11.6 Formulas to be used when changing atmosphere in an tank

Transcription

1 11.6 Formulas to be used when changing atmosphere in an tank There are some formulas that we can use to calculate the consumption of nitrogen or inert for changing atmosphere in tanks and the time to be used for the same purpose. The formula is a bit different, if we use inert versus nitrogen Using nitrogen When we use nitrogen, the oxygen content is 0%. That means we should purge 21% oxygen from the air to a given maximum content of oxygen in the tank e.g. 0,2% Numbers of volume changed is ln (original O 2 content/ desired O 2 content) ln Natural logarithm Original O 2 content The original content of O 2 in the tank that we should purge Desired O 2 content The specified O 2 content given in the charter party Numbers of volume changed The number of times the specified tank capacity needs to be completely purged of nitrogen If we have one tank at 1000 m 3 capacity and the O 2 content, according to the charter party, should be less than 0,2%, we start with air in the tank. The calculation will be as follows: Number of volume changed = ln ( 20,8% / 0,18%) ln ( 115,56) Number of volume changed = 4,75 The desired O 2 content has been set to 0,18%, to be less than 0,2%. The nitrogen consumption will then be 1000m 3 * 4,75 which equals 4750m 3 nitrogen. This is the minimum required nitrogen. When ordering nitrogen add 10% to the minimum needed (5225m 3 ) nitrogen Using inert When using inert we also use the formula with ln natural logarithm but we must calculate the O 2 content in the inert gas also.

2 Numbers of volume changed is ln (original O 2 content/ desired O 2 content) ln Natural logarithm Original O 2 content The original content of O 2 in the tank that we should purge Desired O 2 content The specified O 2 content given in the charter party O 2 content in the inert The O 2 content that we set the inert gas generator to give, never above 5% Numbers of volume changed The number of times the specified tank capacity needs to be completely purged of nitrogen Take an example with the same tank at 1000m 3. The charter party states maximum 2% O 2 and the O 2 content on the inert is set to 0,5%. We start with air in the tank. Number of volume changed = ln ( 20,8% - 0,5% / 2% - 0,5%) ln ( 13,53) Number of volume changed = 2,61 We have to subtract the inert gas O 2 content from the original and desired O 2 content. The total consumption of inert will be 1000m 3 * 2,61 = 2610m Use with allowed vacuum On vessels that have facility to have some vacuum on their tanks they can use their compressors to create the minimum allowed pressure in their tanks. If your vessel can have 30% vacuum in the tanks, it means a 0,7 bar absolute pressure. When you have 30% vacuum you have already quit 30% of the oxygen, which means you have 20,8/100*70 = 14,56% oxygen left. When we come alongside we pressurise the tanks with nitrogen to 1 bar absolute. Then we can continue the purge normally. To pressurise the tanks to 1 bar absolute we need 30% of the total capacity of our tanks. If we should purge one tank on 1000m 3, we need 300m 3 of nitrogen.

3 12- CARGO CALCULATION

4 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 owner s office. When we load and transport liquefied gases there are some variables that we have to have in mind, the setting of the safety valve 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 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 = ρ R / ρ L * 98% ρ R ρ L Density of reference temperature on the relief valve setting 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 tank s 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 35 o C. Relief valve setting 4,5 bar Relief valve setting 0,5 bar Cargo temperature 35 o C 4,5 bar + 1 bar = 5,5 bar 5 o C = 523,3 kg/m 3 0,5 bar + 1 bar = 1,5 bar - = 570,2 32 o C kg/m 3 = 573,7 kg/m 3

5 In our example with 1000 m 3 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 If we are on a gas carrier on m 3 the loss of income will then be USD 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 decklog 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 T 0 = 273 o C 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 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 30 o C 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 30 o C and for 6 o C, this only to get an overview of how mush we can load. At 6 o C ρ is 522,0 kg/m 3 and at 30 o C ρ is 567,9 kg/m 3 Then we get 98% * ρ R /ρ L 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 5 o C 5,61 bar 6 o C 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 5,39 o C and reference density is 522,79 kg/m 3.

