FUNDAMENTALS OF PROPULSION Course Objectives

This course covers the fundamental principles of jet propulsion systems. By the end of the course, you should be able to
1. understand the major elements of a jet engine, and the reasons for their presence
2. calculate the overall performance of a jet engine, given a few critical parameters
3. understand technological limitations on the performance of engines
4. understand the differences in design between engines intended for different applications
We will not attempt to teach you about the detailed parts or workings of any particular engine. Instead we will attempt to ensure that you have the capability to analyze any engine.

The reference book is Hill, and Peterson, "Mechanics and Thermodynamics of Propulsion".
  Course Outline The course will be taught in three subject areas, running in parallel. Each subject area will have its own tests, problem sets, and computer-based assignments. Forty contact hours are planned for each subject area, including tests. The topics in each subject area are listed below; please note that the number of hours for each topic and the synchronization of topics may vary. The course is tailored to the level of advanced seniors in mechanical engineering.
 

    P1: Introduction

In this course, we will learn how jet engines work, and to analyze the performance of any jet engine. We will do this by reducing the complex mechanisms of an engine to a few essential features. Shown above is a cut-away view of a modern turbofan engine. Many of the complex parts have been deleted. Starting from the left, we see a compressor system which does work on the air entering the engine, increasing its pressure. In this engine, the compressor is in two parts, running on two different shafts which can operate at different speeds. Next, we see pipes introducing fuel into the combustion chamber, where the fuel is mixed with air and burned. Thus, heat is added to the air. Next we see a turbine, which takes work out of the air leaving the combustion chamber, and uses it to run the compressor. Finally, the air leaving the turbine blows out at high speed through a nozzle. Upon closer examination, we see that some of the air entering the engine goes through the outer parts of the compressor, called a "fan", and then blow out of the engine directly without going through the combustion chamber or nozzle.

The net effect of the engine is that the momentum of the air leaving the engine is greater than that of the air entering the engine. The reaction to this added momentum is reflected in the thrust of the engine.

In this course, we will consider the features of the overall engine, and then of each of the components in turn. To understand the flow in the compressor, turbines, intakes, and nozzles, we will first learn some concepts in compressible fluid dynamics. To understand the flow in the combustor, we will first learn about combustion and reacting flows. These concepts will then be brought together to develop analysis procedures for complete engines, and to compute their performance over a range of flight conditions.   Thrust and Engine Design

Momentum of incoming air is (u₁) per unit time.

Momentum of outgoing air is (u₂)  per unit time.

By Newton's 2nd Law of Motion, force is equal to the rate of change of momentum. The thrust is the reaction to this force, acting in the direction opposite to the increase in momentum.

 As shown above, the thrust of a jet engine is given approximately by
 

T = (Du),    the increase in momentum of the propellant,

where is the mass of propellant flowing out of the engine per unit time, and Du =  u₂   - u₁   is  the increase in velocity of the propellant through the engine. Thus if a jet engine takes in (and expels) 100 kg/s of air, and increases its velocity from 200 m/s to 300 m/s, its thrust is approximately 10,000 Newtons (1020 Kgf, or 2248 lbf). This equation is surprisingly useful for jet engines, where the contribution from "pressure thrust" is small or zero, as we will see later, and the fuel mass flow rate is only about 1 to 2% of the air mass flow rate.

Thus, we can design engines to either

a) accelerate a small amount of air  through a large velocity increment  (examples: rockets, ramjets, turbojets), or

b) accelerate a large through a small  (examples: turbofan, turboprop, propfan )
 

Approach	Advantages	Disadvantages	Typical Applications
(a) small , large	1. small engine area & weight.   2. better high-speed performance  3. better high-altitude performance	1. less efficient conversion of thermal to kinetic energy  2. noise   3. jet blast	rockets, ramjets, high-altitude flight; turbojets, low-bypass turbofans.
(b) large , small	1.High efficiency  2. Better low-altitude performance  3. Better performance below Mach 1.  4. Low noise	1. Large engine diameter  2. More complex engine  3. Slower acceleration	Commercial aircraft: high-bypass turbofans, propfans, turbojets.

In the following sections, we see the factors which determine the maximum or "ideal" efficiency of a given design: this can be calculated from basic thermodynamics if we know a very few crucial parameters of the design, without worrying about the precise shape or complexity of a particular engine.

