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Gas Turbines Summary
1. Basics of gas turbines
In this first chapter, we’re going to look at the basics of gas turbines. What are they? When were they developed? How do they work? And how do we perform basic calculations?
1.1 Introduction to gas turbines
This is where our journey into the world of gas turbines takes off. And we start with a very important question: what is a gas turbine?
1.1.1 What is a gas turbine?
A gas turbine is a machine delivering mechanical power or thrust. It does this using a gaseous working fluid. The mechanical power generated can be used by, for example, an industrial device. The outgoing gaseous fluid can be used to generate thrust.
In the gas turbine, there is a continuous flow of the working fluid. This working fluid is initially compressed in the compressor. It is then heated in the combustion chamber. Finally, it goes through the turbine. The turbine converts the energy of the gas into mechanical work. Part of this work is used to drive the compressor. The remaining part is known as the net work of the gas turbine.
1.1.2 History of gas turbines
We can distinguish two important types of gas turbines. There are industrial gas turbines and there are jet engine gas turbines. Both types of gas turbines have a short but interesting history.
Industrial gas turbines were developed rather slowly. This was because, to use a gas turbine, a high initial compression is necessary. This rather troubled early engineers. Due to this, the first working gas turbine was only made in 1905 by the Frenchman Rateau. The first gas turbine for power generation became operational in 1939 in Switzerland. It was developed by the company Brown Boveri.
Back then, gas turbines had a rather low thermal efficiency. But they were still useful. This was because they could start up rather quickly. They were therefore used to provide power at peak loads in the electricity network. In the 1980’s, natural gas made its breakthrough as fuel. Since then, gas turbines have increased in popularity.
The first time a gas turbine was considered as a jet engine, was in 1929 by the Englishman Frank Whittle. However, he had trouble finding funds. The first actual jet aircraft was build by the German Von Ohain in 1939. After world war 2, the gas turbine developed rapidly. New high-temperature materials, new cooling techniques and research in aerodynamics strongly improved the efficiency of the jet engine. It therefore soon became the primary choice for many applications.
Currently, there are several companies producing gas turbines. The biggest producer of both industrial gas turbines and jet engines is General Electric (GE) from the USA. Rolls Royce and Pratt & Whitney are also important manufacturers of jet engines.
1.1.3 Gas turbine topics
When designing a gas turbine, you need to be schooled in various topics. To define the compressor and the turbine, you need to use aerodynamics. To get an efficient combustion, knowledge on thermody- namics is required. And finally, to make sure the engine survives the big temperature differences and high forces, you must be familiar with material sciences.
Gas turbines come in various sizes and types. Which kind of gas turbine to use depends on a lot of criteria. These criteria include the required power output, the bounds on the volume and weight of the turbine, the operating profile, the fuel type, and many more. Industrial gas turbines can deliver a power from 200kW to 240M W. Similarly, jet engines can deliver thrust from 40N to 400kN.
1.2 The ideal gas turbine cycle
To start examing a gas turbine in detail, we make a few simplifications. By doing this, we wind up with the ideal gas turbine. How do we analyze such a turbine?
1.2.1 Examining the cycle
Let’s examine the thermodynamic process in an ideal gas turbine. The cycle that is present is known as the Joule-Brayton cycle. This cycle consists of five important points.
We start at position 1. After the gas has passed through the inlet, we are at position 2. The inlet doesn’t do much special, so T 1 = T 2 and p 1 = p 2. (Since the properties in points 1 and 2 are equal, point 1 is usually ignored.) The gas then passes through the compressor. We assume that the compression is performed isentropically. So, s 2 = s 3. The gas is then heated in the combustor. (Point 4.) This is done isobarically (at constant pressure). So, p 3 = p 4. Finally, the gas is expanded in the turbine. (Point 5.) This is again done isentropically. So, s 4 = s 5. The whole process is visualized in the enthalpy-entropy diagram shown in figure 1.1.
Figure 1.1: The enthalpy-entropy diagram for an ideal cycle.