6 Then we use -30 o C and we find density to 567,9 kg/m 3. Filling limit = ρr / ρl * 98% = 522,79 kg/m 3 / 567,9 kg/m 3 x 98% = 90,22% In this example the filling limit will be 90,17% when we load propane at a temperature on -30 o C. If the loading temperature is colder than -30 o C the filling limit will be less than 90,17% and higher if the temperature is above -30 o C Example 1 Cargo Propane Temp in o C -30 o C 567,9 kg/m 3 density Temp. reference rel. valve R 5,39 o C 522,79 kg/m 3 density Tank #1, 100% Volume 1182,18 m 3 Relieve valve set point 4,5 bar Atmospheric pressure 1,013 bar Absolute pressure relieve 5,513 bar valve Filling limit = rr / rl x 98 % Filling limit 522,793 / 567,900 x 98,00 % 90,22 % When we have loaded propane on 30 o C 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 5,39 o C 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.

7 V L = 0,98 * V * ρ R / ρ L V L V ρ R ρ L Volume liquid 100% Volume of the cargo tank Density of reference temperature on the relief valve setting 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 m 3 volume at 100%, we find the maximum volume to be loaded by multiplying with 90,17% filling limit. Cargo tank 100% volume in m 3 Filling limit in % Volume to be loaded in m ,18 90, ,943 When we do this calculation we use the formula: V L = 0,98 x V x rr / rl Example 2 Filling limit 522,793 / 567,900 Filling volume = Filling limit * Cargo tank 100% vol. Filling volume 90,22 % x 1182,18 98 % 90,216 % m 3 = 1065,972 m3 Or Filling Volume V L = 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.

8 Ullage Sounding 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 Example 3 Filling volume = Filling limit * Cargo tank 100% vol. Filling limit 90,17 % 1182,180 m 3 = 1065,943 m 3 sounding in m volume in m 3 8, ,25 m 3 8, ,36 m 3 8, ,94 m 3 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.

9 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/dm 3 ) Corrections in meter

10 If we have cargo density 0,55 kg/dm 3, 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 correction s signs the opposite way Example 4 Corrected sounding 8,1662 Trim correction from table -0,021 List correction from table 0 Sounding w. 20 o C 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 30 o C and density 567,9 kg/m 3 = 0,5679kg/dm 3, aft trim on 0,5 meter zero list and 15 o C 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.

11 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 valve s set point is the same 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 15 o C 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 (m 3 ) or litre (lt) as unit. Unit for density is either kg/m 3 tonn/m 3 or kg/dm 3 kg/lt. Density and specific gravity is often given in vacuum, then we need tables or calculations to convert to weight in air at 15 o C 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 15 o C, as we do in the crude oil trade. We then get either specific gravity 60/60 o F or density at 15 o C from shore and we have to use the ASTM-IP-API tables. In table ASTM-IP no. 21, we find density at 15 o C when the gravity 60/60 o F is given. In table ASTM-IP no. 54, we find the reduction factor to the actual cargo temperature compared with density at 15 o C. 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 -25 o C and specific gravity 0,5075, we will calculate the weight in air at 15 o C.

12 We then start with table ASTM-IP-API no. 21 to find density at 15 o C from specific gravity 60/60 o F 0,5075. We look in the column for Specific gravity 60/60 o F 0,507 and find density at 15 o C to We then look in the column for specific gravity 60/60 o F 0,508 an find density at 15 o C to The density has now increased with Our Specific gravity is 0,5075, we then have to interpolate as follows 0, (0,0010 / 0,001 x 0,0005) that give 0,5073 kg/lt 0,5083 kg/lt 0,0010 kg/lt 0,5078 kg/lt Example on table ASTM IP-API 21 Specific API Gravity Gravity Density 60/60 o F 60 o F 15 o C 0,506-0,5063 0,507-0,5073 0,508-0,5083 Specific gravity 60/60 o F 0,5075 that gives 0,5078 kg/lt We have now find the density at 15 o C 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 25 o C. In table ASTM IP no.54, we look in the column for actual liquid temperature 25 o C. The table is divided in three columns and we have to interpolate between the 0,505 and 0,510 columns.

