m3.

 

12.4.19             
Example of calculations on vapour
density with different tank pressures

 




r v
0,3bar =


288


x


134


x


42,08


 


 


 




 


243


 


101,3


 


23,6382


=


2,791


kg/m3




Volume


12000


x


2,791


kg/m3


33 490


kg


 


 




 


 


 


 


 


 


 


 


 




r v
1,5bar =


288


x


252


x


42,08


 


 


 




 


243


 


101,3


 


23,6382


=


5,249


kg/m3




Volume


12000


x


5,249


kg/m3


62 982


kg


 


 




Difference
in mass


 


29 492


kg


Difference in r


 


2,458


kg/m3




 

With a difference of tank pressure at 1,3 bar
on a 12 000 m3 vessel, we get 29 492 kg in mass difference. It is a good routine to always calculate the
maximum mass of vapour, which we can have as ROB to reach 99,5% of Bill of
Lading before we start discharging.

 

If we are onboard a fully refrigerated gas
carrier, we do not have any problem with high tank pressure when we have
completed discharging.

 

12.3   
Calculation of cargo weight using density tables

When transport of chemical gases and also
sometimes LPG cargoes, we use density tables for the actual cargo. We get the
density tables from the surveyor, the shipper of the cargo or thermodynamic
properties of gases. The weight of cargo is calculated by use of the actual
cargo temperature and the density tables are either in vacuum or in air. On
clean cargoes, such as, ethylene, propylene, butadiene and VCM, we can use the
density tables composed by SGS or thermodynamic properties of gases. We have to
be sure that the density tables we are using are either in vacuum or in air and
it has to be noted on the Bill of Lading. The density table we are using in the
load port has to be used also in the discharge port. The only ASTM table we are
using is ASTM-IP table no.56 for converting weight in air to weight in vacuum
or vice versa.

 

When the calculation is completed, we have to
note that the weight is in vacuum or in air. We always have to calculate the
vapour density because the vapour temperature does not match the cargo tank
pressure. We should use the actual vapour temperature and actual tank pressure
in the calculation of vapour density. First we take a look at how we are
calculating the weight of liquid.

 

First of all we have to find out the maximum
filling volume on the actual cargo tanks that we have to load.

 

Maximum
filling volume is as follows:

 

Maximum filling volume = 0,98 x VT x rR
/ rL

The cargo tank 100% volume is 1182,18 m3,
safety valve set point is 4,5 bar Þ
5,5 bar ata, liquid temperature is
24oC. Liquid density at 5,5 bar
is 523,3 kg/m3 and density at
24oC is 560,6 kg/m3.

 




 


98%


Vt m3


r SV kg/m3


r c kg/m3




Maximum filling volume =


0.98


1182.18


523.3


560.6




Maximum filling volume =


1081.452


m3


 


 




 

We should calculate the weight of liquid
propane, cargo tank pressure is 1,1 bar. We have loaded 1089,556m3
liquid propane, density from density table and
24oC is 560,6 kg/m3.
Cargo tank expansion factor at -24oC is 0,99870. Weight in vacuum will then be 605 477 kg, weight in air 604 115 kg.

 

 

 

 

 

 

 

 

 

 

12.3.1                  
Example on calculation of weight in
air




Loaded volume


1081.452


m3




Correction factor for -24oC


0.99870


 




Corrected volume


1080.052


m3




Density at -24oC from table


560.6


kg/m3




Weight in vacuum at -24oC


605 477


kg




Factor from table 56


0.99775


 




Weight in air at -24oC


604 115


kg




 

12.3.2                  
Calculation of vapour density and
weight

To calculate the weight of vapour, we first
have to calculate density of the vapour on the actual temperature. The actual
vapour temperature has to be in K (Kelvin) and pressures in kPa (kilo Pascal).
Another factor we should use is molar gas constant which is 8,31441 J/(mol x
K). To find the pressure in kPa “kilo Pascal" we multiply the pressure in bar
with 100, that means 1 bar is equal to 100 kPa.

In all calculations in this manual, we use
273K as 0oC. When we do the calculations onboard we use 273,15K as 0oC.

