14 Cooling calculations

The Mollier diagram used applies only to (pure) propane, and is in

fact called a "Log p-h Mollier diagram". "Log P" means that it has a

logarithmic scale for pressure (p), which is the vertical scale, whilst

the "h" stands for enthalpy (or energy) and is the horizontal scale.

Note that when you read values off this diagram, you must use the

scale at the top (enthalpy in kJ/kg) and the scale on the right side

(pressure in MPa). The scales on the left-hand side and along the

bottom of the diagram are English units that we are not going to use

in this course.

The pressure scale is, as mentioned, in MPa (Mega Pascal), and we

ask you to write 1bar where it says 0,l MPa, so that later you will not

have to think about changing the units to "bar" when they are to be

used (0,l MPa = 1bar); e.g. where it says 0,4 MPa will be 4 bar etc.

In the left and lower area of the diagram a "sack" is drawn. This sack

marks the saturation line and the phase change for liquid propane to

propane vapour. The left side of the "sack" shows, in the same way,

the line for boiling liquid propane, and the right hand side shows the

line for saturated propane. The area that is on the left of the sack is

under-cooled (sub-cooled) liquid propane, and the area to the right

of the sack is superheated propane vapour. The phase change from

liquid to vapour is in the sack.

Let us now look at a few problems in connection with what we have

gone through.

Example 1:

Use the Mollier diagram for propane, and draw a line through the

phase change (sack) at 2 bar (0,2 MPa) pressure, as figure 1 shows.

Pressure2,0 (bar) - Enthalpy (kJ/kg)

Let us now find out how large the enthalpy content (heat/energy

content) is for a kilo of boiling liquid propane at 2bar saturation

pressure. Boiling liquid propane at different pressures is found as

mentioned on the left-hand part of the sack. In our case at point "A".

Go from point "A" and vertically up the scale on the top of your

diagram. Here you will find a value of ca. 460 kJ/kg. This means that

each kilo of propane in the given state (boiling liquid propane at 2

bar saturation pressure) has an enthalpy content of 460kJ. 10 kilo

propane will, therefore, contain:

10 Kg x 460kJ/kg = 4600kJ

Next, we shall try to find out how large the enthalpy content is for

saturated vapour at the same saturation pressure (2bar). Now we

have to go vertically up from "B" point which is on the right-hand

side of the sack, to the enthalpy scale, and here we read off

870kJ/kg.

We will now continue with more practice in the use of the Mollier

diagram. As we have already said, the phase change between liquid

and vapour is in the sack. To find out how dry or wet this phase

change is in the different stages, there are a few curves drawn

approximately parallel to the sides of the sack. These lines are

labelled from left to right with 0,1 - 0,2 - 0,3 etc. This is to say that

"0,1" means 0,1 parts of vapour, and, therefore, 0,9 parts of liquid.

Another, and important piece of information that we get in the

phase change is the temperature that the substance (the propane)

has when the liquid changes over to vapour. This temperature is

constant whilst the substance is in the phase change.

If we have the propane in a closed system, and have both liquid and

vapour in this system (as in a cargo tank), we are in a position to

find the liquid temperature by using the vapour pressure, and

conversely we can find the vapour pressure using the liquid

temperature. We say that the pressure and temperature follow each

other when the liquid and vapour are in equilibrium and in a closed

system.

The temperature values in your Mollier diagram are given in 0F

(degrees Fahrenheit), and are given on the horizontal lines in the

phase change. Always remember to change them to 0C when you

are plotting the temperatures on a diagram. Note how the

temperature curve goes approximately vertically up the left-hand

side of the sack, and in a curve downwards on the right-hand side of

the sack.

Example 2: We shall now show on a sketch how we mark off the

values that we take out of the Mollier diagram. Let us now first

consider that we have 1 kilo of propane in a phase change (part is

vapour, and part is liquid), the saturation pressure is 7,5bar, and the

enthalpy of the vapour/liquid mixture is 635kJ. How much of the

propane is liquid, and how much is vapour. Find also the

temperature of the liquid/vapour.

Fig. 2 Pressure (bar) - Enthalpy (kJ/kg)

By "plotting" the pressure at 7.5 bar and the enthalpy at 635 kJ/kg

(see fig. 2), we find that the point of intersection is on the

"dampness curve" where it says 0.2, which means that in the

mixture of liquid/vapour we have 0.2 parts of vapour and so 0.8

parts of liquid. We also find that the temperature line that runs

parallel to the pressure line for 7,5bar saturation pressure is that for

+600F. If you convert 600F to Celsius, you get +150C. To find the

tank pressure (also known as manometer pressure or overpressure)

you must realise that the atmospheric pressure must be subtracted

from the saturation pressure. Normal atmospheric pressure is 1013

millibar (1013mb = 1,0l3bar); and no serious mistake is made if we

subtract 1 (one) from the saturation pressure to get the tank

pressure, for simplicity. If the saturation pressure (often called the

absolute pressure) is 7,5 bar, as in our example, then we can use

the tank pressure as 6,5bar, if nothing else is mentioned about the

atmospheric pressure.

We have previously said that in a phase change both the vapour and

liquid are at the same temperature, and that this temperature

decides which vapour pressure we have in the closed system. This is

correct, but in practice you will see that the vapour in the top of the

tank (in the dome) has often a significantly higher temperature than

the liquid temperature down in the tank. The vapour has, in other

words, been superheated due to the heat leakage into the tank (sun

and warm air). This does not, however, influence the pressure in the

tank that is decided by the temperature of the liquid (really the

temperature at the liquid surface). The vapour over the liquid

surface is at the same temperature as the liquid surface, and it is

here that the phase change takes place.