13 Example from table 54 Table 54C Observed Density 15 o C 0,500 0,505 0,510 temperature, o C Factor to reduce volume to 15 o C When we do the interpolation, we find the reduction factor to 1, 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

14 Example on table ASTM IP-API 54 Table 54C Observed Density 15 o C temperature, 0,500 0,505 0,510 o C Factor to reduce volume to 15 o C -26 1, , ,105-25,5 1, , , , , ,103-24,5 1, , , , , , ,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 20 o C and is less than one when the steel is colder than 20 o C. The shrinkage factor is the correction for the thermal expansion on the cargo tank steel. It is the correction between 20 o C 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 Example on shrinkage factor at different temperatures Temp. Sh.fact. Temp. Sh.fact. Temp. Sh.fact , , , , , , , ,99753

15 17 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 15 o C, which in this example is 0,5078 kg/ltr, and find the factor for propane to 0, 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 15 o C on the actual cargo to find the correct factor Example on table ASTM IP-API 56 Table 56 Density at 15 o C kg/ltr Factor for mass in vacuum to mass in air 0,5000 to 0,5191 0, ,5192 to 0,5421 0, ,5422 to 0,5673 0, ,5674 to 0,5950 0,99805 Factor is 0,99775 We can look at one example where we have loaded 1089,556m 3 propane with specific gravity 60/60F 0,5075 and liquid temperature is 25 o C. From table ASTM-IP-API no. 21 we find the cargo density at 15 o C to 0,5078 kg/ltr. 507,8 kg/m 3 From table ASTM IP no. 54, we find reduction factor from 15 o C to 25 o C to 1, From table ASTM IP no. 56, we find factor from mass in vacuum to mass in air to 0, From the cargo tank shrinkage table, we find shrinkage factor to 0,99868 at 25 o C. The calculation gives us kg in vacuum at 15 o C that gives us 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.

16 Calculation of the liquid s mass Volume loaded 1089,556 m 3 Shrinkage factor for -25 o C 0,99868 Corrected volume at -25 o C 1088,118 m 3 Reduction factor to 15 o C 1,10432 Volume at 15 o C 1201,630 m 3 Density at 15 o C 507,8 kg/m 3 Mass in vacuum at 15 o C kg Factor from table 56 0,99775 Mass in air at 15 o C kg Calculation of vapour We will now calculate mass of the vapour in air at 15 o C. 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 15 o C, 101,325 kpa which is equal to 1,013 bar. Molar volume of ideal gas at 288 K is 23,6382 m 3 /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 18 o C and cargo tank pressure at 1,5 bar. T s is standard temperature 288 K T v is average temperature on vapour in K P v is absolute pressure of vapour in kpa P s is standard pressure kpa 1,013 bar M m is molecular mass of the product in kg/kmol I is molar gas volume at 288 K and standard pressure 1,013bar 23,6382 m 3 /kmol ρv = (T s x P v x M m ) / (T v x P s x I) kg/m 3 When we insert the values in the formula we find the following vapour density. T s = 288 K T v = (-18) = 255 K P v = (P s + P T ) x 100 = (1, ,5) x 100 = kpa P s = kpa M m = 44,1 kg/kmol for propane I = 23,6382 m 3 /kmol

17 Density calculation of vapour T s K P s 1, ,3 kpa P v 1,013 1,5 251,3 kpa M m 44,1 44,1 kg/kmol I 23, ,6382 m 3 /kmol 288 x 251,3 x 44,1 rv = ,3 23,6382 = 5,227 kg/m 3 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 1182,18m 3 and we have loaded 1089,556m 3 liquid. The vapour volume is then 100% cargo tank volume minus liquid volume. That gives us 1182,18m ,556m 3 = 92,624m 3. We have a vapour temperature on 18 o C, which gives us a shrinkage factor (cargo tank expansion factor) on 0,99888 taken from the vessel s shrinkage table. The vapour density is in kg/m 3 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/m 3, which is equal to 0, mt/m 3. In this example, the mass of vapour is 484 kilos Calculation of vapour mass at 15 o C in kilo Cargo tank 100% volume 1182,180 m 3 Liquid volume 1089,556 m 3 Gas volume 92,624 m 3 Shrinkage for - 18 o C 0,99888 Corrected Gas volume 92,520 m 3 Density of gas at 15 o C 5,227 kg/m 3 Mass of gas in vacuum at 15 o C 484 kg