We should now look at one example to find
vapour density on propane with vapour temperature on
25oC, tank
pressure is 1,4 bar and the atmospheric pressure is 1.013 bar. Molecular mass
on propane is 44,1 kg/kmol.

 

Vapour density at actual temperature formula:

(Tank pressure
in kPa + Atmospheric pressure in kPa) x Molecular mass molar gas
constant x (T0 K + Gas temperature in oC)

Tank pressure 1,4 bar is equal to 140 kPa and
the atmospheric pressure 1,013 bar is equal to 101,3 kPa. Vapour temperature DT
in K = 273 + - 25 = 248K Tank pressure plus atmospheric pressure DP
is equal to 241,3 kPa.

 




D P x Molecular mass


 


 


 




Molar gas const. x D T


 


 


 




( 140,0 +


101,30) x


44,1


 


 




8,31441 x


( 273,00 +


-25,00)


5,16075


kg/m3




 

We can take another example with ethylene and
calculate the vapour density, molecular mass is 28,05 kg/kmol, vapour
temperature is
99oC Þ
174K and tank pressure is 0,35 bar Þ 35 kPa. Atmospheric
pressure is 1,012 bar Þ 101,2 kPa.

 




D P x Molecular mass


 


 


 




Molar gas const. x D T


 


 


 




 


 


 


 


 




( 35,0 +


101,20) x


28,05


 


 




8,31441 x


( 273,15 +


-99,00)


2,638


kg/m3




 

Now when we have found the vapour density at
the actual vapour temperature, we can calculate the weight of vapour at the
actual temperature. We have loaded one
tank with ethylene, tank 100% volume is 1182,15 m3 and liquid volume
is 1088,6 m3. Liquid temperature is
100oC and vapour
temperature is
99,5oC shrinkage factor at
99,5oC are
0,99648. Tank pressure is 0,15 bar and the atmospheric pressure is 1.014 bar.

 




Vapour volume


93.55


m3 =


(1182.15 -


1088.60)


m3

 



Vapour density


2.263367


kg/m3 =


(D P x
28,05) / (8,31441 x D T)




Vapour weight at
99,5oC


210.99


kg =


103,55 x 0,99648 x 2,26337


 

 



 

We have now seen how to calculate weight of
liquid and weight of vapour and we should now calculate both liquid and vapour.
We should calculate one tank loaded with ethylene, relief valve set point is
4,5 bar and atmospheric pressure is 1,020 bar. After loading the vessel we have
1 meter by stern trim with the following values:

 




Vapour


-95oC and tank pressure 0,345
bar.




Liquid


-100,5oC, density 563,63 kg/m3,
liquid volume 1313,348 m3




 

Maximum filling limit is 89,45%, which is
equal to 1313,348 m3 with relief valve setting on 4,5 bar. Total
weight of cargo in the tank is 738 009 kg in vacuum.

 

12.3.3                  
Calculation of Ethylene set point 4,5
bar




Liquid volume


1313.35


m3




Shrinkage factor at -100,5oC


0.99645


 




Corrected liquid volume


1308.69


m3




Liquid density at -100,5oC


563.625


kg/m3




Weight of liquid in vacuum at
-100,5oC


737610


kg




Cargo tank 100% volume


1468.18


m3




Vapour volume


154.832


m3




Shrinkage factor at -95oC


0.99661


 




Corrected vapour volume


154.308


m3




Vapour density at -95oC


2.583


kg/m3




Weight of vapour in vacuum at
-95oC


399


kg




 


 


 




 


 


 




TOTAL LOADED IN VACUUM


738009


kg




 

When we change the relief valve set point to
0,5 bar the maximum allowable filling limit then increase to 97,0% that is
equal to 1424,127m3. We then get a total weight of cargo in the tank
on 799 940 kg, which is 61 931 kg more than with set point on 4,5 bar. First,
we have to calculate maximum allowed filling limit.

 

 




Set point is


0.5 bar


 


Absolute pres.


1.520 bar




 


 


 


Ref. temp.


-96.53 oC




 


 


 


Ref. dens.


