HACETTEPE UNIVERSITY

KMU226-23
CHEMICAL ENGINEERING
THERMODYNAMICS I

Hacettepe University
Department of Chemical Engineering
Fall Semester

Selis Önel, PhD

selis@hacettepe.edu.tr
 Syllabus: Weeks 1-3
2

Week 1. Lectures 1-3 Introduction: Chapters 1 & 2

 Introduction to course content: Fundamental concepts of thermodynamics,

Definitions
 Work and heat, The first law of thermodynamics, Internal energy

 Energy balance for closed system

Week 2. Lectures 4-6 Chapter 2: Mass and Energy Balances

 Thermodynamic state and state functions

 Equilibrium, reversible process, constant-V and constant-P processes

 Enthalpy, heat capacity

 Mass and energy balances for open systems: General energy balance, energy

balances for steady-state processes

Week 3. Lectures 7-9 Chapter 3: Volumetric Properties of Pure Fluids

 PVT Behavior of pure substances, the ideal gas state

 Virial equations of state

SelisÖnel©
 Syllabus: Weeks 4-6
3

Week 4. Lecture 10-12 Chapter 3: Volumetric Properties of Pure Fluids

 Cubic Equations of State

 Generalized Correlations for Gases

 Generalized Correlations for Liquids

Week 5. Lecture 13-15 Chapter 5: Second Law of Thermodynamics

 Atomic statement of the second law. Heat Engines.

 Thermodynamic Temperature Scales

 Entropy, Entropy Changes of an Ideal Gas

 Mathematical Statement of the Second Law

Week 6. Lectures 16-18 Midterm 1

SelisÖnel©
 Syllabus: Weeks 7-9
4

Week 7. Lectures 19-21 Chapter 5: Second Law of Thermodynamics

 Mathematical Statement of the Second Law

 Entropy balance for open systems

 Calculation of ideal work and lost work

Week 8. Lectures 22-24 Chapter 6: Thermodynamic Properties of Fluids

 Property relations, Residual properties

 Residual properties from the virial equations of state

Week 9. Lectures 25-27 Chapter 6: Thermodynamic Properties of Fluids

 Thermodynamic Diagrams, Tables of thermodynamic properties

 Generalized Property Correlations for Gases

SelisÖnel©
 Syllabus: Weeks 10-12
5

Week 10. Lectures 28-30 Chapter 7: Application of thermodynamics to flow

processes
 Duct flow of compressible fluids

Week 11. Lectures 30-32 Midterm 2

Week 12. Lectures 33-35 Chapter 7: Application of thermodynamics to flow

processes
 Turbines (expanders)

 Expansion and compression processes

SelisÖnel©
 Syllabus: Weeks 13-16
6

Week 13. Lectures 36-38 Chapter 8: Production of power from heat

 Steam power plant, Rankine cycle

 Internal-Combustion Engines, Jet engines

Week 14. Lectures 39-41 Chapter 9: Refrigeration and Liquefaction

 Carnot refrigerator

 Vapor-compression cycle

 Heat pump

Week 15. Preparation week for final exam

Week 16. Final Exam

SelisÖnel©
 Refrigeration
7

 Maintenance of a T <Tsurroundings
Ex:
-Air conditioning of buildings
-Preservation of foods and chilling of beverages
-Large-scale commercial processes requiring refrigeration:
- manufacture of ice and solid CO2
- dehydration and liquefaction of gases,
- separation of air into oxygen and nitrogen.

SelisÖnel©
 Goals
8

Making operational engineering calculations considering:

 The model refrigerator, operating on a reverse Carnot cycle
 Refrigeration via the vapor compression cycle , as in the common house hold
refrigerator and air conditioner
 The choice of refrigerants as influenced by their properties
 Refrigeration based on vapor absorption, an alternative to vapor compression
 Heating or cooling by heat pumps with heat extracted from or rejected to the
surroundings
 Liquefaction of gases by refrigeration

SelisÖnel©
 9.1: The Carnot Refrigerator
• A refrigerator is a heat pump that
absorbs heat from a region at a
temperature below that of the
surroundings and rejects heat to
the surroundings
• Transferring heat from lower to
higher temperature requires an
input of work
• Maximum efficiency is obtained
for the Carnot refrigeration cycle,
the reverse of the Carnot engine
cycle

