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

36 Enclose 3 Shrinkage factor for exercise Seagull 04 Temp. Sh.fakt. Temp. Sh.fakt. Temp. Sh.fakt , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,99638

37 Enclose 4 Correction for expansion of sounding tape tank #1 Temp. Read Sounding in meter o C ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0,001 ± -0,003-0,003-0,002-0,002-0,001-0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,001-0,001-0,001-0,001-0,001 0, ,001-0,001-0,001-0,001-0,001 0, ,001-0,001-0,001-0,001 0,000 0, ,001-0,001-0,001-0,001 0,000 0, ,001-0,001-0,001-0,001 0,000 0,000

38 Enclose 5 Trim table tank #1, tank 100% volume 1182,18 m 3 Sounding and trim meter in Sounding Stern trim Sounding Stern trim Sounding Stern trim 0,5 1 0,5 1 0,5 1 8,04-0,021-0,043 8,41-0,021-0,043 8,78-0,022-0,045 8,05-0,021-0,043 8,42-0,021-0,043 8,79-0,022-0,045 8,06-0,021-0,043 8,43-0,021-0,043 8,80-0,022-0,045 8,07-0,021-0,043 8,44-0,021-0,043 8,81-0,022-0,045 8,08-0,021-0,043 8,45-0,021-0,043 8,82-0,022-0,045 8,09-0,021-0,043 8,46-0,021-0,043 8,83-0,022-0,045 8,10-0,021-0,042 8,47-0,021-0,043 8,84-0,022-0,045 8,11-0,021-0,042 8,48-0,021-0,043 8,85-0,022-0,045 8,12-0,021-0,042 8,49-0,022-0,043 8,86-0,022-0,045 8,13-0,021-0,042 8,50-0,022-0,043 8,87-0,022-0,045 8,14-0,021-0,042 8,51-0,022-0,043 8,88-0,022-0,045 8,15-0,021-0,042 8,52-0,022-0,043 8,89-0,022-0,045 8,16-0,021-0,042 8,53-0,022-0,043 8,90-0,022-0,045 8,17-0,021-0,042 8,54-0,022-0,043 8,91-0,022-0,045 8,18-0,021-0,042 8,55-0,022-0,043 8,92-0,022-0,045 8,19-0,021-0,042 8,56-0,022-0,043 8,93-0,022-0,045 8,20-0,021-0,042 8,57-0,022-0,043 8,94-0,022-0,045 8,21-0,021-0,042 8,58-0,022-0,043 8,95-0,022-0,045 8,22-0,021-0,042 8,59-0,022-0,043 8,96-0,022-0,045 8,23-0,021-0,042 8,60-0,022-0,043 8,97-0,022-0,045 8,24-0,021-0,042 8,61-0,022-0,044 8,98-0,022-0,045 8,25-0,021-0,043 8,62-0,022-0,044 8,99-0,022-0,045 8,26-0,021-0,043 8,63-0,022-0,044 9,00-0,023-0,045 8,27-0,021-0,043 8,64-0,022-0,044 9,01-0,023-0,045 8,28-0,021-0,043 8,65-0,022-0,044 9,02-0,023-0,045 8,29-0,021-0,043 8,66-0,022-0,044 9,03-0,023-0,046 8,30-0,021-0,043 8,67-0,022-0,044 9,04-0,023-0,046 8,31-0,021-0,043 8,68-0,022-0,044 9,05-0,023-0,046 8,32-0,021-0,043 8,69-0,022-0,044 9,06-0,023-0,046 8,33-0,021-0,043 8,70-0,022-0,044 9,07-0,023-0,046 8,34-0,021-0,043 8,71-0,022-0,044 9,08-0,023-0,046 8,35-0,021-0,043 8,72-0,022-0,044 9,09-0,023-0,046 8,36-0,021-0,043 8,73-0,022-0,044 9,10-0,023-0,046 8,37-0,021-0,043 8,74-0,022-0,044 9,11-0,023-0,046 8,38-0,021-0,043 8,75-0,022-0,044 9,12-0,023-0,046 8,39-0,021-0,043 8,76-0,022-0,045 9,13-0,023-0,047 8,40-0,021-0,043 8,77-0,022-0,045 9,14-0,023-0,047

39 Enclose 6 Sounding table tank #1, tank 100% volume 1182,18 m 3 Sounding Volume m m 3 Sounding m Volume m 3 Sounding m Volume m 3 8, ,77 8, ,05 8, ,33 8, ,91 8, ,08 8, ,20 8, ,04 8, ,10 8, ,05 8, ,17 8, ,12 8, ,90 8, ,30 8, ,14 8, ,74 8, ,33 8, ,15 8, ,58 8, ,55 8, ,16 8, ,41 8, ,67 8, ,17 8, ,23 8, ,79 8, ,17 8, ,05 8, ,91 8, ,17 8, ,85 8, ,03 8, ,16 8, ,65 8, ,14 8, ,15 8, ,44 8, ,25 8, ,14 8, ,23 8, ,36 8, ,12 8, ,01 8, ,46 8, ,10 8, ,78 8, ,56 8, ,07 8, ,54 8, ,66 8, ,04 8, ,30 8, ,76 8, ,00 8, ,06 8, ,86 8, ,96 8, ,80 8, ,95 8, ,92 8, ,54 8, ,03 8, ,87 8, ,27 8, ,12 8, ,82 8, ,00 8, ,20 8, ,76 9, ,71 8, ,28 8, ,70 9, ,43 8, ,36 8, ,63 9, ,13 8, ,43 8, ,56 9, ,83 8, ,50 8, ,48 9, ,53 8, ,57 8, ,40 9, ,21 8, ,63 8, ,32 9, ,89 8, ,69 8, ,23 9, ,56 8, ,75 8, ,14 9, ,23 8, ,80 8, ,04 9, ,89 8, ,85 8, ,94 9, ,54 8, ,90 8, ,83 9, ,19 8, ,94 8, ,71 9, ,82 8, ,98 8, ,59 9, ,46 8, ,01 8, ,47 9, ,08

40 Enclose 7 Correction for expansion of sounding tape tank #2 / #3 Temp. Read sounding in meter o C ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0, ,004-0,003-0,003-0,002-0,001-0,001 ± -0,003-0,003-0,002-0,002-0,001-0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,002-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,002-0,001-0,001-0,001-0,001 0, ,001-0,001-0,001-0,001-0,001 0, ,001-0,001-0,001-0,001-0,001 0, ,001-0,001-0,001-0,001-0,000 0, ,001-0,001-0,001-0,001-0,000 0, ,001-0,001-0,001-0,001-0,000 0,000

41 Enclose 8 Trim table tank #2 and #3, tank 100% volume 1468,18 m 3 Sounding and trim in meter Sounding Stern trim Sounding Stern trim Sounding Stern trim 0,5 1 0,5 1 0,5 1 8,04-0,029-0,059 8,41-0,029-0,059 8,78-0,028-0,057 8,05-0,029-0,059 8,42-0,029-0,059 8,79-0,028-0,057 8,06-0,029-0,059 8,43-0,029-0,059 8,80-0,028-0,057 8,07-0,029-0,059 8,44-0,029-0,059 8,81-0,028-0,057 8,08-0,029-0,059 8,45-0,029-0,059 8,82-0,028-0,057 8,09-0,029-0,059 8,46-0,029-0,059 8,83-0,028-0,057 8,10-0,029-0,059 8,47-0,029-0,059 8,84-0,028-0,057 8,11-0,029-0,059 8,48-0,029-0,059 8,85-0,028-0,057 8,12-0,029-0,059 8,49-0,029-0,059 8,86-0,028-0,057 8,13-0,029-0,059 8,50-0,029-0,059 8,87-0,028-0,057 8,14-0,029-0,059 8,51-0,029-0,059 8,88-0,028-0,057 8,15-0,029-0,059 8,52-0,029-0,059 8,89-0,028-0,057 8,16-0,029-0,059 8,53-0,029-0,059 8,90-0,029-0,057 8,17-0,029-0,059 8,54-0,029-0,059 8,91-0,029-0,058 8,18-0,029-0,059 8,55-0,029-0,059 8,92-0,029-0,058 8,19-0,029-0,059 8,56-0,029-0,059 8,93-0,029-0,058 8,20-0,029-0,059 8,57-0,029-0,059 8,94-0,029-0,058 8,21-0,029-0,059 8,58-0,029-0,059 8,95-0,029-0,058 8,22-0,029-0,059 8,59-0,029-0,059 8,96-0,029-0,058 8,23-0,029-0,059 8,60-0,029-0,059 8,97-0,029-0,058 8,24-0,029-0,059 8,61-0,029-0,059 8,98-0,029-0,058 8,25-0,029-0,059 8,62-0,029-0,058 8,99-0,029-0,058 8,26-0,029-0,059 8,63-0,029-0,058 9,00-0,029-0,058 8,27-0,029-0,059 8,64-0,029-0,058 9,01-0,029-0,058 8,28-0,029-0,059 8,65-0,029-0,058 9,02-0,029-0,058 8,29-0,029-0,059 8,66-0,029-0,058 9,03-0,029-0,058 8,30-0,029-0,059 8,67-0,029-0,058 9,04-0,029-0,058 8,31-0,029-0,059 8,68-0,028-0,058 9,05-0,029-0,058 8,32-0,029-0,059 8,69-0,028-0,058 9,06-0,029-0,058 8,33-0,029-0,059 8,70-0,028-0,058 9,07-0,029-0,058 8,34-0,029-0,059 8,71-0,028-0,058 9,08-0,029-0,058 8,35-0,029-0,059 8,72-0,028-0,058 9,09-0,029-0,058 8,36-0,029-0,059 8,73-0,028-0,057 9,10-0,029-0,058 8,37-0,029-0,059 8,74-0,028-0,057 9,11-0,029-0,058 8,38-0,029-0,059 8,75-0,028-0,057 9,12-0,029-0,058 8,39-0,029-0,059 8,76-0,028-0,057 9,13-0,029-0,058 8,40-0,029-0,059 8,77-0,028-0,057 9,14-0,028-0,058