18 To find the total mass of liquid and vapour in the cargo tank, we have to add mass of liquid kg kg = kg. Then we use ASTM-IP table 56 and find the conversion factor to mass in air. Cargo density at 15 o C is 507,8 kg/m 3 with a factor of 0, Then we multiply total mass in vacuum kg with 0,99775 which gives us kg in air Calculation of total mass in air at 15 o C Mass of liquid in vacuum at 15 o C Mass of gas in vacuum at 15 o C Total mass in vacuum at 15 o C kg 484 kg kg Factor from ASTM-IP 56 table 0,99775 Total mass in air at 15 o C 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 Vapour temperature -20 o C o C 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 -24 o C 560,6 kg/m 3 Density at Relief valve setting 522,756 kg/m 3 Trim by stern 1 meter Sounding 8,152 meter 100 % Volume of cargo tank 1468,180 m 3 ROB before loading 3114 kg

19 With a set point on the relief valve at 4,5 bar we can load maximum 91,38% with liquid temperature 24 o C. Maximum filling volume is, as follows: Maximum filling volume = 0,98 x V T x ρr / ρl Maximum filling volume = 0,98 x 1468,18 x 522,7 / 560,6 = 1341,69 m 3 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 Example on a full calculation on mass at 15 o C PROPAN Tank # %Volume cargo tank 1468,180 m 3 2 Liquid temperature -24,0 o C 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 20 o C 8,250 m 9 Liquid volume at 20 o C 1341,373 m 3 10 Shrinkage factor tank steel at 24 o C 0, Corrected liquid volume 1339,643 m 3 12 Reductions factor from table 54C 1, Liquid volume at 15 o C 1476,287 m 3 14 Liquid density at 15 o C table ,5 kg/m 3

20 15 Mass of liquid in vacuum at 15 o C kg 16 Uncorrected vapour volume 126,807 m 3 17 Shrinkage factor vapour phase 20 o C 0, Corrected vapour volume 126,657 m 3 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 o C 23 Vapour density at 15 o C 5,382 kg/m 3 24 Mass of vapour in vacuum at 15 o C 682 kg 25 Total mass of cargo in the tank in vacuum kg Mass in air kg x 0,99775 ROB in air Total loaded in air at 15 o C kg kg 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 surveyor s figure. In our example, we have loaded kg in air at 15 o C in the actual cargo tank. It must be specified on the Bill of Lading that the mass is in air at 15 o C. 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 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 27 o C. 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.

21 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 vessel s 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 12000m Example of calculations on vapour density with different tank pressures r v 0,3bar = 288 x 134 x 42, ,3 23,6382 = 2,791 kg/m 3 Volume x 2,791 kg/m kg r v 1,5bar = 288 x 252 x 42, ,3 23,6382 = 5,249 kg/m 3 Volume x 5,249 kg/m kg Difference in mass kg Difference in r 2,458 kg/m 3 With a difference of tank pressure at 1,3 bar on a m 3 vessel, we get 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.

22 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 V T x ρr / ρl The cargo tank 100% volume is 1182,18 m 3, safety valve set point is 4,5 bar 5,5 bar ata, liquid temperature is 24 o C. Liquid density at 5,5 bar is 523,3 kg/m 3 and density at 24 o C is 560,6 kg/m 3. 98% Vt m 3 r SV kg/m 3 r c kg/m 3 Maximum filling volume = Maximum filling volume = m 3 We should calculate the weight of liquid propane, cargo tank pressure is 1,1 bar. We have loaded 1089,556m 3 liquid propane, density from density table and 24 o C is 560,6 kg/m 3. Cargo tank expansion factor at -24 o C is 0, Weight in vacuum will then be kg, weight in air kg.

