Example 9.6.4

Problem Statement: A closed Brayton cycle has a two-stage compressor, with each stage having an efficiency (compared to an isentropic compressor) of 95 percent. The working fluid is argon. The compressor inlet is at 1 atm and 27°C. The compressor outlet is at 10 atm. Between the compressor stages, the argon is cooled at P = 5 atm to 100°C and then enters the second compressor stage. The single-stage turbine has an efficiency of 90 percent and an inlet temperature of 1100 K.

(a) Draw a T-s diagram for the cycle.
(b) Find the cycle efficiency (percentage).
(c) Find the cycle efficiency without intercooling (percentage).

Solution:

Problem Interpretation: (Note)

• Identify the type of the cycle and whether a control mass or control volume analysis is used: Answer
• What is the difference between an open and closed Brayton cycle? Answer
• Describe any significant characteristics of the working fluid for the cycle: Answer
• Note any assumptions that are required to solve this problem: Answer

Diagram:

Given:

Cycle: Closed Brayton Cycle Working Fluid: Argon (monatomic with constant specific heats) The given property and efficiency data are summarized in the state and process tables below. The tables for the intercooling cycle are given first followed by the cycle with intercooling, which we will refer to as the simple cycle. Note that values in green are assumed. State Table for Intercooling Cycle Mass Flow T P State (K) (atm) 1 300 1 2   5 3 373 = P3 4   10 5 1100 = P4 6   = P1 Process Table for Intercooling Cycle Process Component q w Isentropic Efficiency 1->2 Compressor 0 wIN,1-2 95% 2->3 Intercooler qOUT,2-3 0 - 3->4 Compressor 0 wIN,3-4 95% 4->5 Heat Exchanger qIN,4-5 0 - 5->6 Turbine 0 wOUT,5-6 90% 6->1 Heat Exchanger qOUT,6-1 0 - State Table for Simple Cycle Mass Flow T P State (K) (atm) 1 300 = P6 2   = P3 3 1100 10 4   1 Process Table for Simple Cycle Process Component q w Isentropic Efficiency 1->2 Compressor 0 wIN,1-2 95% 2->3 Heat Exchanger qIN,2-3 0 - 3->4 High Pressure Turbine 0 wOUT,3-4 90% 4->1 Heat Exchanger qOUT,4-1 0 - Note that states 1, 2 and 3 and the processes 1->2 and 2->3 are identical for the cycle with intercooling and the cycle without intercooling.

Find:

(a) Draw a T-s diagram for the cycle
(b) Find the cycle efficiency (percentage)
(c) Find the cycle efficiency without intercooling (percentage)

Assumptions:

• NKEPE, SSSF
• All values in the State and Process Tables in green are assumed.

Governing Relations:

The cycle efficiency is defined as

where the subscript INTERCLG is used to denote the cycle with intercooling. The required heat transfers can be calculated from the following energy balance, where each device is a control volume (CV) operating at SSSF and argon has constant specific heats:

Combining the information in the Process Table in the Given section with EB.1 yields the following for the cycle with intercooling:

Similarly, combining the information in the Process Table in the Given section with EB.1 yields the following for the simple cycle (no intercooling):

Isentropic relations can be combined with the isentropic efficiencies to fix the outlet states of the compressor and turbine. We will use the subscript S to denote the isentropic outlet state. Recall that the isentropic efficiencies are defined based on the isentropic outlet state having the same pressure as the actual outlet state.

Note that Eqn.11 can be applied to the processes 1->2 and 3->4 for the intercooling cycle and the process 1->2 for the simple cycle. Eqn.12 can be applied to process 5->6 for the intercooling cycle and process 3->4 for the simple cycle.

Property Data: Argon is a monatomic ideal gas with R = 0.208 kJ/(kg K). For all monatomic ideal gases k = 5/3 = 1.667 and c_P is constant and equal to

To draw the T-s diagrams we will calculate the change in entropy for each process using

Quantitative Solution:

a) To draw the T-s diagram we need to calculate T₂, T₄ and T₆ for the cycle with intercooling using Eqn.11 and Eqn.12:

We will assume s₁ = 0 and calculate the other entropy values relative to this value. The T and P data are substituted into Eqn.13 for each process and the results are summarized in the table below:

Plotting the T and s data,

b) The cycle efficiency with intercoolingis

c) For the cycle without intercooling, T₂ is

The cycle efficiency without intercooling is

Discussion of Results: Why is the efficiency without intercooling (36.7%) higher than that with intercooling (30.8%)? Note that (w_OUT-w_IN)_INTERCLG = 97 kJ/kg is higher than (w_OUT-w_IN)_SIMPLE = 61 kJ/kg). However, the net heat added to the cycle is also larger when intercooling is added (you can see this by comparing the T-s diagram for this problem with that without intercooling), and the ratio of |wnet|/qin is lower with intercooling. That is indeed the result of this problem. In practice, however, intercooling provides other benefits that we have not considered. The presence of intercooling reduces the temperature and therefore the specific volume of the Argon entering the second compressor stage. That stage can therefore be designed to be smaller and more efficient than a stage operating at higher inlet T, although the problem statement does not allow this to be considered. This increase in compressor efficiency and resulting reduced compressor work often outweighs the increase in heat addition necessary in the cycle. Also, the compressor can be physically smaller.