©McGraw-Hill Education.
 The Carnot Refrigerator (2)
• Operating in a cycle for which ∆U = 0, the 1st law becomes:
𝑊𝑊 = − 𝑄𝑄𝐶𝐶 + 𝑄𝑄𝐻𝐻 (9.1)
• Here, 𝑄𝑄𝐶𝐶 > 0 and 𝑄𝑄𝐻𝐻 < 0 (but larger in absolute value)
• Measure of effectiveness of a refrigerator = coefficient of
performance, 𝜔𝜔
heat absorbed at the lower temperature 𝑄𝑄𝐶𝐶
𝜔𝜔 ≡ = (9.2)
net work 𝑊𝑊
• Combining Eqn. (9.2) with Eqn. (5.4), the coefficient of
performance for a Carnot refrigerator becomes:
𝑇𝑇𝐶𝐶
𝜔𝜔 =
𝑇𝑇𝐻𝐻 − 𝑇𝑇𝐶𝐶
• Refrigeration effect per unit work decreases as TC decreases
and as TH increases
©McGraw-Hill Education.
 9.2: The Vapor-Compression Cycle
• Practical factors like those that prevent use of a Carnot cycle in a
steam power plant also make its use for refrigeration impractical.
• The vapor-compression cycle is a more practical refrigeration cycle

©McGraw-Hill Education.
 The Vapor-Compression Cycle (2)
• (1 ➝ 2) Evaporator: liquid
refrigerant evaporates at constant
T & P, absorbing heat & producing
the refrigeration effect
• (2 ➝ 3’) Compressor (ideal): vapor
is compressed isentropically
• (2 ➝ 3) Compressor (real):
compression with inherent
irreversibilities & ∆S > 0
• (3 ➝ 4) Condenser: Vapor cooled
& condensed to reject heat
• (4 ➝ 1) Throttle valve: expands
liquid to original pressure with
partial evaporation
©McGraw-Hill Education.
 The Vapor-Compression Cycle (3)
Per unit mass of fluid:
𝑄𝑄𝐶𝐶 = 𝐻𝐻2 − 𝐻𝐻1 & 𝑄𝑄𝐻𝐻 = 𝐻𝐻4 − 𝐻𝐻3
Work of compression:
𝑊𝑊 = 𝐻𝐻3 − 𝐻𝐻2
Coefficient of Performance:
𝐻𝐻2 − 𝐻𝐻1
𝜔𝜔 = (9.4)
𝐻𝐻3 − 𝐻𝐻2
Rate of circulation of refrigerant:
𝑄𝑄̇𝐶𝐶
𝑚𝑚̇ = (9.5)
𝐻𝐻2 − 𝐻𝐻1

©McGraw-Hill Education.
 The Vapor-Compression Cycle (4)
• Refrigerant data is commonly
found in tables or in the form
of PH diagrams
• PH diagrams directly show all
required enthalpies
• Note: small pressure drops
do occur in the (ideally
isobaric) condenser and
evaporator due to fluid
friction
• For a given TC & TH, the
vapor-compression cycle
results in lower 𝜔𝜔 than the
Carnot cycle
©McGraw-Hill Education.
 Calculation Protocol for Vapor-Compression Cycle
• Typical parameters known:
necessary refrigeration
temperature, temperature of
surroundings, compressor
efficiency, minimum-temperature
differences for heat transfer
• Must choose refrigerant (Sec. 9.3)
• Minimum-temperature difference
in evaporator and condenser is
the difference between the
temperature of the process fluid
and the surroundings or the
refrigerated area

©McGraw-Hill Education.
 Comment About Heat Transfer
• The reason a minimum-
temperature difference must be
set (or calculated) is because
there must always be some
temperature gradient to
promote heat transfer
• If the ∆T between the fluid and
system or surroundings is too
small, the necessary heat
transfer area becomes
impractically large
• See a Heat Transfer course/text
for further details

©McGraw-Hill Education.
 Vapor-Compression Cycle: Fluid Temperatures
Define the following parameters:
• Minimum temperature
difference in evaporator:
∆𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚,𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
• Minimum temperature
difference in condenser:
∆𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚,𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
• Temp. of Surroundings: 𝑇𝑇𝑆𝑆
• Refrigerator Temperature: 𝑇𝑇𝑟𝑟
• Fluid operating temperatures:
𝑇𝑇𝐶𝐶 = 𝑇𝑇2 = 𝑇𝑇𝑟𝑟 − ∆𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚,𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
𝑇𝑇𝐻𝐻 = 𝑇𝑇4 = 𝑇𝑇𝑠𝑠 + ∆𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚,𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
©McGraw-Hill Education.
 Vapor-Compression Cycle: Refrigerant Choice