42 Enclose 9 Sounding table tank #2 and #3, tank 100% volume 1468,18 m 3 Sounding m Volume m 3 Sounding m Volume m 3 Sounding m Volume m 3 8, ,35 8, ,91 8, ,51 8, ,81 8, ,21 8, ,56 8, ,27 8, ,50 8, ,59 8, ,72 8, ,79 8, ,61 8, ,18 8, ,07 8, ,63 8, ,63 8, ,35 8, ,64 8, ,08 8, ,62 8, ,64 8, ,52 8, ,88 8, ,64 8, ,96 8, ,14 8, ,62 8, ,40 8, ,39 8, ,60 8, ,84 8, ,64 8, ,57 8, ,27 8, ,87 8, ,52 8, ,70 8, ,11 8, ,47 8, ,13 8, ,33 8, ,42 8, ,55 8, ,55 8, ,35 8, ,97 8, ,76 8, ,27 8, ,38 8, ,97 8, ,18 8, ,80 8, ,17 8, ,09 8, ,20 8, ,36 8, ,98 8, ,60 8, ,54 8, ,86 8, ,00 8, ,71 8, ,74 8, ,29 8, ,88 8, ,60 8, ,77 8, ,04 9, ,46 8, ,15 8, ,19 9, ,31 8, ,53 8, ,33 9, ,14 8, ,89 8, ,46 9, ,97 8, ,26 8, ,59 9, ,79 8, ,62 8, ,70 9, ,59 8, ,97 8, ,82 9, ,39 8, ,32 8, ,92 9, ,18 8, ,66 8, ,02 9, ,95 8, ,00 8, ,11 9, ,72 8, ,33 8, ,20 9, ,47 8, ,66 8, ,28 9, ,22 8, ,98 8, ,35 9, ,95 8, ,29 8, ,41 9, ,67 8, ,60 8, ,47 9, ,38

43 Enclose 10 Correction on the float Specific gravity (kg/dm 3 ) Correction in meter 0,40 + 0,220 0,45 + 0,187 0,50 + 0,170 0,60 + 0,150 0,70 + 0,136 0,80 + 0,127 0,90 + 0,121 1,00 + 0,114 1,10 + 0,110 Cargo calculation ASTM table LPG Seagull Port: Date: Last: Skip: Tank no. # 1 P 1 S 2 P 2 S 3 P 3 S 1 Volume of tank 100 % m 3 2 Liquid temp. o C 3 Sounding read m 4 Float correction m 5 Correction for vapour m temperature 6 List correction m 7 Trim correction m 8 Sounding at 20 o C m 9 Volume at 20 o C m 3 10 Shrink. factor tank steel 11 Corrected liquid volume m 3 12 Reduction factor from table 54C 13 Volume 15 o C m 3 14 Density on liquid at 15 o C Mt/m 3

44 tab Mass of liquid in vacuum at Mt 15 o C 16 Uncorrected vapour volume m 3 17 Shrinkage factor vapour phase 18 Corrected Vapour volume m 3 19 Tank pressure bar 20 Atmosphere pressure bar 21 Molecular weight 22 Temperature on vapour 23 Density on vapour kg/m 3 24 Mass of vapour in vacuum Mt 25 Total mass in vacuum Mt o C Total weight in air ROB in air Total loaded in air Mt Mt Mt

45

46 13- Cooling Processes and Calculation

47 13 COOLING PROCESSES AND CALCULATIONS 13.1 MOLLIER DIAGRAM INTRODUCTION The Mollier diagram is an invaluable in helping to understand refrigeration calculations. In a cooling plant, the cooling media will constantly change its state. Vapour is compressed and gets a higher pressure and temperature, vapour condenses and liquid vaporises. In the cooling process, the aggregate state, pressure and temperature change continuously. To perform calculations of the cooling process, one must know the enthalpy changes taking place. If one is to make enthalpy-tables for a cooling media in all possible states, you will find a large and unpractical table. Diagrams for simply cooling media where the cooling media s enthalpy under the actual aggregate states, are therefore developed. Such a diagram contains infinitely much more information than a table can have, and gives, in addition, possibility to make a heat-technical picture of the cooling process. Before making use of the Mollier diagram one has to learn how the diagram is built, and how the different lines lie in the diagram. This will make it easier to find information in the diagram. Log p-h diagram or called the Mollier diagram, has a vertical logarithmic scale for pressure (p) and a horizontal scale for enthalpy (h). In the Mollier diagram, one will often think of the two different units for the different qualities. We will consistently make use of the SI-units and refer to these scales only.

48 The pressure of the Mollier diagram is, as in all heat technical tables, given as absolute pressure. If nothing else is given, we have chosen to set the atmospheric pressure to equal 1 bar, which corresponds to 0,1 MPa. In cooling technical calculations one is only interested in enthalpy changes and not the absolute enthalpy values. The values for enthalpy are therefore chosen from a random reference state. One must take note of this, if one wants to compare enthalpy values from the different tables or diagrams. In the Mollier diagram, the sack is the most bearing curve. The bent line that goes from the lower left corner and upwards the KP (the kinetic point) is called the liquid line. If this line is lengthways, the liquid is always in its boiling point. The line bends from KP and is almost vertically down the middle of the diagram. This line is called the saturation curve. On this line, the state will be saturated gas. At the right of the saturation curve, the gas is super-heated. To the left of the liquid line the liquid is super-cooled. The area between the liquid and the saturation line specifies a mixture of boiling liquid and saturated gas. The dashed line indicates the proportion of mixture between liquid and gas. Halfway between the liquid line and the saturation line there will be equal parts of gas as liquid. In the sketch above, one line is marked 0,1. Along this line there is 90% liquid and 10% saturated gas. The distance between the liquid and the saturation line indicates how large the vaporisation is. Notice that the vaporisation varies with the pressure and is lessened, the higher the pressure is. In the critical point, liquid can not appear. The Mollier diagram also has lines that indicate density (or specific volume), temperature and entropy. The lines for entropy indicate how pressure, temperature and heat content change in an adiabatic state of proportion. An adiabatic state of proportion is an alteration without heat exchanging with the surroundings. The real compression progress in a compressor will of course deviate some from these lines because of loss in the compressor and heat exchange with the environment.

49 The lines for density (kg/m 3 ) or specific volume (m 3 /kg) indicate density with varying pressure and temperature. The line for constant temperature is vertical from the top of the diagram down to the liquid line. The temperature goes from the liquid line horizontal to the saturation line, thereby to bend vertically towards the enthalpy from the saturation line. To maintain necessary training in using the Mollier diagram, we will see in some examples how to obtain useful information from the diagram. This course provides a Mollier diagram for propane utilises the diagram in the following advice. Take note that there are often two scales for pressure and enthalpy. Note which units to utilise for temperature, density and entropy. We will only utilise SI-units in our calculations. In the course enclosure, tables for conversion between the most common units are enclosed. When you plot and draw in the diagram it is recommended to always utilise a soft pencil. The diagram can be used several times Example 1 The ship is loaded with propane and the tank manometer pressure is 2 bar. Find the physical state for the liquid in the loading tank. Any state of proportion can settle in the diagram if we can identify two actual crossing lines. As the cargo in the loading tank always lies in its boiling point, the state of proportion must lie somewhere among the liquid line. It is also known that the pressure in the tank and above the liquid is read off at 2 bar on the manometer. This pressure is equivalent 3 bar absolute (if we assume the atmospheric pressure to 1 bar) or 0.3 MPa. The state of proportion here is the point of intersection between the liquid line and the pressure line of 0,3 MPa.

50 Example 2 The ship is loaded with propane and the tank pressure is read off on the manometer at 2 bar. Plot the state of proportion for the gas above the liquid in the cargo tank. As the gas above the liquid in the loading tank is, at all times, saturated, the state of proportion must lie somewhere on the saturation line. The pressure in the tank and above the liquid is read off to 2 bar on the manometer. This pressure is equivalent to 3 bar absolute (if we assume the atmospheric pressure to 1 bar) or 0,3 MPa. The state of proportion here will be the state of intersection between the saturation line and the pressure line of 0,3 MPa.

51 Example 3 The ship is loaded with propane and the tank pressure is read off on the manometer to 2 bar. Utilise the Mollier diagram to find how much heat one must supply the tank to evaporate (boil off) 1 kg propane. In example 1 and 2, the state of proportions for the liquid and the gas in the loading tank is stated at a tank pressure of 2 bar. The difference between the liquid enthalpy h 1 and the saturated gas enthalpy h 2 is the heat quantity that is needed to evaporate 1kg propane at 2 bar pressure. Calculation of evaporation heat Gas : Propane Pressure (abs.) = 3 bar Enthalpy for saturated gas, h 2 = 880 kj/kg - Enthalpy for saturated liquid, h 1 = 490 kj/kg Latent heat of evaporation = 390 kj/kg The latent heat of evaporation for propane can also be found in a heat thermal property table for propane. Find this table in the course enclosure, and check that the latent heat of evaporation is the same.

52 Example 4 The gas carrier is loaded with propane and the tank pressure is read of on the manometer of 2 bar. On a cargo compressor in operation the suction pressure is read off to 1,5 bar, and the suction temperature to 10 o C (14 o F). Plot the state of proportion for the gas on the compressor. As the pressure in the gas in to the compressor is 1,5 bar, the state of proportion must lie on a pressure line equal 0,25 MPA (1,2 bar + 1 bar) in the diagram. The exact state of proportion is plotted where the temperature line of 10 o C crosses the pressure line of 0,14 MPa. When the point is plotted, the density of the gas into the compressor can be defined from the density lines that run sideways out to the right in the diagram. A cooling plant s net cold capacity is expressed as: Q nett = m x Dh where m =the mass of gas that streams through the cooling plant per hour (kg/s) Dh = the difference between enthalpy on the gas that abandons the tank and enthalpy on the condensate that returns back to the tank (kj/kg) Notice that the density of the gas increases at higher gas pressure and lower temperature. Larger density gives more mass per hour that will flow through the plant. More mass involves larger cold capacity for the plant.

