ANALYSIS OF COOLING CAPACITY AND OPTIMIZATION OF COMPRESSOR OUTLET REFRIGERATOR/LIQUEFIER

ANALYSIS OF COOLING CAPACITY AND OPTIMIZATION OF COMPRESSOR OUTLET REFRIGERATOR/LIQUEFIER
ANALYSIS OF COOLING CAPACITY AND
OPTIMIZATION OF COMPRESSOR OUTLET
PRESSURE FOR kW CLASS HELIUM
REFRIGERATOR/LIQUEFIER
A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE
REQUIREMENTS FOR THE DEGREE OF
Master of Technology
In
Mechanical Engineering
By
Nishigandha Jadhav (212ME5326)
Department of Mechanical Engineering
National Institute of Technology
Rourkela
2014
ANALYSIS OF COOLING CAPACITY AND
OPTIMIZATION OF COMPRESSOR OUTLET
PRESSURE FOR kW CLASS HELIUM
REFRIGERATOR/LIQUEFIER
A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE
REQUIREMENTS FOR THE DEGREE OF
Master of Technology
In
Mechanical Engineering
By
Nishigandha Jadhav (212ME5326)
Under the Guidance of
Mr. A. K. Sahu.
(Institute for Plasma Research, Bhat, Gandhinagar),
Prof. Ranjit Kumar Sahoo.
Department of Mechanical Engineering
National Institute of Technology
Rourkela
2014
National Institute of Technology
Rourkela
CERTIFICATE
This is to certify that the thesis entitled, “Analysis of Cooling Capacity and Optimization of
Compressor Outlet Pressure for kW Class Helium Refrigerator / Liquefier” submitted to
National Institute of Technology, Rourkela by Nishigandha Jadhav, Roll No. 212ME5326 for the
award of the Degree of Master of Technology in Mechanical Engineering with specialization in”
Cryogenic and Vacuum Technology” is a record of bonafide research work carried out by her under
my supervision and guidance. The results presented in this thesis have not been, to the best of my
knowledge, submitted to any other University/ Institute for the award of any degree or diploma.
The thesis, in my opinion, has reached the standards fulfilling the requirement
for the award of degree of Master of Technology in accordance with regulations of the Institute.
Mr. A. K. Sahu.
Prof. R. K. SAHOO
Scientist / Engineer – SF, Division Head,
Professor,
Large Cryogenic Plant and Cryosystem,
Department of Mechanical Engineering,
Institute for Plasma Research,
National Institute of Technology,
Gandhinagar, Gujarat
Rourkela.
i
CERTIFICATE
This is to certify that the dissertation, entitled
“Analysis of Cooling Capacity and Optimization of Compressor Outlet Pressure for kW Class
helium Refrigerator / Liquefier”
Is a bonafide work done by
Nishigandha Jadhav
Under my close guidance and supervision in the Large Cryogenic Plant and Cryosystem Group
Of
Institute for Plasma Research, Gandhinagar, Gujarat
For the partial fulfillment of the award of the Degree of Master of Technology in Mechanical
Engineering with specialization in “Cryogenic and vacuum Technology”
at
National Institute of Technology, Rourkela.
The work presented here, to the best of my knowledge, has not been submitted to any university
For the award of similar degree.
GUIDE:
Mr. A. K. Sahu
Scientist / Engineer – SF, Division Head
Large Cryogenic Plant and Cryosystem,
Institute for Plasma Research,
Gandhinagar, Gujarat
ii
ACKNOWLEDGEMENT
I avail this opportunity to express my indebtedness to my guide Mr.A.K.Sahu, Divisional Head of
large cryogenics plant and Cryo system, Institute for Plasma Research, Bhat, Gandhinagar, Gujarat,
and co-guide Prof. R.K.Sahoo, Mechanical Engineering Department, National Institute of
Technology, Rourkela, for their valuable guidance, constant encouragement and kind help at various
stages for the execution of this dissertation work. I consider myself fortunate to have worked under
their supervision. I also express my sincere gratitude to Dr. Sarangi, Director of NIT Rourkela. Prof.
Maithy, Head of The Department of mechanical Engineering at NIT Rourkela for providing valuable
department facilities.
.
Submitted By:
Nishigandha Jadhav
Roll No: 212ME5326
Department of Mechanical Engineering
National Institute Of Technology, Rourkela
Rourkela-769008.
iii
Abstract
The main components of a helium liquefier which determines the performance of the HRL for a given
compressor flow rate are Turbine, Heat exchanger and JT valve. Turbine and JT valve produces cooling
effect of helium gas by isentropic and isenthalpic expansion process respectively. Different arrangement
of components can be made to have different thermodynamic cycle configuration. For each configuration
main components can have different operating process parameters leading to different performance of
HRL. This project involves the analysis and optimization of compressor outlet pressure parameter for a
given configuration. Normally JT valve is kept at the lowest temperature to get the highest liquefaction
and this lowest temperature depends on the performance of other components and hence optimization of
process parameter of JT valve is not considered here. One of the different cycle configurations is analyzed
here and this cycle is often used in helium refrigeration and liquefaction plant. This configuration, planned
to use for the indigenous helium plant, has three turbines and eight heat exchangers which produces liquid
helium at 4.5 K. 1st and 2nd turbines operates at warmer temperature compared to 3rd turbine and s process
flow paths of these warmer turbines are connected in series. Helium stream coming out of the 1st turbine
passes to a heat exchanger which will reduce its temperature before entering the 2nd turbine. The nominal
helium mass flow rate supplied by the compressor system is 210 g/s at pressure of 14 bars and 310 k
temperature considering available standard compressors and capacity (~2 kW at 4.5 K) of the indigenous
plant. Effects of compressor mass flow rate and pressure on the cooling capacity of the plant have been
analyzed in this project. A part of this mass flow rate passes through a 1st and 2nd turbine for isentropic
expansion to 1.2 bar and then this low pressure helium stream comes back to compressor suction through
different heat exchangers to transfer cooling effect to the hot stream coming from the compressor. 3 rd
turbine will expand the remaining part of the main helium stream to 4 bar and this stream further passes
through a heat exchanger before entering the JT valve for liquid helium production. This analysis and
optimization work involves different practical factors and in efficiencies of main components. The
analysis result for compressor delivery flow 140 g/s at 14 bar pressure is further compared with the
performance of existing helium plant at IPR which has same compressor flow parameter. The results are
also compared with that of the aspen tech software.
Keywords: Liquefaction, Helium, process parameters, Optimization, Turbine, heat exchanger.
iv
CONTENTS
Certificate
I
Acknowledgement
III
Abstract
IV
Contents
V
List of Figures
X
List of Tables
XIII
Chapter-1
Introduction
1
1.1
Liquefaction of gas
2
1.2
Helium
2
1.3
Helium liquefier/refrigerator
3
1.4
The thermodynamically ideal system
5
1.5
Production of low temperature
6
1.5.1
Joule – Thompson effect
6
1.5.2
Adiabatic expansion
7
1.6
Thermodynamic Configuration for helium plant
8
1.6.1
Collins helium liquefaction system
8
1.6.2
Assumptions in Collins helium liquefaction system
9
1.6.3
Analysis and Performance of the system
9
Chapter-2
Literature Review
10
Chapter-3
Methodology
15
3.1
Different methods to analyze the process
16
Parameters of heat exchangers and turbine
3.2
3.2.1
(a)
Transient approach
17
One turbine and one heat exchanger
17
At a particular effectiveness different turbine inlet
Temperatures
(i)
3.2.1.2
At different effectiveness:
Flow chart for one turbine and one heat exchanger
v
19
3.2.2
One turbine and three heat exchangers
(a)
Turbine in a loop
(I)
20
At a particular effectiveness calculated UA
from converged temperatures
(i)
3.2.2.1
At different turbine inlet temperatures
Flow chart for one turbine and three heat exchanger
21
using effectiveness and turbine is in loop
(ii)
3.2.2.4
Different valves of UA for heat exchanger
23
Flow chart for one turbine and three heat exchanger
24
using UA and Turbine in loop
1.
3.2.2.6
Turbine is not in a loop
26
Flow chart for one turbine and three heat exchanger
26
using UA and Turbine is not in loop
3.2.3
One turbine and three heat exchangers with a JT
28
Valve
3.3
3.3.1
Steady state approach
29
Effectiveness based method for all three heat
29
exchangers: (nitrogen)
(i)
At a particular mass flow rates different turbine
inlet temperatures
(a)
3.3.2
At different mass flow rates
UA based method for heat exchanger
vi
32
3.3.3
Effectiveness based method only for middle heat
33
exchanger
3.4
Effect of compressor outlet pressure on liquefaction
33
and refrigeration capacity
(1)
Three compressors and compressor outlet mass flow
34
rate is 210 g/s
(a)
3.4.1
With T III
Flow chart explanation for two compressor system
34
with 3rd turbine
3.4.2
Plant layout for given configuration with 3rd
37
turbine
3.4.3
TS Diagram of a given configuration
38
3.4.4
Flow chart for 2 compressors, 140.7 g/s with 3rd
39
Turbine
(b)
3.4.5
Without T III
43
Plant layout of given configuration without 3rd
44
turbine
3.4.6
Flow chart for 2 compressors, 140.7 g/s without
45
3rd turbine
(2)
Three compressors and compressor outlet mass
flow rate is 210 g/s
(a)
With T III
vii
49
(b)
Without T III
Chapter-4
Results and Discussion
50
4.1
Effect of compressor outlet pressure on a given
51
configuration
(1)
Three compressor system with compressor outlet mass
flow rate is 210 g/s
(a)
With T III
(b)
Without T III
52
(2)
Three compressor system with compressor
53
outlet mass flow rate is 210 g/s
(a)
With T III
(b)
Without T III
54
Effect of compressor outlet mass flow rate on
55
4.2
a given configuration
(1)
Two compressors system without 3rd turbine
Chapter-5
Validation using Aspen HYSYS
57
5.1
Introduction to Aspen HYSYS
58
5.2
Entering the simulation environment
59
5.3
Process design procedure in Aspen HYSYS
60
5.4
Input parameters in a PFD
62
Process flow diagram of helium liquefier in
66
5.4.1
viii
Aspen HYSYS
5.4.2
5.5
Material streams
67
Comparison of analytical and aspen HYSYS results
68
with existing plant
5.6
Behavior of heat exchangers in given configuration
70
Chapter-6
Conclusion and future work
73
6.1
Conclusion
74
6.2
Future Work
74
References
75
ix
LIST OF FIGURES
Figure 1.3.1:
Typical Schematic of the cold box along with
4
the warm and cold end components for Helium
plant of Tokamak
Figure1.4.1:
(a) Thermodynamic cycle T-S plane (b) Apparatus Set-up
5
Figure1.5.1.1:
Isenthalpic expansion of a real gas
7
Figure 1.5.2.1:
Isentropic expansion of a Turbine
7
Figure1.6.1.1:
Collins Helium (a) Liquefaction Cycle (b) T-S diagram
8
Figure 3.1.1:
Different methods to analyze the process parameters of
16
Heat Exchangers and Turbine
Figure 3.2.1.1:
One Turbine and One Heat Exchanger
17
Figure 3.2.1.2:
Flow chart for one turbine and one heat exchanger.
19
Figure 3.2.1.3
Plot of converged temperatures of one Turbine and One
20
Heat Exchanger
Figure 3.2.2.1:
Schematic diagram of one turbine and one heat exchanger.
21
Figure 3.2.2.2:
Flow chart for one turbine and one heat exchanger.
21
Figure 3.2.2.3:
Plot of converged temperatures of one Turbine and Three
23
Heat Exchanger using different effectiveness of all Heat
Exchangers.
Figure 3.2.2.4:
Flow chart for one Turbine and Three Heat Exchanger
24
using calculated UA and turbine is in a loop.
Figure 3.2.2.5:
Plot of converged temperatures of one Turbine, Three Heat
Exchanger using calculated UA and Turbine is in loop.
x
26
Figure 3.2.2.6:
Flow chart for one Turbine and Three Heat Exchanger
26
using calculated UA and turbine is not in a loop.
Figure 3.2.3.1:
One Turbine and Three Heat Exchangers with a JT valve:
28
Transient approach
Figure 3.2.3.2:
Plot of converged temperatures for one Turbine and Three
29
Heat Exchangers with a JT valve: Transient approach
Figure 3.3.1.1:
One Turbine and Three Heat Exchangers with a JT valve:
30
Steady state approach
Figure 3.3.1.2:
Plot of LN2 production and refrigeration for one Turbine
32
and Three Heat Exchangers with a JT valve: Steady state
approach
Figure 3.4.1:
Effect of compressor outlet pressure on liquid formation at
33
JT outlet and refrigeration capacity
Figure 3.4.2:
Plant layout of given configuration with 3rd turbine
37
Figure 3.4.3:
TS diagram of a given configuration
38
Figure 3.4.4:
Flow chart for 2 compressors, 140.7 g/s with 3rd turbine
39
Figure 3.4.5:
Plant layout of given configuration without 3rd turbine
44
Figure 3.4.6:
Flow chart for 2 compressors, 140.7 g/s without 3rd turbine
45
Figure 4.1.1:
Liquid formation at JT outlet, Refrigeration capacity VS
51
pressure for 2 compressor system with 3rd turbine
Figure 4.1.2:
Liquid formation at JT outlet, Refrigeration capacity VS
xi
52
pressure for 2 compressor system without 3rd turbine
Figure 4.1.3:
Liquid formation at JT outlet, Refrigeration capacity VS
53
pressure for 3 compressor system with 3rd turbine
Figure 4.1.4:
Liquid formation at JT outlet, Refrigeration capacity VS
54
pressure for 3 compressor system without 3rd turbine
Figure 4.2.1:
Liquid formation at JT outlet, Refrigeration capacity at
56
different compressor outlet mass flow rate for 2
compressor system without 3rd turbine
Figure 5.1.1:
Aspen ONE Engineering Family
58
Figure 5.2.1:
Aspen HYSYS Simulation Environment
60
Figure 5.4.1:
PFD for a given configuration in a simulation
62
environment
Figure 5.4.1.1:
PFD of Helium Liquefier
67
Figure 5.6.1:
Plot of Temperature VS Heat Flow for HE I
71
Figure 5.6.2:
Plot of Temperature VS UA for HE I
72
Figure 5.6.3:
Plot of Delta Temperature VS UA for HE I
72
xii
LIST OF TABLES
Table1.2.1:
Thermodynamic properties of Helium
Table 4.1.1:
Liquid formation at JT outlet, Refrigeration
2
51
capacity and JT inlet temperature at different
compressor outlet pressure for 2 compressor system
with 3rd turbine
Table 4.1.2:
Liquid formation at JT outlet, Refrigeration
52
capacity and JT inlet temperature at different
compressor outlet pressure for 2 compressor system
without 3rd turbine
Table 4.1.3:
Liquid formation at JT outlet, Refrigeration
53
capacity and JT inlet temperature at different
compressor outlet pressure for 3 compressor system
with 3rd turbine
Table 4.1.4:
Liquid formation at JT outlet, Refrigeration
54
capacity and JT inlet temperature at different
compressor outlet pressure for 3 compressor system
without 3rd turbine
Table 4.2.1:
Liquid formation at JT outlet, Refrigeration capacity at
55
different compressor outlet mass flow rate for 2
compressor system without 3rd turbine
Table 5.4.2.1:
Material streams in Helium Liquefier
67
Table 5.5.1:
Comparison of UA values
68
Table 5.5.2:
Comparison of Refrigeration and Liquefaction
Capacity
xiii
69
Table 5.5.3:
Comparison of turbine inlet outlet temperatures
69
Table 5.5.4:
Comparison of Heat Exchangers hot and cold
70
Stream inlet outlet temperatures
xiv
Chapter 1
INTRODUCTION
1
1.1 LIQUEFACTION OF GAS
Liquefaction is nothing but physical conversion of gas into liquid state and used for analyzing the
fundamental properties of gas molecules, for storage of gases and in refrigeration and air conditioning.
Many gases can be converted into gaseous state by simple cooling at normal atmospheric pressure and
some other requires pressurization like carbon dioxide. Liquefaction is the process of cooling or
refrigerating a gas to a temperature below its critical temperature so that liquid can be formed at some
suitable pressure which is below the critical pressure. Using an ambient-temperature compressor, the
gas is first compressed to an elevated pressure. This high-pressure gas is then passed through a
counter-current heat exchanger to a throttling valve (Joule-Thompson valve) or an expansion engine.
Upon expanding to a certain lower pressure below the critical pressure, cooling takes place and some
fraction of gas is liquefied. The cool, low-pressure gas returns to the compressor inlet through a
recycle stream to repeat the cycle. The counter-current heat exchanger warms the low-pressure gas
prior to recompression, and simultaneously cools the high-pressure gas to the lowest temperature
possible prior to expansion.
1.2 HELIUM
Air contains 78% of Nitrogen and 0.0005% of helium whereas both are used as cryogenic refrigerants.
Thermodynamic properties of helium are given below:
Property Data / Fluid
4He
Normal boiling point (K)
4.22
Critical temperature (K)
5.20
Critical pressure (Bar)
2.3
Table1.2.1: Thermodynamic properties of Helium
2
The critical temperature of the fluid refers to the temperature of the critical point where the saturated
liquid and saturated vapor states are identical. Liquefaction of helium (4He) with the Hampson-Linde
cycle led to a Nobel Prize for Heike Kamerlingh Onnes in 1913. At ambient pressure the boiling point
of liquefied helium is 4.22 K (-268.93°C). Below 2.17 K liquid 4He has many amazing properties,
such as exhibiting super fluidity (under certain conditions it acts as if it had zero viscosity) and
climbing the walls of the vessel. Liquid helium (4He) is used as a cryogenic refrigerant; it is produced
commercially for use in superconducting magnets such as those used in superconducting Tokamak,
MRI or NMR. Cryogenic technology is the study of production of very low temperature (below 1500C or 123 K) and the behavior of materials at those temperatures. For the liquefaction process,
development of such low temperature working device, air separation and fundamental principles and
procedures have been discussed in well-known text books of cryogenics [1-5].This chapter discusses
several of the systems used to liquefy the cryogenic fluids.
1.3 HELIUM LIQUEFIER/REFRIGERATOR
Helium liquefier as the name suggest is used for the liquefaction process of Helium gas. The cold box
shown below is used for the cool down and liquefaction purpose of Helium gas coming out of the
Tokomak. Cold box contains total 8 heat exchangers and 3 turbines which expand isentropic ally and
one JT valve which expands isenthalpic ally. Process parameters of heat exchanger are effectiveness or
UA, mass flow rate, Temperatures and for turbine are temperatures, mass flow rate, inlet outlet
pressure; efficiency has to be optimized to get maximum liquefaction of LHe with minimum
refrigeration load.
3
Figure 1.3.1: Typical Schematic of the cold box along with the warm and cold end components for
Helium plant of Tokamak
4
1.4 THE THERMODYNAMICALLY IDEAL SYSTEM
Thermodynamically ideal liquefaction system is firstly used for the comparison of liquefaction
systems through the figure of merit. This system is ideal in the thermodynamic sense, but it is not ideal
as far as practical system is concerned. Carnot cycle is the perfect cycle in which ideal liquefaction is
having a reversible isothermal compression followed by a reversible isentropic expansion. The gas to
be liquefied is compressed reversibly and isothermally from ambient conditions to some high pressure.
This high pressure is selected so that gas will become saturated liquid upon reversible isentropic
expansion through the expander. The final pressure is taken as the same as the initial pressure. The
pressure attained at the end of isothermal compression is extremely high in the order of 70 GPa and it
is not an ideal process for a practicable system as it is impracticable to handle such a pressure. [1].
Figure1.4.1: (a) Thermodynamic cycle T-S plane (b) Apparatus Set-up [1].
5
The First law of thermodynamic for steady flow may be written as:
Qnet – Wnet = Outlet mh - Inlet mh
Applying the First law to the system shown in figure:
QR – W1 = m (hf – h1)
The heat transfer process is reversible and isothermal in the Carnot cycle. Thus, from the second law
of Thermodynamics:
QR = mT1 (S2 – S1) = - mT1 (S1 – Sf)
Because the process from point 2 to point f is isentropic, S2 = S3, where S is the entropy of the fluid.
Substituting QR, we may determine the work requirement for the ideal system:
- (Wi/m) = T1 (S1 – Sf) – (h1 – hf)
1.5 PRODUCTION OF LOW TEMPERATURE
1.5.1 Joule–Thompson effect
Most of the practical liquefaction systems produce low temperatures using either an expansion valve
or a Joule Thomson valve. Applying the first law for steady flow to the expansion valve, for zero heat
and work transfer and for negligible kinetic and potential changes, we find h1= h2. Flow within the
valve is irreversible and is not an isenthalpic process; the inlet and the outlet do lie on the same
enthalpy curve. We note that there is a region in which an expansion through the valve produces an
increase in temperature; while in another region the expansion results in a decrease in temperature.
Obviously we should operate the expansion valve in a liquefaction system in the region where there is
a net decrease in temperature results. The curve that separates two regions is called the inversion
curve. The effect of change in temperature for an isenthalpic change in pressure is represented by the
Joule-Thompson coefficient [1].
6
Figure1.5.1.1: Isenthalpic expansion of a real gas [1].
1.5.2 Adiabatic expansion
The second method of producing low temperatures is the adiabatic expansion of the gas through a
work producing device, such as an expansion engine. In the ideal case, the expansion would be
reversible and adiabatic and therefore isentropic. In this case we can define the isentropic coefficient
which expresses the temperature change due to a pressure change at constant entropy [1]. Isentropic
outlet temperature (T6s) is calculated for the turbine from isentropic relation for an ideal gas is T6s =
T2*(P6/P2) (r-1)/r and then actual temperature (T6a) is found out from the turbine isentropic efficiency
using formula: ɳ = (T2-T6a) / (T2-T6s)
Figure 1.5.2.1: Isentropic expansion of a Turbine
7
1.6 THERMODYNAMIC CONFIGURATION FOR HELIUM PLANT
1.6.1 Collins helium liquefaction system:
The Collins cycle or the modified Claude cycle is the one which is normally used for helium
liquefaction. Figure 1.6.1.1 (a) gives a schematic diagram of the Collins cycle and (b) gives its process
representation on the T-S diagram. HX1, HX2… HX6 are the nomenclature for the six heat
exchangers used in this liquefaction system and EX1 and EX2 are the two reciprocating expanders as
shown in the schematic diagram below. m is the total mass flow rate of the helium gas through the
compressor while me1 and me2 are the mass flow rates diverted through the expansion engine number
1 and 2, respectively. mf is the liquefaction yield. Ph and Pl represent discharge and suction pressure
of the compressor [6].
a)
b)
Figure1.6.1.1: Collins Helium (a) Liquefaction Cycle (b) T-S diagram
8
1.6.2 Assumptions in Collins Helium Liquefaction system:

The maximum pressure (Ph) in the system is 15 bar and the minimum pressure (Pl) is 1 bar.

The temperature of the gas after compression is 300 K which the ambient temperature and
the return stream temperature of the helium gas after liquefaction is at its boiling point, i.e.
4.21 K.

The pressure drop in the heat exchangers is negligible.

The J-T expansion is a perfect isenthalpic expansion process.

Heat in-leak in the system is negligible.

Effectiveness of heat exchangers and efficiencies of expanders are assumed to be constant
and their dependence on pressure, temperature and mass flow rate is ignored.
1.6.3 Analysis and Performance of the system:
The thermo physical properties of the helium gas, at different temperatures and pressures, are taken
from Van Sciver [6]. For any intermediate temperatures, the values for enthalpy, entropy, etc. are
linearly interpolated. Applying the first law of thermodynamics to the system, excepting the
compressor, for the steady state condition, the ratio of liquid yield to the total mass flow rate, y, is
given as follows: where, x1 = me1 / m and x2 = me2 / m. del he1 and del he2 are the net enthalpy
changes in helium occurring in EX1 and EX2 respectively. h represents enthalpy at the respective
points. Different parameters like heat exchanger effectiveness (ε), expander efficiencies (n1 and n2),
temperatures of gas before expansion, total mass flow rate (m ), mass flow fraction through expanders
(me1 + me2) etc. affect the performance of the liquefier. The cold produced in the expanders is directly
proportional to the mass flow rate diverted through them and the liquefaction yield is proportional to
the remaining mass flow rate that passes through the J-T valve.
9
Chapter 2
LITERATURE
REVIEW
10
M.D. Atrey [7], suggest the effect of expander efficiency and heat exchanger effectiveness on
the performance of the liquefier in a Collins helium liquefaction cycle with two reciprocating
expanders. It states that for a given efficiency of expanders and effectiveness of heat exchangers, there
exists an optimum mass flow fraction of total helium gas mass flow rate that should be diverted
through the expanders for which liquid yield is maximum and net power input is minimum. It gives
final steady state temperature distribution across the cycle, which is necessary for carrying out the
preliminary design of various components in the cycle.
G. Cammarata., A. Fichera., D. Guglielmino, [8] gave an optimization methodology for
liquefaction/refrigeration systems in the cryogenic field with Figure of Merit as the evaluation index,
and genetic algorithms as evaluation criteria. This methodology has been applied to an existing helium
liquefaction system that works according to a Collins cycle which allows optimizing the system by
taking suitable number of independent variables that are sufficient to characterize the plant. Optimized
the liquid helium production in considered application with maximum mass flow rate conditions, gives
an improvement of 10% in the FOM.
D. Henry, J.Y. Journeaux, P. Roussel, F. Michel, J.M. Poncet, A. Girard, V. Kalinin, P. Chesny
[9], states that CEA is carrying out an analysis of the various ITER cryoplant operational modes. ITER
has designed to be operated 365 days per year to optimize the available time of the Tokamak.
Operation is running for a long time but separated by a maintenance period with annual or bi-annual
major shutdown periods of a few months. Auxiliary subsystems like the cryoplant and the
cryodistribution have to cope with different heat loads which depend on the different ITER operating
states. Cryoplant consists of four identical 4.5K refrigerators and two 80K helium loops coupled with
two LN2 modules. All these systems are operated in parallel to remove the heat loads from the
magnet, cryopumps and other small users. A new design consists of updated layout of the
cryodistribution system, refrigeration loop for the HTS current leads and revised strategy for
operations of the cryopumps. Plasma operation state, short term stand by, short term maintenance, or
test and conditioning state are normal operating scenarios of the cryoplant which are checked for the
typical ITER operating states. Last part of the paper presents the abnormal operating modes of the
magnets and generated by the cryoplant. The occurrence of a fast discharge or a quench of the magnets
generates large heat loads disturbances and produces exceptional high mass flow rates which have to
11
be managed by the cryoplant, while a failure of a cryogenic component induces a major disturbance
for the magnet system. Because of this analysis modifications are made in the present PFD to match
the technical specifications of the cryogenic system with the ITER operation requirements.
R. J. Thomas, P. Ghosh, K. Chowdhury [10], stated that efficiency of helium liquefiers used in
fusion reactors can be calculated by the performance of their constituting components like heat
exchanger. On simulating with Aspen HYSYS V7.0, the effects of heat exchanger process parameters
in a helium liquefier can be understood. Independent parameter UA (product of overall heat transfer
coefficient U, heat transfer surface area A and deterioration factor F) which takes into account all
thermal irreversibility and configuration effects. Rate of liquefaction is directly proportional to UA,
saturates at limiting UA and shows the linear variation with the effectiveness of heat exchangers. Also
has influence on performance of those heat exchangers that determine the inlet temperatures to
expanders. Variation of sizes of heat exchangers does not affect the optimum mass flow rate through
expanders. When effectiveness remains equal for all heat exchangers gives the maximum liquefaction.
R. J. Thomas, P. Ghosh, K. Chowdhury [11], performed the parametric studies using Aspen
HYSYS® 7.0 and results are extrapolated to understand the behavior of large scale helium liquefiers.
It shows that the maximum liquid production is obtained when 80% of the compressor flow is diverted
through the expanders and it is equally distributed between the two expanders in a Collins cycle
analysis. Liquid production and the isentropic efficiency of expanders show the linear relationship
which is same for the higher and lower temperature expanders.
Rijo Jacob Thomas, Parthasarathi Ghosh, Kanchan Chowdhury [12], suggest that in a helium
liquefiers/ refrigeration expanders connected in parallel (reverse Brayton stage) or in series and also in
series with heat exchangers between them (modified Brayton stages). Using exergy analysis the
options of splitting and combining Brayton stages into modified Brayton stages are evaluated. Results
show that the performance is not good when two Brayton stages are combined to make two modified
Brayton stages. When one Brayton stage is split into two modified Brayton stages, the performance
shows improvement with the total heat exchanger surface area remaining unchanged. Splitting led to
more improvement when the stage operates at lower temperatures. Each stages either Brayton or
modified Brayton has been found to behave as independent refrigeration stage allowing more additions
12
of heat exchanger area. At any temperature of operation, brayton stage has been found to be superior
to a modified Brayton stage while doing one to one comparison. When heat exchangers in the
configuration are less balanced in mass flow the impact of replacing Brayton stage with modified
Brayton stage has been found to be more pronounced.
Partho S. Roy and Ruhul Amin M. [13], states increasing demand of gas production against
low production rate at the time of energy crisis effects the domestic and industrial operations as natural
gas is major power source. There is a dwindling situation in gas production as almost all plants are
operating beyond limits. Establishment of a new gas plant and other power sources has made the
situation complicated. In such a case optimization of the gas plant is the only better way. This paper
presents the steady state simulation of Bakhrabad gas processing plant (at Sylhet) using the Aspen
HYSYS shall be performed based on both the design and physical property data of the plant.
Rijo Jacob Thomas, Parthasarathi Ghosh, Kanchan Chowdhury [14], Suggest that exergy is a
useful tool for analyzing and optimizing the design and operation of systems. There are some literature
serves available on helium refrigerators and liquefiers based on exergy. This paper evaluates the
operating and geometric parameters to determine the exergy destructions in components as well as in
the entire cycle of Collins helium liquefiers. Grassmann diagram of exergy flow helps in
understanding relative importance of different components used in the system. Compressor pressure,
expander flow rates, heat exchanger surface area are some of the parameters optimized considering
both presence and absence of pressure drop in the heat exchangers. For a plant of any capacity results
are applicable using Non-dimensionalization of parameters. Exergy-analysis based on Second Law is
far superior to the First Law based energy analysis in designing of the helium plant and capable of
deriving some additional conclusions. Derived results from a Collins cycle may be applicable in largescale helium liquefiers by giving basic knowledge of the components on the plant performance and
reasonable initial guess values in their design and simulation.
QIU Lilong., ZHUANG Ming., MAO Jin., HU Liangbing., SHENG Linhai [15], Suggest a
designed steady-state program to simulate the primary cycle of the EAST cryogenic system and
compressor is tested to obtain the best isothermal efficiency. The actual UA values of the heat
13
exchanger, turbine efficiencies and fraction of the mass flow rate have been analyzed to optimize the
cycle. An equivalent refrigeration capacity is used to evaluate the refrigerator in different operation
modes. Finally, an upgraded mode is proposed on the basis of the calculations and estimate of future
heat loads from the tokamak device.
Rijo Jacob Thomas, Rohan Dutta , Parthasarathi Ghosh, Kanchan Chowdhury[16], gives the
proper design of helium systems with large number of components and involved configurations such
as helium liquefiers/refrigerators requires the use of tools like process simulators. Simulations results
are accurate as per the accuracy of given data. The 32-parameter MBWR equation of state proposed by
McCarty and Arp [19] for computation of thermodynamic properties of helium is widely used. It is
computationally involved which makes the simulation process more time-consuming and sometimes
leads to computational difficulties such as numerical oscillations, divergence in solution especially,
when the process operates over a wide thermodynamic region and is constituted of many components.
Substituting MBWR EOS by simpler equations of state (EOS(s)) at selected thermodynamic planes,
where the simpler EOS(s) have the similar accuracy as that of MBWR EOS may enhance ease of
computation. This paper has been adopted with the above mentioned methodology with an example of
steady as well as dynamic simulation of helium liquefier/refrigerator based on Collins cycle. The
above concept can be applied to the computation of fluid property which involves the thermodynamic
analysis of other process cycles.
14
Chapter 3
METHODOLOGY
15
3.1 DIFFERENT METHODS TO ANALYZE THE PROCESS PARAMETERS
OF HEAT EXCHANGERS AND TURBINE:
Indigenous helium plant of about 2 KW cooling capacity is planned to be built at Institute for Plasma
Research (IPR). Existing helium plant has about 1.3 kW cooling capacity and similar arrangement of
heat exchangers and turbines. The planned indigenous helium plant’s process flow diagram and T-S
diagram are shown in Fig 3.1.1 before going to do thermodynamic analysis of such bigger system,
simpler systems have been tried to develop certain analysis methods. These methods are further
analyzed to produce a hybrid method as per the requirement for the convergence of all temperatures of
heat exchangers. Then it is easy to optimize using that method on various process parameters such as
Effectiveness, UA, mass flow rate for heat exchanger and pressure, mass flow rate, efficiency for a
turbine.
Different methods to analyse the
process parameters of heat Exchangers
and Turbine
One Turbine
and One Heat
Exchanger:
At a perticular
Effectiveness different
Turbine inlet
Temperatures.
At different
Effectiveness
Transcient
Approach:
Steady State
Approach:
Iterative
Methods
Effectiveness based
method for all
Three Heat
Exchangers:
One Turbine and
Three Heat
Exchangers with a
JT valve:
One Turbine
and Three Heat
Exchangers:
Turbine in a
loop:
Turbine is not
in a loop:
Effectivenes
s based
method
only for
middle Heat
Exchanger.
UA based
method
for Heat
Exchanger
At a particular
mass flow rates
different Turbine
inlet
Temperatures.
At different mass
flowrates
At a particular
effectiveness
calculated UA
from converged
temperatures
At different turbine
inlet temperatures
Figure 3.1.1: Different methods to analyze the process parameters of Heat Exchangers and Turbine.
16
3.2 TRANSIANT APPROACH:
3.2.1 ONE TURBINE AND ONE HEAT EXCHANGER: (HELIUM)
1) At a particular Effectiveness different Turbine inlet Temperatures
a) At different Effectiveness:
Figure 3.2.1.1: One Turbine and One Heat Exchanger

Let turbine inlet temperature (Thi) =300K fixed for all the iterations and Effectiveness of a
heat exchanger (E) =0.9.Turbine expands isentropic ally from 14 bar (P2) to 1 bar (P6) with a
turbine isentropic efficiency = 70%

Total mass flow rate through a compressor (m1) =100g/s of which 50% is diverted to the
turbine (m2) and rest (m3) passes through the hot stream of the heat exchanger. At the inlet of
the cold stream of heat exchanger, mass flow rates add up and gives the total mass flow rate
same as through the compressor (m1).