557.87 kg/m3




Filling limit


 


 


 


 




rR/rL x 98%


 


97.00 %


 


 




 

 

12.3.4 Calculation of
Ethylene set point 0,5 bar




Liquid volume


1 424,13


m3




Shrinkage factor at -100,5oC


0,99645


 




Corrected liquid volume


1 419,08


m3




Liquid density from table at -100,5oC


563,625


kg/m3




Weight of liquid in vacuum at -100,5oC


799 827


kg




Cargo tank 100% volume


1 468,18


m3




Vapour volume


44,053


m3




Shrinkage factor at -95oC


0,99661


 




Corrected vapour volume


43,904


m3




Vapour density at -95oC


2,581


kg/m3




Weight of vapour in vacuum at -95oC


113,326


kg




 


 


 




TOTAL LOADED IN VACUUM


799 940


kg




When we are loading on ROB from previous
cargo, the total loaded cargo is total weight of liquid and vapour in the tank
minus ROB. If we, in this example, had ROB before loading and we surveyed the
tank atmosphere at
87oC and tank pressure 0,02 bar, atmospheric
pressure 1,019 bar, the ROB will then be 2758 kg in vacuum. Total loaded will
then be 799 940 kg
2 758 kg = 797 182 kg in vacuum.

 

12.3.4                  
Weight of ROB before loading at
temperature -87oC and tank pressure 0,02 bar




Vapour density


1,885


kg/m3


 


 


 




Tank volume 100%


1468,18


m3


D T =


186


K




Weight of vapour


2 758


kg


D P =


104


kPa




Shrinkage factor


0,99685


 


 


 


 




To find the weight in air we can either use
table ASTM-IP-API 56 if we know the density at 15oC or we have to
calculate a factor. The factor is, as follows:

 

(1
(r
air/ r cargo liquid)) / (1
( r
air/ r Brass)

(1
(1,2 kg/m3 /r
cargo liquid)) / ( 1
(1,2 kg/m3 / 8100kg/m3))

 

In our example we will get a factor, as
follows:

 

(1 - (1,2
kg/m3 /563,625 kg/m3)) / ( 1
(1,2 kg/m3 /
8100kg/m3)) = 0,997985

 

Then we have to multiply mass in vacuum with
the factor:

 

797 182 kg x 0,997985 = 795 575 kg in air

 

On a full-loaded tank, we can use the
following formula:

Mass in vacuum loaded - (Mass in vacuum
loaded x r
air / r liquid)
When we use the values from our last example it will be, as follows

 




Weight in air = Mass in vacuum - (Mass in
vacuum x 1,2/ r liquid)




Weight in air = 797 182 - (797 182 * 1,2/
563,625)


= 795 485


kg




 

Before we commence with cargo calculations,
we have to be sure that the density given is in air or in vacuum. With most
chemical gases, we get the density on the actual liquid temperature in vacuum.
Always note on the Bill of Lading that the quantity is either in vacuum or air.
On the calculation forms, we calculate both in vacuum and in air.

We should now do a full cargo calculation. We
start to calculate ROB before loading. Then we do calculations after we have
completed loading.

The vessel has three twin tanks numbered as
follows 1P, 1S, 2P, 2S, 3P and 3S. Cargo tanks 2 and 3 are equal and tank 1 is
a bit smaller.

 

12.3.5                  
Calculation of ROB before loading

 






Loading report




 


 


 


 


 


 


 


 




Cargo


Propylene


 


Port


Al Jubail


 


 




Molecular mass


42,08


 


Date


17.05.


1994


 


 




Atm.press.