53 Example 5 A gas cylinder is filled with floating propane. Temperature of the air and propane liquid and gas is 15 oc. We say that the gas over the liquid is saturated. The valve opens and floating propane flows over in an open container. Plot the state of proportion for the liquid in the bottle, before the valve was opened, and for the liquid in the opened container later. The liquid lies on the liquid line in the diagram. The point (1) is defined either from the temperature line of 15 o C (59 o F) or equivalent pressure line 0,73 Mpa (7,3 bar). When the liquid is let out of the bottle, the pressure of the liquid to the atmospheric pressure (1 bar) lowers. The fast reduction of the pressure involves a powerful boiling of the liquid because of an unbalance between the liquid s temperature and gas pressure of the liquid. The heat of the boiling is taken from the liquid itself and the surroundings and the liquid gets colder. An enthalpy change during the process will not take place. We can draw the process line (from point 1 to point 2) for any change to the liquid, as a vertical line from the cross-point through the liquid line to the pressure line. Notice that the new state of proportion (2) is inside the sack and that a precise share of the liquid has evaporated because of pressure reduction.

54 13.2 THE COOLING PLANTS COMPONENT A good processor should have a construction that secures from gas leakage, be applicable for the different media qualities, have large regulation opportunity, obtain least space and give as little noise and vibrations as possible. We separate between four types of compressors in the cooling plant: oil lubricated pistons, oil free pistons, oil lubricated screw compressors and oil free screw compressors. Both types of pistons and oil free screw compressors are used on the cargo side in the cooling plant. On the Freon side, oil lubricated piston and screw compressors are used Pistons An oil-lubricated piston has piston rings made of cast-iron. It is therefore necessary to lubricate the cylinder walls. A part of this oil will be lead out of the compressor. An oil separator in the pressure pipeline separates most of the oil. But some of the oil is lead further out in the system in form of oil vapour. The consequence is that one can not have too large demands for defilement of cargo on ships with such cargo compressors. The state of the piston rings and the cylinder liner is conclusive for how much oil leads out of the compressor. In time, mud will extend from the oil settler. This reduces the settler s capacity. Cleaning of this is therefore one of the assumptions to maintain the oil consumption and the pollution at a fair level. The working valves are usually a plate type valve and are placed inside the compressor. The working valves have work-over intervals from 2000 to 5000 hours. The capacity regulation by the suction valve plates is gradually lifted from the seats and only the gas is pumped in and out of the cylinder. The lifting arrangement is normally performed hydraulically by oil taken from the pressure side of the oil pump. The lubricate oil pump is normally the gear type and normally placed in the extension of the crankshaft. The oil lubricates the bearing, shaft and the cylinder walls. One supplies the crankcase with a coil for heating and cooling. The lubricated piston compressors are normally built as V or W machines. This construction is less space demanding, but also less friendly working-wise. The compressor case is sterling in cast-iron and the valves normally of aluminium. The casings are loose and cast-iron. The principal for oil free piston compressors is that no parts, which are in contact with the gas, are lubricated. The piston runs dry in the cylinder. These construction problems are solved, as follows: The sealing device between piston and cylinder liner is performed by piston rings with self-lubricated qualities. Piston rings of teflon material are often used in these compressors. A labyrinth gland performs the sealing device between piston and cylinder walls. Piston/cylinder is made by a small groove. The clearance is as small as possible. As the piston is in touch with the cylinder walls, there is minimal wear on compressors with this construction. Loss of energy, which is due to friction in a normal

55 compressor, is about the same as the leakage loss for a compressor with labyrinth gland. Oil free piston compressors are at all times built as double acting. That involves that they are supplied with a cross-head, and the working valves are placed outside the cylinders. The compressor is also built as two and three stage compressors. This means that one cylinder is used as 1 st stage, the next to 2 nd stage and the third as 3 rd stage. The individual stages here will have different cylinder diameters where the first is the largest and the others gradually smaller. The first stage is also referring to as a low-pressure (LT) cylinder and the highest as high-pressure (HT) cylinder Piston compressors and lubricate oil The oil will, at all times, be led out with the gas, from the compressor in plants, by oil lubricated piston compressors. Oil that departs from the oil remover is lead back to the crank room or to an exhaust tank. Where is decided by what cargo we have. One can lead ammonia back to the crank room because this is not soluble in the oil. LPG gas is soluble in oil and should therefore first boil out in an exhaust tank. The lubricant free piston compressors have, under normal situations, no draft of oil. Some gas will however leak down in the crankcase on the piston compressors and there mix with the oil. High pressure in the crankcase is an indication of inferior sealing. This indicates that the viscosity of the oil is less. If sealing and oil is in bad shape, oil will also go up in to the cylinder liner and go with the vapour out of the compressor. Pollution of cargo and reduced compressor capacity are the consequences of this. One recommends using a mineral oil for the butadiene, while synthetic oil is used for the rest of the cargo. Many of the synthetic oils are hygroscopic and will therefore accumulate dampness if they are exposed to damp air. When changing oil types, one

56 must be sure not to mix the oil types. To be sure that all remnants of the old oil are removed, washing the crank room with the new oil before refilling is recommended Linde compressor Linde is a type of oil free piston compressor with Teflon sealing rings that is used as a cargo compressor in the cooling plant. These are built as V or W machines with a number of revolutions of about 1200 rpm. The piston is kept central in the cylinder by the help of steering on the topside of the piston. A packing seal and oil blocks the connection between cylinder and crank room. Cylinder, cylinder cover and crank house are cooled and heated by glycol. The working valves are plate type and are capacity regulated at 50% and 100%. They are generally performed with a hydraulic lifting of the suction valves. It is important to consider that the length of lifetime of the Teflon material depends upon at which working temperature the compressor is operating. Too high operating temperature will lead to higher wear and thereby higher leakage loss in the compressor Sulzer Compressor Sulzer produces oil free piston compressors with labyrinth sealing for use as cargo compressors in the cooling plant. The cylinders are arranged in series and can be used as one to three stage compressors with the number of revolutions from 600 to 1000 rpm. The pistons are kept central in the cylinder by the help of a piston steering and oil lubricated drawback. Labyrinth sealing and an arrangement of oil scraper rings fence the compound between cylinder and crank room. Gas that flows through the labyrinth sealing is lead back to the suction side. The working valves are made of the ring flapper type. The capacity can be regulated to 50% and 100% and is performed with a hydraulic lifting of the suction valve

57 flapper. Cylinder, labyrinth sealing, crank room and piston rod steering is cooled and heated by glycol Delivery rate A compressor delivery rate is the difference between the suction volume and the stroke volume. A high delivery rate is thereby an important factor for the cooling plant capacity. If we compare the piston compressors, the delivery rate to an oillubricated compressor will be clearly better than an oil free compressor. The volume difference is lowest at low working pressure. The pressure ratio over the compressor or a cylinder in the compressor is the ratio between delivery pressure and suction pressure. The placing of the working valves externally and oil free compressor makes the damaging room larger and is the cause for a different delivery rate. If the diagram above is actually for an oil free compressor with a read off suction pressure of 0,5 bar and a delivery pressure of 5 bar, the pressure ratio is: 6 / 1,5 = 4 A pressure ratio of 4 gives a delivery rate of about 0,60. If the compressor has a gross capacity of 400 m 3 /h, net capacity is: 400 m 3 /h x 0,60 = 240 m 3 /h 240 m 3 /h gas flow through the compressor in this working situation. Comparison between oil free piston compressors with labyrinth sealing and Teflon sealing indicates that the delivery volume is a bit larger for a compressor with Teflon sealing. However, the power consumption rises at a higher delivery pressure percentage, more than occurs with delivery volume.

58 Screw compressors The screw compressor is divided into two groups: oil lubricated and oil free. The difference is the same as for the piston compressors. In the oil free screw compressors, the parts that are in contact with the gas are not lubricated with oil. This type of compressor works by the displacement principal. The working parts of the compressor are two rotating screws, parallel placed in a compressor case. The screws, or the rotors, are shaped with male and female profiles that mesh into on another.

59 The impetus is transferred to the male rotor. The most common combination is four and six lobes for the male and the female rotor. Four lobes on the male rotor mesh into 6 lobes on the female rotor. If the male rotor has a number of revolutions of 3600 rpm, the female rotor number of revolutions is: The compressors working progress is, as follows: 4 / 6 x 3600 rpm = 2400 rpm The lobe on the female rotor reveals the die orifice. When the rotors turn, the chamber gets longer and gas sucks into the compressor The chamber passes the inlet opening. The gas is now closed between the end casing Gradually as the rotor twists additionally, the volume is less and the gas compress against the outlet wall An additional twist of the rotor uncovers the outlet wall and the gas is pressed out of the compressor. In the most common screw compressors on board, the confined gas will compress to a precise volume before blowing down to the pressure side. One says that the compressor has a constant embedded volume proportion. The embedded volume proportion is determined from the proportion: V 1 / V 2 where V 1 is the maximum induction volume per rotor per rotation. V 2 is the volume of reticent gas after compression, but before blow down to pressure side. The pressure proportion is not constant, but dependent from ploy for the employed gas and rate of cooling. The connection between the volume proportion and the pressure proportion can therefore be expressed, as:

60 (V 1 / V 2 ) k = Designed compression proportion = Delivery pressure/suction pressure Practically, this means that when the drift parameters change, the screw compressors with a built-in volume proportion will compress the gas needlessly too much or too little. Too low compression increases back flow of gas and hence reduced efficiency. The oil free screw compressors are used on board as cargo compressors. A synchronising gear is used to keep the rotors from wearing one another. The axial strength recovers when placed on the outlet side. Bearings and gear lubricate by a self-pressure oil system. To prevent oil from forcing into the compressor room, axial sealing is installed between the bearings and the gas area inside the compressor. The compressor case is cooled and heated by glycol. The number of revolutions is kept relatively high, 5000 to rpm, to keep the leakage loss as low as possible. This means that if electric motors are used as; the gear must be used. Oil lubricated screw compressors are used on board on the Freon side in cascade plants. The female rotor is driven directly by the male rotor in this type of compressor. Ample oil must therefore be supplied to the rotors. The oil has the following purpose: Lubricate the bearings Lubricate, cool and seal between rotors and cases Lubricate, cool and seal between the rotors Cool and lubricate the axial sealing Give hydraulic energy to the capacity regulation Cool the gas The screw compressors power consumption is unfavourable compared with the twostage piston compressor. Installation of a superfeed arrangement compensates some in this condition.