Jot down the valves of Cp from the fluid database software (NIST). Cph at 14 bar pressure, 300
K temperature and Cpc at 1 bar pressure, 150 K temperature.

Isentropic outlet temperature (T6s) is calculated for the turbine from isentropic relation for an
ideal gas is
17
P2
T2
P6
T6s = T2*(P6/P2) (r-1)/r
T6s

T6a
Then actual temperature (T6a) is found out from the turbine isentropic efficiency using
formula:
ɳ = (T2-T6a) / (T2-T6s)

For the initial case let Tho = Thi = 300K and inlet cold stream temperature of a heat
exchanger can be calculated using formula:
T6 = ((T6a*m2) + (Tho*m3)) / m1

Assign heat exchanger cold stream temperature (Tci) = T6.

Calculate the heat exchanger hot stream outlet temperature (Tho) and cold stream outlet
temperature (Tco) using effectiveness formula:
E = ((m*Cp) h *(Thi – Tho)) / ((m*Cp) min * (Thi-Tci)) OR
E = ((m*Cp) c *(Tco – Tci)) / ((m*Cp) min * (Thi-Tci))

Now, in the second iteration Thi = 300 K which is fixed and Tci= T6 can be calculated
using Tho from previous iteration and T6a. Then with the use of effectiveness formula new
Tho and Tco can be calculated. Again the same procedure repeats till Tho, Tci and Tco will
get converged.
18
3.2.1.2 Flow chart for one turbine and one heat exchanger:
START
Thi =300 K, ɛ = 0.9, P2 = 14 bar, P6 = 1 bar,
ɳ = 70 %, m1 = 100 g/s, m2 = 50 g/s
m3 = m1 – m2; Cph = at 14 b and 300 K; Cpc = at 1 b and
150 K; T6s = T2*(P6/P2) ^
(r-1)/r
; T6a = t2 - ɳ * (T2 - T6s)
Tho =300 K
T6 = [(T6a * m2) + (Tho * m3)] / m1
Tci = T6
Tho = Thi – {[(m*Cp) min*(Thi - Tci) * ɛ] / (m*Cp) h}
Tco = Tci + {[(m*Cp) min*(Thi - Tci) * ɛ] / (m*Cp) c}
No
Is solution for
Tho, Tco, Tci
converged?
Yes
STOP
Figure 3.2.1.2: Flow chart for one turbine and one heat exchanger.

Above method can be carried out for
 heat exchanger effectiveness ( E ) = 0.9 at different fixed Turbine inlet temperatures
(Thi) = 300K, 200K, 150K
19
 Heat exchanger effectiveness (E) = 0.8 at different fixed Turbine inlet temperatures
(Thi) = 300K, 200K, 150K and results are plotted on a graph to see whether it has been
converged or not.
Figure 3.2.1.3: Plot of converged temperatures of one Turbine and One Heat Exchanger

Conclusion from the above plots:
 all the temperatures are converged with this method and
 Higher value of effectiveness gives the better heat transfer.
3.2.2 ONE TURBINE AND THREE HEAT EXCHANGERS:
1.
Turbine in a loop:
(a) At a particular effectiveness calculated UA from converged temperatures
(i) At different turbine inlet temperatures:
20
Figure 3.2.2.1: Schematic diagram of one turbine and one heat exchanger.
3.2.2.1 Flow chart for one turbine and three heat exchanger using effectiveness and turbine is in
loop:
START
Thi =T2 = 300 K, T1 = 300 K, ɛ I= 0.96, ɛ II=
0.9, ɛ III= 0.96, P2 = 14 bar, P6 = 1 bar, ɳ = 70
%, m1 = 100 g/s, m2 = 50 g/s
m3 = m1 – m2; Cph = at 14 b and 300 K; Cpc = at 1 b
and 150 K; Tco HE III = T2
A
21
A
T6s = T2*(P6/P2) ^(r-1)/r; T6a = t2 - ɳ * (T2 - T6s)
HE II:
T6 = [(T6a * m2) + (Tco HE III * m3)] / m1
Tci HE II = T6
Tho HE II = Thi – {[(m*Cp) min*(Thi - Tci) * ɛ II] / (m*Cp) h}
Tco HE II = Tci + {[(m*Cp) min*(Thi - Tci) * ɛ II] / (m*Cp) c}
HE I:
Tci HE I = T7
Tho HE I = Thi – {[(m*Cp) min*(Thi - Tci) * ɛ I] / (m*Cp) h}
Tco HE I = Tci + {[(m*Cp) min*(Thi - Tci) * ɛ I] / (m*Cp) c}
HE III:
Thi HE III = Tho HE II
Tco HE III = Tci HE II
Tho HE III = Thi – {[(m*Cp) min*(Thi - Tci) * ɛ III] / (m*Cp) h}
Tci HE III = Thi - {[(m*Cp) h*(Thi - Tho)] / [(m*Cp) c * ɛ III]}
T2 = Tho HE I
Is solution for
No
Tho, Tco, Tci
converged?
Yes
STOP
Figure 3.2.2.2: Flow chart for one turbine and one heat exchanger.
22

Above method can be carried out for:

Heat exchanger (I, II, III) effectiveness (E) = 0.96, 0.9, and 0.96 respectively at
different fixed Turbine inlet temperatures (Thi II) = 300K...
Figure 3.2.2.3: Plot of converged temperatures of one Turbine and Three Heat Exchanger
using different effectiveness of all Heat Exchangers.
ii. Different valves of UA for Heat Exchanger:

From this plot UA values can be calculated at the converged points and those are UA (I, II,
III) values =218.684 W/K, 13.211 W/K, 87.4738 W/K respectively.
23
3.2.2.4 Flow chart for one turbine and three heat exchanger using UA and Turbine in loop:
STAR
Thi =T2 = 300 K, T1 = 300 K, UA I= 218.684 W/K, UA II=
13.211 W/K, UA III= 87.4738 W/K, P2 = 14 bar, P6 = 1 bar, ɳ =
70 %, m1 = 100 g/s, m2 = 50 g/s
m3 = m1 – m2; Cph = at 14 b and 300-110 K; Cpc = at 1 b and 300-77
B
T6s = T2*(P6/P2) ^(r-1)/r; T6a = t2 - ɳ * (T2 - T6s)
T6 = [(T6a * m2) + (Tco HE III * m3)] / m1
Tho HE II = Assumed; Tci HE II = T6
Tco HE II = Tci + {(m*Cp) h*(Thi - Tho) / (m*Cp) c}
Q theoretical = (m*Cp) h*(Thi - Tho)
I = Thi – Tco; θII = Tho – Tci; LMTD = [(θI – θII)/ ln (θI /θII)]
UA = Q theoretical / LMTD
Is (Q theoretical
No
– Qcalculated)
= 0?
Yes
Tho HE I = Assumed; Tci HE I = T6
Tco HE I = Tci + {(m*Cp) h*(Thi - Tho) / (m*Cp) c}
Q theoretical = (m*Cp) h*(Thi - Tho)
θI = Thi – Tco ; θII = Tho – Tci ; LMTD = [ (θI – θII)/ ln (θI /θII) ]
UA = Q theoretical / LMTD
No
Is (Q theoretical
– Qcalculated) =
0?
Yes
A
24
A
Tho HE III = Assumed; Tci HE III = T6
Tco HE III = Tci + {(m*Cp)h*(Thi - Tho) / (m*Cp)c}
Q theoretical = (m*Cp)h*(Thi - Tho)
θI = Thi – Tco ; θII = Tho – Tci ; LMTD = [ (θI – θII)/ ln (θI /θII) ]
UA = Q theoretical / LMTD
No
Is (Q theoretical
– Qcalculated) =
0?
Yes
T2 = Tho HE I
B
STOP
Figure 3.2.2.4: Flow chart for one Turbine and Three Heat Exchanger using calculated UA and turbine
is in a loop.

Same method follows till the temperatures will get converged.
25
Figure 3.2.2.5: Plot of converged temperatures of one Turbine, Three Heat Exchanger using
calculated UA and Turbine is in loop.

Conclusion from the above plots:
 All the temperatures are converged with this method. These methods can be used for
the optimization.
1.

Turbine is not in a loop:
This method does not give a good cooling effect as turbine is expanding isentropic ally only
once.

Same procedure repeats for several times, till get the converged results.
3.2.2.6 Flow chart for one turbine and three heat exchanger using UA and Turbine is not in loop:
STAR
Thi =T2 = 300 K, T1 = 300 K, UA I= 218.684 W/K,UA II=
13.211 W/K, UA III= 87.4738 W/K, P2 = 14 bar, P6 = 1 bar, ɳ =
70 %, m1 = 100 g/s, m2 = 50 g/s
A
26
A
m3 = m1 – m2; Cph = at 14 b and 300-110 K; Cpc = at 1 b and 300-77
T6s = T2*(P6/P2)^(r-1)/r; T6a = t2 - ɳ * (T2 - T6s )
T6 = [(T6a * m2) + (Tco HE III * m3)] / m1
Tho HE II = Assumed; Tci HE II = T6
Tco HE II = Tci + {(m*Cp)h*(Thi - Tho) / (m*Cp)c}; Q theoretical = (m*Cp)h*(Thi - Tho)
θI = Thi – Tco ; θII = Tho – Tci ; LMTD = [ (θI – θII)/ ln (θI /θII) ]; UA = Q theoretical / LMTD
Is (Q theoretical –
No
Qcalculated) = 0?
Yes
Tho HE I = Assumed; Tci HE I = T6
Tco HE I = Tci + {(m*Cp)h*(Thi - Tho) / (m*Cp)c}; Q theoretical = (m*Cp)h*(Thi - Tho)
θI = Thi – Tco ; θII = Tho – Tci ; LMTD = [ (θI – θII)/ ln (θI /θII) ]; UA = Q theoretical / LMTD
No
Is (Q theoretical –
Qcalculated) = 0?
Yes
Tho HE III = Assumed; Tci HE III = T6
Tco HE III = Tci + {(m*Cp)h*(Thi - Tho) / (m*Cp)c}; Q theoretical = (m*Cp)h*(Thi - Tho)
θI = Thi – Tco ; θII = Tho – Tci ; LMTD = [ (θI – θII)/ ln (θI /θII) ]; UA = Q theoretical / LMTD
No
Is (Q theoretical –
Qcalculated) = 0?
Yes
STOP
27
Figure 3.2.2.6: Flow chart for one Turbine and Three Heat Exchanger using calculated UA and turbine
is not in a loop.
3.2.3
ONE TURBINE AND THREE HEAT EXCHANGERS WITH A JT VALVE:
Figure 3.2.3.1: One Turbine and Three Heat Exchangers with a JT valve: Transient approach

In this method exact same approach need to be used like one turbine with three heat
exchangers but in addition to that some more points to be noted as follows:
 Enthalpy can be calculated at 40 bar pressure and Tho III temperature.
 As JT valve expands isenthalpic ally so h4 = h9, assuming some value of temperature at
T9 and by using goal seek enthalpy difference (h4-h9) tends to zero gives the corrected
T9 temperature.
28
Figure 3.2.3.2: Plot of converged temperatures for one Turbine and Three Heat Exchangers with a JT
valve: Transient approach

Conclusion from the above plots:
 All the temperatures for heat exchanger I and II are converged but the temperature for
heat exchanger III and JT outlet are still decreasing and reaching to 84 K after 20th
iteration. These methods can be used for the optimization.
3.3 STEADY STATE APPROACH:
3.3.1. EFFECTIVENESS
BASED
METHOD
FOR
ALL
THREE
EXCHANGERS: (Nitrogen)
(1) At a particular mass flow rates different Turbine inlet Temperatures:
(a) At different mass flow rates:
29
HEAT
Figure 3.3.1.1: One Turbine and Three Heat Exchangers with a JT valve: Steady state approach

This is a steady state approach in which effectiveness of each heat exchanger is 0.95, 0.9, and
0.9 respectively. Assuming different turbine inlet temperatures as 140 K, 160 K, 180 K, 200 K,
220 K, and 240 K at a fixed mass flow rate from a turbine (m2).