• Considerations: toxicity, flammability, cost, corrosion properties, and vapor

pressure
• Environmental concerns (ozone depletion potential, global warming potential,
regulatory status) strongly constrain the range of compounds to be considered
• To avoid air leakage INTO the system, the vapor pressure of the refrigerant at the
evaporator temperature should be greater than atmospheric pressure (i.e. P2 >
1.01 bar)
• The vapor pressure at the condenser temperature should not be unduly high
because of the initial cost and operating expense of high-pressure equipment (i.e.
P3 can’t be too large)
• Common choices: ammonia, methyl chloride, carbon dioxide, propane, 1,1,1,2-
tetrafluoroethane (HFC-134a), 2,3,3,3-tetrafluoropropene (HFO-1234yf)

©McGraw-Hill Education.
 9.3: The Choice of Refrigerant
• See Table 9.1, Figure F.2 or the NIST Chemistry Webbook for
thermodynamic tables for most common refrigerants

©McGraw-Hill Education.
 Vapor-Compression Cycle: Operating Pressures
• From T2, constraint that
P2 > 1.01 bar at T2, & all other
considerations, choose a
refrigerant
• For refrigerant and known T4,
determine P4 (= P3)
• From these pressures,
determine enthalpies and
entropies from PH diagrams or
data tables (or, for an exotic
refrigerant, generalized
correlations)

©McGraw-Hill Education.
 Vapor-Compression Cycle: Compressor

• Knowing T2, P2, find H2 and S2

(from tables or diagrams)
• Initially assume isentropic
compression, and set 𝑆𝑆3′ = S2
• From known P4 = P3 and 𝑆𝑆3′ , find
𝑇𝑇3′ and 𝐻𝐻3′ (may require
interpolation)
• Ideal work: 𝑊𝑊𝑠𝑠 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 =
∆𝐻𝐻 𝑆𝑆 = 𝐻𝐻3′ − 𝐻𝐻2
• Real work:
𝑊𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = ∆𝐻𝐻 = ∆𝐻𝐻 𝑠𝑠 /𝜂𝜂𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

©McGraw-Hill Education.
 Vapor-Compression Cycle: Compressor (2)

• Find real state 3 enthalpy, 𝐻𝐻3 :

𝐻𝐻3 = 𝐻𝐻2 + 𝑊𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
• If necessary, from known
pressure and enthalpy,
determine 𝑇𝑇3 by interpolation
• Note: if 𝑇𝑇3 or 𝑃𝑃3 is found to be
too high, a different refrigerant
must be chosen or capital costs
for refrigeration unit will be too
high

©McGraw-Hill Education.
 Vapor-Compression Cycle: Rest of Cycle

• Knowing T4, P4, find H4

• Determine condenser duty:
𝑄𝑄𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑄𝑄𝐻𝐻 = 𝐻𝐻3 − 𝐻𝐻4
• Throttling process is enthalpic,
∆H = 0, hence H1 = H4
• If necessary, use saturated
liquid/vapor values at T2 to
determine vapor fraction at
State 1
• Determine evaporator heat:
𝑄𝑄𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑄𝑄𝐶𝐶 = 𝐻𝐻2 − 𝐻𝐻1

©McGraw-Hill Education.
 Vapor-Compression Cycle: Final Calculations

• Find coefficient of performance:

𝑄𝑄𝐶𝐶 𝐻𝐻2 − 𝐻𝐻1
𝜔𝜔 = = (9.4)
𝑊𝑊 𝐻𝐻3 − 𝐻𝐻2
• If refrigeration capacity is
known, determine necessary
refrigerant flow rate:
𝑄𝑄̇𝐶𝐶
𝑚𝑚̇ = (9.5)
𝐻𝐻2 − 𝐻𝐻1

©McGraw-Hill Education.
 Example 9.1

A refrigerated space is maintained at -20°C and cooling water is available at

21°C. The refrigeration capacity is 120 000 kJ·h-1.
The evaporator and condenser are of sufficient size that a 5°C minimum-
temperature difference for heat transfer can be realized in each.
The refrigerant is 1,1,1,2-tetrafluoroethane (HFC-134a), for which data are
given in Table 9.1 and Fig. F.2.
a) What is the value of 𝜔𝜔 for a Carnot refrigerator?
b) Calculate 𝜔𝜔 and 𝑚𝑚̇ for a vapor-compression cycle if the compressor
efficiency is 0.80.