61 The condensate is super-cooled in a coil placed in a combined evaporator and intercooler. The boil off in the inter-cooler is led back to the compressor (S). The plants cooling capacity will rise ( Q). The power consumption will also rise, but less than the increased cooling capacity.

62 The capacity regulation on oil lubricated screw compressors takes place by help of a hydraulic operated drawback that opens a gate to the inlet side. The capacity if often infinitely variable between 10 % and 100 %. The advantages with screw compressors are: no suction or pressure valves few mobile parts lower cost at purchase, installation and maintenance little vibrations and simple foundation even gas flow, no pulsating no sensitiveness for liquid in the gas flow high volumetric efficiency of the actual working area possibilities for infinitely variable capacity regulation low pressure pipe temperature for oil lubricated screw compressors

63 Lubricating oil system The sketch below indicates a normal oil system for cooling plants with soil lubricated screw compressors. Despite the fact that the oil separator is installed after the compressor, a larger share of oil is constantly in the freon consequently this leads to poor plant cooling capacity. It is not possible to drain off pure oil on the liquid side and a system is therefore installed as an oil-restoring device in these plants. The oil-restoring device system is often used in these systems. Some of the cold and the oil-rich freon liquid is taken from the pressure side and delivered to the lower inlet of the heat exchanger. A relatively warm freon liquid from the liquid collector meets it here. The oil-rich freon liquid evaporates inside the pipes. Because of the large velocity of the gas out of the heat exchanger, the oil particles are carried with it, as well.

64 An independent regulation valve provides that a suitable amount of freon is let into the heat exchanger when the liquid is taken from the pressure side. A constant overflow pipe provides that the level in the heat exchanger is constant Thermostatic expansion valve Thermostatic expansion valves are used in the cooling plant to regulate the liquid injection in the heat exchangers. The evaporation pressure and the overheating in the heat exchanger control the valve. For the valve to function satisfactorily the valve must be installed as recommended by the supplier and the valve is dimensioned to the plant. In choosing a valve, one must consider the operation terms and the capacity need for the plant.

65 13.3 ONE-STAGE DIRECT OPERATION Cooling plants with 1-stage direct operation are the simplest cooling plants with the least number of components. The description 1-stage refers to the compression that occurs in only one step. Direct operation means that the cargo vapour is sucked from the cargo tank and compressed directly either against seawater or Freon in an cascade plant. In this chapter we have chosen to use one-sage direct cooling for our examples, which is used on board many semi-pressurised gas carrier. Propylene is chosen as cargo and cooling media. The simplified sketch of the plant indicates the main components in a cooling plant with possibility for motoring both with 1 stage and 2 stage direct operation. With 1 stage direct operation cargo tank(s), liquid separator, compressor, condenser, liquid collector and regulation valves are utilised. Notice that the compressor has two or more cylinders with different diameters. This is normal when the plant is also designed for 2 stage operation. Following, there is also an intermediate cooler with coil. This is not in use in 1 stage operation and supply lines are therefore drawn as a broken dashed line. On the following log sheet, an actual operation situation for this plat is registered. The cooling process itself is drawn in a sketch of a Mollier diagram for propylene.

66 Date: Time: Cargo Sea water Pressure Temp. Temp Air Power Tank 1.suct 1.deliv. 1.suct. 1.deliv. In Out o C Amp. % bar bar bar o C o C o C o C load 21/3 12: ,0 2,8 10, The loading tank is filled with propylene. The temperature of the liquid in the tank is 12 o C, this gives a tank pressure of 3 bars. Heat will be transferred to the tank and the cargo because of the difference in temperature between cargo and the surroundings. The condition of the liquid in the tank is in the boiling point line at a pressure of 3 bars. When liquid boils, the gas over the liquid is regarded as saturated. The condition of the gas at this point is, therefore, in the Mollier diagram on the saturation line at an absolute pressure of 4 bar. The point is marked as A. The piston compressor sucks the gas from the tank via a liquid separator. Because of heat leakage, the gas is heated between the tank and compressor. In addition, the pressure in the gas will reduce some because of friction in the pipeline. When gas sucks into the compressor, both temperature and pressure will be changed. If the compressor s suction temperature is 5 o C and read-off suction pressure is 2,8 bars, the condition of the gas (when it sucks into the compressor) can be plotted into the diagram. The condition is in an overheated area where the temperature line of 5 o C and the pressure line of 3.8 bars (2,8 bars + 1 bars) crosses. The point is marked as B. The gas is compressed parallel over both cylinders and delivered to the seawatercooled condenser. The compressor performs a job on the gas and the gas is supplied with energy. Pressure, temperature and enthalpy increases. The compressing process is traced into the diagram as the line from B to C.

67 The heat is transferred from the gas to the seawater in the condenser. The gas into the condenser now has a temperature of 80 o C. The temperature of the seawater into the condenser is read-off as 23 o C. The gas cools and condenses against the relative cold seawater under constant pressure. The pressure is maintained by the compressor, which continuously compresses new vapour. The process line in the Mollier diagram is traced from point C to D. Notice that the gas s overheating at first is removed, and then the gas is saturated. Then the evaporation heat is removed and the gas condenses. When all gas is condensed in the condenser, the state is on the boiling point line. The relative cold gas of about 12 o C that was sucked out from the tank has, through the compression, increased the temperature to a relative warmer state than the seawater. This renders a possibility of heat transmission from an originally colder to a warmer medium. One can see that the gas s enthalpy reduces in the process. A heat quantity is transferred to the seawater that is corresponding to the difference in enthalpy of the gas that sucks from the tank and the condensate enthalpy after the condenser. It is the temperature of the seawater that circulates through the condenser that determines the condensation pressure. The heat exchanger is normally designed so that the temperature difference between seawater and condensate is between 3 o C and 8 o C. In our example, the read-off condensation pressure is 10,5 bars. In the heat technical table for propylene, one can see that this corresponds to a condensation temperature of 25 o C, which is within the normal area. A change in the temperature of seawater and gas over the condenser indicates something is wrong. If for example, the seawater side becomes dirty, the temperature increase of the seawater over the condenser is smaller and the condensation pressure higher. One should notice that such changes of the cooling plant mainly occur over time, something that makes it difficult to intercept the signals. The only useful method to reveal such a development is to log the drift parameter regularly. This routine is of great importance. The suction of the gas is now condensed and the condensate collected in a liquid collector. A regulation valve between the liquid collector and the cargo tank regulates the level in the liquid collector. One can see the process line between point D and E in the Mollier diagram. Notice that there is no enthalpy change. Temperature and pressure is what changes over the valve. When the pressure over the condensate is reduced from condensation pressure of 10,5 bars to tank pressure of 3 bars, the liquid will boil energetically. Boiling requires heat. This heat is mainly taken from the liquid itself. The liquid cools and some of the gas evaporates. A mixture of liquid and gas with the same temperature as the liquid and the gas in the tank is what returns into the cargo tank. The temperature must be the same for the mixture, because the pressure is the same. One can locate the shares of liquid and saturated gas in the Mollier diagram as point E. The regulation valve s job is to let the condensate back to the cargo tank in a constant and controlled process. The valve should simultaneously keep the liquid locked to secure the compressor s maintenance of the compression pressure. It is of great importance that the regulation valve has satisfactory activity. If the liquid level and thereby the liquid lock disappears, uncondensed gas will blow through the

68 condenser and reduce the capability of the plant. If the liquid level rises, so the condenser refills this result in an inferior heat exchange of the condenser and a higher condensation pressure. If the compressor has a capability of 350 m 3 /h, the net cooling capacity of the plant, Q net is: Q net = m x (h 2 - h 1 ) The capacity of the compressor (V) and the density (p) of the gas into the compressor determines the mass (m), which flows through the plant. The gas s specific volume (v) is point B in the Mollier diagram to 0,12 m 3 /kg. This corresponds to a density of (l/0,12 m 3 /kg) or 8,33 kg/m 3. Q net = {V /3600 s} x r x (h 1 - h 2 ) = {(350 m 3 /h) / 3600 s} x 8,33 kg/m 3 x (240 kj7kg kj/kg) = 243 kw Notice the factor, which expresses the cooling plant s capability to influence and to control, thereby, the cooling capacity of the plant. Its construction and condition determine the capacity of the compressor. Good control and maintenance influence the capability of the cooling plant. Pressure drops determine the density of the gas into the compressor and temperature increases between tank and compressor. Needless throttling of valves on the suction side or tightened filters, gives a needless low pressure and thereby, a lower density. Inferior isolation of suction pipe or tank gives a larger heat leakage, higher temperature and thereby lower density. Lower density gives less mass through the compressor hence with reduced capability. The enthalpy difference of the plant is determined by the cooling media and other outer conditions like the seawater temperature. These factors are therefore less influential than the others are. By studying the process of the Mollier diagram, one can see that the condition for the temperature of the liquid in the tank and h 2 determines the gas at the top of the tank. The other enthalpy point, h 1, is determined by the condensation pressure. Notice that a higher condensation pressure removes this point to the right and will give less enthalpy difference than with a low condensation pressure. A lowest possible condensation pressure gives therefore the best capacity in addition to the best operating condition for the compressor. If the capability of the cooling plant is larger than the total heat leakage, the temperature of the cargo will be lower over time. When the temperature sinks, the tank pressure sinks. If the seawater temperature and other operating conditions are unaltered, the process lines EAB are displaced parallel downward with the tank pressure against the process lines E 1 A 1 B 1.