Outlet temperature and pressure of a compressor is 300 K and 40 bar. Assuming inlet
temperature of turbine is 140 K at 40 % mass flow rate from turbine (m2=0.04 kg/s).

Thi I = 300 K, Tho I = 140 K, calculating Tci I and Tco I using effectiveness of heat exchanger
I.
E = ((m*Cp) h *(Thi – Tho)) / ((m*Cp) min * (Thi-Tci)) OR
E = ((m*Cp) c *(Tco – Tci)) / ((m*Cp) min * (Thi-Tci))

Turbine expands from 40 bar to 1 bar at inlet temperature of 140 K and found out actual
turbine outlet temperature using isentropic expansion and adiabatic efficiency of a turbine.

Isentropic outlet temperature (T6s) is calculated for the turbine from isentropic relation for an
ideal gas is
T6s = T2*(P6/P2) (r-1)/r
30

And then actual temperature (T6a) is found out from the turbine isentropic efficiency using
formula:
ɳ = (T2-T6a) / (T2-T6s)

Assigning Tho I = Thi II and Tci I = Tco II, assuming some arbitrary value to Tho II
calculating Tci II using effectiveness formula.
E = ((m*Cp) h *(Thi – Tho)) / ((m*Cp) min * (Thi-Tci))

Temperature difference (Tho II – Tci II) has to be managed using goal seek to get Tci II = T6a
and corrected Tho II which was assumed.

Calculated Tci III using energy balance,
(T6* m1)= (T6a*m2) + (Tco III *m3)

Assigning Tho II = Thi III, assuming Tho III and knowing Tco III found out Tci III.
Maintaining temperature difference (Tho III – Tci III) using goal seek so that Tci III should
reach to a boiling point temperature of nitrogen at 1 bar which is 77 K and corrected Tho III.

Calculating enthalpy at Tho III at 40 bar ( h4) which will be same as h9 as helium gas expands
isenthalpic ally using JT valve, calculating vapor and liquid fraction from h9 and refrigeration
load and LN2 production as follows:
h4 = h9 = hf +x (hg - hf)
hf = Saturated liquid at 1 bar, hg = Saturated vapor at 1 bar, (hg-hf) = Latent heat, x = Vapor fraction,
(1-x) = Liquid fraction.
LN2 Production = (1-x) * m3
Refrigeration load = (1-x) * hfg * m3
31
Figure 3.3.1.2: Plot of LN2 production and refrigeration for one Turbine and Three Heat Exchangers
with a JT valve: Steady state approach

Above plots are plotted for the LN2 production and refrigeration load at m2 = 40% at turbine
inlet temperatures at 140 K, 160 K, 180 K, 200 K, 220 K, 240 K, 260 K.