1,015


 


Vessel


LPG Seagull


 


 




 


 


 


 


 


 


 


 




 


 


 


 


 


 


 


 




Liquid


 


 


 


 


 


 


 




Tank #


Sounding in meter


Volume from tab. in m3


Temp.
in oC


Pressure in bar


r liquid in kg/m3


Shrinkage factor


Mass of liquid in kg




1P


 


 


 


0,02


 


 


0




1S


 


 


 


0,02


 


 


0




2P


 


 


 


0,02


 


 


0




2S


 


 


 


0,02


 


 


0




3P


 


 


 


0,02


 


 


0




3S


 


 


 


0,02


 


 


0




 


 


 


 


 


Total
mass of Liquid


0




 
 
Vapour


 


 


 


 


 


 


 
 




Tank #


!00% vol.
in m3


Vapour volume in m3


Temp.
in oC


r vapour in kg/m3


Shrinkage factor


Mass of vapour in kg




1P


1182,18


1182,18


-27


 


2,129


0,99862


2 514




1S


1182,18


1182,18


-27


 


2,129


0,99862


2 514




2P


1468,18


1468,18


-27


 


2,129


0,99862


3 122




2S


1468,18


1468,18


-27


 


2,129


0,99862


3 122




3P


1468,18


1468,18


-25


 


2,112


0,99868


3 097




3S


1468,18


1468,18


-25


 


2,112


0,99868


3 097




 


 


 


 


Total
mass of vapour


17 465




 


 


 


 


Total
mass in vacuum


17 465




 


 


 


 


ROB


 


 


 




 


 


 


 


Total
loaded in vacuum


 




 


 


 


 


Total
loaded in air


 






 

12.3.6                  
Calculation of mass after loading






 


 


 


 


 


 


 


 




Loading report

 



 


 


 


 


 


 


 


 

 



Cargo


Propylene


 


Port


Al Jubail


 


 

 



Molecular mass


42,08


 


Date


18.05.


1994


 


 

 



Atm.press


1,020


 


Skip


LPG Seagull


 


 

 



 


 


 


 


 


 


 


 

 



 


 


 


 


 


 


 


 

 



Liquid


 


 


 


 


 


 


 

 



Tank #


Sounding in meter


Volume from tab. in m3


Temp.
in oC


Press in bar


r liquid in kg/m3


Shrinkage factor


Mass of liquid in kg

 



1P


8,74


1123,83


-39


0,6


601,2


0,99826


674 471

 



1S


8,76


1125,55


-39


0,6


601,2


0,99826


675 503

 



2P


8,72


1400,11


-39


0,6


601,2


0,99826


840 281

 



2S


8,73


1401,20


-39


0,6


601,2


0,99826


840 936

 



3P


8,76


1404,41


-38


0,6


600,0


0,99829


841 205

 



3S


8,75


1403,35


-38


0,6


600,0


0,99829


840 570

 



 


 


 


 


 


Total
mass of Liquid


4 712 967

 



 
 
 
 
 
 
 
Vapour


 


 


 


 


 


 


 

 



Tank #


100% vol.
in m3


Vapour volume in m3


Temp.
in oC


r vapour in kg/m3


Shrinkage factor


Masse of vapour in kg

 



1P


1182,18


58,35


-35


 


3,445


0,99838


201

 



1S


1182,18


56,63


-35


 


3,445


0,99838


195

 



2P


1468,18


68,07


-36


 


3,459


0,99835


235

 



2S


1468,18


66,98


-36


 


3,459


0,99835


231

 



3P


1468,18


63,77


-34


 


3,431


0,99841


218

 



3S


1468,18


64,83


-34


 


3,431


0,99841


222

 



 


 


 


 


Total
mass of


vapour


1 302

 



 


 


 


 


Total
mass in vacuum


4 714 269

 



 


 


 


 


ROB


 


 


17 465

 



 


 


 


 


Total
Loaded in vacuum


4 696 803

 



 


 


 


 


Total
loaded in air


4 687 973

 
























 

 

 

 

 

 

12.4 Calculation of liquid to
be used for gassing up

There are some parameters we have to have in
mind to find out how much liquid we need to take onboard for gassing up our
cargo tanks. The first is the temperature of the liquid we will take onboard
then the temperature of the cargo tank steel and what volume we should gas up.

