gas thermodynamics
DESCRIPTION
ThermodynamicsTRANSCRIPT
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CONCEPTS OF THERMODYNAMICS
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A part of a universe set aside for studies
Must have enclosing boundaries outside which must be the surrounding
Examples include: reservoir gas flowing to the
wellbore,
gas undergoing a compression in a mechanical compressor
Gas flowing vertically in a tubing etc.
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It states that energy can neither be created nor destroyed but can be transformed from one form to the other.
Accounts for transfer of energy to and from a system and changes of energy within a system
Two accounting procedures are: Control mass and Control volume
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It is a selected piece of matter (gas)
Any assembly is a control mass
This method calculates the properties of given pieces of matter as a function of time
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It is any defined region in space
This method gives the properties of whatever piece of matter happens to be in a given region of space at any given instant
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The general methodology of energy balance analysis is:
Define the control-mass or control-volume, indicating its boundaries on a sketch
Indicate what flows of energy across the boundaries will be considered and set up the sign conventions in the drawing
Indicate the time basis for the energy balance
Write the conservation of energy in general terms
make appropriate idealizations and bring in equation of state or other information necessary to allow solution
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Energy carried with the fluid(gas) include:
The internal energy(U)
The energy of motion (kinetic energy, )
Energy of position (Potential energy, )
The pressure energy (PV), carried by the system because of its introduction into or exit from flow under pressure
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Energy transferred between a fluid or system and its surrounding is of two kinds:
1. Heat absorbed by the flowing material or system as a result of temperature difference between the system and the surrounding (+ve)
2. The work done (w) by the system on the surrounding (shaft work). It does not include lost work due to friction
Heat and work are the only means of transferring energy between the system and the surrounding.
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An energy balance of flow of a system between point1 to point2 and the surroundings assuming no accumulation of material at any point in the system is given by the equation:
By definition
Therefore,
(1)
(2)
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Enthalpy is defined as:
Change in enthalpy is given as:
Hence equation (2) becomes:
Where H is the total heat content (BTU)
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Between two thermodynamic states (pressure and temperature), the change in enthalpy can be defined as:
Or
Where: m= pound-mass of gas n = pound-mole of gas h= specific enthalpy of the fluid (Btu/lbm) h = molal specific enthalpy of the fluid (Btu/lb-mol)
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For natural gas, the molal specific enthalpy is a function of pressure, temperature and composition i.e:
For a given gas composition,
Taking the differential
And integrating gives:
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Considering a functional relation, ,the differencein energy between two states separated byinfinitesimal temperature and specific volumedifferences, dT and dv, is:
The derivative, is called specific heat at constantvolume
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While the derivative is called specific heat atconstant pressure
The specific heat at constant volume is related to thespecific heat at constant pressure by the relation:
The ratio of the specific heats is very usefulin adiabatic compression of gases for which idealgases follow the relationship:
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The specific heats of gasesand liquids are determinedexperimentally in acalorimeter usually at apressure of 1 atm.
For natural gases, specificheat at 1atm pressure is afunction of temperature andgas gravity or molecularweight
Specific heat of hydrocarbon gases at 1atm (after Brown)
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Entropy of a system is defined as:
But the change in entropy is analogous to the change in energy; hence
Where
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Example 1: calculate the molal enthalpy and molal entropy change, h and s when a given quantity of 0.7 gravity natural gas is compressed from 14.7psia and 100oF to 800psia and 300oF
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From Thomas, Hankinson and Phillips equation:
But change in enthalpy can be defined as:
From this equation, change in enthalpy can be divided into two processes:
Change in enthalpy at constant pressure and
Change in enthalpy at constant temperature
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Considering change in enthalpy at constant pressure of 14.7psia from 100oF to 300oF will give:
Average temperature = 200oF
The specific heat at constant pressure at average pressure is obtained from chart
= 11.1 Btu/ib-mol oF
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Hence change in enthalpy at constant pressure is:
Again, lets consider the enthalpy change at constant temperature. That is from 14.7psia to 800psia at 300oF
or
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Calculate the reduced properties
From table, the z-factor equations for 0.2ppr1.2 and 1.4 Tpr 3.0 is:
Therefore,
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Hence, change in enthalpy at constant temperature will be:
Total enthalpy change will be:
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Change in entropy the change in entropy equation can be modified to
reflect 100oF instead of 32oF as:
= 11.1 Btu/ib-mol oF at 200 oF
The average z-factor from z-factor chart is:
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Averaging gives:
Therefore,
Hence, change in entropy is:
=-4.99Btu/lb-mol oF
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Very useful in the analysis of steady-state process
Enthalpy is the important thermodynamic property in the steady-flow energy balance
Entropy is the principal property of concern with respect to the second law
Hence, the coordinates of h-s in the diagram represents the two major properties of interests in the both laws
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Only applicable to gases
If a gas is cooled below its dew point, condensation occurs and heat removal cannot be determined directly from the charts
These charts are made for different range of specific gravity gases (from 0.6-1.0)
These charts are useful in finding the temperature change on expanding or compressing gases
They are also used for finding the reversible work of compression or expansion
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Using Enthalpy-entropy diagrams (MollierDiagrams)
From browns h-s diagram of 0.7 gravity natural gas
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Determination of enthalpy change
At 14.7psia and 100oF ,
h1=680Btu/lb-mol
At 800Psia and 300oF
h2 = 2600Btu/lb-mol
Therefore,
Enthalpy change from 1 -2 will be :
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Determination of Entropy Change
At 14.7psia and 100oF ,
s1=1.2 Btu/lb-mol oF
At 800Psia and 300oF
S2 = -3.7Btu/lb-mol oF
Therefore,
Entropy change from 1 -2 will be :
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Example 2
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From the diagram, read the intersection of 600oF line and the 200-psia line
h1= 6100Btu/lb-mol
Then follow the 200-psia line to its intersection with the 100oF line and read the enthalpy
h2 = 500Btu/lb-mol
Heat removed = h1-h2
= 6100- 500
= 5600 Btu/lb-mol
(a)
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(b)
At 200 psia and 300oF the enthalpy is
h3 =2600Btu/lb-mol
Heat added = h3-h2
= 2600-500
= 2100 Btu/lb-mol