P2: Thermodynamics Review

Fundamental Concepts

1. Objects are at equilibrium with each other only when they have the same "degree of hotness" : concept of Temperature:

2. The impossibility of perpetual motion machines : concept of Internal Energy.

3. The impossibility of reversing any natural process in its entirety : concept of Entropy:

Closed System: A collection of matter of fixed identity. The boundaries may change, but there can be no mass transfer across them. Example: a nuclear submarine moving under the sea.

Control Volume: A region of fixed shape, location, and volume. Matter can enter and leave the boundaries. Example: the inside of a classroom, or a jet engine.

Note: The laws of thermodynamics are usually found written for closed systems. However, jet engines are more easily modeled as control volumes, since we don't usually care which particular molecules of air are inside the engine at any given time.
  First Law of Thermodynamics Q = DE + W for a closed system, where Q is the heat transferred into the system, W is the work done by, or taken out, of the system, and DE is the increase in the internal energy E of the system.

Note that this precludes the possibility of "perpetual motion" machines, whose work output is always greater than the work obtained by converting the heat put into them. At some point, there will be no more internal energy available to convert to work. Derivation of the Energy Equation for a Control Volume

We can also relate the rates of heat transfer, energy change, and work as:

Consider each of these rates for a control volume in a fluid:

where  is the mass flow rate out of the control surface through its surface, and e is the energy per unit mass of the fluid.

where  is specific volume (volume per unit mass), and  is the power taken out through shear forces and through rotating shafts.

The energy per unit mass of the fluid can be written as

where eint is internal energy, u is speed, and P.E. is potential energy per unit mass.Thus,

where the integral is taken over the surface of the control volume. Now

, the enthalpy per unit mass.

can be written as  , and  can be written as 

where  is the component of the velocity directed outward from the surface of the control volume.

Thus,

 

Special Cases

Adiabatic: q = 0

Steady:  ( ) = 0

No body forces: P.E. = 0

No work done except that of expansion:  = 0

 If all of the above are true,

so

is a constant.

Stagnation enthalpy is constant along a streamline in steady adiabatic flow, where body forces are negligible and the only work is that of expansion.
  Energy Equation for Steady One-Dimensional Flow Consider an engine represented by a control volume as shown below. Fluid enters from the left at the rate of  per unit time, and leaves from the right boundary. Work  is extracted from the fluid per unit mass flow rate, and heat  is added to the fluid per unit mass flow rate.

 

Neglect changes in potential energy of the fluid. Assume that steady-state conditions exist, so that time derivatives are zero in the frame of reference moving with the control volume. Ignore variations across the streamlines, so that integrals such as  can be replaced by . Thus, the energy equation reduces to

or

Note:
Stagnation enthalpy increases when heat is added, and decreases when work is taken out.   Second Law of Thermodynamics

For any process in a closed system,

where dQ is the heat transferred at temperature T to the system, and ds is the resulting increase in entropy of the system.

 For a control volume, the entropy inequality may be written as

 For a change in an isolated system (no energy or mass transfer),

 Reversible Process

 Rearranging terms,

  Applicable to a closed system in internal equilibrium with the only work being that due to volume change against a pressure.

Equations of State

 For a system composed of a pure substance at equilibrium, only two independent static properties need be specified to describe the thermodynamic state of the system. The relation between any two properties of such a system is called an equation of state.

 Thermal Equation of State

For a thermally perfect gas,

where p is absolute pressure, r is density, T is absolute temperature, and R is the gas constant for the particular gas.

where  is the "universal gas constant", a constant for all gases, and  is the molecular weight of the gas.  has the value of 8314 Joules per Kelvin per kilogram-mole.

 Note: At extremely high temperatures, such as those encountered in flames, and in hypersonic flight (> 2500K) nitrogen and oxygen start dissociating into single atoms, so that the effective molecular weight of air decreases.

 Example using the perfect gas relation:

Calculate the density of air at a pressure of 1 atmosphere and a temperature of 25 deg. C.