Next look at the curves on the Mollier diagram for the density

(earlier called gravity). These curves go gradually up towards the

right and are given in the unit lb/ft3 (pounds per cubic foot). We shall

in the meantime convert all the values we find from lb/ft3 to kg/m3

(kilos per cubic meter). On the curve for 1,0 lb/ft3 we find a

conversion factor for this number that is 16,02. That is to say that 1

lb/ft3 is the same as 16.02 kg/&. If we find the density at, for

instance, 0.3 lb/ft3, it will become 4,806 kg/m3 (0.3 16.02 = 4,806).

Example 3:

We will now show an example of the use of the density curve. Let us

find the density for a superheated propane vapour with a saturation

pressure of 0,95bar and a temperature of +490C (+1200F). The

specific enthalpy for this propane vapour will also be found.

Pressure (bar) Fig. 3 Enthalpy (kJ/kg)

By plotting the points of intersection (see figure 3) between the

pressure line for 0,95bar and the temperature curve for +490C

(+1200F), we find that the point of intersection lies in the density

curve for 0.1 lb/ft3. The density will, therefore, be 1,6kg/m3 for the

superheated vapour in our example (0.1 x 16,02kg/ m3 = 1,6 kg/m3).

The enthalpy at this point is found as previously (about l000 kJ/kg).

As you see, we always use propane in our examples. The reason for

this is that if you have learnt the thermodynamics for one gas, you

understand the thermodynamics for all the other gases. Concerning

the thermodynamic properties, it is only the values that are changed

from gas to gas. Another important reason for using only one gas in

this course is that fewer curves and diagrams are needed. In Letter

3 under the title "USE OF DATA PAGES AND CURVES", a complete

set of diagrams/tables is given for propane only.

We have now looked at all the information that our Mollier diagram

gives, except for one value. This one value is a curve known as

"entropy" (5). The unit for specific entropy (5) is J/kg·K (Joule per kilo

and Kelvin). We shall, for the meantime, not bother ourselves further

about this value on this course, except to mention that an ideal

compression of a gas will follow the curve (adiabatic) that goes

steeply up to right of the diagram.

You have now used the Mollier diagram for a while, and you are to

continue to use it later in the course, but it is now time to introduce

more of the values you can find in the thermodynamic tables. Find

the tables "THERMODYNAMIC PROPERTIES FOR SATURATED

PROPANE" in Letter 3.

Most of the information you have extracted from the Mollier diagram

can be found in the table with great accuracy. You might now ask

why it is necessary to use the diagram. The answer is simple. We

can only take out values from the saturation line (the sack) in a

table, and, therefore, no information from the under-cooled (sub-

cooled) or the superheated area (to the left and right of the sack) is

given. Also it is extremely difficult to "see" the cooling process

without such a diagram. On the other hand, the table is very useful

when we require accurate values, and we are, therefore, going to

use the table when it can be used, to correct the values that we get

from the diagram.

In the next example we are going to use the table for accurate

values and put them on a log p-h sketch in order calculate an

answer of the highest accuracy.

Example 4:

We have propane in a tank at a temperature of -200C. We wish to

find the saturation pressure and tank pressure as accurately as

possible at this temperature. Further, we will find the density and

enthalpy of the liquid and vapour as well as the enthalpy difference

(Δh) between the liquid and vapour. We assume that the vapour

above the liquid surface is at the same temperature as the liquid in

this example, and that the atmospheric pressure is 980 mb. Draw a

sketch and mark the values on it. Compare the values from the table

with those you would get from the Mollier diagram.

Enter the table using the liquid temperature (in this case and read

off the value 2,44526bar under the saturation pressure.

The tank pressure is calculated thus:

Saturation pressure - Atmospheric pressure = Tank pressure

2,44526bar - 0,980bar = 1,46526bar

Under the column for the liquid density we find 0.55448 kg/1, and

under the vapour density we find 560kg/m3. Enthalpy for the liquid is

476.2 kJ/kg and for vapour is 876.8 kJ/kg.

The enthalpy difference (_h) = (876.8 – 476.2) kJ/kg = 400,6 kJ/kg.

Pressure (bar) Fig.4 Enthalpy (kJ/kg)

14.1 Exercises

l. a) What is the enthalpy content (specific enthalpy) for one kilo of

boiling liquid propane at 3 bar saturation pressure?

b) What is the specific enthalpy for saturated propane vapour at

3bar saturation pressure?

c) What is the saturation pressure of boiling liquid propane when the

specific enthalpy for the liquid is 510kJ/kg?

d) What is the saturation pressure of saturated propane vapour with

a specific enthalpy content of 900kJ/kg?

e) What is the enthalpy content for 7 kilos of boiling liquid propane

at a saturation pressure of 5 bar?

