Example 9.6.3

Problem Statement: A closed Brayton cycle has a two-stage turbine. The high-pressure (first) stage has inlet T = 1100 K at 10 atm and an efficiency of 90 percent. Argon is the working fluid. After leaving the first stage at 5 atm, the argon is reheated to 1100 K and enters the second stage, which also has an efficiency of 90 percent. The exhaust from the turbine is at 1 atm. The compressor in the cycle has an efficiency of 95 percent, and the compressor inlet temperature is 27°C.

(a) Draw a T-s diagram for the cycle
(b) Find the cycle efficiency (percentage)
(c) Find the cycle efficiency without reheat (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 reheat cycle are given first followed by the cycle with reheat, which we will refer to as the simple cycle. Note that values in green are assumed. State Table for Reheat Cycle Mass Flow T P State (K) (atm) 1 300 = P6 2   = P3 3 1100 10 4   5 5 1100 = P4 6   1 Process Table for Reheat 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->5 Reheater qIN,4-5 0 - 5->6 High Pressure 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,6-1 0 - Note that states 1, 2 and 3 and the processes 1->2 and 2->3 are identical for the cycle with reheat and the cycle without reheat.

Find:

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

Assumptions:

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

Governing Relations:

The cycle efficiency is defined as

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:

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.2 can be applied to the processes from 3->4 for both the reheat and simple cycles and from 5->6 for the reheat 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 reheat using Eqn.1 and Eqn.2:

We will assume s₁ = 0 and calculate the other entropy values relative to this value. The T and P data are substituted into Eqn.3 for each process and the results are summarized in the table below:

Plotting the T and s data,

b) The cycle efficiency with reheat is

c) For the cycle without reheat, T₄ is

The cycle efficiency without reheat is

Discussion of Results: Reheating normally increases overall cycle efficiency. Usually, a two-stage turbine with reheat will have a slightly lower isentropic efficiency in the low-pressure (second) stage than will a two-stage turbine with no reheat, because with reheat the mass passing through the second stage will have higher specific volume. This causes higher velocities and greater frictional losses. We have ignored this effect in the problem by using ideal relationships in the cycle analysis, and by assuming all turbine stages have the same efficiency in all cases.