©McGraw-Hill Education.
 Cascade Cycles
• Limits on the operating
pressures of the evaporator
and condenser also limit the
temperature difference, TH –TC
• With TH fixed by the
temperature of the
surroundings, a lower limit is
placed on the temperature
level of refrigeration
• This can be overcome by the
operation of 2 or more cycles
employing different
refrigerants in a cascade

©McGraw-Hill Education.
 Cascade Cycles (2)

• Here, heat absorbed in the

interchangers by the
refrigerant of the higher-T
cycle 2 serves to condense
the refrigerant in the lower-
T cycle 1
• The two refrigerants, again,
are chosen such that each
cycle operates at
reasonable condenser and
evaporator pressures

©McGraw-Hill Education.
 9.4: Absorption Refrigeration

• In vapor-compression refrigeration, the work of compression

is usually supplied by an electric motor, but the ultimate
source of the electric energy is most likely a heat engine
(power plant) – suggesting a heat engine and refrigeration
cycle could be combined
• Heat can be taken in at both an elevated temperature, TH ,
and at a reduced temperature, TC , rejecting heat to the
surroundings at TS
• This is known as an absorption-refrigeration process

©McGraw-Hill Education.
 Absorption Refrigeration (2)
• Work required by a Carnot refrigerator absorbing heat at
temperature TC and rejecting at the temperature of the
surroundings TS, from Eqns. (9.2) & (9.3):
𝑇𝑇𝑆𝑆 − 𝑇𝑇𝐶𝐶
𝑊𝑊 = 𝑄𝑄𝐶𝐶
𝑇𝑇𝐶𝐶
• If work is provided by a Carnot engine absorbing heat at TH >
TS, from Eqn. (5.5) we can show that:
−𝑊𝑊 𝑇𝑇𝑆𝑆 𝑇𝑇𝐻𝐻
= − 1 𝑜𝑜𝑜𝑜 𝑄𝑄𝐻𝐻 = 𝑊𝑊
𝑄𝑄𝐻𝐻 𝑇𝑇𝐻𝐻 𝑇𝑇𝐻𝐻 − 𝑇𝑇𝑆𝑆
• Elimination of W yields minimum ratio of absorbed heats:
𝑄𝑄𝐻𝐻 𝑇𝑇𝐻𝐻 𝑇𝑇𝑆𝑆 − 𝑇𝑇𝐶𝐶
= (9.6)
𝑄𝑄𝐶𝐶 𝑇𝑇𝐻𝐻 − 𝑇𝑇𝑆𝑆 𝑇𝑇𝐶𝐶

©McGraw-Hill Education.
 Figure 9.4: Absorption-Refrigeration Unit

©McGraw-Hill Education.
 Absorption-Refrigeration Unit (2)

• Difference between a vapor-compression and absorption

refrigerator is in the different means employed for compression
• Section to the right of dashed line in Fig. 9.4 is the same as in a
vapor-compression refrigerator; section to the left accomplishes
compression effectively via heat engine
• Refrigerant as vapor from the evaporator is absorbed in a
relatively nonvolatile liquid solvent at the evaporator pressure
and relatively low T
• Heat given off in this process is discarded to surroundings

©McGraw-Hill Education.
 Absorption-Refrigeration Unit (3)

• Liquid solution from absorber (with high concentration of

refrigerant) passes to pump which raises the pressure of the
liquid to that of the condenser
• Heat from the high T source evaporates the refrigerant from the
solvent
• Vapor passes from the regenerator to the condenser & solvent
returns to absorber by way of a heat exchanger, conserving
energy & optimizing stream temperatures
• Low-pressure (low level) steam is the usual source of heat for
the regenerator

©McGraw-Hill Education.
 Absorption-Refrigeration Unit (4)
• Most commonly, absorption-refrigeration systems use water as
the refrigerant and lithium bromide solution as the absorbent
– Limits refrigeration to temperatures above the freezing point of water