1

.

a

)

Specific enthalpy for liquid propane at 3bar

saturation pressure:Αbt. 490kJ/kg

  b

)

Specific enthalpy for saturated propane vapour at

3bar saturation pressure:Αbt. 883kJ/kg

  c)Saturation pressure for boiling liquid propane when

h = 510kJ/kgΑbt. 4bar

  d

)

Saturation pressure for saturated propane vapour

when

h = 900kJ/kg

Αbt. 5bar

  e

)

Enthalpy content for 7 kilo boiling liquid propane at

5bar saturation pressure = 529kJ/kg · 7kg3703kJ

2. a) What is the saturation pressure and tank pressure in a tank

containing propane where the liquid temperature is _400C?

b) What is the temperature of liquid propane in a tank where the

tank pressure reading is l,5bar?

c) The temperature of a liquid/vapour mixture of propane in phase

change is _100C. The mixture consists of 0,3 parts of vapour (gas)

and the rest is liquid. What is the specific enthalpy for the

liquid/vapour mixture?

d) If we put a manometer in the vapour pocket over the liquid (in the

closed system) in question 2c, what will the manometer read?

2.

a)Saturation pressure for propane at _40_C 1,1 Bar

    _ Atmospheric pressure 1,0 bar

    Tank pressure 0,1 bar

      (= 100mb)

  b)Tank pressure 1,5 bar

    +Atmospheric pressure 1,0 bar

    Saturation pressure 2,5 bar

    Temperature of the liquid propane at 2,5bar saturated pressure

Αbt. 19 o C

  c)Specific enthalpy Αbt. 618kJ/kg

  d)Saturation pressure for propane at _10_C 3,5bar

    _ Atmospheric pressure 1,0bar

    Tank pressure (Manometer pressure) 2,5bar

3. In a propane tank the liquid has a temperature of +150C. The

atmospheric pressure is 1030mb. Make a log p-h sketch and find out

as accurately as possible the answers to the following questions:

a) What is the saturation pressure for the propane?

b) What is the tank pressure?

c) What is the liquid density?

d) What is the vapour density above the liquid surface?

e) What is the vapour density in the dome, when the temperature of

this vapour is +300C? (The answer must be taken from the Mollier

diagram)?

f) What is the liquid enthalpy?

g) What is the vapour enthalpy just above the liquid surface?

h) What is the vapour enthalpy in the dome?

i) What is the enthalpy difference (Δh) between the liquid and the

vapour just above the liquid surface?

3

.

a

)Saturation pressure 7,32921bar

 

b

)

Tank pressure

= Saturation pressure _ Atmospheric pressure

= 7,32921bar _ 1,030bar

6,29921bar

  c)Liquid density 0,50753kg/1

  d

)Vapour density (saturated propane vapour) 15,89kg/m 3

 

e

)

Density of overheated propane vapour at 7,3bar and

+30_C

= (0,9 · 16,02)kg/ m3

14,4kg/m 3

  f) Liquid enthalpy (specific) 562,3kJ/kg

  g

)Vapour enthalpy (specific) 913,2kJ(kg)

  h

)Vapour enthalpy in the dome 942kJ/kg

 

i)

Enthalpy difference (_h) between the liquid and

vapour

(913,2 _ 562,3)kJ/kg

350,9kJ/kg

14.2 Single Stage Direct Refrigeration

We have now come so far that we can begin to look at the simplest

of re-liquefaction plants, so-called single stage direct refrigeration.

The plant is called »single stage direct refrigeration" because the

vapour is compressed in one stage directly against sea water. Fig. 5

shows the most important components in such a system.

Fig. 5

The points A to F on the log p-h sketch are found in this way:

Pressure (bar) Fig. 6 Enthalpy (kJ/kg)

The process is as follows; Due to the heat leaking into the cargo

tank, some of the liquid (A) will boil off to saturated vapour (B). This

boil-off (vapour) uses heat that is taken from the liquid, so that when

we remove this vapour, the liquid in the tank will remain at the same

temperature. If we suck out more vapour than generated by heat

leakage, the temperature in the tank will decrease. Basically, this is

the cooling process. We suck vapour into the compressor at point

(C). The vapour has, in the meantime, been superheated. It has also

slightly decreased in pressure in relation to the tank pressure

because it has passed through pipes, valves, bends, liquid separator

etc. before it is sucked into the compressor.

In the compressor the vapour is pressurised (given energy) , and the

vapour comes out from the compressor at (D) with a higher pressure

and greater enthalpy value than at point (C). From point (D) the

vapour goes into the sea water-cooled condenser. Note that the high

vapour pressure achieved in the compressor remains all the way

until it reaches the regulating valve.

The vapour at a high pressure (and temperature), condenses to

liquid when cooled in the condenser. It is the sea water temperature

that decides what temperature and pressure the propane-vapour

condenses at.

The condensed liquid propane collects in the liquid receiver (E).

From the liquid receiver (whose level is controlled by the regulating

valve) this relatively warm liquid expands from a high pressure to a

low pressure through the regulating valve (F). In this instance the

condensing pressure (right side of F) decreases to the tank pressure

(left side of F), and some of the liquid boils off due to the reduction

in pressure. This boil-off consumes heat that is removed from the

liquid, and the liquid temperature decreases from that in the liquid

receiver to that of the tank. Remember that pressure and

temperature follow each other.

As described, the condensate (that runs through the regulating

valve back to the tank) consists primarily of liquid, but some will

boil-off to vapour on its way through the regulating valve. The

amount that vaporises can be between 10% and 30% of the liquid

amount, depending upon several different conditions that we shall

look at later.

As we understand, we have sucked 100% vapour from the tank into

the compressor, whilst we deliver this back as 70 to 90% liquid and

at the same temperature as the liquid in the tank. This is the so-

called "reliquefaction process". It is not the return of the condensate

that gives the cooling effect (reduction of tank temperature), but on

the contrary, it is the vapour that we suck off with the compressor

that gives the cooling effect.