23 Example on calculation of weight in air Loaded volume m 3 Correction factor for -24 o C Corrected volume m 3 Density at -24 o C from table kg/m 3 Weight in vacuum at -24 o C kg Factor from table Weight in air at -24 o C kg 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 0 o C. When we do the calculations onboard we use 273,15K as 0 o C. We should now look at one example to find vapour density on propane with vapour temperature on 25 o C, tank pressure is 1,4 bar and the atmospheric pressure is 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 (T 0 K + Gas temperature in o C) 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 T in K = = 248K Tank pressure plus atmospheric pressure P is equal to 241,3 kpa. D P x Molecular mass Molar gas const. x D T ( 140, ,30) x 44,1 8,31441 x ( 273, ,00) 5,16075 kg/m 3 We can take another example with ethylene and calculate the vapour density, molecular mass is 28,05 kg/kmol, vapour temperature is 99 o C 174K and tank pressure is 0,35 bar 35 kpa. Atmospheric pressure is 1,012 bar 101,2 kpa. D P x Molecular mass Molar gas const. x D T

24 ( 35, ,20) x 28,05 8,31441 x ( 273,15-99,00) 2,638 kg/m 3 + 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 m 3 and liquid volume is 1088,6 m 3. Liquid temperature is 100 o C and vapour temperature is 99,5 o C shrinkage factor at 99,5 o C are 0, Tank pressure is 0,15 bar and the atmospheric pressure is bar. Vapour volume m 3 = ( ) m 3 - Vapour density kg/m 3 = (D P x 28,05) / (8,31441 x D T) Vapour weight at 99,5 o C 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 -95 o C and tank pressure 0,345 bar. Liquid -100,5 o C, density 563,63 kg/m 3, liquid volume 1313,348 m 3 Maximum filling limit is 89,45%, which is equal to 1313,348 m 3 with relief valve setting on 4,5 bar. Total weight of cargo in the tank is kg in vacuum Calculation of Ethylene set point 4,5 bar Liquid volume m 3 Shrinkage factor at -100,5 o C Corrected liquid volume m 3 Liquid density at -100,5 o C kg/m 3 Weight of liquid in vacuum at -100,5 o C kg Cargo tank 100% volume m 3 Vapour volume m 3 Shrinkage factor at -95 o C Corrected vapour volume m 3 Vapour density at -95 o C kg/m 3 Weight of vapour in vacuum at -95 o C 399 kg TOTAL LOADED IN VACUUM kg

25 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 1424,127m 3. We then get a total weight of cargo in the tank on kg, which is 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. Ref. temp. Ref. dens. Filling limit rr/rl x 98% % bar o C kg/m Calculation of Ethylene set point 0,5 bar Liquid volume 1 424,13 m 3 Shrinkage factor at -100,5 o C 0,99645 Corrected liquid volume 1 419,08 m 3 Liquid density from table at -100,5 o C 563,625 kg/m 3 Weight of liquid in vacuum at -100,5 o C kg Cargo tank 100% volume 1 468,18 m 3 Vapour volume 44,053 m 3 Shrinkage factor at -95 o C 0,99661 Corrected vapour volume 43,904 m 3 Vapour density at -95 o C 2,581 kg/m 3 Weight of vapour in vacuum at -95 o C 113,326 kg TOTAL LOADED IN VACUUM 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 87 o C 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 kg kg = kg in vacuum.

26 Weight of ROB before loading at temperature -87 o C and tank pressure 0,02 bar Vapour density 1,885 kg/m 3 Tank volume 100% 1468,18 m 3 D T = 186 K Weight of vapour 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 15 o C or we have to calculate a factor. The factor is, as follows: (1 (ρ air/ ρ cargo liquid)) / (1 ( ρ air/ ρ Brass) (1 (1,2 kg/m 3 /ρ cargo liquid)) / ( 1 (1,2 kg/m 3 / 8100kg/m 3 )) In our example we will get a factor, as follows: (1 - (1,2 kg/m 3 /563,625 kg/m 3 )) / ( 1 (1,2 kg/m 3 / 8100kg/m 3 )) = 0, Then we have to multiply mass in vacuum with the factor: kg x 0, = 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 = ( * 1,2/ 563,625) = 795 kg 485 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.