• For lower temperatures, ammonia can serve as the refrigerant

with water as the solvent. Such a system can achieve refrigerant
effect up to –25°C
• Alternatively, methanol as the refrigerant and polyglycoethers as
the absorbents
• Actual absorption-refrigeration systems have heat ratios of
𝑄𝑄𝐻𝐻 ⁄𝑄𝑄𝐶𝐶 near 2.5

©McGraw-Hill Education.
 9.5: The Heat Pump
• Heat pump (reversed heat engine) is a device for heating
houses and commercial buildings during the winter and
cooling them during the summer
• In the winter: absorbs heat from surroundings and rejects
heat into the building; the refrigerant evaporates in coils
underground or in outside air; during condensation, heat is
transferred to air or water – used to heat the building
– Compression must be to a pressure such that the condensation
temperature of the refrigerant is higher than the building temperature

• In the summer: heat pumps serves for air conditioning; flow

of refrigerant is reversed; heat is absorbed from the building
and rejected through underground coils or to the outside air

©McGraw-Hill Education.
 Example 9.2

A house has a winter heating requirement of 30 kW and a summer cooling

requirement of 60 kW.
Consider a heat-pump installation to maintain the house temperature at 20°C
in winter and 25°C in summer.
This requires circulation of the refrigerant through interior exchanger coils at
30°C in winter and 5°C in summer. Underground coils provide the heat source
in winter and the heat sink in summer.
For a year-round ground temperature of 15°C, the heat-transfer
characteristics of the coils necessitate refrigerant temperatures of 10°C in
winter and 25°C in summer.
What are the minimum power requirements for winter heating and summer
cooling?

©McGraw-Hill Education.
 9.6: Liquefaction Processes
• Liquefied gases are used for a variety of purposes:
– Domestic fuel, like liquid propane or liquefied petroleum gas
– Oxidant for rocket fuel, like liquid oxygen
– Easier transport, as done for natural gas
– Low-temperature refrigeration, like liquid nitrogen
– Separation of air constituents by distillation

• Liquefaction results when a gas is cooled to a temperature in

the two-phase VLE region, in processes such as:
1) By heat exchange at constant pressure (condenser)
2) By an expansion process from which work is obtained (turbine)
3) By a throttling process (valve)

©McGraw-Hill Education.
 Cooling Processes
1) By heat exchange at constant
pressure (condenser)
2) By an expansion process from
which work is obtained (turbine)
3) By a throttling process (valve)
From Figure 9.5:
• Note: if starting at A, process (3)
does not approach VLE region
• Initial state must be at a point
such as A' for (3) to achieve
liquefaction

©McGraw-Hill Education.
 Cooling Processes (2)
From Figure 9.5:
• To get from A to A' , typically a
gas is compressed to B then
isobaric cooling to A'
• PH diagram for air shows that at
a temperature of 160 K,
P > 80 bar for liquefaction along
any isenthalpic path
• Thus, if air is compressed to at
least 80 bar, then cooled below
160 K, it can be partially liquefied
by throttling

©McGraw-Hill Education.
 Linde Liquefaction Process
• Depends solely on throttling expansion, shown on the next
slide
• After compression, the gas is precooled to ambient
temperature, or further cooled by refrigeration
• Additional cooling can take place in a counter-current heat
exchanger with post-throttled, cold (not liquefied) vapor
• The lower the temperature of the gas entering the throttle
valve, the greater the fraction of gas that is liquefied
• Example: a refrigerant evaporating in the cooler at -40°C
provides a lower temperature at the valve than if water at
20°C is the cooling medium

©McGraw-Hill Education.
 Linde Liquefaction Process (2)

©McGraw-Hill Education.
 Claude Liquefaction Process

• More efficient that Linde process, replaces throttle valve with

an expander (but operation in VLE is impractical)
• Gas at an intermediate temperature is extracted from the
heat-exchange system and passed through an expander from
which it exhausts as a saturated or slightly superheated vapor
• Remaining gas is further cooled and throttled through a valve
to produce liquefaction, as in the Linde process
• Unliquefied portion, which is saturated vapor, mixes with the
expander exhaust and returns for recycle through the heat-
exchange system

©McGraw-Hill Education.
 Claude Liquefaction Process (2)