The simplest form for cooling of a propane cargo would be to suck

off (or really release) the vapour to the atmosphere. We would, in

this way, come down to atmospheric pressure on the cargo, and in

the case of propane it would correspond to a temperature of under

_420C (check with the tables for propane!). We can, of course, not

use this method except in an emergency, as it could be rather

expensive to arrive at the discharging port with too little cargo.

We shall once more look at the two last figures, but this time we are

to use the temperature and pressure we would read off the process

plant in real life on board.

Example 5:

Use the following data:

  Cargo Propane

 

  Tank temperature: -100C

 

  Tank pressure: 2.45bar (3.45 abs.)

  Compressor suction

pressure:

2.40bar (3.40 abs.)

  Compressor suction

temperaure:

+50C 

  Compressor delivery

pressure:

7,60bar (8,60 abs.)

  Compressor delivery

temperature:

+600C 

First draw a diagram of the process plant and plot in the values

given. In addition, using the condensation pressure, (8,60bar) find

the condensation temperature (+21oC), which is the temperature of

the liquid in the liquid receiver.

Fig. 7

Thereafter plot all these values on a log p-h diagram.. See figure 8.

Pressure (bar) Fig. 8 Enthalpy (kJ/kg)

As you understand there is much more information we could find

indirectly, such as enthalpy values, densities etc. but we shall leave

these to you as exercises.

From example 5 it can be seen that the sea water temperature is

about +110C, since the condensation temperature is +210C. We can

say this because most condensers of this type are made so that the

temperature difference between the cooling medium (sea water)

and the condensation (propane) is at about 10%.

The condensate is, in other words, 10% higher in temperature than

the cooling media after having gone through the heat exchanger.

14.2.1 Calculation of Time Used

Example 6:

We shall now look at an example where we wish to calculate the

time used to decrease the temperature a set number of degrees for

a gas cargo. For the sake of simplicity we shall ignore heat leakage

into the tank as well as cooling of the tank shell, insulation and tank

atmosphere. We shall only use one compressor to cool the cargo

tank.

The following data are known:

Process plant: Single stage direct cooling.

Cargo: Propane (pure)

Cargo quantity: 1,000.00 MTons

Tank temperature: -50C

Compressor suction pressure: 3,00bar

Compressor suction

temperature:

+6oC

Compressor delivery pressure: 6, 95 bar

Compressor delivery

temperature:

+55%

Compressor sucking volume: 350m3/hour

Atmospheric pressure: 1000mb

Temp. Cargo is to be cooled

down to:

-100C

We first find that the saturation pressure of the cargo is 4.06 bar

(tank pressure is 3,06 bar) by using the cargo temperature. The

condensation temperature is +180C at the corresponding

condensation pressure of 7.95 bar, (we can, therefore, assume that

the sea water temperature is about +80C).

Thereafter we take out the enthalpy values for boiling liquid at -

100C, -50C and +180C as well as the enthalpy values for saturated

vapour at -50C We also need the density of the superheated vapour

at +60C/4,0bar (abs) which we take out of the Mollier diagram.

Pressure (bar) Fig. 9 Enthalpy (kJ/kg)

Note that we have marked off certain points of intersection on the

diagram with h1, h2 etc. This is so that these positions can be

identified for various calculations etc.

T1 stands for the cargo temperature before cooling, and T2 stands for

the temperature we will have after cooling off the cargo.

When we are to calculate the cooling time to cool the cargo from -

50C to -10oC, we must first calculate how much heat (enthalpy) we

are to remove. The enthalpy difference between T1 and T2 shows us

how much heat we must remove from each kilo of propane cargo, to

reduce the temperature from -50C to -100C. We also know how many

kilos of cargo we are to cool down (1, 000 MT = 1,000,000 kilo).

The calculation will be:

Heat to be removed (Q) = (511.6 – 499.5) kJ/kg x 1000000kg =

12 , 100 , 000 kJ

We now know how much heat that has to be removed to reduce the

temperature from -50C to -10oC (l2,l00,000 kJ), and have now to

determine how much heat we can remove per time unit (per hour),

by using our cooling plant. The formula for net cooling capacity

looks like this:

= (Vs ) x (s) x (δh)

Stands for net cooling capacity

Vs stands for compressor volume

s stands for density of the vapour on suction side

δh stands for the enthalpy difference between saturated vapour

after the phase change and boiling liquid before the expansion valve

(h1-h4)

We will calculate the net cooling capacity in our example:

= Vs _s · _h

= 350 m3/hour · 8.3 kg/m3 · (893,2-570,2)kJ/kg

= 938 , 315 kJ/hour

We have now come far enough to calculate the cooling time:

Cooling time

=

= 12,9

hours

Before we go further, we shall look at a simpler and more accurate

method to find the density for superheated vapour. Instead of using

the Mollier diagram for this purpose, you can, from now on, use the

curve page in Letter 3 called "The density for superheated vapour -

propane".

You use the curve by entering it at the lower scale with the

temperature of the superheated vapour (+60C in our case), and then

go vertically up to the curve that corresponds to the saturation

pressure for the superheated vapour (4,00bar). Make a mark here,

and then go horizontally from this to the scale on the left-hand side

where you will find the density (8,25kg/m3).

At the end we must mention a small mistake that we do in the

written method to calculate the cooling time. The mistake appears

during the calculation of " " and is insignificant when we are using

small temperature intervals (e.g. 5 - 10 degrees). The reason for the

mistake is that both the suction volume, the density of the suction

vapours and δh change value as the temperature in the tank

decreases. If we were to increase the accuracy of the calculations,

we would have to use average values. The heat leakage to the tanks

also changes with decreasing temperature of the cargo.