69 One can see that a lower tank pressure results in a lower cooling capacity. The mass through the compressor sinks because the density of the vapour is lower with a lower pressure. In addition, the point h 2 is displacing gently against the left; this results in some lower enthalpy difference. One must notice that the saturation line of the different material has different inclines. Therefore the influence of this change varies. If the condensation pressure is kept unaltered, the process will be longer for the compression and the outlet temperature from the compressor will be higher. The condition for the gas after compression is removed from C to C 1. This has in itself no direct influence on the capacity, as long as the condenser can transfer the necessary heat amount from the gas to the seawater. Previously, we have seen how the piston compressor delivery extent depends on the pressure condition of the compressor. The pressure condition in our example is, as follows: Pressure condition = Delivery pressure/suction pressure = 11,5 / 3,8 = 3 According to the fall of the tank pressure, the pressure condition will rise and result in a constant decreasing delivery extent. A reduction of the suction pressure with 2 bars and the same condensation pressure gives a pressure condition of: 11,5 / 1,8 = 6,4 This increased pressure condition gives an essential reduction in delivery extent and thereby cooling capacity. The influence of reduced delivery extent of the cooling capacity is often larger than the summary of the other factors. The explanation of why the cooling capacity is at the lowest by low tank pressure is some complex. In addition to above-mentioned condition, the heat leakage will, of course, be largest at a low tank pressure.

70 Example 1 The cooling capacity of the vessel is at the moment 234 kw. We want to find the time to cool down the cargo of 1000 MT propylene, from 12 o to 30 o. At first we set an expression for the heat balance: Net heat abducted in the cooling plant = heat-leakage + heat abducted from the cargo In the vessel s operation manual, we find that the heat leakage to the cargo tanks at presence air and seawater temperature is 36 kw. The necessary heat amount removed from the cargo to cool the cargo is located in the Mollier diagram or in a heat technical table: Enthalpy for propylene at-12 o C = 61,5 kcal/kg = 257,1 kj/kg - Enthalpy for propylene at -30 o C = 52,1 kcal/kg = 217,8 kj/kg =Enthalpy difference = 39,3 kj/kg Here one must eliminate 39,3 kj from one kilo propylene to lower the heat temperature from -12 o C to -30 o C. As we have 1000 MT propylene, the heat amount that must be abducted is: (1000 x 103) kg x 39,3 kj/kg = 39,3 x 106 kj We insert above-mentioned values of the heat balance and get: (234 kw x T) = (36 kw x T) + 39,3 x 10 6 kj where T = cooling time in seconds T = (39,3 x 106 / 198)s = s T = ( / 3600) hrs. = 55 hrs. The calculated cooling time here at unaltered heat leakage and cooling capacity is about 55 hours. In practice, the cooling time will be a bit longer because the cooling capacity and the heat leakage is not constant in the period Example 2 The vessel is loaded with propane and the cargo tank pressure is read to 2 bars. The suction pressure of a cargo compressor is read to 1,5 bars and the suction temperature is 10oC (14oF). Plot the state-point for the gas into the compressor. As the pressure of the gas into the compressor is 1,5 bars, the state-point must lie in a pressure line equal 0,25 MPa (1,5 bars + 1 bars) in the diagram. The exact statepoint is plotted where the temperature line of 10 o C crosses the pressure line of 0,25 MPa.

71 When the point is plotted, the density of the gas into the compressor is determined from the density lines aslant towards right in the diagram. Q net = m x h Where m = the mass of gas flowing through the cooling plant per time (kg/s) h = the difference between enthalpy of the gas that leaves the tank and enthalpy on the condensate that is returned back to the tank (kj/kg) Notice that the density of the gas increases at higher gas pressure and lower temperature. Larger density gives more mass per hour that flows through the plant. More mass involves larger cold capacity for the plant Example 3 A gas bottle is filled with liquid propane. The temperature of the air and propane liquid and gas is 15oC. The gas above the liquid is regarded as saturated. The valve opens and liquid propane runs into an open container. Plot the statepoints for the liquid in the bottle before the valve was opened and for the liquid in the open container afterwards. The liquid lies on the liquid line and the point (1) is determined either from the temperature line of 15 o C (59 o F) or corresponding pressure line of 0,73 MPa (7,3 bars). When the liquid is discharged from the bottle, the pressure lowers above the liquid to the atmosphere pressure (1 bars). The quick reduction of the pressure involves powerful boiling in the liquid because of an unbalance between the liquid temperature and the gas pressure above the liquid. The heat from boiling is taken from the liquid itself and the surroundings and the liquid gets colder. No enthalpy change will take place during the process. One can trace the process line (from point 1 to point 2) for the alteration that occurs with the liquid, as a vertical line from the point of the liquid line to the atmosphere pressure line. Notice that the new state-point (2) lies inside the sack and that a precise share of the liquid has vaporised because of the pressure reduction STAGE DIRECT OPERATION One can see how the cooling plant s capacity was reduced at larger pressure conditions in the chapter concerning 1 stage direct operation. At 2 stage operation, stated pressure conditions will improve the capacity of the plant. This plant is the same as used in the one-stage lesson. The difference is that the plant now is altered to 2 stage operation with intermediate cooling. Notice how the volume flow now compares to 1stage operation and which valves that regulates this.

72 The compressor is the same, but only one of the cylinders is used to suck from the tank. The compressor is traced with different diameters of the cylinders. Most of the piston compressors that are used on the cargo-side have this construction. The largest cylinder is used as 1 st stage in the compression and is called the low-pressure cylinder. The smallest cylinder is used in 2 nd stage compression and is called the high-pressure cylinder. There is also installed an intermediate pressure container with a coil for sub-cooling of the condensate in this plant. The following table indicates a summary from the cooling plant s log sheet in an actual stage situation. Cargo Sea water Pressure Temp. Temp. Air Tank 1.suct 1.deliv 2.deliv 1.suct 1.deliv 2.suct 2.deliv In Out o C bar bar bar bar o C o C o C o C o C o C 27 0,7 0,5 3,5 10, The compressor s low pressure cylinder or 1 st stage suck vapour from the cargo tank. The vapour leaves the tank with state A and are sucked into the compressor s 1 st stage with state B. The vapour compresses in the compressor s 1 st stage and is delivered to the intermediate cooler (MT) with state C. The intermediate cooler is

73 partly filled with cargo liquid. The superheated gas is normally led down in the liquid where it is cooled by the relatively cold liquid. The compressor s 2 nd stage suck in saturated vapour from the top of the MT cooler and compresses this vapour and delivers it to the cargo condenser. The process line is marked DE. The heat is transferred from the vapour to the seawater in the cargo condenser, the vapour condensate and is collected in the liquid collector. The process line is marked EF. The condensate is led via a coil in the MT cooler through the regulation valve and back to the cargo tank. The process line is marked FG. Notice that the state-point is below the sub-cooled area. The pressure in the intermediate cooler determines the temperature of the liquid in it. A pressure of 3,5 bars for propylene corresponds to a temperature of 8 o C. The vapour that is sucked inn to the 2 nd stage has a temperature of 7 o C in the log. The small difference of temperature may be caused by heat leakage in the suction line or reading error. The state-point G is therefore determined from the temperature line corresponding to a pressure of 0,45 MPa (3,5 bars + 1 bars). A to low liquid level in the MT cooler results in a sub-cooling of the condensate and thereby reduces cooling capacity. The heat supplied to the liquid in the MT cooler from the warm vapour from the 1 st stage and the relatively warm condensate leads to a lot of liquid evaporation. An independent regulation valve regulates the liquid level in the MT cooler. The vapour is led from 1 st stage and down in the liquid of the MT cooler in our 2- stage plant. Many plants that also have the possibility for flash cooling where the vapour is lead into the top of the MT cooler and give operational advantages. After the coil in the MT cooler the condensate is released back to the cargo tank through the regulation valve. Notice that there is a larger share of liquid now than at 1-stage operation. The state-point for the condensate that returns back to the tank is further to the left in the diagram.

74 The cooling plants net cold capacity is calculated, as previously, by the expression: Qnet = m x (h2 - h1) The enthalpy difference is larger at 2-stage operation and sub-cooling, which is something that increases the result. But the mass of gas flowing through the plant is also altered when the plant is reorganised from 1-stage to 2-stage operation. A lower pressure condition gives a better delivery extent and thereby larger mass. With 2- stage operation, vapour from the cargo tank is sucked with one cylinder only compared with 1-stage where all cylinders is used to suck in the vapour. This has a negative guided influence so that the total increase of the cooling capacity is thereby reduced Example 1 To better visualise the difference between 1-stage operation and 2-stage operation, we can look at an example where the vessel has loaded warm propylene and delivered the cargo fully cooled. The plant is started with 1-stage operation because of a very high tank pressure. Subsequent to the cargo cooling, the cargo tank pressure is reduced. The result is increased differential pressure between 1 st and 2 nd stage and reduced delivery extent. When the temperature of the cargo is at 30 o C, one decides to reorganise the operation to a 2-stage operation with intermediate cooling. The following working parameters are registered before and after the re-adjustment: Cargo Sea water Pressure Temp. Temp Tank 1.suct 1.deliv. 2.deliv. 1.suct 1.deliv. 2.suct. 2.deliv. In Out bar bar bar bar o C o C o C o C o C o C 1,1 1,0 3,7 10, ,1 0,9 10, The compressors capacity at 1-stage operation and a pressure condition of 6 (11,5/1,9) are calculated to 275 m 3 /h. At 2-stage operation and a pressure condition of 2,4 (4,7/2), this is calculated to 240 m 3 /h. The cooling capacity at 1-stage operation is calculated from the following values located in the Mollies diagram and technical table: Density on gas at -22 o C and 0,9 bar: 4,17 kg/m 3 Enthalpy of gas in a cargo tank: 230 kj/kg Enthalpy of the condensate after condenser: -60 kj/kg Qnett = {V /3600 s} x r x (h1 - h2) = {(275 m3/h) / 3600 s} x 4,17 kg/m3 x (230 kj7kg kj/kg)