Conclusion from the above plots:
 Mass flow rate through a turbine is 40%, at lower turbine inlet temperature (140 K)
maximum LN2 production occurs and high refrigeration load is needed and at higher
temperatures (260 K) minimum LN2 production and lowest refrigeration load is
required.
 A point has to be selected where LN2 production increases and refrigeration load
decreases.
3.3.2
UA BASED METHOD FOR HEAT EXCHANGER:
This will be based on UA rather than effectiveness otherwise same procedure will
be followed as above.
32
3.3.3
EFFECTIVENESS
BASED
METHOD
ONLY
FOR
MIDDLE
HEAT
EXCHANGER:
This will be based on effectiveness of middle heat exchanger otherwise same
procedure will be followed as above.
3.4 EFFECT OF COMPRESSOR OUTLET PRESSURE ON LIQUEFACTION
AND REFRIGERATION CAPACITY
Compressor outlet pressure is one of the important operating parameter in a given configuration which
has to be optimized. A procedure has been developed analytically at given conditions. Developed
analytical procedure has been calculated at different compressor outlet pressures e.g. 10 b, 12 b, 14
b,.22b to see its effect on liquefaction at the outlet of JT and refrigeration capacity.
Effect of compressor outlet pressure on
Liquefaction and Refrigeratin Capacity
2 Compressors,
140.7 g/s
With T III
3 Compressors,
210 g/s
Without T III
With T III
With out T III
Figure 3.4.1: Effect of compressor outlet pressure on liquid formation at JT outlet and refrigeration
capacity
33
(1) TWO COMPRESSORS AND COMPRESSOR OUTLET MASS FLOW RATE IS
140.7 g/s
(a) With T III :
To achieve cooling capacity of about 2 kW at 4.5 K, total flow rates will be supplied by
3 standard compressors each having delivery helium flow capacity of about 70 g/s. For
different operational configurations of having 1, 2 and 3 compressors are analyzed. 2
compressor systems can provide flow rate of 140.7 g/s which is same as that of existing
helium plant’s nominal compressor flow rate. The cyclic configuration of 2 compressor
system is shown below which contains total 8 Heat Exchangers HE 1, HE LN2, … HE
VII respectively, Three Turbines in series combination named T I, T II, T III, JT Valve
and a Liquefier / Separator where liquefaction rate is 7 g/s. In this 2 compressor system
outlet mass flow at the compressor is 140.7 g/s at 14 bar which passes through the hot
stream which is denoted as mh and some amount of liquid is formed at the outlet of JT
out of which liquefaction rate is removed out for some minor applications and rest is
returned back to cold stream as mc. TS diagram of this configuration has been plotted
below and the procedure has been described below with the help of flow chart.
3.4.1 FLOW CHART EXPLANATION FOR TWO COMPRESSOR SYSTEM
WITH 3RD TURBINE:
HE II: It is a two stream heat exchanger in which heat has been transferred from hot
helium stream to cold helium stream. Thi and Tho is known and minimum approach
(θ2) has been assumed such a Tco is calculated from θ2 and Tci from energy balance.
By knowing all temperatures UA has been calculated.
HE I: It is a three stream heat exchanger which contains hot and cold stream of helium
and another cold stream of GN2. Minimum approach (θ1) has been assumed to find out
helium cold stream outlet temperature which is same as the GN2 outlet temperature.
34
HE LN2: In this HE phase change occur from LN2 to GN2 and that heat has been
transferred to cool down the helium hot stream temperature to 80 K. Thi is known from
HE I calculations. Using all temperatures LN2 mass flow rate has been found out.
T I: Turbine 1 inlet temperature is user defined which is given as 35.3 K. Turbine
expands adiabatically with adiabatic efficiency 76%
HE III: This is a two stream helium heat exchanger in which Thi and Tco is known
from HE II. Minimum approach (θ3) has been assumed and Tho, Tci is calculated using
energy balance. Knowing all temperatures UA has been found out to know the size of
the heat exchanger.
HE IV: It is a three stream helium heat exchanger in which Tci has been calculated
from total heat load on hot streams of helium as T II inlet temperature is given as 15.62
K. UA has been calculated from all temperatures.
T II: Turbine 2 expands adiabatically with adiabatic efficiency of 72% with its user
defined inlet temperature as 15.62K.
T III: To provide extra cooling effect 3rd turbine is included in this modified Claude
cycle which expands adiabatically with adiabatic efficiency of 64% and turbine inlet
temperature as a 7.5 K.
HE VII: It is a two stream helium heat exchanger in which Thi is equal to the T III
outlet temperature. Minimum approach (θ7) has been assumed to calculate Tco. Tci is
given as 4.408 K which is the boiling point of helium. From energy balance Tho has
been calculated.
JT Valve: It gives the cooling effect by isenthalpic expansion. Liquid has been formed
at the outlet of JT.
HE VI: Minimum approach (θ6) has been assumed between helium hot stream of HE
VI and T II outlet stream. It is a two stream helium heat exchanger in which Tco has
been calculated from energy balance.
35
Mixer: Two different inlet streams of helium are mixed together and a product stream’s
temperature has been calculated using energy balance.
HE V: This is a two stream helium heat exchanger in which all temperatures has been
found out and UA is being calculated.
Liquefier / Separator: It is connected after JT valve so that liquid formed at the outlet
of JT is accumulated in a liquefier and 7 g/s which are the liquefaction rate is taken out
and rest is heated. GHe is sent back to cool the helium hot stream of HE VII.
Refrigeration Capacity of total plant has been calculated.
36
3.4.2 PLANT LAYOUT FOR GIVEN CONFIGURATION WITH 3RD TURBINE:
Figure 3.4.2: Plant layout of given configuration with 3rd turbine
37
3.4.3 TS DIAGRAM OF A GIVEN CONFIGURATION:
Figure 3.4.3: TS diagram of a given configuration
38
3.4.4 FLOW CHART FOR 2 COMPRESSORS, 140.7 g/s WITH 3RD TURBINE
START
HE II:
UA, Heat Leak, LN2 HE Tho = 80 K, TI inlet = 35.3 K
Assume θ2
Thi = HE (LN2) Tho ; hThi = f( Thi,Ph2 )
Tco = Thi - θ2 ; hTco = f( Tco,Pc2 )
Tho = T inlet of T I ; hTho = f( Tho,Ph2 )
From Energy balance,
hTci = hTco -{[ mh*( hThi - hTho ) + HL ]/mc } ; Tci = f( hTci,Pc2 )
θI = Thi – Tco ; θII = Tho – Tci; LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ; QCalculated = LMTD * UA
Is
No
QCalculated = QTheorotical
?
Yes
HE I:
UA, Heat Leak
Assume θ1
Thi = 310 K ; hThi = f( Thi,Ph1 )
Tco (He)= Thi – θ1 ; hTco(He) = f( Tco(He),Pc1 )
Tco (GN2)= Tco (He) ; hTco (GN2)= f( Tco(GN2),Pc1 )
Tci = Tco of HE II ; hTci = f( Tci,Pc1 )
From Energy balance,
Qc(He)=mc * (hTco - hTci ) ; Qc(GN2)=mc * (hTco - hTci ) ; Qc (total) = Qc(He)+ Qc(GN2) = Qh
hTho = hThi - (Qh/ mh) ; Tho = f( hTho,Ph1 ); θI = Thi – Tco ; θII = Tho – Tci ; LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ; QCalculated = LMTD * UA
No
Is
QCalculated = QTheorotical
?
Yes
A
39
A
HE LN2:
Tho = 80 K, Phase change of LN2
Thi = Tho of HE I; hThi = f( Thi,Ph LN2 ); Tho = 80 K ; hTho = f( Tho,Ph LN2 )
From Energy balance,
mc = [ mh*( hThi - hTho )/ Latent Heat ]
T I:
TI inlet = 35.3 K, ɳa = 76%, P I inlet, PI outlet
hI1 = f( T I inlet ,P I inlet ) ; SI1 = f( T I inlet ,P I inlet )
SI1 = SI2s ; hI2s = f( S I2s ,P II inlet )
hI2a = hI1 – [ɳa * (hI1 - hI2s)] ; TI2a = f( hI2a,P I outlet ); θ T I = 27.32 - T I outlet
HE III:
θ3, Heat Leak, TI inlet = 35.3 K
Assume UA
Thi = TI inlet ; hThi = f( Thi,Ph3 )
Tco = Tci of HE II ; hTco = f( Tco,Pc3 )
Tho = T I outlet ; hTho = f( Tho,Ph3 )
From Energy balance,
hTci = hTco -{[ mh*( hThi - hTho ) + HL ]/mc } ; Tci = Tho – θ3
θI = Thi – Tco ; θII = Tho – Tci; LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ; QCalculated = LMTD * UA
Is
No
QCalculated = QTheorotical
?
Yes
B
40
B
HE IV:
Heat Leak
Assume UA
Thi 1 = Thi 2 = T I outlet; hThi = f( Thi,Ph4 )
Tho 1 = Tho 2 =T II inlet; hTho = f( Tho,Pc4 ); Tco = Tci HE III ; hTco = f( Tco,Pc4 )
From Energy balance,
Qh1= mh TII* (hThi - hTho ) ; Qh2= mh * (hThi - hTho ) ; Qh (total) = Qh1 + Qh2 = Qc
hTci = hTco - (Qc/ mc) ; Tci = f( hTci,Ph4 )
θI = Thi – Tco ; θII = Tho – Tci ; LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ; QCalculated = LMTD * UA
Is
No
QCalculated = QTheorotical
?
Yes
T II:
TII inlet = (15.62 – θ T I ) K, ɳa = 72%, P II inlet, PII outlet
hII1 = f( T II inlet ,P II inlet ) ; SII1 = f( T II inlet ,P II inlet ); SII1 = SII2s ; hII2s = f( S II2s ,P II inlet )
hII2a = hII1 – [ɳa * (hII1 - hII2s)] ; TII2a = f( hII2a,P II outlet ); θ T II = 10.32 - T II outlet
T III:
TIII inlet = (7.5 – θ T II ) K, ɳa = 64%, P III inlet, PIII outlet
hIII1 = f( T III inlet ,P III inlet ) ; SIII1 = f( T III inlet ,P III inlet )
SIII1 = SIII2s ; hIII2s = f( S III2s ,P III inlet )
hIII2a = hIII1 – [ɳa * (hIII1 - hIII2s)] ; TIII2a = f( hIII2a,P III outlet ); θ T III = 5.982 - T III outlet
C
41
C
HE VII:
UA, Heat Leak, Tci = 4.408 K
Assume θ7
Thi = T III outlet; hThi = f( Thi,Ph7 )
Tci = 4.408 K ; hTci = f( Tci,Pc7 ); Tco = Thi – θ7 ; hTco = f( Tco,Pc7 )
From Energy balance,
hTho = hThi -{[ mc*( hTco - hTci ) - HL ]/mh } ; Tho = f( hTho,Pc7 )
θI = Thi – Tco ; θII = Tho – Tci; LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ; QCalculated = LMTD * UA
Is
No
QCalculated = QTheorotical
?
Yes
HE VI:
UA, Heat Leak
Assume θ6 = T II outlet - Thi
Thi = T II outlet + θ6 ; hThi = f( Thi,Ph6 )
Tho = T III inlet; hTho= f( Tho,Pc6 ); Tci = Tco of HE VII ; hTci = f( Tci,Pc6 )
From Energy balance,
hTco = hTci +{[ mh*( hThi - hTho ) + HL ]/mc } ; Tco = f( hTco,Pc2 )
θI = Thi – Tco ; θII = Tho – Tci ; LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ; QCalculated = LMTD * UA
Is
No
QCalculated = QTheorotical
?
Yes
D
42
D
HE V:
Heat Leak
Thi = Tho HE IV ; hThi = f( Thi,Ph5 )
Tho = Thi HE VI ; hTho = f( Tho,Pc5 )
Tci = Tco HE VI ; hTci = f( Tci,Pc5 )
From Energy balance in mixer,
hTci = {[( mhT * hII2a) + (mc* hTco HE VI)] / mc } ; Tci = f( hTci,Ph4 )
θI = Thi – Tco ; θII = Tho – Tci ;
LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ;
UA = QTheorotical / LMTD
JT Valve:
Liquefaction rate = 7 g/s = 0.007 Kg/s , hfg at Pc
Vapor Fraction at JT outlet ( x) = f( hTho HE VII, P c )
Liquid Fraction = 1 – x
Liquefaction at JT outlet (y) = (1 – x)*mh JT
Refrigeration Capacity = (y – 0.007) * hfg
STOP
Figure 3.4.4: Flow chart for 2 compressors, 140.7 g/s with 3rd turbine
(b) WITHOUT T III :
This is the same procedure as above but 3rd turbine (T III) has been removed to see
the effect of compressor outlet pressure on the liquid formation at the outlet of JT
43
and the refrigeration capacity of plant. The procedure is shown below with the help
of flow chart and plant layout diagram.
3.4.5 PLANT LAYOUT OF GIVEN CONFIGURATION WITHOUT 3RD TURBINE:
Figure 3.4.5: Plant layout of given configuration without 3rd turbine
44
3.4.6 FLOW CHART FOR 2 COMPRESSORS, 140.7 g/s WITHOUT 3RD TURBINE
START
HE II:
UA, Heat Leak, LN2 HE Tho = 80 K, TI inlet =
35.3 K
Assume θ2