14.3 Exercises

1. You are on board a semi-pressurised LPG ship, and with help of all

three cooling plants on board you are to lower the temperature of

the cargo of (pure) propane. Remember that all the manometer

pressure readings (suction pressure etc.) are over-pressures.

The following data is given:

Process plant: Single stage direct plant

Cargo quantity: 2400MT

Tank temperature: +40C

Compressor suction press: 4,3bar

Compressor suction temp: +110C

Compressor delivery press: 8.l bar

Compressor delivery temp.: +800C

Cargo to be cooled down to: -2οC

Sucking vol. per compressor: 290m3/time

Atmospheric pressure: 1000mb

a) Write all the values on a log p-h diagram (sketch).

b) What is the condensation temperature of the plant?

c) What would you assume the sea water temperature to be?

d) What is the specific liquid enthalpy in the tank before cooling?

e) What will the specific liquid enthalpy be in the tank after cooling?

f) What is the vapour enthalpy after the phase change in the tank?

g) How much heat must be removed from the tank (Q)?

h) What will the net cooling capacity be ( )? Remember to multiply

the suction volume with the correct number of compressors in use!

i) How many hours will be needed for cooling, when we ignore the

heat leakage into the tank etc.?

j) What was the tank pressure before cooling?

k) What will the tank pressure be after cooling?

1a)

Pressure (bar)5, Enthalpy (kJ/kg)

b) Condensation temperature: +23_C

c) Sea water temp. (Approx): (+23 -10) oC = +13_C

d) Liquid enthalpy before cooling: 534,1kJ/kg

e) Liquid enthalpy after cooling: 519,0kJ/kg

f) Vapour enthalpy after phase change: 902,2kJ/kg

g) Heat that must be removed (Q) = (m) x (δh)

= 2.4 · 106kg · (534,1 _ 519,0)kJ/kg = 36,23 · 10 6 kg

h) = Vs · _s · _h = 3,290m3/h · 11, 2kg/ m3 (902.2 - 583,8)kJ/kg

=3102489,6kJ/h

= 3,1 · 10 6 kg kJ/h

i) Cooling time = = = 11 . 7 hours

j) Tank pressure before cooling: (5,358 - 1,000) bar = 4,358bar

k) Tank pressure after cooling: (4,466-1,000) bar = 3,466bar

14.4 Two Stage Direct Refrigeration Plant

We shall now look at a more advanced reliquefication plant than the

one in the last section. The plant is called a "two-stage direct plant

with intermediate cooler". Figure 10 shows the most important

components in such a plant:

Fig. 10

Points A to J are again found on a log p-h diagram in this way (see

figure 11):

Pressure (bar) Fig. 11 Enthalpy (kJ/kg)

The process works in the following way. Some of the liquid (A) in the

tank is boiled to vapour (B) due to the heat leaking in. We suck the

vapour into the first stage of the compressor (C), and deliver the

superheated vapour with a higher pressure (D). This vapour is

pressed down into the liquid in the intermediate cooler where the

vapour is cooled to the same temperature as the liquid. The liquid in

the intermediate cooler has a temperature corresponding to the

intermediate pressure, which again corresponds to the delivery

pressure from the first stage.

The saturated vapour in the intermediate cooler (above the liquid) is

sucked into the second stage of the compressor (E) and is delivered

with a high pressure and temperature (F) to the sea water cooled

condenser. In the condenser the superheat is first removed from the

vapour, and thereafter begins its phase change from vapour back to

liquid, as the excess enthalpy is removed from the vapour.

The liquid collects in the liquid receiver (C) with a comparative high

temperature (normally ca. 100C higher than the sea water

temperature). The main flow of liquid from the liquid receiver goes

first through the coil in the intermediate cooler, where the liquid is

sub-cooled to a lower temperature than the pressure should indicate

(H). Thereafter the sub-cooled liquid goes through the regulating

valve and decreases to tank pressure (I), and thereafter also to tank

temperature.

Some of the liquid flow from the liquid receiver goes through a

regulating valve into the intermediate cooler (3). The pressure over

this valve will fall from high pressure to intermediate pressure, and

the temperature will follow the pressure (closed liquid/vapour

system in equilibrium.

It is the liquid level in the liquid receiver that controls the regulating

valve for return of liquid back to the tank, and it is the liquid level in

the intermediate cooler that controls the regulating valve to this

receiver.

Example 7:

We shall look at both our figures once more, and we shall now use a

realistic temperature and pressure that we would get in an actual

situation on board.

We shall use the following data:

Cargo : Propane

Cargo temperature: -400C Suction pressure stage 1: 0.l bar

Suction temperature stage 1: 250C

Delivery pressure stage 1: 4.5bar

Delivery temperature stage 1: +500C

Delivery pressure stage 2: 10.9bar

Delivery temperature stage 2: +780C

Temperature out from the intermediate pressure cooler coil:

+11oC

We first plot the tank temperature of -400C on the simplified drawing

of the process plant (fig. 11) and on the log p-h diagram (fig. 12).

The tank temperature corresponds to a saturation pressure of abt.

