# Thermodynamic Cycles

This is a topic from Thermodynamics:

## Introduction

A basic understanding of various forms of thermodynamic cycles is necessary for solving thermodynamic problems of a practical nature. The cycles covered in these notes can be broken down into either air-standard cycles (where the ideal gas laws apply) or vapour cycles (where the pure substance curves must be considered and the steam tables must be used).The Carnot Cycle is covered in the previous section of the notes.

## The Compression Ratio

For certain piston cycles, where there is a swept and clearance volume, the compression ratio is defined as:

rv = (Vswept + Vclearance) / Vclearance

## Air-standard Cycles

### The "Otto" Cycle

The cycle is as follows:

4→1: Constant volume heat rejection

The Otto cycle on a PV diagram: Analysis:

• The volume in states 2 and 3 is the 'clearance' volume, such that the piston is almost at the top of the cylinder
• The volume in states 1 and 4 is the 'swept' volume , such that the piston is as far down the cylinder as it reaches during the cycle
• The work done for the isochoric processes (2→3 and 4→1) is equal to zero, and so by the first law: QH = m(u3/1 - u2/4) = mCV(T3/1 - T2/4)
• Recalling the definition of engine efficiency:
ηth = Wnet / Qin = (QH - QL) / QH,
ηth = (mCV(T3 - T2) - mCV(T4 - T1))/mCV(T3 - T2) = ((T3 - T2) - (T4 - T1)) / (T3 - T2)
• And so:
ηth = 1 - ((T4 - T1) / (T3 - T2)
T2 / T1 = (v1/v2)k-1 and T3 / T4 = (v2/v3)k-1, and so
T2 / T1 = T3 / T4
• The compression ratio here would be given by:
rv = v1/v2
• And thus the expression for efficiency can be rewritten:
ηth = 1 - T1 / T2 = 1 - rv1-k
...we can see therefore, that increasing the compression ratio is best for efficiency.

### The Diesel Engine Cycle

The cycle is as follows:

4→1: Constant volume heat rejection

Note that the second process is isobaric rather than isochoric.

The Diesel cycle on a PV diagram: Analysis:

• The 2→3 process is isobaric, and so applying a common rearrangement of the first law: QH = mCP(T3 - T2)
• The 4→1 process is isochoric, and so there is no work, and so applying the first law: QL = ΔU41 = mCV(T4 - T1)
• Recalling the definition of engine efficiency:
ηth = Wnet / Qin = (QH - QL) / QH,
ηth = (mCP(T3 - T2) - mCV(T4 - T1))/mCP(T3 - T2)
= 1 - (T4 - T1) / k(T3 - T2)
• The cutoff ratio, rc is defined as v3/v2
• Meanwhile for the constant pressure process: v3/v2 = T3/T2
• For the isentropic compression process: T2/T1 = (v1/v2)k-1 = rvk-1
• Therefore:
ηth = 1 - (T4 - T2/rv1-k) / k(rcT2 - T2)
ηth = 1 - (T4/T2 - 1/rv1-k) / k(rc - 1)
• And thus the expression for efficiency can be rewritten:
ηth = 1 - rv1-k(rck - 1) / k(rc - 1)

### The Gas Turbine Cycle (Brayton Cycle)

The gas turbine cycle applies to airplane propulsion and involves the generation of thrust through the expulsion of air. Air is compressed as it enters the turbine, it is heated during combustion, is expands through the entrance the turbine, losing heat within the turbine itself. The cycle is as follows:

4→1: Constant pressure heat rejection

Note that both heat processes are isobaric rather than isochoric.

The Diesel cycle on a PV diagram: Analysis:

• Open system analysis is necessary in this case, with the first law for an open system simplifying to: q - w = hout - hin, where the q and w refer to specific heat and specific work flow rates.
• Thus during the constant pressure heat addition: qH = q23 = h3 - h4 = CP(T3 - T2)
• And during the constant pressure heat rejection: qL = q41 = h4 - h1 = CP(T4 - T1)
• Recalling the definition of engine efficiency:
ηth = Wnet / Qin = (QH - QL) / QH,
ηth = (CP(T3 - T2) - CP(T4 - T1))/mCP(T3 - T2)
= 1 - (T4 - T1) / (T3 - T2)
• The pressure ratio, rp is defined as P2/P1
• For the isentropic compression process: T2/T1 = (P2/P1)(k-1)/k = rp(k-1)/k
• Also: T3/T4 = (P3/P4)(k-1)/k = rp(k-1)/k
• Therefore combining the two gives: T2/T1 = T3/T4
• Substituting into our efficiency equation:
ηth = 1 - (T1T3/T2 - T1) / (T3 - T2)
ηth = 1 - T1(T3/T2 - 1) / T2(T3/T2 - 1)
• And thus the expression for efficiency can be rewritten:
ηth = 1 - rpk/(k-1)

### Other Cycles

The following cycles are more difficult to analyse, and so for this course you only need to be somewhat familiar with them.

#### Dual Cycle

The dual cycle combines elements of the Otto and Diesel cycles and proceeds as follows: #### The Atkinson Cycle

The Atkinson cycle can be considered to be the inverse of the Diesel cycle and proceeds as follows: #### The Stirling Cycle

The Stirling cycle is like a cross between the Carnot cycle and the Otto cycle and proceeds as follows: #### The Compression Refrigeration Cycle

For the refrigeration cycle work is put into the system to push heat out. Curve 12 is typically an isentrope (the compressor has 100% efficiency)/ ## Cycle PV Diagram Summary

• The Carnot cycle has 4 curves, 2 isothermal and 2 adiabatic isentropic curves, while the other cycles covered in this section have 2 curves, both adiabatic and isentropic and 2 or more straight lines. The heat transfers occur during the isothermal processes for the Carnot cycle and during the straight line processes for the other cycles.
• The Otto cycle PV diagram has two vertical lines (isochoric processes), The Gas Turbine cycle PV diagram is the opposite, with two horizontal lines (isobaric processes).
• The Gas Turbine has a step 3' between 3 and 4 to separate the output work used to drive the compressor (33') from the useful output work (3'4).
• The Diesel cycle PV diagram is a cross between the Otto and the Gas Turbine, with one vertical line in the bottom right, and one horizontal line in the top left. The Atkinson cycle is the opposite of the Diesel cycle, with one vertical line in the top left, and one horizontal line in the bottom right.
• The Dual cycle PV diagram has the a right angled corner in the top left, and a vertical line (like the Otto cycle) in the bottom right. The heat addition occurs equally during both of the straight line processes making up the right angle.
• The Stirling cycle PV diagram has the two vertical lines of the Otto cycle, and the two isotherms of the Carnot cycle. There are no isentropic or adiabatic processes.