ap3290_chapter_5_ii_2009

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Thermodynamics Y Y Shan

AP3290 75

2.3. Cooling by isentropic expansion (adiabatic expansion) process:

S P

T

∂

∂

It has been pointed out earlier that thermodynamic processes can be mathematically

described by a partial derivative under specific conditions.

“isentropic” above indicates no change of entropy ( 0=dS , i.e. 0=dQ ) happened in

the process. “isentropic” process is equivalent to “adiabatic” process (since 0=dS means

0=dQ ).

“expansion” means when volume expands and pressure will change accordingly.

“cooling” means the temperature decreases.

Therefore, cooling by isentropic expansion is dealing with the T~P relationship, or is

aking how temperature changes with respect to the decrease of pressure (when volume

expands) under the isentropic condition?

Thus, the study of “cooling by isentropic expansion” is to answer the following

questions: ?=

∂

∂

S P

T

How does the temperature of a gas change when it expands?

Derivation ofS P

T

∂

∂

:

Entropy S is a state function, for ),( PT S , mathematically we have

),(TP

dPP

S dT

T

S PT dS

∂

∂+

∂

∂=

For isentropic process (constant entropy process), 0=dS :

dPP

S dT

T

S

TP

0

∂

∂+

∂

∂=∴

thus

P

T

∂

∂

∂

∂

−=

∂

∂

T

S

P

S

P

T

S

Eq5-6

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AP3290 76

-----------------------------------------------------------------------------------------

*Or Eq5-6 can also be obtained directly by applying the “triple product rule”:

For ),( PT S :

1−=

∂

∂

∂

∂

∂

∂

S T P P

T

S

P

T

S

Therefore:

P

T

T P

S

T

S

P

S

S

P

T

S P

T

∂

∂

∂

∂

−=

∂

∂

∂

∂−=

∂

∂ 1

--------------------------------------------------------------------------------------------

Applying the Maxwell’s relation (from VdPSdT dG +−= ):

PT

-

∂

∂=

∂

∂

T V

PS ,

Since the volume expansion coefficient:

P

1

∂

∂=

T

V

V α , or V

T

V

P

α =

∂

∂. The specific heat

P

∂

∂=

T

S T C P , or T C

T

S P /

P

=

∂

∂is the,

Therefore Eqn5-6 becomes:

0

P

T>=

∂

∂

∂∂

−=

∂

∂

PS C

TV

T

S

PS

P

T α Eq5-7

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AP3290 77

Positive slope means that when pressure P decreases (volume expands), temperature T will

drop. It can be concluded that isentropic expansion always causes a decrease in temperature.

This process can be used for gas cooling.

2.4.

Joule-Thomson effect(Throttling process): Cooling by isenthalpic expansion

H P

T

∂

∂

Let us perform a thermodynamic process: a high-pressure gas is forced through a

constriction, such as a partially opened valve (a throttle) or a porous wall. The throttle valve

is well thermally insulated so that no heat is transferred during the process.

The gas initially has a pressure P1, temperature T1 and volume V1. After it passes

through the valve, its final pressure is P2 and the volume is V2

The first law for this system can be written as: W QU U U +=−=∆12

Since the system is thermally insulated, Q=0, so W U U =− 12 ,

2211

0

2

0

121

2

1

V PV PdV PdV PW W W V

V

−=−−=+= ∫∫

so we obtain:

221112 V PV PU U −=− , i.e., 111222 V PU V PU +=+ ,

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AP3290 78

Since the enthalpy is defined as PV U H += , so we get

0or1212

=−=∆= H H H H H

Thus throttling process is a constant enthalpy process (isenthalpic process).

Given a particular gas and the initial conditions P1 and T1, and final pressure P2 , we

wish to know what the final temperature T2 will be. In other words, we want to know the

change of temperature of the gas (from T1 to T2) with respect to the change of its pressure

(from P1 to P2) in this isenthalpic process:

?=

∂

∂=

H

JT P

T µ Eqn5-8

This thermodynamic quantity µJT is known as the Joule-Thomson coefficient, which is named

after James Joule and William Thomson, who together studied this process in the 1850s.