1,ll bar (tank pressure 0,l1bar). The suction pressure for the first

stage is 1,l0 bar abs, and the suction temperature for this stage is

for this stage is -250C. The delivery pressure at stage 1 (5,5bar)

corresponds to a liquid temperature in the intermediate cooler of

+50C. Hence the suction pressure and temperature for the 2nd stage

is known (5,5bar abs. at +50C). The delivery temperature for stage 1

(+500C) is also plotted.

Thereafter we note the delivery pressure and temperature for the

2nd stage (11,9bar abs. at +78%). The condensate temperature that

we have in the liquid receiver will be +34oC, which corresponds to

the condensation pressure (l1.9 bar abs.).

Fig. 11

Pressure (bar) Fig. 12 Enthalpy (kJ/kg)

By looking at the condensation temperature, (+34oC) we can say

that the sea water temperature is abt. +240C (t = 100C).

The liquid temperature after the intermediate-cooler coil (Pt. h7) is

+11oC and the pressure is the same as in the condenser (ll.9 bar

abs.). Note how we plot the temperature for the sub-cooled liquid

after the coil.

The temperature line rises vertically on the left side of the phase

change in the Mollier diagram, and the "little trick" makes it easier

to take out the enthalpy in pt. h7 which has the same enthalpy as

pt. h9. We can, therefore, take out the enthalpy from the table for

pt. h9, and regard it as the enthalpy in pt. h7.

As you see, it would have been theoretically possible to sub-cool the

liquid in the coil down to +5% (with an infinitely long coil), which is

the temperature of the liquid in the intermediate cooler. We have,

however, in our example managed to sub-cool the liquid from +340C

down to +110C, and that is about the limit that can be managed with

such a plant ( t is abt. 5 - 60C between sub-cooled liquid in the coil

and liquid in the intermediate cooler).

To calculate we take the density of the suction vapour in Pt. h2

(2,35 kg/m3), and the enthalpy in pt. h1 and h7 (h9) which are

respectively 853,3 kJ/kg and 551,8 kJ/kg.

If the sucking volume for this compressor had been 500m3/hour

under these conditions, the calculation of net cooling capacity would

be as follows:

= Vs · _s · _h = 500m3/hour · 2,35kg/ m3 · (853.3 _ 55l.8)

kJ/kg = 354,62kJ/hour

When we load a gas ship, it is the cooling capacity that normally

limits the loading speed. The heat that must be removed from the

loaded cargo, is calculated in the same way as we have done up to

now. Let us look at an example where we are to load propane that

keeps a temperature over the manifold (the shore tank temperature

plus the eventual heat leakage into the loading pipes) of -320C. The

cargo is to be cooled down to -390C during loading by using all the

three cooling plants on board. There is no vapour return to shore.

We shall calculate how much heat that must be removed from the

on board flowing cargo, using a log p-h sketch for help. We are to

load 1000 MT.

Pressure (bar) Fig. 13 Enthalpy (kJ/kg)

Pt. "M" stands for the manifold liquid condition (X-over), and "T"

stands as earlier for the tank liquid condition. The calculations will

be as follows:

Heat to be removed (Q) = m · _h = 1000000 kg · (448.5 – 432.4)

kJ/kg = 16l00000kJ

= 16100 MJ

This heat quantity (16100 MJ) must be removed with the ship's own

cooling plant, if there is no facility to vapour return to shore. With

vapour return we mean a return to shore of the cargo vapour that is

generated in the ship's cargo tanks during loading. The vapour is

returned from the ship back to the shore plant through the vapour

manifold connection.

It is worth noting that even when the cargo comes on board at a

temperature of -32%, the tank temperature at any time will be -390C

if the loading rate is not higher than that which the cooling plant can

remove of excess heat (the enthalpy difference).

14.4.1 Exercises

1. The following data is given:

Cargo: Propane

Process plant: Two stage direct, intermediate, cooler

Cargo quantity: 6000 MT

Tank temperature: _370C

Suction pressure stage 1: 0,2bar

Suction temperature stage 1: -220C

Delivery pressure stage 1: 5,2bar

Delivery temperature stage 1: +620C

Delivery pressure stage 2: 11,5bar

Delivery temperature stage 2: +840C

Temperature out from the intermediate cooler (inter cooler):

+150C

Suction volume per compressor: 520m3/time

a) Draw all the values on a log p-h sketch.

b) What is the temperature in the intermediate cooler?

c) What is the condensation temperature?

d) What do you reckon the sea water temperature will be?

e) How much heat must be removed from the cargo, if we are to

reduce the temperature to _410C?

f) How long will it take to do this cooling if all three plants are in

operation?

Pressure (bar) - Enthalpy (kJ/kg)

b) Temperature in the intermed. Cooler +9_C (corresponds to 6,2bar

saturation pressure)

c) Condensation temperature +36_C (corresponds to 12,5bar

saturation pressure)

d) Sea water temperature ca.26_C (condensation temperature minus

10_C)

e) Q = m · _h = 6 · 106kg · (437,0 _ 427,8) kJ/kg = 55,2 · 10 6 kg

f) net = Vs · _s · _h = 3,520m3/h · 2,55kg/ m3 · (856,9 _ 562,5)

kJ/kg = 1,17 · 106 kJ/h

g) Cooling time =

= = 47.2 hours

2. The following data is given:

Process plant: Two stage w/intercooler

Cargo: Propane

Cargo quantity: 8200 MT

Atmospheric pressure: 1000mb

Tank temperature: -39_C

Temperature at X-over: -33_C

Suction pressure stage 1: 150mb

Suction temperature stage 1: -34_C

Delivery pressure stage 1: 4,5bar

Delivery temperature stage 1: +590C

Delivery pressure stage 2: See sea water temp. (t = 10oC)

Delivery temperature stage 2: +85oC

Sea water temperature: +210C

Temperature after coil: +100C

Suction volume per compressor: 920m3/hour

Vapour return to shore: None

a) Draw all the values on a log p-h sketch.

b) What is the delivery pressure for stage 2 (the condensation

pressure)?

c) Calculate the loading time when all three process plants are to be

run for the whole loading period.