Thi = HE (LN2) Tho ; hThi = f( Thi,Ph2 )
Tco = Thi - θ2 ; hTco = f( Tco,Pc2 ); Tho = T inlet of T I ; hTho = f( Tho,Ph2 )
From Energy balance,
hTci = hTco -{[ mh*( hThi - hTho ) + HL ]/mc } ; Tci = f( hTci,Pc2 )
θI = Thi – Tco ; θII = Tho – Tci; LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ; QCalculated = LMTD * UA
Is
No
QCalculated = QTheorotical
?
HE I:
Yes
UA, Heat Leak
Assume θ1
Thi = 310 K ; hThi = f( Thi,Ph1 )
Tco (He)= Thi – θ1 ; hTco(He) = f( Tco(He),Pc1 )
Tco (GN2)= Tco (He) ; hTco (GN2)= f( Tco(GN2),Pc1 ); Tci = Tco of HE II ; hTci = f( Tci,Pc1 )
From Energy balance,
Qc(He)=mc * (hTco - hTci ) ; Qc(GN2)=mc * (hTco - hTci ) ; Qc (total) = Qc(He)+ Qc(GN2) =
Qh
hTho = hThi - (Qh/ mh) ; Tho = f( hTho,Ph1 )
θI = Thi – Tco ; θII = Tho – Tci ; LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ; QCalculated = LMTD * UA
No
Is
QCalculated = QTheorotical
?
Yes
A
45
A
HE LN2:
Tho = 80 K, Phase change of LN2
Thi = Tho of HE I; hThi = f( Thi,Ph LN2 ); Tho = 80 K ; hTho = f( Tho,Ph LN2 )
From Energy balance,
mc = [ mh*( hThi - hTho )/ Latent Heat ]
T I:
TI inlet = 35.3 K, ɳa = 76%, P I inlet, PI outlet
hI1 = f( T I inlet ,P I inlet ) ; SI1 = f( T I inlet ,P I inlet );SI1 = SI2s ; hI2s = f( S I2s ,P II
inlet )
hI2a = hI1 – [ɳa * (hI1 - hI2s)] ; TI2a = f( hI2a,P I outlet )
θ T I = 27.32 - T I outlet
HE III:
θ3, Heat Leak, TI inlet = 35.3 K
Assume UA
Thi = TI inlet ; hThi = f( Thi,Ph3 )
Tco = Tci of HE II ; hTco = f( Tco,Pc3 )
Tho = T I outlet ; hTho = f( Tho,Ph3 )
From Energy balance,
hTci = hTco -{[ mh*( hThi - hTho ) + HL ]/mc } ; Tci = Tho – θ3
θI = Thi – Tco ; θII = Tho – Tci; LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ; QCalculated = LMTD * UA
Is
No
QCalculated = QTheorotical
?
Yes
B
46
B
HE IV:
Heat Leak
Assume UA
Thi 1 = Thi 2 = T I outlet; hThi = f( Thi,Ph4 )
Tho 1 = Tho 2 =T II inlet; hTho = f( Tho,Pc4 )
Tco = Tci HE III ; hTco = f( Tco,Pc4 )
From Energy balance,
Qh1= mh TII* (hThi - hTho ) ; Qh2= mh * (hThi - hTho ) ; Qh (total) = Qh1 + Qh2 = Qc
hTci = hTco - (Qc/ mc) ; Tci = f( hTci,Ph4 )
θI = Thi – Tco ; θII = Tho – Tci ; LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ; QCalculated = LMTD * UA
Is
No
QCalculated = QTheorotical
?
Yes
T II:
TII inlet = (15.62 – θ T I ) K, ɳa = 72%, P II inlet, PII
outlet
hII1 = f( T II inlet ,P II inlet ) ; SII1 = f( T II inlet ,P II inlet )
SII1 = SII2s ; hII2s = f( S II2s ,P II inlet )
hII2a = hII1 – [ɳa * (hII1 - hII2s)] ; TII2a = f( hII2a,P II outlet )
θ T II = 10.32 - T II outlet
HE VI +
VII:
UA ( VI + VII ), Heat Leak
C
47
C
Assume θ6 = T II outlet Thi
Thi = T II outlet + θ6 ; hThi = f( Thi,Ph6 )
Tho = T III inlet; hTho= f( Tho,Pc6 )
Tci = Tco of HE VII ; hTci = f( Tci,Pc6 )
From Energy balance,
hTco = hTci +{[ mh*( hThi - hTho ) + HL ]/mc } ; Tco = f( hTco,Pc2 )
θI = Thi – Tco ; θII = Tho – Tci ; LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ; QCalculated = LMTD * UA
Is
No
QCalculated = QTheorotical
?
Yes
HE V:
Heat Leak
Thi = Tho HE IV ; hThi = f( Thi,Ph5 )
Tho = Thi HE VI ; hTho = f( Tho,Pc5 )
Tci = Tco HE VI ; hTci = f( Tci,Pc5 )
From Energy balance in mixer,
hTci = {[( mhT * hII2a) + (mc* hTco HE VI)] / mc } ; Tci = f( hTci,Ph4 )
θI = Thi – Tco ; θII = Tho – Tci ;
LMTD = [ (θI – θII)/ ln (θI /θII) ]
QTheorotical = mh*( hThi - hTho ) + HL ;
UA = QTheorotical / LMTD
JT Valve:
Liquefaction rate = 7 g/s = 0.007 Kg/s , hfg at Pc
Vapor Fraction at JT outlet ( x) = f(hTho HE VII, P c ) ; Liquid Fraction = 1 – x
Liquefaction at JT outlet (y) = (1 – x)*mh JT; Refrigeration Capacity = (y –0.007) * hfg
STOP
Figure 3.4.6: Flow chart for 2 compressors, 140.7 g/s without 3rd turbine
48
(2) THREE COMPRESSORS AND COMPRESSOR OUTLET MASS
FLOW RATE IS 210 g/s
(a) With T III :
Same procedure has been adopted as above with 3rd turbine using compressor outlet
mass flow rate of 210 g/s instead of 140.7 g/s and the whole cycle calculations has
been done analytically at different compressor outlet pressures.
(b) Without T III :
The only change in compressor outlet mass flow rate has been done from 140.7 g/s
to 210 g/s as compressor system contains 3 compressors at different compressor
outlet pressures and the results has been plotted to see how refrigeration capacity of
plant varies.
49
Chapter-4
RESULTS AND
DISCUSSION
50
4.1 EFFECT OF COMPRESSOR OUTLET PRESSURE ON A GIVEN
CONFIGURATION:
(1) TWO COMPRESSOR SYSTEM WITH OUTLET MASS FLOW RATE IS 140.7 g/s:
(a) With T III:
Analytically developed procedure for 2 compressors system with outlet mass flow rate of 140.7 g/s is
at 14 bar compressor outlet pressure. At different compressor outlet pressures liquid formation at JT
outlet and refrigeration capacity is calculated and tabulated below. Graph of liquid formation at JT
outlet and refrigeration capacity against compressor outlet pressure has been plotted below which
shows that both liquid formation and refrigeration capacity increases as compressor outlet pressure is
increasing.
Pressure
Pa
JT Inlet
k
liquefaction
g/s
Refrigeration load
W
1200000
1400000
1600000
1800000
2000000
2200000
5.761282527
5.255432267
4.885541404
4.687424143
4.583452649
4.525941217
29.33742534
42.91480427
49.2079059
52.03264347
53.40708397
54.13970546
436.55
701.90
824.89
880.10
906.96
921.28
Table 4.1.1: Liquid formation at JT outlet, Refrigeration capacity and JT inlet temperature at different
compressor outlet pressure for 2 compressor system with 3rd turbine
Figure 4.1.1: Liquid formation at JT outlet, Refrigeration capacity VS pressure for 2 compressor
system with 3rd turbine
51
(b) Without T III:
As per discussion in 3.4.5, at different compressor outlet pressures liquid formation at JT outlet
and refrigeration capacity is calculated and tabulated below. Graphs has been plotted against
compressor outlet pressure. Both the plots shows increasing nature till pressure reaches to 17.1 bar
and then decreases. Maximum value of liquid formation at JT outlet is 36.7 g/s and Refrigeration
capacity is 582.12 W. at 17.1 bar compressor outlet pressure.
Pressure
Pa
JT Inlet
k
liquefaction
g/s
Refrigeration load
W
1200000
1400000
1600000
1650000
1680000
1700000
1705000
1710000
1715000
1720000
1750000
1800000
1900000
2000000
2200000
6.74630532
5.787061827
4.894978169
4.749956438
4.68102954
4.641969391
4.633107624
4.624486816
4.616176123
4.608161368
4.565839724
4.514004042
4.456524464
4.431219959
4.414242356
16.98811153
28.97207206
35.90612238
36.53881825
36.72474613
36.77876641
36.78343028
36.78567726
36.7849161
36.78129384
36.70448253
36.4018711
35.37925816
34.07452
31.17803007
195.20
429.41
564.93
577.29
580.93
581.98
582.07
582.12
582.10
582.03
580.53
574.62
554.63
529.13
472.53
Table 4.1.2: Liquid formation at JT outlet, Refrigeration capacity and JT inlet temperature at different
compressor outlet pressure for 2 compressor system without 3rd turbine
Figure 4.1.2: Liquid formation at JT outlet, Refrigeration capacity VS pressure for 2 compressor
system without 3rd turbine
52
(2) THREE COMPRESSOR SYSTEM WITH OUTLET MASS FLOW RATE IS 210 g/s:
(a) With T III:
Below plot shows that liquid formation at JT outlet and refrigeration capacity is directly
proportional to pressure. It increases as compressor outlet pressure increases.
pressure
JT Inlet
liquefaction
Pa
k
g/s
Refrigeration
load
W
1100000
1200000
1300000
1400000
1500000
1600000
1700000
1800000
1900000
2000000
2100000
2200000
5.927323236
5.708255454
5.441330461
5.182552054
4.974039679
4.821330897
4.713808133
4.638507297
4.585195802
4.546826422
4.51873756
4.497854014
32.67443164
46.71577453
58.11690347
66.10965052
71.40626042
74.85983351
77.12370443
78.63827585
79.67848098
80.41167099
80.94054319
81.32956259
501.77
776.19
999.01
1155.21
1258.73
1326.22
1370.47
1400.07
1420.40
1434.73
1445.06
1452.67
Table 4.1.3: Liquid formation at JT outlet, Refrigeration capacity and JT inlet temperature at different
compressor outlet pressure for 3 compressor system with 3rd turbine
Figure 4.1.3: Liquid formation at JT outlet, Refrigeration capacity VS pressure for 3 compressor
system with 3rd turbine
53
(b) Without T III:
Refrigeration capacity and liquid production shows a peak value at 16 bar and then
decreases.Maximum values of refrigeration capacity is 983.64 W and liquid formation at JT outlet is
57.33 g/s at 16 bar compressor outlet pressure.
Pressure
Pa
JT Inlet
k
liquefaction
g/s
Refrigeration load
W
1200000
1400000
1500000
1600000
1700000
1800000
1900000
2000000
6.380853736
5.381981306
4.927399245
4.624491574
4.482354176
4.432108822
4.416155075
4.410984197
34.06260995
50.21483081
55.24397429
57.3304184
56.9745864
55.36258769
53.3020345
51.10249639
528.90
844.57
942.86
983.64
976.68
945.18
904.91
861.92
Table 4.1.4: Liquid formation at JT outlet, Refrigeration capacity and JT inlet temperature at different
compressor outlet pressure for 3 compressor system without 3rd turbine
Figure 4.1.4: Liquid formation at JT outlet, Refrigeration capacity VS pressure for 3
compressor system without 3rd turbine
54
4.2 EFFECT OF COMPRESSOR OUTLET MASS FLOW RATE ON A GIVEN
CONFIGURATION:
(1) TWO COMPRESSOR SYSTEM WITHOUT 3RD TURBINE:
A procedure has been developed such that process fluid temperature for component entry and exit
points have been kept constant for different compressor flow rates (however they will change a bit)
and change the UA of each Heat Exchanger with different compressor outlet mass flow rates. This
procedure has been developed for 2 compressor system without 3rd turbine. Below table shows the
variation of liquid formation at JT outlet and refrigeration capacity with the different compressor
outlet mass flow rate. Plot shows as a compressor outlet mass flow increases which give the highest
liquid formation at JT outlet and highest refrigeration capacity at that point. Bigger is the system,
higher efficiency with higher mass flow rate at compressor outlet.
g/s
liquid formation at
JT outlet
g/s
Refrigeration
Capacity
W
210
180
140.7
100
70
68.52496711
43.38479234
28.78552213
15.87540648
8.262593718
1202.42
711.09
425.77
173.46
24.68
Compressor outlet mass flow rate
Table 4.2.1: Liquid formation at JT outlet, Refrigeration capacity at different compressor outlet mass
flow rate for 2 compressor system without 3rd turbine
55
Figure 4.2.1: Liquid formation at JT outlet, Refrigeration capacity at different compressor outlet mass
flow rate for 2 compressor system without 3rd turbine
56
Chapter-5
VALIDATION USING
ASPEN HYSYS
57
5.1 INTRODUCTION TO ASPEN HYSYS:
It belongs to aspen one engineering family. Aspen one is a model processing simulation software by
Aspen Tech. Its main purpose is to simulate the optimized configuration with operational excellence.
Aspen
APLE
Aspen
HTFS
Aspen
Plus
Aspen
Aspen
ACOL
Aspen
FIHR
FRAN
ASPENONE
ENGINEERING
Aspen
TASC
Aspen
Aspen
Dynamics
HYSYS
Aspen
Aspen
PIPE
MUSE
Figure 5.1.1: Aspen ONE Engineering Family
Aspen HYSYS consists of three process modeling regions: Gas processing, Refining, Chemicals
1. Gas Processing