75 = 92 kw The cooling capacity at 2-stage operation and intermediate cooling are calculated from the following values located in the Mollier diagram and technical table: Density of gas at -20 o C and 1 bar: 4,17 kg/m 3 Enthalpy of gas in a cargo tank: 230 kj/kg Enthalpy of the condensate after condenser: -140 kj/kg Qnett = {V /3600 s} x r x (h1 - h2) = {(240 m3/h) / 3600 s} x 4,17 kg/m3 x (230 kj7kg kj/kg) = 103 kw The two calculated cooling capacities in the example must be evaluated from the accuracy in our measures. With normal measure divergence, one can conclude that the cooling capacities are about equal for this plant at 30 o C for propylene and a condensation pressure of 10,5 bars. The decision of reorganising the plant to 2-stage operation at this moment looks correct, if one only evaluates the cooling capacities. One must however notice that normally it refers, in the operating manuals, only to maximum delivery-pressure/maximum differential-pressure over the compressor and not pressure condition. These parameters are simple practical expressions for the compressor s constructional limitations, which influence the operation. If it is specified, in the operating manual, that maximum delivery pressure at 1-stage is 12 bars and maximum differential pressure is 8 bars, alteration from 1-stage to 2- stage operation is determined, as follows: At unchanged seawater temperature, the condensation pressure is 10,5 bars. Limitation of maximum delivery pressure is thereby kept regardless of operations form. The differential pressure at 1-stage will rise at falling tank pressure. If the condensation pressure is a constant 10,5 bar, 1-stage will have a differential pressure of 8 bars when the suction pressure p 0 is: p o = (10,5-8) bar = 2,5 bar If the pressure drop of the suction line is constant at 0,1 bars, the plant must alter to 2-stage operation at a tank pressure of 2,6 bars, which corresponds to a cargo temperature of 15 o C. From constructional limitations one can see that the cooling plant should be altered to 2-stage operation at 15 o C and not at 30 o C Example 2 When we are loading cargo that is warmer than compared with the safety valve s set point, the loading time is determined by the capacity of the cooling plant. To maintain the tank pressure during the opening pressure for the valves, the cargo that is loaded must be cooled down. Bad plant condition and poor plant operation reduces the cooling capacity and results in longer loading time. The same gas tanker, as in the example above is to load 4000 MT hot propylene of +15 o C. The loading tank s safety valves have an opening pressure of 4,5 bars. It is determined to maintain the tank pressure of 4 bars during the loading.

76 In the following log two different operational parameters are registered from the cargo cooling plant. In the first alternative, the compressor s suction valves are throttled to a suction pressure of 1,5 bars. In the second alternative, the suction valve is less throttled and the suction pressure is 2,5 bars. The following operational parameters is registered before and after the alteration: Cargo Pressure Temp. Tank 1.suct 1.deliv. 1.suct. 1.deliv. Alt. bar bar bar o C o C 1 4,0 1,5 8, ,0 2,5 8, The two cooling processes are plotted into a Mollier s diagram for propylene. Notice that the enthalpy difference is the same regardless of suction pressure. In the Mollier diagram, we find specific volume and density for the gas into the compressors at the two suction pressures: Specific volume for propylene: v/2,5 bar and 0 o C: 0,14 m 3 /kg = 7,143 kg/m 3 Specific volume for propylene v/1,5 bar and 5 o C: 0,22 m 3 /kg = 4,546 kg/m 3 As the cooling capacity expresses: Q net = m x (h 2 - h 1 ) One can see that the difference in the cooling capacity will be proportional with the change in the mass flowing through the plant. The percentage difference in the cooling capacity thereby is expressed, as follows: {(7,143-4,546) x 100 / 7,143}% = 36 % The cooling capacity is reduced by about 36% if the suction pressure is reduced from 2,5 bars to 1,5 bars for this re-condensation plant during the above-mentioned conditions. This also means that loading time will increase with 36%.

77 To visualise this influence better, we calculate the loading time for the vessel. We assume that the cargo tanks with ROB, steal and isolation are cooled down to -5 o C before commence loading. The heat leakage to the cargo tanks (Q Tr ) is stated in the vessel s operational manuals to 144 kw and three identical cargo cooling units drive in the re-condensation plant, where the compressor s capacity in each cooling unit is set to 400 m 3 /h. We locate how much heat has to be removed from each kilo propylene liquid to cool this down to +15 o C to 5 o C in the heat technical table: Enthalpy for propylene v/15 o C: 76,8 kcal/kg = 321,5 kj/kg - Enthalpy for propylene v/-5 o C: 65,7 kcal/kg = 275,0 kj/kg = Necessary abducted heat, h = 46,5 kj/kg Necessary heat that must be removed from the cargo (Q L ) is: Q L = m x Dh = (4000 x 10 3 )kg x 46,5 kj/kg = 186 x 10 6 kj We find the enthalpy values of the gas out of the tank (h 2 ) and the enthalpy of the mixture that is returned to the tank (h 1 ), in the Mollier diagram: h2 = 240 kj/kg h1 = -80 kj/kg The cooling capacity for the entire re-condensation plant at a suction pressure of 2,5 bars and 0 o C is: Qnetto = m x (h2 - h1) = {V /3600 s} x r x (h1 - h2) = {3 x 400 m3/h / 3600} x 7,143 kg/m3 x (240 80)kJ/kg = 762 kw

78 The loading time T 1, at this operation situation is then: T 1 = Q L / (Q netto - QT r ) = 186 x 10 6 kj / ( ) kw = s = 83.6 hours The cooling capacity for the entire re-condensation plant at a suction pressure of 1,5 bar and 5 o C is: Qnetto = m x (h2 - h1) = {V /3600 s} x r x (h1 - h2) = {3 x 400 m3/h / 3600} x 4,546 kg/m3 x (240 80)kJ/kg = 485 kw The loading time, T 2, at this operation situation is then: T 2 = Q L / (Q netto - QT r ) = 186 x 10 6 kj / ( ) kw = s = 152 hours We can see that the influence of unnecessary throttling on the suction side of the compressor gives large deflection of the cooling capacity and thereby loading time. The tank pressure during the loading period also has influence on the total loading time. If a high tank pressure is kept close up to the safety valve s opening pressure, less heat will be abducted from the loaded cargo and loading time will be shorter CASCADE PLANT A cascade plant onboard a gas tanker is a cooling plant composed by two coolant circuits working in serial with each other. Both circuits are complete cooling plants that are built in many different configurations. Both 1-stage and 2-stage operations are used depending on the cargo that should be cooled and choice of components in the cooling systems. One of the cooling circuits consist in a closed cooling process where the cargo directly condenses contra freon or other cooling media. The other cooling circuit consists in a closed cooling process where the freon or another cooling medium condenses contra seawater. The cargo side of the cascade plant is the same as in the 2-stage plant example, with exception for the condenser. The seawater-cooled condenser is now replaced with a freon-cooled condenser. On most cascade plants we have both a freon condenser and seawater cooled condensers in the cargo cooling plant. This gives

79 flexibility and good operation economy. The freon side of this plant is a simple 1- stage plant with screw compressor. Freon compressors are of oil-lubricated type. Large amounts of oil will follow the compression gas out of the compressor. If this is not separated and returned to the oil receiver, the share of oil in the freon circuit will be too high. The oil will reduce heat transfer in the heat exchangers and create operation interruptions. The oil separator has two functions where it separates oil from the gas and at the same time is a system tank for the oil system. The following summary from a cooling plant s log indicates an actual operation situation for this plant. The cargo that is cooled is ethylene. Ethylene cycle Pressure Temp. Tank 1.suct 1.deliv. 2.deliv. 1.suct. 1.deliv. 2.suct. 2.deliv. bar bar bar bar o C o C o C o C 0,6 0,5 5,3 17, Freon cycle Sea water Pressure Temp. Temp 1.suct 1.deliv. 1.suct. 1.deliv. In Out bar bar o C o C o C o C 0,2 12,

80 The cooling process of the cargo side is plotted into a Mollier diagram for ethylene and is, as follows: The cooling process on the freon side of the cascade plants is plotted equal. The condensation pressure of the loading side depends on the temperature of the freon liquid circulating through the loading condenser. The pressure in the liquid separator again depends on the suction pressure of the freon compressor and determines the freon temperature. (We assume that the freon side has the right filling). Notice that the condensation pressure for ethylene lies at 17,3 bars, which corresponds to a temperature of 27 o C. The pressure in the liquid separator is 0,2 bars, which corresponds to a temperature of 37 o C. This gives a temperature difference of 10 o C in the cargo condenser, usually too high for this type of heat exchanger. If the freon pump don t deliver sufficient liquid to the cargo condenser, the cause is probably one or a combination of the following: reduced heat transmission caused by incrustation in the loading condenser reduced heat transmission caused by too large share of oil in the Freon liquid too high condenser pressure caused by unknown gas on the loading side Cascade plants are used both for LPG and ethylene. The plant exists both on atmosphere gas carrier LPG and for semi pressured gas carriers LPG/ ethylene. This

81 plant is an example of a usual cascade plant that can also be used to re-condense ethylene. Normally screw compressors are used both on the loading side and on the freon side in the cooling plant onboard big atmosphere pressure gas carrier LPG. There are many different configurations of cascade plants. The freon side is frequently equipped with super-feed or intermediate pressure container with subcooling. Some plants have freon pumps, which pump freon through the cargo condenser, where others use thermal expansion valves. Freon compressors can be piston compressors or screw compressors. Screw compressors can be built with one or two stages. Plants with and without MT-containers, with and without flash cooling, with and without de-super-heaters, 2-stage compression with and without sub-cooling, exist on the loading side. The compressors on the cargo side have to be oil free piston or screw compressors. The configuration possibilities are many and the variation in plants from vessel to vessel is what one faces onboard. The understanding of the cooling process and knowledge to your vessel s special plant is a basic assumption for safe and economic operation of the plant.

82 14- Insulation and Heat Transfer

83 14.1 INSULATION AND INSULATION MATERIAL There are three different methods in transporting heat: thermal conductance, convection and radiation. The insulation material s main task is to reduce heat transmission from thermal conductance. Most insulation material s insulating qualities emerge from stationary gases, bad thermal conductance capability and thereby good insulation capability. The thermal conductance capability is expressed by a material s thermal conductance number (thermal conductivity) and states the heat quantity measured by Watt, which is transported through to surface of 1 meter thickness when the temperature difference is 1K. Thermal conductivity Natural pure metals From 8 to 400 W/m K Iron 80 W/m K Ice at -20 o C 2,4 W/m K Natural liquids From 0,10 to 0,60 W/m K Wood From 0,1 to 0,3 W/m K Glass 1 W/m K Polypropylene 0,12 W/m K Sand 0,35 W/m K Natural gases From 0,008 to 0,048 W/m K Stationary air 0,024 W/m K Freon 22 0,012 W/m K One can see that the best heat conductor or worst insulation materials are pure metals. The worst heat conductors or best insulation materials are stationary gases. It is of most importance that the gases are kept as stationary as possible, because the total thermal transmission is higher if convection also arises. This is in practice solved by trying to catch the gas inside the smallest possible cavity, or by keeping the gas inside a net of thin fibres. The thermal conductance figure will mainly increase at higher temperature because of larger convection. The best insulation material regarding thermal transmission is a composition of a firm material and a gas with the separately lowest thermal conductance figures. First of all, it is the thermal conductance figure that is of interest when looking at the important qualities for an insulation material. Lower thermal conductance figures render possible thinner insulation and thereby place saved.

