# Isobaric Process

An isobaric process is a thermodynamic process, in which the pressure of the system remains constant (p = const). The heat transfer into or out of the system does work, but also changes the internal energy of the system.

Since there are changes in internal energy (dU) and changes in system volume (∆V), engineers often use the enthalpy of the system, which is defined as:

H = U + pV

In many thermodynamic analyses it is convenient to use the enthalpy instead of the internal energy. Especially in case of the first law of thermodynamics.

The enthalpy is the preferred expression of system energy that changes in many chemical, biological, and physical measurements at constant pressure. It is so useful that it is tabulated in the steam tables along with specific volume and specific internal energy. It is due to the fact, it simplifies the description of energy transfer. At constant pressure, the enthalpy change equals the energy transferred from the environment through heating (Q = H2 – H1) or work other than expansion work. For a variable-pressure process, the difference in enthalpy is not quite as obvious.

There are expressions in terms of more familiar variables such as temperature and pressure:

dH = CpdT + V(1-αT)dp

Where Cp is the heat capacity at constant pressure and α is the coefficient of (cubic) thermal expansion. For ideal gas αT = 1 and therefore:

dH = CpdT

For an ideal gas and a polytropic process, the case n = 0 corresponds to an isobaric (constant-pressure) process. In contrast to adiabatic process , in which n =  and a system exchanges no heat with its surroundings (Q = 0; ∆T≠0), in an isobaric process there is a change in the internal energy (due to ∆T≠0) and therefore ΔU 0 (for ideal gases) and Q ≠ 0.

In engineering, both very important thermodynamic cycles (Brayton and Rankine cycle) are based on two isobaric processes, therefore the study of this process is crucial for power plants.

## Isobaric Process and the First Law

The classical form of the first law of thermodynamics is the following equation:

dU = dQ – dW

In this equation dW is equal to dW = pdV and is known as the boundary work.

In an isobaric process and the ideal gas, part of heat added to the system will be used to do work and part of heat added will increase the internal energy (increase the temperature).  Therefore it is convenient to use the enthalpy instead of the internal energy.Since H = U + pV, therefore dH = dU + pdV + Vdp and we substitute dU = dH – pdV – Vdp into the classical form of the law:

dH – pdV – Vdp = dQ – pdV

We obtain the law in terms of enthalpy:

dH = dQ + Vdp

or

dH = TdS + Vdp

In this equation the term Vdp is a flow process work. This work,  Vdp, is used for open flow systems like a turbine or a pump in which there is a “dp”, i.e. change in pressure. There are no changes in control volume. As can be seen, this form of the law simplifies the description of energy transfer. At constant pressure, the enthalpy change equals the energy transferred from the environment through heating:

Isobaric process (Vdp = 0):

dH = dQ          Q = H2 – H1

At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system.

Isentropic process (dQ = 0):

dH = Vdp     →     W = H2 – H1

It is obvious, it will be very useful in analysis of both thermodynamic cycles used in power engineering, i.e. in Brayton cycle and Rankine cycle.

Example: Frictionless Piston – Heat – Enthalpy

A frictionless piston is used to provide a constant pressure of 500 kPa in a cylinder containing steam (superheated steam) of a volume of 2 m3  at 500 K. Calculate the final temperature, if 3000 kJ of heat is added.

Solution:

Using steam tables we know, that the specific enthalpy of such steam (500 kPa; 500 K) is about 2912 kJ/kg. Since at this condition the steam has density of 2.2 kg/m3, then we know there is about 4.4 kg of steam in the piston at enthalpy of 2912 kJ/kg x 4.4 kg = 12812 kJ.

When we use simply Q = H2 − H1, then the resulting enthalpy of steam will be:

H2 = H1 + Q = 15812 kJ

From steam tables, such superheated steam (15812/4.4 = 3593 kJ/kg) will have a temperature of 828 K (555°C). Since at this enthalpy the steam have density of 1.31 kg/m3, it is obvious that it has expanded by about 2.2/1.31 = 1.67 (+67%). Therefore the resulting volume is 2 m3 x 1.67 = 3.34 m3 and ∆V = 3.34 m3 – 2 m3 = 1.34 m3.

The p∆V part of enthalpy, i.e. the work done is:

W = p∆V = 500 000 Pa x 1.34 m3 = 670 kJ

## Isobaric Process – Ideal Gas Equation

Let assume an isobaric heat addition in an ideal gas. In an ideal gas, molecules have no volume and do not interact. According to the ideal gas law, pressure varies linearly with temperature and quantity, and inversely with volume.

pV = nRT

where:

• p is the absolute pressure of the gas
• n is the amount of substance
• T is the absolute temperature
• V is the volume
• R  is the ideal, or universal, gas constant, equal to the product of the Boltzmann constant and the Avogadro constant,

In this equation the symbol R is a constant called the universal gas constant that has the same value for all gases—namely, R =  8.31 J/mol K.

The isobaric process can be expressed with the ideal gas law as:

or

On a p-V diagram, the process occurs along a horizontal line (called an isobar) that has the equation p = constant.