27 Calculation of ROB before loading Loading report Cargo Propylene Port Al Jubail Molecular 42,08 Date mass Atm.press. 1,015 Vessel LPG Seagull Liquid Tank # Sounding in meter Volume from tab. in m 3 Temp. in o C Pressure in bar r liquid in kg/m 3 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 0 Liquid Vapour Tank #!00% vol. in m 3 Vapour volume in m 3 Temp. in o C r vapour in kg/m 3 Shrinkage factor Mass of vapour in kg 1P 1182, , ,129 0, S 1182, , ,129 0, P 1468, , ,129 0, S 1468, , ,129 0, P 1468, , ,112 0, S 1468, , ,112 0, Total mass of vapour Total mass in vacuum ROB Total loaded in vacuum Total loaded in air

28 Calculation of mass after loading Loading report Cargo Propylene Port Al Jubail Molecular 42,08 Date mass Atm.press 1,020 Skip LPG Seagull Liquid Tank # Sounding in meter Volume from tab. in m 3 Temp. in o C Press in bar r liquid Shrinkage in kg/m 3 factor Mass of liquid in kg 1P 8, , ,6 601,2 0, S 8, , ,6 601,2 0, P 8, , ,6 601,2 0, S 8, , ,6 601,2 0, P 8, , ,6 600,0 0, S 8, , ,6 600,0 0, Total mass of Liquid Vapour Tank # 100% vol. in m 3 Vapour volume in m 3 Temp. in o C r vapour Shrinkage in kg/m 3 factor Masse of vapour in kg 1P 1182,18 58, ,445 0, S 1182,18 56, ,445 0, P 1468,18 68, ,459 0, S 1468,18 66, ,459 0, P 1468,18 63, ,431 0, S 1468,18 64, ,431 0, Total mass of vapour Total mass in vacuum ROB Total Loaded in vacuum Total loaded in air

29 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 don t 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 Volume of liquid to be used for gassing up We have a cargo tank with volume 1182,18 m 3 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 m 3 x 2 = 2364,36 m 3. 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 40 o C and our cargo tank steel 20 o C. The formula is mass = ρ vapour x total volume. From the table thermodynamic properties for propylene superheated vapour, we find the vapour ρ to 1,812 kg/m 3 on 20 o C and P=1 bar. Then we take the total volume 2364,36 m 3 and multiply with vapour ρ 1,812 kg/m 3 = 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 Calculation of volume liquid we have to order Cargo Propylene 100% Tank volume 1182,18 m 3 r vapour at atmospheric pressure from 20 o C 1,812 kg/m 3 table r liquid from table -40 o C 602,4 kg/m 3 Number of changes 2 Total volume to be changed 2364,36 m 3 Mass volume = Volume x r for vapour Mass total volume = 4283,26 kg Volume liquid to be loaded = Mass volume / r liquid

30 Volume to be loaded = 7,110 m 3 We have to load 7,11 m 3 propylene at 40 o C from shore tank to change the vapour atmosphere at 20 o C 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 vessel s 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 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 10 o C, total tank volume is 2564,36 m 3 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 Mass of vapour after gassing up Cargo Propylene Tank volume #1 P/S 2364,36 m 3 r for vapour at atmospheric -10 o C 1,953 kg/m 3 pressure Mass of vapour in the tank kg We have now calculated that we should have 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 kg from the amount of liquid we need to gas up the whole vessel. We can continue with the example and have 15 m 3 liquid propylene at 10 o C, vapour temperature 0 o C and the pressure 3,3 bar in tank #1 P/S. Total volume of the vessel is 8237 m 3 and atmospheric pressure is 1,015 bar. That means we have to gas up vessel s total volume volume of tank #1 P/S, which is equal to 8237 m ,36