84 The upper temperature limit for the material s relevance normally has no importance for an insulation material that should be used for cool insulation, but one prefers that the material tolerate highest possible temperature considering fire. The insulation material s lower temperature is of major importance. In the plastic insulation condensation can occur at a lower temperature and thereby increase the thermal inductance figure for the material. Thermal expansion and elasticity are two qualities of great importance for plastic insulation materials, for example cargo tanks on gas tankers. Changing temperature for tank shells and insulation can lead to periodical expansions and compressions. As the thermal expansion co-efficiency for the insulation material can be 4 to 5 times larger than the steel, cracks may easily arise if the insulation material don t have good elasticity. A plastic insulation material like polyurethane has good adhesion firmness to the steel priming and good elasticity and is thereby resistant to cracks. An insulating problem area is around the loose tank cradle of cylinder cargo tanks. All of the expansion and compression movement of longitudinal direction of the tank takes place here. Large demands are made both to material and for the insulation to be good and long performance. A special developed insulation with especially good elasticity is suitable for such areas. All insulation materials that are used onboard must be fire resistant. Insulation materials made of plastic are added or built up in such a way that in case of fire, fireextinguishing vapour are released, and they are self-quenching. The insulation material will only burn as long as a foreign fire source is present. One must not underestimate the risk of fire in the plastic insulation and the consequences by such a fire. Fire can easily arise in such materials in connection with weld work. Ignition of the insulation on the cargo tanks has occurred with fatal consequences. The large amount of thick, black and poisonous smoke that is formed by such a fire has prevented escape from hold space and serious poisoned injures. Where the material is exposed to strains the material s compressive-strength is important. The strain points will, for example, be large in the support points for an insulated pipeline. To evade lasting compression or crumbling of the insulation material, special compressive-strengthen materials are used in these areas. Of the chemical qualities for insulation materials, it is the corrosive qualities that often are underestimated. When the insulation material of glass or mineral wool got a high humidity, it will have a strong corroding effect on metals. It is difficult over time to avoid this onboard. Even stainless steel pipes corrode in such an environment and if this is allowed to proceed, an expensive replacement is soon the result. The capability to resist humidity or diffusion resistance is an important quality that must be evaluated in choosing insulation material, as well as, in planning maintenance. When insulating between a warm and cold side, for example, a cargo tank, the cold side will be tight (tank shell) and the warm side will be surrounded by air with high humidity. Because of a higher saturation pressure on the warm side in proportion to the cold side, moist air will be forced through and will condense against the cold side. The thermal conductance figure will, because of the humidity, increase and the water will freeze and destroy the insulation. The damage extent will accelerate if such a process is continuously unchanged. If the diffusion resistance for stationary air is set to 1, the proportional diffusion resistance for mineral wool will be about 1,5 and for polyurethane it will be about 60.

85 This indicates that insulation of polyurethane is about 40 times more resistant against moist air to leak through than an insulation of mineral wool. Regardless of which insulation material that is used, moist air will penetrate into the material and destroy it. A diffusion-tight damp-latch on the warm side is, at all times, imperative on a chill insulation. The most commonly used materials and methods are: Thin aluminium foil glued on the insulation. This method is suitable for insulation non-exposed for mechanical wear and tear, for example, some loading tank constructions. Galvanised and stainless steel sheet that is fastened by pop cones and the joint seals with jointing compound. This method is expensive, but strong against mechanical wear and tear and necessary for foaming. Glass-fibre armed polyester. This method is more moderate, but gets easily fragile. Most suitable for repair One or more layers with asphalt emulsion armoured with multiple layers of glass fibre fabric. Moderate method and easy to maintain, but weak for mechanical wear and tear. Sprayed mastic with or without armouring. Moderate method, but week against mechanical war and tear and requires more maintenance on exposed places. Most suitable for repair and sprinkling on insulation not exposed to rough weather. The insulation material on gas ships may be divided into the following three different groups based on structure and material: Cellular plastic, which is expanded plastic raw material, built up in a cellular structure. Wallboard, which is built up of a net with thin fibres. Expanded volcanic perlite, which is built up on a cellular structure Polyurethane There are a number of insulation materials, which are built up of raw plastic materials. The most used is polyurethane. Mixing isocynate and polyole, normally in the proportion 1:1 makes polyurethane. Isocyanate has a resemblance to thick oil and polyole has a resemblance to clear liquid. A chemical reaction that is exothermic is actuated when mixing the liquids. If one adds some water, carbon dioxide is formed, which because of the reaction-heat evaporates and blows up the material. Foam with about 90% closed cells and a very low thermal conductance number appear. Polyurethane-foam based on CO 2 has a relatively long time of expansion.

86 The qualities of the polyurethane-foam can be improved by using different freon materials as a blowing agent. Because of the freon material s lower boiling point, the expansion is quicker. R11 was used earlier as a blowing agent. The thermal conductance capability for R11 is only half of CO 2. If applying polyurethane foam with a sprayer, or frothing, freon is used as the blowing agent. Freon evaporates with speed in normal surrounding temperature and one obtains pre-expansion, when the mixture leave the spray. One can mix polyurethane foam for smaller repairs in a bucket before pouring into a mould. As there are different suppliers on the market, it is recommended to check with the supplier about the composition and if water has to be used when mixing. There are also a number of machines for spraying on the market, also disposable spray equipment with smaller containers for iciyanat and polyol. Disposal-spray equipment of this type or a simple modifying of the paint sprayer onboard, is suitable for smaller repairs to the insulation. Polyurethane is also available as half-cups for insulation of pipes. It is important that the dimension of the cup fit to the outer diameter of the pipe so that air leak is avoided. The polyurethane cups can easily be cut for matching the pipe bend, bend and valves. It is recommended to lay two layers with half cups on larger pipes, so that the connections are displaced in proportion to each other. The half cups and the connections are glued, and secure the density and strength and a water barrier is laid either with thin metal plates or mastic Polystyrene Polystyrene is produced in two stages. At the first stage polystyrene is pre-expanded with vapour where blisters from 1 to 6 mm are formed. The pre-expanded material is sent up into large silos for de-aeration. After 2 to 4 day in the silo, the pre-expanded material is filled in forms for further expansion and compression of the grains, to shape blocks. Heat from steam or electrical elements are used in this last expansion. The finished blocks are cut up into plates and pipe cups Isolation of LNG ships with spherical tanks Cellular plastic is used as insulation material for Moss Rosenberg's spherical cargo tanks. Dow Chemicals in USA and Technical Isolation in Norway developed a new method of mounting insulation, because the insulation materials contraction is 2,5 times more than aluminium and 5 times more than steel at -163 o C. Poles of Styrofoam of about 3 meters long are welded together and set continuously around the spherical tank, from equator against the bottom and on top of the tank. An external aluminium foil of 0,25 mm is laid on also. The room around the loading tank is filled with nitrogen with a dew point down to 40 o C. The purpose is partly to protect the tanks from corrosion and to reduce the pressure of humidity against the insulation.

87 Mineral wool Mineral wool is a collective term of different fibre rich insulation materials. Rock wool and glass wool are two types of insulation material that are used. The production method and user area is equal. A mixture of several types of stones is used to produce rock wool. The stones are melted in a temperature up to 1600 o C and are dispatched over a wheel with very high rotation. The melted mixture of stones is hurled out and chilled in long thin fibres. Cementing agent based on plastic is added and hardened with hot air. The amount of cementing agent varies and is determined by the material s purpose of use. Rock wool plates are elastic and a normal density of about 45 kg/m 3. The plates are also delivered with larger density and firmness. Glass wool is produced like rock wool by hurling and chilled melted glass to very fine fibres, only about 0,0025 mm diameter Expanded Perlite Perlite is made of a volcanic rock species with perlite structure. The main component is about 71% SiO 2 and about 16% Al 2 O 3. The raw material has some water content, which by heating to about 1200 o C evaporates and blows up the material. One obtains a 10 to 20 times expansion with numerous closed airtight cells. Simultaneously the individual corns loosen form each other, the material explodes and forms sharp-edged corns with sizes from 0 to 3 mm with very large mechanical strength. The density is about 60 kg/m 3. One can easily fill the whole room around a loading tank with perlite. It is first of all used on atmosphere pressure gas carriers. Maintenance of insulation The cooling plant on a gas carrier is constructed and calculated for thermal leakage from cargo tanks and system when the ship is new. Insulation is exposed for ageing, wear and tear and will in time be reduce if maintenance of the insulation is not kept. If the insulation on cargo tanks and pipelines reduces, increased thermal leakage will occur. Increased thermal leakage involves removal of more heat from the cargo. Load time and time used to cool the cargo increase. Preserving the insulation is good economy. Regular control and systematic maintenance of this from day one will save large future expenses. Protection of water and humidity is of high importance and that s why we purge hold spaces with dry inert/ nitrogen. With exception of mechanical wear and tear, there is nothing more destroying for the insulation then the humidity. The only way to protect the insulation from humidity is to assure that the water barrier is intact. Re-insulation of the insulation on cargo tanks is especially expensive. The best and cheapest way to preserve this is to be sure that the atmosphere around the loading tanks is dry. It is important to consider the different material s capability to absorb the heat of radiation when working with external insulation materials. A light water barrier absorbs less heat than dark. In practice, this means that white pipe insulation absorbs less heat than one with red or orange colour on the pipe insulation. One must also notice that pipe insulation faced with stainless steel plates radiates less

88 heat and thereby is warmer than a galvanised plate. The thermal conductance is thereby larger. Five good advice for maintaining the cargo tank insulation: 1. Held the atmosphere in hold spaces dry by drying the atmosphere regularly, use dry inert gas or nitrogen, if possible. 2. Control the cargo tank insulation regularly. Areas with ice or humidity indicate thermal loss. Note these areas with spray paint to easier locate the areas when these need repairing. 3. Control external insulation regularly and repair wrecked water barrier plates immediately. Areas with ice or humidity indicate thermal loss. Note the areas with spray painting in order to easily locate these when need of repairing. 4. At all times, have necessary materials to repair wrecked insulation onboard, minimum materials to repair damage on water barrier. As some of the insulation materials have limited operating time, the stock onboard must be adjusted to the expected consumption the next month. 5. Before adding new insulation in place, corrosion and pitting must be controlled. The steel must be protected from corrosion before new insulation is put to place.