©McGraw-Hill Education.
 Claude Liquefaction Process (3)
• Energy balance applied to the part of the process to the right
of the dashed vertical line yields:
𝑚𝑚̇9 𝐻𝐻9 + 𝑚𝑚̇15 𝐻𝐻15 − 𝑚𝑚̇4 𝐻𝐻4 = 𝑊𝑊̇𝑜𝑜𝑜𝑜𝑜𝑜
• Ifs the expander operates adiabatically, then from Eqn. (7.13):
𝑊𝑊̇𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑚𝑚̇12 (𝐻𝐻12 − 𝐻𝐻5 )
• By mass balance, 𝑚𝑚̇15 = 𝑚𝑚̇4 − 𝑚𝑚̇9 , so the energy balance
becomes:
𝑚𝑚̇9 𝑚𝑚̇4 − 𝑚𝑚̇9 𝑚𝑚̇12
𝐻𝐻9 + 𝐻𝐻15 − 𝐻𝐻4 = (𝐻𝐻12 − 𝐻𝐻5 )
𝑚𝑚̇4 𝑚𝑚̇4 𝑚𝑚̇4
• Defining 𝑧𝑧 ≡ 𝑚𝑚̇9 ⁄𝑚𝑚̇4 and x ≡ 𝑚𝑚̇12 ⁄𝑚𝑚̇4 , solving for z :
𝑥𝑥 𝐻𝐻12 − 𝐻𝐻5 + 𝐻𝐻4 − 𝐻𝐻15
𝑧𝑧 = (9.7)
𝐻𝐻9 − 𝐻𝐻15
©McGraw-Hill Education.
 Claude Liquefaction Process (4)
𝑥𝑥 𝐻𝐻12 − 𝐻𝐻5 + 𝐻𝐻4 − 𝐻𝐻15
𝑧𝑧 = (9.7)
𝐻𝐻9 − 𝐻𝐻15
• z is the fraction of the stream entering the heat-exchanger system
that is liquefied
• x is the fraction of the stream that is drawn off between the heat
exchangers and passed through the expander (must be specified
to solve Eqn. 9.7)
• The Linde process results when x = 0:
𝐻𝐻4 − 𝐻𝐻15
𝑧𝑧 = (9.8)
𝐻𝐻9 − 𝐻𝐻15
• Thus the Linde process is a limiting case of the Claude process
• Both eqns assume no heat flows in from surroundings (never
exactly true)
©McGraw-Hill Education.
 Example 9.3

Natural gas, assumed here to be pure methane, is liquefied in a

Claude process. Compression is to 60 bar and precooling is to
300 K. The expander and throttle exhaust to a pressure of 1 bar.
Recycle methane at this pressure leaves the exchanger system
(point 15 in Fig. 9.7) at 295 K. Assume no heat leaks into the
system from the surroundings, an expander efficiency of 75%,
and an expander exhaust of saturated vapor. For a draw-off to
the expander of 25% of the methane entering the exchanger
system (x = 0.25), what fraction z of the methane is liquefied,
and what is the temperature of the high-pressure stream
entering the throttle valve?

©McGraw-Hill Education.
 Example 9.3 Data

From the NIST WebBook:

@ 300 K & 60 bar: 𝐻𝐻4 = 855.3 kJ·kg−1

@ 295 K & 1 bar: 𝐻𝐻15 = 903.0 kJ·kg−1

For saturated liquid and vapor at a pressure of 1 bar:
𝑇𝑇 𝑠𝑠𝑠𝑠𝑠𝑠 = 111.5 K

Saturated liquid: 𝐻𝐻9 = −0.6 kJ·kg−1

Saturated vapor: 𝐻𝐻12 = 510.6 kJ·kg−1

Saturated vapor: 𝑆𝑆12 = 4.579 kJ·kg−1 ·K−1

©McGraw-Hill Education.
 Now you should be able to:
47

 Compute the coefficient of performance for a Carnot refrigeration cycle and

recognize that this represents an upper limit for any real refrigeration process
 Carry out a thermodynamic analysis of a vapor compression refrigeration cycle
 Describe a practical absorption refrigeration process and explain why its use
might be advantageous
 Sketch a cascade refrigeration system, explain why one might use such a system,
and understand how to approach the selection of refrigerants for such a system
 Carry out a thermodynamic analysis of a Linde or Claude liquefaction process

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