# Example of Isobaric Process – Isobaric Heat Addition

## 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

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

References:
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