To find out

H

JT P

T

∂

∂= µ , we proceed by starting from the relationship:

dPV P

S T dT

T

S T VdPdP

P

S dT

T

S T dH

T PT P

+

∂

∂+

∂

∂=+

∂

∂+

∂

∂=∴

Applying the Maxwell’s relation:PT T

V

P

S

∂

∂=

∂

∂- , and

P

PP

S T C

∂

∂= ,

dH becomes:

dP

T

V T V dT C dH

P

P

∂

∂−+=

For isenthalpic process (constant enthalpy process): 0=dH , so

dPV T

V T dT C

P

P

−

∂

∂=

therefore we obtain the expression for the Joule-Thomson coefficient:

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AP3290 79

−

∂

∂=

∂

∂= V

T

V T

C P

T

PP H

JT

1 µ Eqn5-10

For an ideal gas, ,,T

V

P

nR

T

V nRT PV

P

==

∂

∂→= , it is straightforward to show that

0= JT

µ , the temperature of an ideal gas does not change after throttling expansion.

However, when a real gas expands freely at constant enthalpy, the final temperature

( 2T ) may either decrease or increase, depending on the initial temperature ( 1T ) and

pressure( 1P ). For any given pressure, a real gas has a Joule-Thomson (Kelvin) inversion

temperature InvT . (i) When initial temperature of the gas InvT T >1 , throttling expansion

causes the temperature to rise ( 12 T T >

). (ii) When InvT T <

1 isenthalpic expansion causes

cooling ( 12 T T < ).

For most gases at atmospheric pressure, the inversion temperature InvT is fairly high

(above room temperature), and so isenthalpic expansion can be applied for gas cooling at

those temperature and pressure conditions.

The value of µ JT , representing the change of temperature with respect to a change of pressure,

depends on the specific gas, as well as the initial temperature and pressure of the gas before

expansion. For all real gases, µ JT will equal zero at some temperature point called the Joule-

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AP3290 80

Thomson inversion temperature or inversion point. It is the temperature where the

coefficient changes sign (i.e., where the coefficient equals zero).

In any gas expansion, the gas pressure decreases, meaning P∂ is always negative

( 0<∂P ). With that in mind ( P

T JT

∂

∂

= µ ),the following table explains when the Joule-

Thomson effect cools or heats a real gas.

If the gas temperature is then µ JT is since P∂ is thus T ∂ must be so the gas

below the inversion temperature >0 always “-“ negative cools

above the inversion temperature <0 always “-“ positive heats

Helium and hydrogen are two gases whose Joule-Thomson inversion temperatures at

one atmosphere are very low (e.g., about −222 °C for helium). Thus, helium and hydrogen

will warm when expanded at constant enthalpy at typical room temperatures. So hydrogen

expansion at room temperature is very dangerous.

3. Gas cooling, liquefaction, and low temperature achievement

The Joule-Thomson effect is applied in the Linde technique as a standard process in

the petrochemical industry, where the cooling effect is used to liquefy gases (for the

production of liquid O2, N2, H2, and He), so that extremely low temperature can be achieved.

From air liquefaction to air separation: with air liquefaction, German engineer Carl

von Linde(1891–1934), created the conditions needed to produce pure gases using low-

temperature processes. These gases include not only oxygen and nitrogen, but also hydrogen

and inert gases - a technology for which the future has only just begun.

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AP3290 81

Sketch of the first air liquefaction plant from 1895.

A gas must be below its inversion temperature to be cooled and liquefied. Once a gas has

been pre-cooled to a temperature lower than the maximum inversion temperature, the

optimum pressure, from which to start throttling, corresponds to a point on the inversion

curve, where the Joule-Thomson coefficient 0= JT

µ . Starting at this pressure and ending at

a lower pressure (after expansion), the process produces a large temperature drop.

Consequently, the gas that has been cooled by throttling is used to cool the incoming gas,

which after throttling becomes still cooler. After successive cooling processes, the

temperature of the gas is lowered to such a temperature that , after throttling, it becomes

partly liquefied.

Gas Inversion temperature TInv Liquefying temperature TLiq

At 1atm

O2 620oC -183

oC

N2 352oC -196oC

H2 -71oC -253oC

Ne -239oC -269

oC

ap3290_chapter_5_ii_2009

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