Pressure (bar) Enthalpy (kJ/kg)

b) Delivery pressure 2nd step: 10,1bar (11,1bar abs.)

Corresponds to a condensation temp of +31oC (sea temp. +10oC)

c) Loading time = = =

= = 52.7 hours

15 Choice Single or Two-Stage Operation

We have now looked at the operation of both a single and a two

stage cooling plant. Many plants can be run in both modes, and

there is often the need for a decision to be made concerning the

type of mode that is to be used before the operation. It is always an

advantage to run the plant in the one-stage mode, if the conditions

are right for it (the suction volume will be greater if two cylinders

suck at the same time in parallel, instead of one cylinder delivering

to the other in series). The conditions that must be satisfied for such

a decision are:

Cargo type Tank pressure (the cargo temperature)

Sea water temperature

Maximum delivery pressure of compressor (1st stage)

Maximum differential pressure of compressor (1st stage)

Let us make it clear with an example. We shall load propane with a

temperature of -20oC (corresponds to a tank pressure of 1,4 bar).

The sea water temperature at the loading area is +70C. In the

instruction book (the ship's manual) we find that the maximum

delivery pressure for the first stage of the compressor is 7bar (8bar

abs.), and the maximum differential pressure for the first stage is

given as 6,5 bar (the difference between the suction pressure and

the delivery pressure). We assume the condenser temperature

difference (δt) to be 100C, and use a log p-h sketch as follows:

Pressure (bar) Fig. 14 Enthalpy (kJ/kg)

With a sea water temperature of +7oC we can expect a

condensation temperature of +17οC, which correspond to a

saturation pressure of 7.7 bars. Our compressor is able to deliver up

to 7 bar manometer pressure (8 bar saturation pressure), so this is

sufficient. We get the pressure difference as 5,3bar in our example

(7.7 – 2.4 = 5.3), and that is well under the maximum pressure

difference which was given as 6,5 bar. We can, in other words, "line

up" and get ready for a single-stage operation in our example.

Let us now consider that after the ship is loaded, we shall, during

the journey, cool the cargo down from -20oC to -41oC. We wish to

find out at what tank temperature and pressure we must change to

two-stage cooling. We assume the same sea water temperature as

at the loading area (+70 C). A new log p-h sketch will show this:

Pressure (bar) Fig. 15 Enthalpy (kJ/kg)

The sea water temperature gives the condensation temperature at

+17_C and, therefore the condensation pressure at 7,7bar (abs.).

We can cool the cargo down until the tank pressure is 0,2bar (1,2bar

abs.), which corresponds to a tank temperature of -38oC. The answer

to the question is that we can run a single-stage operation from the

loading port until the tank temperature drops to -38oC. From this

temperature and further down to -41oC, we have to use a two-stage

operation.

15.1 Exercises

1. A semi-pressurised LPG ship is at the port of loading and ready to

commence loading. The following data are given:

Cargo: Propane

Tank pressure during loading: 2,2bar

Sea water temperature: +100C

Max. Delivery pressure of compressor: 7,8bar

Max. Differential pressure of compressor: 6bar

a) What tank temperature will the ship keep during loading?

b) What will the condensation temperature be? (We assume _t for

the condenser to be 100C)

c) What condensation pressure will this result in?

d) Can the plant be run at single-stage operation during loading?

Explain why, and show it on a log p-h sketch.

=========================================

=====

a) Tank temperature -12 o C - (Corresponds to a saturation pressure of

3,2 abs)

b) Condensation temperature: +20 o C (Sea temperature + _t = 10oC

+10oC)

c) Condensation pressure: 7,4bar (8,4abs)

d) The plant may be operated as a single stage. The delivery

pressure will be 7,4 bar (max. 7,8bar), and the pressure difference

will be 5,2 bar (max. 6 bar).

16 Complete Loading Time Calculation

Smaller loading terminals have normally no possibility to receive

vapour in return during loading. If the ship is to load from shore

tanks with a cargo that has a higher temperature than that accepted

on board, due to the max allowable pressure in the cargo tank, the

cargo must be cooled during loading. Besides the cargo type, the

following parameters are important for the loading rate:

a. The cargo temperature at the ship's cross over.

b. The tank atmosphere temperature and condition before

loading.

c. The pressure in the cargo tanks during loading.

d. The capacity of the reliquefication plants.

e. The surrounding temperature (sea, air).

The heat balance for the cargo tanks during the loading period can

be described in the following way: Cooling of the tank shell,

insulation, cargo and tank atmosphere plus heat leakage from the

surroundings = heat removed by the reliquefaction plant.

Loading time can be calculated with the following formula:

Loading time =

Where

= Amount of heat removed from the tank shell.

= Amount of heat removed from the insulation.

= Amount of heat removed from the cargo liquid.

= Amount of heat removed from the tank atmosphere.

= Amount of heat leaked into the tanks per time unit.

= Amount of heat that is removed by the reliquefaction plant per

time unit.

With "·" above the symbol the values are given in units of time.

