CONCEPTS OF THERMODYNAMICS

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.

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

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

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

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

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

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.

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)

Enthalpy is defined as:

Change in enthalpy is given as:

Hence equation (2) becomes:

Where H is the total heat content (BTU)

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)

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:

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

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:

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)

Entropy of a system is defined as:

But the change in entropy is analogous to the change in energy; hence

Where

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

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

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

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

Calculate the reduced properties

From table, the z-factor equations for 0.2ppr1.2 and 1.4 Tpr 3.0 is:

Therefore,

Hence, change in enthalpy at constant temperature will be:

Total enthalpy change will be:

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:

Averaging gives:

Therefore,

Hence, change in entropy is:

=-4.99Btu/lb-mol oF

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

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

Using Enthalpy-entropy diagrams (MollierDiagrams)

From browns h-s diagram of 0.7 gravity natural gas

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 :

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 :

Example 2

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)

(b)

At 200 psia and 300oF the enthalpy is

h3 =2600Btu/lb-mol

Heat added = h3-h2

= 2600-500

= 2100 Btu/lb-mol