Steady State: It models a gas refrigeration plant consisting of compressor, expander, gas heat
exchanger, chiller, low-temperature separator etc.

Dynamics: All models are built up in dynamic processing mode.
2. Refining

Steady State: This modeling includes a crude oil processing facility consisting of a pre-flash
drum, crude furnace and an atmospheric crude column.

Dynamics: Models the Refining problem in Dynamic mode.
58
3. Chemicals

Steady State: It contains modeling of a propylene glycol production process consisting of a
continuously-stirred-tank reactor and a distillation tower.

Dynamics: Models the Chemical problem in Dynamic mode.
Helium liquefaction system comes under gas process modeling in a steady state.
5.2 ENTERING THE SIMULATION ENVIRONMENT:
For entering into the Simulation environment, detailed instructions for choosing a property package
and components, installing and defining streams, unit operations, and using various aspects of the
HYSYS interface to examine the results while you are creating the simulation. The gas processing
simulation is built using following basic steps:
1. Create a unit set.
2. Choose a property package.
3. Select the components.
4. Create and specify the feed streams.
5. Install and define the unit operations prior to the Heat Exchanger / Compressor / Turbine.
6. Install and define the Heat Exchanger / Compressor / Turbine.
59
Figure 5.2.1: Aspen HYSYS Simulation Environment
5.3 PROCESS DESIGN PROCEDURE IN ASPEN HYSYS:
For creating a new case in Aspen HYSYS, select New from file menu and Case from submenu.
Then Simulation Basis Manager Window will appear. The Simulation Basis Manager is the main
property view of the Simulation environment. You can access information in Simulation Basis
manager while the other areas of HYSYS are kept on hold avoiding unnecessary Flow sheet
calculations. Any changes made in the Simulation Basis environment will take effect at the same time.
At the same time thermodynamic data has been fixed and cannot be manipulated in the flow sheet of
simulation environment.
The minimum input criteria of Simulation Basis manager are to select a Fluid Package with an
attached Property Package and At least one component in the Fluid Package. In a simulation basis
manager all components are present in a component manager tab which contains all chemical
information of respective component. This information is stored as component list from the collection
of library. The Components Manager always contains a Master Component List that cannot be deleted.
60
This master list contains every component available from all component lists. If you add components
to any other component list, they automatically get added to the Master Component List. Also, if you
delete a component from the master, it also gets deleted from any other component list that is using
that component. Fluid Package contains all necessary information of a component which is required in
Calculation. There are four key advantages to this approach:

All associated information is defined in a single location, allowing for easy creation and
modification of the information.

Fluid Packages can be exported and imported as completely defined packages for use in any
simulation.

Fluid Packages can be cloned, which simplifies the task of making small changes to a complex
Fluid Package.

Multiple Fluid Packages can be used in the same simulation.
In simulation Basis Manager contains fluid package tab on fluid package manager. In this tab
many other fluid packages can be created as well as manipulated. Once we go to fluid package
manager then we can choose one appropriate fluid package according to our requirement of properties
of given configuration. Selected fluid package from fluid package manager in Simulation Basis
Environment is listed in the Current Fluid packages group with the following information: name,
number of components attached to the fluid package, and property package attached to the fluid
package. To see the list of all fluid packages click view from the Fluid Package tab of the Simulation
Basis Manager and Add button to add respective Fluid Package to the Environment. Select the proper
fluid package and from the Component List Selection drop-down list, select the required components
for simulation of given configuration.
From Fluid package manager, active the Aspen properties tab and select the required Fluid
Package. Here RefProp fluid package has been selected for a given configuration and from Aspen
properties database two components i.e. Helium-4, Nitrogen is selected.
After selecting fluid packages and components, click on Enter Simulation Environment button so
that a process flow sheet window will appear. From Menu bar, set the preferences in Tools option.
61
Choose a Unit set from variables in the preferences or add a new user defines unit set. In process flow
sheet unit operations can be installed in many ways. There are many unit operations available on a
palette. As soon as we double click on the unit operations to be installed tab opens. There we can input
all the connections and values. In worksheet of that unit operation we can see the material streams
information which is automatically calculated as soon as we enter some input values. We can
reposition streams and operations. In steady state analysis recycler unit operations can be used to
calculate the unknown parameters in the process flow diagram. The process flow diagram (PFD)
provides the best representation of the flow sheet. Using the PFD gives you immediate reference to the
progress of the simulation currently being built, such as what streams and operations are installed,
flow sheet connectivity, and the status of objects. In addition to graphical representation, we can build
our flow sheet within the PFD.
5.4 INPUT PARAMETERS IN A PFD:
From simulation basis manager, aspen properties database two components Nitrogen and Helium-4 are
taken as material stream. From fluid package, Aspen properties, Riprap is selected as a fluid package.
Then enter into the simulation environment. All unit operations are arranged in order and linked by
material streams. PFD is shown below for a given configuration in HYSYS:
Figure 5.4.1: PFD for a given configuration in a simulation environment
62
For each unit operations following input values are given.
1. Cooler ( C1 )
Outlet temperature = 300 K
Pressure drop = 0.15 bar
2. Mixer ( MIX 100 )
Inlet streams:

De Furties:
Inlet temperature = 300 K
Inlet pressure = 1.05 bar
Mass flow rate = 0.7 g/s

De Paliers:
Inlet temperature = 300 K
Inlet pressure = 1.05 bar
Mass flow rate = 8 g/s

Injection of GHe:
Inlet temperature = 300 K
Inlet pressure = 1.05 bar
Mass flow rate = 7 g/s
3. Compressor system
Mass flow rate = 140.7 g/s
Inlet temperature = 300 K
Inlet pressure = 1.05 bar
Outlet pressure =14 bar
4. Cooler ( C2 )
Outlet temperature = 310 K
Pressure drop = 0.2 bar
5. Tee ( Tee 100 )
Mass flow rate to Paliers = 8 g/s
6. Cooler ( C3 )
Outlet temperature = 300 K
Pressure drop = 12.75 bar
63
7. Valve ( V1 )
Pressure drop = 0.5 bar
8. Heat exchanger 1 ( HE I )
Pressure drop in hot stream = 0.1 bar
Pressure drop in Cold stream = 0 bar
Temperature at 8 = 307 K
Minimum Approach (θ1) = 3 K
Heat Leak = 0.07 KW
9. Heat exchanger LN2 ( HE LN2 )
Pressure drop in hot stream = 0 bar
Pressure drop in Cold stream = 0 bar
LN2 inlet vapor fraction = 0.00
LN2 outlet vapor fraction = 1.00
Inlet temperature = 79.19 K
LN2 mass flow rate = 23.96 g/s
10. Valve ( V2 )
Pressure drop = 0.15 bar
11. Heat exchanger 2 ( HE II )
Pressure drop in hot stream = 0.06 bar
Pressure drop in Cold stream = 0 bar
Minimum Approach (θ2) = 1.344 K
Heat Leak = 0.05 KW
12. Tee ( Tee 101 )
Mass flow rate to T I = 74.92 g/s
13. Turbine 1 ( T I )
Efficiency of turbo expander = 76 %
Outlet pressure = 5.4 bar
14. Heat exchanger 3 ( HE III )
Pressure drop in hot stream = 0 bar
Pressure drop in Cold stream = 0 bar
Minimum Approach (θ3) = 0.527 K
64
Heat Leak = 0.015 KW
UA = 1372 W/C
15. Tee ( Tee 102 )
Mass flow rate to De Furties = 0.2 g/s
16. Valve ( V3 )
Pressure drop = 0.1 bar
17. Heat exchanger 4 ( HE IV )
Pressure drop in hot stream = 0 bar
Pressure drop in Cold stream = 0 bar
Heat Leak = 0.04 KW
Temperature at 29 and 30 = 15.62 K
18. Valve ( V4 )
Pressure drop = 0.15 bar
19. Turbine 2 ( T II )
Efficiency of turbo expander = 72 %
Outlet pressure = 1.2 bar
20. Tee ( Tee 103 )
Mass flow rate to De Furties = 0.2 g/s
21. Heat exchanger 5 ( HE V )
Pressure drop in hot stream = 0 bar
Pressure drop in Cold stream = 0 bar
Heat Leak = 0.025 KW
Temperature at 36 = 10.67 K
22. Mixer ( MIX 101 )
Outlet mass flow rate = 125 g/s
23. Heat exchanger 6 ( HE VI )
Pressure drop in hot stream = 0 bar
Pressure drop in Cold stream = 0 bar
Minimum Approach (θ6) = 0.296 K
Heat Leak = 0.025 KW
Temperature at 40 = 7.5 K
65
24. Turbine 3 ( T III )
Efficiency of turbo expander = 64 %
Outlet pressure = 4 bar
25. Tee ( Tee 104 )
Mass flow rate to De Furties = 0.3 g/s
26. Mixer ( MIX 102 )
Outlet mass flow rate = 0.7 g/s
27. Heater ( H1 )
Outlet temperature = 300 K
Pressure drop = 0.15 bar
28. Heat exchanger 7 ( HE VII )
Pressure drop in hot stream = 0 bar
Pressure drop in Cold stream = 0 bar
Minimum Approach (θ7) = 0.317 K
Heat Leak = 0.025 KW
Temperature at 47 = 4.416 K
29. JT Valve
Pressure drop = 2.8 bar
30. Separator
Outlet pressure = 1.2 bar
If we change any input value mentioned above all other information is updated automatically in
given cyclic configuration.
5.4.1 PROCESS FLOW DIAGRAM OF HELIUM LIQUEFIER IN ASPEN
HYSYS:
Figure shows the process flow diagram that drawn in HYSYS. Table shows the steady state properties
of all the streams in the PFD, material stream 49 gives the helium liquefaction rate.
66
Figure 5.4.1.1: PFD of Helium Liquefier
5.4.2 MATERIAL STREAMS:
Table 5.4.2.1: Material streams in Helium Liquefier
67
5.5 COMPARISION OF ANALYTICAL AND ASPEN HYSYS RESULTS:
Process parameters like temperatures of heat exchanger and turbine, UA of Heat Exchangers
calculated using analytical method and Aspen HYSYS results are compared.

UA values of all Heat Exchangers are tabulated below :
 There is hardly 17% variation between UA values calculated analytically for HE
IV and 14% for HE VII.
 Others are below 10% variation.
UA Value using
Analytical
Method (W/C)
UA Value
using Aspen
Hysys (W/C)
HE I
30037.522
30000
HE II
11448.32773
11930
HE III
1281.415262
1371
HE IV
7734.034239
6393
HE V
2045.627648
2125
HE VI
1612.463348
1491
HE VII
887.6467783
758
Table 5.5.1: Comparison of UA values

Refrigeration and Liquefaction capacity are tabulated below:
 Liquefaction capacity is kept constant and the values of refrigeration capacity
calculated using analytical method and by using Aspen are matching with each other.
68
UA Value
using
Analytical
Method
(W/C)
UA Value UA Values
using Aspen for Existing
Hysys
Plant
(W/C)
(W/C)
Liquifaction
Capacity (g/s)
7
7
7
Refrigeration
Capacity (W)
701.9034928
707.01
650
Table 5.5.2: Comparison of Refrigeration and Liquefaction Capacity

Turbine inlet outlet temperatures are tabulated below:
 Turbine inlet outlet temperatures are almost matching so error is considered to be
negligible.
Inlet temperatures
Using Analytical
Method:
Using Aspen
HYSYS:
Outlet
Temperature
TI
T II
T III
35.3
15.62
7.5
27.32205431 10.31772247
5.981976441
Inlet temperatures
35.29
15.62
7.5
Outlet
Temperature
27.32
10.32
5.982
Table 5.5.3: Comparison of turbine inlet outlet temperatures

Heat Exchangers inlet outlet temperatures are tabulated below:
 Hot and cold stream inlet outlet temperatures of all Heat Exchangers are matching with
each other. Error between those is less than 10%.
69
Heat Exchangers hot
stream
temperatures(K)
Using Analytical
Method:
Heat Exchangers cold
stream
temperatures(K)
HE I (Cold
stream of N2)
HE II
HE III
HE IV (Hot
stream of He)
HE V
HE VI
HE VII
Inlet (Thi)
310
80.00714974
35.3
27.32205431
15.60513935
10.67
5.981976441
Outlet (Tho)
86.7683045
35.3
27.32205431
15.62
10.67
7.5
5.255432267
Inlet (Tci)
78.65714974
30.44353092 26.77205431
13.47358299
10.33908191 5.66197644
Outlet (Tco)
307.016
78.65714974 30.44353092
26.77205431
13.47358299
10.3707
5.661976441
Inlet (Thi / Tci)
3rd hot or cold stream
for 3 stream HE
Outlet (Tho /
Tco)
Heat Exchangers hot
stream
temperatures(K)
Using Aspen
HYSYS:
Heat Exchangers cold
stream
temperatures(K)
79.19
27.32205431
307.016
15.62
4.408
Inlet (Thi)
310
80.01
35.29
26.92
15.61
10.67
5.982
Outlet (Tho)
86.8
35.29
26.92
15.62
10.67
7.5
5.255
Inlet (Tci)
78.66
30.43
26.37
13.27
10.34
5.66
4.416
Outlet (Tco)
307
78.66
30.43
26.37
13.27
10.37
5.66
Inlet (Thi / Tci)
3rd hot or cold stream
for 3 stream HE
Outlet (Tho /
Tco)
79.19
27.32
307
15.62
Table 5.5.4: Comparison of Heat Exchangers hot and cold stream inlet outlet temperatures
All Heat Exchangers hot and cold stream inlet outlet temperatures, Turbines inlet outlet temperatures
UA values of all Heat Exchangers are matching with each other with 17% error between analytical
calculations and values calculated using Aspen HYSYS. This validates the analytical optimization
methodology.
5.6 BEHAVIOR OF HEAT EXCHANGERS IN GIVEN CONFIGURATION:
Different graphs have been plotted for all heat exchangers using Aspen HYSYS. Behavior of all plots
for Heat Exchanger I have been discussed below:
I.
Plot of Temperature VS Heat Flow for HE I:
70

Y axis shows the temperature change along the length of the Heat Exchanger. This plot
signifies how the heat is being transferred from hot stream of helium to the cold stream.

This graph shows the linear variation of properties like Cp.

For HE VII, Cp values are varying so it does not show a linear relation.
Figure 5.6.1: Plot of Temperature VS Heat Flow for HE I
II.
Plot of Temperature VS UA for HE I:

Along the length of the Heat Exchanger how UA is changing with its Temperature is
shown in this plot.

Gradually delta UA is decreasing as temperature is decreasing because at higher
temperatures velocity as well as heat transfer coefficient is high which ultimately
increases UA.
71
Figure 5.6.2: Plot of Temperature VS UA for HE I
III.
Plot of Delta Temperature VS UA for HE I:

This plot shows how UA is varing as Delta Temperature is changing.

As Delta T is decreasing Delta UA is increasing.
Figure 5.6.3: Plot of Delta Temperature VS UA for HE I
72
Chapter-6
CONCLUSION
73
6.1 CONCLUSION:
Among all analyzed methods steady state approach is effectively used for the optimization of
process parameters of turbine and heat exchanger such as mass flow rate, temperature, inlet outlet
turbine pressures, effectiveness or UA. Using steady state approach analytical method has been
developed for 2 compressor system with 3rd turbine and without 3rd turbine for maximizing
refrigeration capacity. Same method has been used to analyze the refrigeration capacity and the effect
of compressor outlet pressure for 3 compressor system.
Analytically developed procedure for 2 compressor system with 3rd turbine has been validated
using Aspen HYSYS. All temperatures and UA values of Heat Exchangers, turbine inlet outlet
temperatures calculated using analytical method is found to be matching with the HYSYS results.
Effect of compressor outlet pressure is directly proportional to liquid formation at JT outlet and
refrigeration capacity with 3rd turbine and without 3rd turbine it increases till certain value then
decreases.
6.2 FUTURE WORK:
Process parameters like turbine mass flow rate have to be varied to optimize the refrigeration capacity
and liquid formation at JT outlet.
74
REFERENCES
75
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(1986)
[4] Ventura, G., Risegari, L., The Art of Cryogenics, Low-temperature experimental Elsevier (2008)
[5] Richard, T. Jacobsen, Steven, G.Penoncello and Eric, W.Lemmon, Thermodynamic Properties of
Cryogenic Fluid Plenum Press
[6] Van Sciver, S.W., Helium cryogenics Plenum Press, New York, USA, 1986.
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Kharagpur, West Bengal 721302, India (2011)
76
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[18] Aspen tutorial # 1: Aspen Basic.
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[20] AspenHYSYSV7_1-User guide
77
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