Further:

QT = mT · cT · (t1 - t2)

QI = mI · c1 · 0,5 · (t1 - t2)

QA = VT · _A · _hA

The symbols stand for:

mT = Weight of tank material.

CT = Specific heat of the tank material.

mI = Weight of insulation.

c1 = Specific heat of the insulation.

VT = Volume of tank.

_hL = Enthalpy difference between entering cargo at the cross over

and the cargo in the tanks at the end of loading.

_A = Cargo density in vapour form at the start of loading.

_hA = The enthalpy difference between the tank atmosphere at the

start of loading and the cargo in liquid form at the end of loading.

t1 = The temperature of the tank shell before loading.

t2 = The temperature of the tank shell at the end of loading.

The amount of heat that must be removed from the tank shell (QT) is

found by multiplying the mass of the tank material with the specific

heat capacity of the tank material, and then multiplying it with _T,

which is the difference in the tank shell temperature before and

after loading.

QI (the amount of heat that must be removed from the insulation) is

found by multiplying the mass of the insulation with the specific

heat capacity for the insulation, and then multiplying it with _T

(which is the difference in the insulation temperature before and

after loading). Finally we multiply the result by 0,5 to get an average

value, since the inside insulation layer is at the cargo temperature,

whilst the outside insulation layer is at the surrounding temperature.

The amount of heat that must be removed from the cargo (QL), has

been calculated earlier in this course. The value is found by

multiplying the mass of the cargo with the enthalpy difference that

is to be removed (m ·_hL).

The amount of heat that is to be removed from the vapour phase

(QA) is probably the value that at first seems the most complex. We

can simplify the problem by putting the actual values on a log p-h

sketch. Let us now look at an example. We consider that the tank

atmosphere contains propane (vapour) before loading. This vapour

has a temperature of +15_C before loading, and a tank pressure of

0,1bar. After loading, this vapour is re-condensed to liquid propane

at a tank pressure of 1,19bar and a temperature of -23oC (see figure

16).

Pressure (bar) Fig. 16 Enthalpy (kJ/kg)

We find the enthalpy difference between pt. "A" (tank atmosphere

before loading) and pt. "T" (liquid propane in the tank after loading):

_hA = (940 _ 469,2) kJ/kg = 470,8 kJ/kg

The density of the propane vapour at pt. "A" is read off as

1,95kg/m3.

If the vapour volume (= tank volume) is 1000m3, the calculation will

be as follows:

QA = VT ·_A · _hA = 1000m3 · 1,95kg/m3 · 470,8kJ/kg = 918060kJ=

918,06MJ

The cooling capacity ( net) is calculated as we did earlier in the

course:

Qnet = Vs ·_s ·_h

TR (transmission heat or heat leakage into the tanks) is taken from

the curve in kW, (see figure 17) and multiply it with 0.5 to get the

average heat leakage into the liquid in the tank(s) during loading.

Remember that 1W = 1J/s.

HEAT TRANSMISSION TO CARGO TANKS

16.1 Exercises

1. The following data are given:

Tank atmosphere before loading: Propane vapour

Cargo type: Propane

Vapour return to shore: None

Cargo tank volume pcs (6 tanks) 5200m3

Loaded mass: 2707MT

Sea water temperature: +110C

Tank pressure before loading: 0,lbar

Tank temperature before loading: +180C

Cargo temperature at the manifold: +100C

Tank temperature during loading: _4_C

Weight of cargo tanks: 740MT

Spec. heat capacity for steel: 0,419kJ/kg·K

Weight of insulation: 14,4MT

Spec. heat capacity of the insulation: 0,84kJ/kg·K

Process plant (3 identical): Single stage dir.

Compressor suction pressure: 3,l5 bar

Compressor suction temperature: +2_C

Compressor delivery pressure: 7,6bar

Compressor delivery temperature: +580C

Suction volume per compressors: 410m3/time

Atmospheric pressure: 1000mb

a) Calculate the heat quantity that is to be removed from the tanks

shell (QT)

b) Calculate the heat quantity that is to be removed from the

insulation (QI)

c) Calculate the heat quantity that must be removed from the cargo

(QL)

d) Calculate the heat quantity that must be removed from the

vapour phase (QA)

e) Find the heat leakage into the tanks using the curve (fig. 17).

f) Calculate the cooling capacity for one plant.

g) Calculate the loading time using all the three (identical)

reliquefication plants during the loading period.

a) QT = mT · cT · (t1 _ t2) = 740 · 103kg · 0,419kJ/kg · K · (291-269) K =

6.82 · 10 6 kJ

b) QI = mI · cI · (t1 _ t2) · 0.5 = 14,4 · 103kg · 0,84kJ/kg · K · (291 -

269)K · 0,5 = 0,13 · 10 6 kJ

c) QL = m ·_h = 2,707 · 106kg · (549.3 - 514,0)kJ/kg = 95,56 · 10 6 kJ

d) QA = VT · _A · _hA = 5200m3 · 1,95kg/m3 · (945 _ 514)kJ/kg = 4,37 ·

10 6 kJ

e) TR = 19kw · 0,5 = 9,5kw = 9,5kJ/s = 9,5kJ/s · 3600s/h = 0,034 ·

10 6 kJ/h

f) net = Vs · _s · _h = 410m3/h · 8,8kg/m3 · (894,2 _ 578,3)kJ/kg =

1,14 · 10 6 kJ/h

g) Loading Time = = = 31,6

hours