Pressure-volume work by the closed system is defined as:

Assuming that the quantity of ideal gas remains constant and applying the ideal gas law, this becomes

According to the ideal gas model, the internal energy can be calculated by:

∆U = m cv ∆T

where the property cv (J/mol K) is referred to as specific heat (or heat capacity) at a constant volume because under certain special conditions (constant volume) it relates the temperature change of a system to the amount of energy added by heat transfer.

Adding these equations together, we obtain the equation for heat:

Q =  m cv ∆T + m R ∆T = m (cv +R)∆T = m cp ∆T

where the property cp (J/mol K) is referred to as specific heat (or heat capacity) at a constant pressure.

## Charles’s Law

Charles’s Law is one of the gas laws. At the end of the 18th century, a French inventor and scientist Jacques Alexandre César Charles studied the relationship between the volume and the temperature of a gas at constant pressure. The results of certain experiments with gases at relatively low pressure led Jacques Alexandre César Charles to formulate a well-known law. It states that:

For a fixed mass of gas at constant pressure, the volume is directly proportional to the Kelvin temperature.

That means that, for example, if you double the temperature, you will double the volume. If you halve the temperature, you will halve the volume.

You can express this mathematically as:

V = constant . T

Yes, it seems to be identical as isobaric process of ideal gas. These results are fully consistent with ideal gas law, which determinates, that the constant is equal to nR/p. If you rearrange the pV = nRT equation by dividing both sides by p, you will obtain:

V = nR/p  .  T

where nR/p is constant and:

• p is the absolute pressure of the gas
• n is the amount of substance
• T is the absolute temperature
• V is the volume
• R  is the ideal, or universal, gas constant, equal to the product of the Boltzmann constant and the Avogadro constant,

In this equation the symbol R is a constant called the universal gas constant that has the same value for all gases—namely, R =  8.31 J/mol K.

## Example of Isobaric Process – Isobaric Heat Addition

Let assume the ideal Brayton cycle that describes the workings of a constant pressure heat engineModern gas turbine engines and airbreathing jet engines also follow the Brayton cycle.

Ideal Brayton cycle consist of four thermodynamic processes. Two isentropic processes and two isobaric processes.

1. isentropic compression – ambient air is drawn into the compressor, where it is pressurized (1 → 2). The work required for the compressor is given by WC = H2 – H1.
2. isobaric heat addition – the compressed air then runs through a combustion chamber, where fuel is burned and air or another medium is heated (2 → 3). It is a constant-pressure process, since the chamber is open to flow in and out. The net heat added is given by Qadd = H– H2
3. isentropic expansion – the heated, pressurized air then expands on turbine, gives up its energy. The work done by turbine is given by WT = H4 – H3
4. isobaric heat rejection – the residual heat must be rejected in order to close the cycle. The net heat rejected is given by Qre = H– H1

Assume an isobaric heat addition (2 → 3) in a heat exchanger. In typical gas turbines the high-pressure stage receives gas (point 3 at the figure; p3 = 6.7 MPa; T3 = 1190 K (917°C)) from a heat exchanger. Moreover we know, that the compressor receives gas (point 1 at the figure; p1 = 2.78 MPa; T1 = 299 K (26°C)) and we know that the isentropic efficiency of the compressor is ηK = 0.87 (87%).

Calculate the heat added by the heat exchanger (between 2 → 3).

Solution:

From the first law of thermodynamics, the net heat added is given by Qadd = H3 – H2 or Qadd = Cp.(T3-T2s), but in this case we do not know the temperature (T2s) at the outlet of the compressor. We will solve this problem in intensive variables. We have to rewrite the previous equation (to include ηK) using the term (+h1 – h1) to:

Qadd = h3 – h2 = h3 – h1 – (h2s – h1)/ηK

Then we will calculate the temperature, T2s, using p, V, T Relation for adiabatic process between (1 → 2).

In this equation the factor for helium is equal to =cp/cv=1.66. From the previous equation follows that the compressor outlet temperature, T2s, is:

From Ideal Gas Law we know, that the molar specific heat of a monatomic ideal gas is:

Cv = 3/2R = 12.5 J/mol K and Cp = Cv + R = 5/2R = 20.8 J/mol K

We transfer the specific heat capacities into units of J/kg K via:

cp = Cp . 1/M (molar weight of helium) = 20.8 x 4.10-3 = 5200 J/kg K

Using this temperature and the isentropic compressor efficiency we can calculate the heat added by the heat exchanger:

Qadd = cp(T3-T1) – (cp(T2s-T1)/ηK) = 5200.(1190 – 299) – 5200.(424-299)/0.87 = 4.633 MJ/kg – 0.747 MJ/kg = 3.886 MJ/kg

References:
Nuclear and Reactor Physics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
6. Kenneth S. Krane. Introductory Nuclear Physics, 3rd Edition, Wiley, 1987, ISBN: 978-0471805533
7. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
8. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
9. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.