Solution:

Pressure: 1 atmosphere = 101300 Newtons per square meter

Temperature: 25 deg. Celsius = 273.15 + 25

= 298.15 deg. Kelvin

The thermal equation of state relates the pressure P, absolute temperature T, and density r of a gas:

P = r RT, where

R =  / MW,

being the Universal Gas Constant (8314.3 in SI units ), and MW the molecular weight of the gas. Air is composed of 79% Nitrogen, and 21% Oxygen. The molecular weight of Nitrogen (N2) is 28, and that of Oxygen (O2) is 32. Thus the mean molecular weight is

MW = 0.79 * 28 + 0.21*32 = 28.84

Thus, the gas constant for air, R = 8314.3 / 28.84 = 288.29 mK^-1s^-2

r = 101300 / (298.15 * 288.29) = 1.1785 kg/m³.

(A more exact representation of air at sea level is: 79% nitrogen, 20% oxygen, and 1% argon (MW=44), giving a molecular weight of 28.96. This gives R for air to be 287.04  mK^-1s^-2

It is useful to remember that in SI units, atmospheric pressure at sea-level is approximately 100,000, temperature is 300, and density is 1.2.

 
Caloric Equations of State

For a calorically perfect gas, internal energy per unit mass depends only on temperature.

e = e(T)

Enthalpy per unit mass

h = e + p/r,

so that h = h(T) as well.
 

Specific Heats

Specific heat at constant volume

    or 

Specific heat at constant pressure

  Now,

, and Define Enthalpy: h = e + p/r measure of heat conatined in a flowing fuild,    so that

At constant pressure, dp = 0, so that dq = dh,    or

Ratio of Specific Heats

         , so that 

 Define      Thus, 

Note: At low to moderate temperatures, (200 to 700K), the ratio of specific heats of diatomic gases such as nitrogen and oxygen is equal to 1.4. In the high temperature regions of engines, this value changes because the specific heats change and the gas composition is different from that of cold air. For most calculations of flows in jet engines, it is usual to assume g = 1.4 in the cooler regions, and g = 1.33 in the hot gases in the turbine and hot nozzle.
 

Variation of specific heats over a large range of temperatures

Where large changes in temperature occur, such as those encountered across a shock wave in hypersonic flow, or through the combustion chamber and turbine of a jet engine, cp, cv and g cannot be assumed to be equal to their low-temperature values.

  P3: Isentropic Flow Relations for a Perfect Gas Assuming reversible processes (no losses), and with the only work being that due to volume change against a pressure,

, so that 

Integrating from one state to another,

Define         Then 

If  is constant  = 

Isentropic Process: s₂ = s₁. Therefore,

  or 

Note: This is the relation between the pressure ratio and temperature ratio for an isentropic change from one state to another at constant specific heats.
  P5: The Brayton Cycle The operation of ramjet and gas turbine ("jet") engines can be expressed, in its most basic form, as a "Brayton Cycle" or "Gas Turbine Cycle" . There are four steps involved:

Note: Step 4 occurs in the atmosphere after the engine has passed. We don't worry about it because we don't need to re-use the same air, or pay to cool it.
  Brayton Cycle Analysis

Net work output = work done by the system in Step (3) minus the work put into the system in Step (1).

 Net heat input = heat put in during Sep (2), assuming that we don't pay to cool the air, and cannot recover the heat that is lost in the exhaust gases.

 Cycle Efficiency = (Net work output) / (Net heat input)

 Net work output per unit mass flow rate = 

Net heat input = 

 Efficiency 

Dividing throughout by specific heat ,

Now, the process from (A) to (B) is isentropic, as is (C) to (D). Also, pressure is constant from B to C, and from D to A. Thus,

and

. Thus,

 These are extremely useful results. Note:

1. As the engine "overall pressure ratio" increases, efficiency increases towards 1. This means that to get high efficiency, we must use a high pressure ratio.

 2. Efficiency = (heat added - heat wasted) / (heat added)

 Thus, to increase efficiency,

a). Increase Tc, the highest temperature in the engine

b). Bring TD down as close to TA as possible: expand through the nozzle, and extract the maximum work possible.

c). Reduce TB: take out heat from the compressor using heat transfer to the fuel (this is considered in supersonic-combustion engines which use cryogenic fuel).

 3. If TD <TA, you get efficiency > 1: a perpetual motion machine. Thus, we see that in reality, the exhaust temperature cannot be lower than the ambient temperature.