To change cargo and gas up costs lot money,
to minimise the cost we have to use all the available cargo equipment onboard
in the most efficient way. We have to be sure that the amount of liquid we
order for gassing up is enough to gas up and to commence cooling down the cargo
tanks. If we have some ROB in one tank, we can begin gassing up at sea if the
tanks are surveyed and approved by a surveyor. If we donłt have any ROB or not
enough, we have to order liquid to gas up the rest of the volume to be gassed
up. To minimise the consumption of cargo for gassing up, we need to heat the
cargo, as mush as possible. The amount of cargo lost when gassing up depends on
the people onboard, cargo equipment and the time we use for gassing up. For
cargoes with a heavy vapour, such as VCM, propane butane and propylene, the
loss of cargo is near to 0 when gassing up correctly. The only way to reduce
the loss of cargo is to control tank pressure when loading coolant, measure and
check when commence heating the coolant for gassing up.

 

12.4.1                  
Volume of liquid to be used for
gassing up

We have a cargo tank with volume 1182,18 m3
that we have to gas up. Our experience is that we need two times the tank
volume for gassing up and commence cooling the tank. We then have to order the
following amount cargo, 1182,18 m3 x 2 = 2364,36 m3. We
will take onboard propylene liquid for gassing up and it is two vital
temperatures we must recognise, tank steel temperature and liquid temperature
on the coolant. Liquid temperature on shore tank is
40oC and our
cargo tank steel 20oC. The formula is mass = r
vapour x total volume. From the table thermodynamic properties for propylene
superheated vapour, we find the vapour r
to 1,812 kg/m3 on 20oC and P=1 bar. Then we take the
total volume 2364,36 m3 and multiply with vapour r
1,812 kg/m3 = 4 283 kg, which is the minimum we need for gassing up
and commence cooling the tank. The loss of cargo and number of changes is
individual for each vessel and it is our duty to reduce the loss of cargo down
to a minimum.

 

12.4.2                  
Calculation of volume liquid we have
to order




Cargo


Propylene


 


 


 




100% Tank volume


1182,18


m3


 


 




r vapour at
atmospheric pressure from table


20


oC


1,812


kg/m3




r liquid from table


-40


oC


602,4


kg/m3




Number of changes


2


 


 


 




Total volume to be changed


2364,36


m3


 


 




Mass volume = Volume x r for vapour


 


 


 


 




Mass total volume =


4283,26


kg


 


 




Volume liquid to be loaded = Mass volume / r liquid


 


 


 




Volume to be loaded =


7,110


m3


 


 




 

We have to load 7,11 m3 propylene
at
40oC from shore tank to change the vapour atmosphere at 20oC
two times. This was a calculation for
one tank, if we gas up all tanks, the calculation has to be on the total volume
of the vesselłs cargo tanks.

After completion of the loading two Bill of
Lading will be made, one for what we have used for gassing up and one for the
quantity we have loaded.

 

12.4.3                  
Number of changes with a given amount
of liquid

To find the number of vapour changes with a
given amount of liquid in either a deck tank or a cargo tank, we then have to
know the liquid temperature and the temperature of the cargo tanks we have to
gas up. Then we have to calculate the mass of the liquid we have. When we know
the mass of liquid and the volume to be gassed up, we know if we then need to
order more liquid or if we can complete to gas up and commence cooling tanks
with the amount of liquid we have onboard. A cargo tank is completely gassed up
when we have more than 97% hydrocarbons in the vapour atmosphere.

We must remember that the tank we use for
gassing up will have a given amount of mass vapour left ROB, which we are
unable to get out. First, we have to calculate the mass of vapour we will have
ROB in our deck tank/cargo tank after we have gassed up the other tanks. When
we have calculated the mass of vapour we have left, we must subtract it from
the amount of liquid we have. How many changes we need depends on the cargo,
the cargo handling equipment we have onboard, temperature of the liquid and
temperature of the atmosphere that we should gas up. If we are able to heat the
vapour, we should have it as hot as possible to use as less liquid as possible.

We can use an example on the calculation of
vapour after gassing up. Average
temperature on the vapour is
10oC, total tank volume is 2564,36 m3
and tank pressure is 0 bar. We then find the density of the vapour, either
calculate the density or use the thermal property table to find it. When we
have found the vapour density, we have to multiply it with the tank shrinkage
factor and the tank volume.