31 m 3 = 5872,64 m 3 with an average temperature of 25 o C. Our experience is that we need 2,5 volume changes to reach 97% hydrocarbons in the vapour atmosphere, 2,5 changes is 5872,64 m 3 x 2,5 = 14681,6 m 3. 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 Mass we can use for gassing up Cargo Propylene Tank volume #1 P/S 2364,36 m 3 Volume liquid in tank #1 P/S 15,00 m 3 Mass of liquid in tank #1 P/S kg Volume of vapour in tank #1 P/S 2349,36 m 3 Mass of vapour in tank #1 P/S kg Total mass in tank #1 P/S kg Mass of vapour in tank #1 P/S after gassing kg up Usable mass in tank #1 P/S kg We have kg available in tank #1 P/S, but when we are completed gassing up, we have 4618 kg vapour left, that means we have kg available for gassing up. Total volume to be gassed up is vessel s total volume minus volume of tank #1 P/S multiplied with 2,5. That gives 8237 m ,36 m ,64 m 3 x 2,5 = m 3. We have to find the vapour density equal to tank steel temperature 25 o C, which is 1,724 kg/m Calculation of volume needed Volume vapour at 25 o C of available m 3 mass r for vapour with atmospheric pressure 25 o C 1,724 kg/m 3 at Total volume vapour 2,5 times tank m 3 volume Volume to be ordered m 3 In this example, we do not have enough liquid to reach 2,5 times for gassing up. There was 1581 m 3 vapour short, so we have to order that 1581 m 3 x 1,724 kg/m 3 = 2726 kg Calculation of mass vapour at a given temperature Cargo Propylene Vessel s total volume 8237 m 3 Tank #1 P/S volume 2364,36 m 3 Volume to be gassed up 5872,64 m 3 Amount changes 2,5

32 Tank steel temperature 25 Atmospheric pressure 1,015 bar Vapour r at 25 o C 1,724 kg/m 3 Total volume to gas up 5872,64 x 2, m 3 Mass of total volume to gas up kg Available mass kg Mass in kg to load to complete gassing kg up 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 25 o C to 60 o C, we do not need to supply any extra from shore Calculation of vapour at 60 o C Vapour r at 60 o C 1,543 kg/m 3 Total volume to gas up 5872,64 x 2, m 3 Mass total volume to gas up kg Available mass 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. o C

33 ENCLOSES Enclose 1 Table 21 0,500-0,510 Specific API Gravity Gravity Density 60/60 o F 60 o F 15 o C 0,500-0,5004 0,501-0,5014 0,502-0,5023 0,503-0,5033 0,504-0,5043 0,505-0,5053 0,506-0,5063 0,507-0,5073 0,508-0,5083 0,509-0,5093 0,510-0, Enclose 2 Table 54C Observed Density 15 o C temperature, 0,500 0,505 0,510 o C Factor for reduction of volume to 15 o C -43 1, , ,146-42,5 1, , , ,15 3 1, ,143-41,5 1, , , , , ,141-40,5 1, , , , , ,139-39,5 1, , , , ,14 3 1,137-38,5 1, , , , , ,134-37,5 1,14 3 1, , , , ,132

34 -36,5 1, , , , , ,13-35,5 1, , , , , ,128-34,5 1, ,13 3 1, , , ,125-33,5 1,13 3 1, , , , ,123-32,5 1, , , , , ,12-31,5 1, , , , , ,118-30,5 1, ,12 3 1, , , ,115-29,5 1,12 3 1, , , , ,113-28,5 1, , , , , ,11-27,5 1, , , , , ,108-26,5 1, ,11 3 1, , , ,105-25,5 1, , , , , ,103-24,5 1, , , , , ,101-23,5 1, , , , , ,098-22,5 1, ,1 3 1, , , ,096-21,5 1, , , , , ,094-20,5 1, , , , , ,092-19,5 1, , , , , ,089-18,5 1, ,09 2 1, , , ,087-17,5 1, , , , , ,084-16,5 1, , , , , ,082-15,5 1, , , , , ,08

35 -14,5 1, ,08 2 1, , , ,077-13,5 1, , , , , ,075-12,5 1, , , , , ,072-11,5 1, , , , , ,07-10,5 1, , , , ,07 2 1,068-9,5 1, , , , , ,065-8,5 1, , , , , ,063-7,5 1, , , , , ,06-6,5 1, , , , ,06 2 1,058-5,5 1, , , , , ,055-4,5 1, , , , , ,052