89 The qualities of the insulation material compared with other materials Thermal Density Pressure Heat Fire Diffusion conductance firmness capacity qualities resistance W/m K kg/m3 kg/cm2 J/kg K Stationary air 0,024 1, on/20 o C CO 2 0,015 2,0 840 Freon 22 0,012 4, Polyurethane foam, R11 Polystyrene foam (Isophor) Carbamide foam Phenol foam (Bakelite) Glass wool plates Rock plates Expanded perlite wool 0,023 40,0 2, Selfquenching 0,033 25,0 0, Selfquenching 0,035 10, ,041 32,0 1, ,035 20,0 840 Flammable above 700 o C 0,035 45,0 840 flammable above 700 o C 0,035 50,0 840 Non flammable ,4 1,4 1,2 Iron Steel (12 Cr) Stainless 17, steel (19 Cr/10Ni) Water of/ 20 0, o C Ice of/ -20 o C 2,

90 14.2 CALCULATION OF THERMAL TRANSFER The cooling plant on a gas carrier is dimensioned by calculated heat transfer to cargo tanks and systems when the ship is new The insulation is exposed for wear and tear and is on many gas carriers partly strong reduced. When the insulation on cargo tanks gets inferior, it will have influence on the capacity of the vessel. Its construction and choice of components give the cooling plant s capacity. Systematic maintenance will hold this capacity. The amount of heat transferred to cargo tanks and cargo pipes dependent on the insulation s state, surrounding temperature, heat radiation and movements. Before taking a closer look at the condition around heat transfer on gas carrier, it is useful to form a picture of the heat balance. It is indicated easily by following illustration: Heat is transferred from the surroundings to the cargo and systems for cargo because of the temperature difference. The transmission heat to the cargo tanks, Q tr.tank is the total transferred heat to the cargo tanks with cargo, steel and insulation. Transmission heat to cargo pipelines Q Tr.pipe is the total transferred heat to cargo in the pipes, pipe and insulation around the pipes. The heat of compression, Q Tr.Compr is the heat supplied to the gas in the compressor and the heat of condensation Q Cond is the heat transferred to the seawater in the loading condenser. When the cooling plant is driven to keep the temperature of the cargo constant, the heat balance is expressed as: Q Tr.tank + Q Tr.pipe + Q Tr.Compr = Q cond or as: Q tr.tank = Q from cargo tank - Q return to cargo tank The actual amount of heat transferred from the surroundings to a cargo tank or a system can be quantified in several methods. At first we will look on how this can be

91 done onboard, how to evaluate the result and what results the eventual effectuated effort will have. Most gas carriers are equipped with a graphic description of the calculated heat transfer to the cargo tanks. This is a theoretical calculated description that does not necessarily give the right image of the heat transfer. The older the ship is the larger probability that the calculated heat transfer DOES NOT coincide with reality. Control of the reel heat transfer can be executed onboard. If a loaded cargo tank is closed and isolated from the cooling plant over a period of time, the transferred heat from the surroundings can be measured. The heat transfer is thereby quantified and is comparable with what it was or should be. Before looking on a concrete example, it is of importance to emphasise that when accomplish such measuring, one must evaluate the results from the accuracy of the instruments. As the measuring instruments onboard has normally no more accuracy than + 10%, the period of measure should be as long as possible. Further it is important that if comparing repeated actual measurements, the measures has to be made at the best possible equal condition. We will now take a look at different examples and what we can do Example 1 A 12 year old smaller intermediate pressure gas carrier with 6 cargo tanks is loaded with ethylene and has just moved the sailing area from Europe to SEA. The captain rapport that the ship capability to cool down ethylene is perceptible inferior. The cooling rate is now at the lower edge of 0,3 o C per day when the temperature of the cargo gets lower than minus 102 o C. Inspection of the loading tanks indicate ice more than usual round all of the tank foundation and tank no.2 and 4 has many ice spots. Several ice spots than before are also observed on suction lines and condensate lines on deck. The cooling plant is checked and driven at optimum, but the cooling rate is more than halved at the same pressure in proportion to when the ship sailed in European waters. As the cooling plants condition is verified good and the plant is verified optimum driven, the bad cooling rate must be the result of the insulation has been worse during the years. The heat transfer has probably increased gradually through the years, but the influence has not been operational visual before the ship altered sailing area. It is obvious that the insulation on tank no. 2 and 4 plus the insulation on the lines on deck are mostly reduced. Repair of the insulation is necessary, but the question is which areas have most influence on the cooling capacity. Clarification of this is of importance when planning and priority of the insulation repair. The amount of heat transfer to the cargo tanks and pipelines must be concretised and compare with the repair costs before making the right decision. Cargo tank no. 2, 3 and 4 are the same type and size. To find out how much heat that is transferred to the bad loading tanks in proportion to one of one of the good, the tanks are shut for 24 hours.

92 The measuring instruments that are utilised during the experiment is calibrated and the following sketch for heat transfer is utilised and filled in: Heat transfer for Cargo tank no.2 Date/time / 10: / 10:30 Cargo tank filling ratio 97,0 % Mass cargo in MT 592,837 Ambient temperature in o C Sea water temperature in o C Average Hold space temperature in o C Ship's movement Calm sea Calm sea Weather condition Cloudy Cloudy Average liquid temperature in o C -103,1-102,2 Cargo tank pressure in mbar Enthalpy liquid in kj/kg 27,0 29,3 As the weight of gas is relative much less than the liquid weight, only the enthalpy change of the liquid is measured. The enthalpy values exist in heat technical table and the heat transfer to the cargo in tank no. 2 is calculated to: [(mass cargo x enthalpy-change) / (time in seconds)] [(592,837 x 10 3 x (29,3-27,0)) / (24 x 60 x 60)] kw = 15,8 kw Corresponding, the heat transfer is measured and calculated to the cargo in tank no. 3 and 4 to respectively 13,9 kw and 22,2 kw. In the technical description of the ship the calculated heat transfer when the ship was built to 13,1 kw for each tank at the same surrounding temperature and seawater temperature is located. A comparable table can be made and indicates as follows: Calculated transfer before: Calculated transfer now: Percentage change: Cargo tank 2 13,1 kw 15,8 kw 20 %

93 Cargo tank 3 Cargo tank 4 13,1 kw 13,9 kw 6% 13,1 kw 22,2 kw 70% The calculations confirm the observations and presumptions made before in connection with the inspection of the tank insulation. It is rather no doubt that the insulation on tank no. 4 is essential much more deteriorated than the remaining two tanks. The priority at en eventual re-insulation of cargo-tanks can thereby well substantiate. By comparing present operational parameters for the cooling plant with earlier registered operational data, the temperature increase on the vapour from the cargo tank to compressor has increased essential. Six years ago, during almost the same condition, the vapour temperature rise from 100 o C to 60 o C from the cargo tank to compressor. Present observed temperature increase is from 100 o C to 40 o C. The vapour is essential now more over heated than earlier. A compressor s cold capacity is expressed as: Q netto = m x Dh Higher temperature on the vapour into the compressor involves lower density and thereby reduced amount of vapour through the compressor per time. The influence of inferior insulation on the suction lines to the cooling plant will have direct influence on the cooling capacity. A comparable table indicates as follows: Density on gas v/1 bars and -60 o C 1,7699 kg/m 3 Density on gas v/1 bar and -40 o C 1,6181 kg/m 3 Percentage reduction in density 9% The reduction of the suction line's insulation involves a direct reduction of the cooling rate of about 10% in this temperature area. The size of the heat loss through condensate lines and liquid lines cannot be measured directly because there will at all times be and unknown and varying mixture of liquid and gas in the pipes. Judgement must at all times be adjusted and from the rapport from this ship, one must assume that the insulation here has the same condition as the suction lines. If this is the case, the influence of bad insulation on the condensate lines will have maximum consequence. The liquid lines are only utilised in a short period (during loading and discharging), while the condensate lines are utilised during all of the cooling period. The increase of the relative heat loss through the insulation on loading tank no. 2 and 4 is now indicated. Likewise is the reduction of cooling capacity because of increased heat transfer to the suction gas established. There is no doubt that both conditions have influence on the operational situation on the ship. But it is difficult to compare these two directly, for thereby to establish which one of them that has the

94 strongest effect on the ship s possibility to execute the transport commission. To make the data s comparable the alteration in the heat transfer to the suction line is quantified. The ship s three loading compressors have stated a capacity of 680 m 3 /h at the same operational condition. The alteration of the heat transfer to the suction line in proportion to earlier years is calculated to: Vapour Ethylene Temperature at inlet compressor -60 o C before, T 1 Enthalpy, h 1 572,7 kj/kg Temperature at inlet compressor -40 o C now, T 2 Enthalpy, h 2 607,4 kj/kg Density, r 2 1,6181 kg/m 3 Number of compressors 3 Capacity per compressor, V 680 m 3 /h Difference in heat transfer, DF 31,8 kw DF = ((3 x V) / 3600) x r 2 x (h 2 - h 1 ) One can see that about 30 kw more heat to the suction cables is supplied now compared to earlier year. Simultaneously the measures indicate that the heat transfer to the cargo is about 15 kw more than earlier. (One must here emphasise that the total heat transfer to the cargo tanks will be larger because the steel in the cargo tanks with insulation is supplied heat). Heat transfer to the suction vapour in the suction pipe has a direct influence on the cooling plant's capacity. In addition, too high suction temperature will have a bad influence on the plant s operational conditions by for example that the pressure pipe temperature may be too high.