 

12.4.4                  
Mass of vapour after gassing up




Cargo


Propylene


 


 


 




Tank volume #1 P/S


2364,36


m3


 


 




r for vapour at
atmospheric pressure


-10


oC


1,953


kg/m3




Mass of vapour in the tank


4 618


kg


 


 




 

We have now calculated that we should have 4
618 kg vapour left in tank #1 P and S when we are not able to get out any more
from the tanks. Before we order any liquid, we have to subtract 4 618 kg from
the amount of liquid we need to gas up the whole vessel.

We can continue with the example and have 15
m3 liquid propylene at
10oC, vapour temperature 0oC
and the pressure 3,3 bar in tank #1 P/S. Total volume of the vessel is 8237 m3
and atmospheric pressure is 1,015 bar. That means we have to gas up vesselłs
total volume
volume of tank #1 P/S, which is equal to 8237 m3

2364,36 m3 = 5872,64 m3 with an average temperature of 25oC.
Our experience is that we need 2,5 volume changes to reach 97% hydrocarbons in
the vapour atmosphere, 2,5 changes is 5872,64 m3 x 2,5 = 14681,6 m3.
It is always stated in the charter party how clean the atmosphere has to be
before loading and it depends on which cargo we have to load.

 

 

12.4.5                  
Mass we can use for gassing up




Cargo


Propylene


 




Tank volume #1 P/S


2364,36


m3




Volume liquid in tank #1 P/S


15,00


m3




Mass of liquid in tank #1 P/S


8 420


kg




Volume of vapour in tank #1 P/S


2349,36


m3




Mass of vapour in tank #1 P/S


18 783


kg




Total mass in tank #1 P/S


27 202


kg




Mass of vapour in tank #1 P/S after gassing
up


4 618


kg




Usable mass in tank #1 P/S


22 584


kg




 

We have 27202 kg available in tank #1 P/S,
but when we are completed gassing up, we have 4618 kg vapour left, that means
we have 22584 kg available for gassing up. Total volume to be gassed up is
vesselÅ‚s total volume minus volume of tank #1 P/S multiplied with 2,5. That gives Þ
8237 m3 - 2349,36 m3 Þ
5872,64 m3 x 2,5 = 14682 m3. We have to find the vapour
density equal to tank steel temperature 25oC, which is 1,724 kg/m3.

 

12.4.6                  
Calculation of volume needed




Volume vapour at 25oC of
available mass


13 101


m3


 


 




r for vapour with
atmospheric pressure at


25


oC


1,724


kg/m3




Total volume vapour 2,5 times tank volume


14 682


m3


 


 




Volume to be ordered


1 581


m3


 


 




 

In this example, we do not have enough liquid
to reach 2,5 times for gassing up. There was 1581 m3 vapour short,
so we have to order that 1581 m3 x 1,724 kg/m3 = 2726 kg.

 

12.4.7                  
Calculation of mass vapour at a given
temperature




Cargo


Propylene


 




Vesselłs total volume


8237


m3




Tank #1 P/S volume


2364,36


m3




Volume to be gassed up


5872,64


m3




Amount changes


2,5


 




Tank steel temperature


25


oC




Atmospheric pressure


1,015


bar




Vapour r at 25oC


1,724


kg/m3




Total volume to gas up 5872,64 x 2,5


14 682


m3




Mass of total volume to gas up


25 309


kg




Available mass


22 584


kg




Mass in kg to load to complete gassing up


2 725


kg




 

To hold the temperature of the vapour we use
for gassing up, we have to use either the compressors or heaters. If we are
able to increase the temperature on the vapour from 25oC to 60oC,
we do not need to supply any extra from shore.

12.4.8                  
Calculation of vapour at 60oC

 




Vapour r at 60oC


1,543


kg/m3




Total volume to gas up 5872,64 x 2,5


14 682


m3




Mass total volume to gas up


22 649


kg




Available mass


22 584


kg




Mass in kg difference


65


kg




 

It is important that we continue to heat the
tank we are taking the vapour from to hold a positive pressure.