We can make a distinction between open and closed cycles. In an open cycle, atmospheric air is used. The exhaust gas is released back into the atmosphere. This implies that p 5 = p 1 = patm. In a closed cycle, the same working fluid is circulated through the engine. After point 5, it passes through a cooler, before it again arrives at point 1. Since the cooling is performed isobarically, we again have p 5 = p 1.
When examining the gas turbine cycle, we do make a few assumptions. We assume that the working fluid is a perfect gas with constant specific heats cp and cv. Also, the specific heat ratio k (sometimes also denoted by γ) is constant. We also assume that the kinetic/potential energy of the working fluid
1.2.5 The optimum pressure ratio
Let’s suppose we don’t want a maximum efficiency. Instead, we want to maximize the power Ws,gg. The corresponding pressure ratio ε is called the optimum pressure ratio εopt. To find it, we can set dWs,gg /dε = 0. From this, we can derive that, at these optimum conditions, we have
T 3 = T 5 =
T 2 T 4. (1.2.7)
It also follows that the optimum pressure ratio itself is given by
εopt =
T 3
T 2
) (^) kk− 1
T 4
T 2
) (^) 2(kk−1)
. (1.2.8)
The corresponding values of Ws,gg and η are given by
Ws,ggmax cpT 2
T 4
T 2
and η = 1 −
T 2
T 4
Note that we have used a non-dimensional version of the specific gas power Ws,gg.
1.3 Enhancing the cycle
There are various tricks, which we can use to enhance the gas turbine cycle. We will examine the three most important ones.
1.3.1 Heat exchange
The first enhancement we look at the heat exchanger (also known as a recuperator). Let’s suppose we’re applying a heat exchanger. After the gas exits the turbine (point 5), it is brought to this heat exchanger. The heat of the exhaust gas is then used, to warm up the gas entering the combustor (point 3). Let’s call the point between the heat exchanger and the combustion chamber point 3.5.
It is important to note that a heat exchanger can only be used if T 5 > T 3. (Heat only flows from warmer to colder gasses.) And we only have T 5 > T 3 if ε < εopt. So a heat exchanger is nice for turbines with low pressure ratios.
Now let’s examine the effects of a heat exchanger. The gas entering the combustor (at point 3.5) is already heated up a bit. So the combustor needs to add less heat. The heat input Qs, 3 − 4 is thus reduced, increasing the efficiency. In an ideal case, we have T 3. 5 = T 5. In this case, the heat input is given by
Qs, 3 − 4 = cp (T 4 − T 3. 5 ) = cp (T 4 − T 5 ) = cpT 4
ε
k− 1 k
The thermodynamic efficiency of the cycle is now given by
ηth =
Ws,gg Qs, 3 − 4
cpT 4
ε k− k 1
− cpT 2
ε k− k 1 − 1
cpT 4
ε k− k 1
T 2
T 4
ε
k− k 1
. (1.3.2)
In this case, the efficiency increases for decreasing temperature ratios. This is contrary to the case without heat exchange. There is a simple reason for this: The lower the pressure ratio, the more heat can be exchanged, and the more the efficiency can be improved by this heat exchange.
1.3.2 Intercooling
We can also try to increase the specific gas power Ws,gg. One way to do this, is by making sure the compressor requires less power. This is where the intercooler comes in.
Let’s suppose we’re using an intercooler in the compressor. First, the compressor compresses the gas a bit (point 2.3). The applied pressure ratio is ε 1 = p 2. 3 /p 2. Then, the gas is cooled by the intercooler (point 2.5). (This is usually done such that T 2. 5 = T 2. Intercooling is also performed isobarically, so p 2. 3 = p 2. 5 .) Finally, the gas is compressed again, until we’re at point 3. The applied pressure ratio during this step is ε 2 = p 3 /p 2. 5. The total applied pressure ratio is εtot = ε 1 ε 2 = p 3 /p 2. The entire process of intercooling has been visualized in figure 1.2 (left).
Figure 1.2: The enthalpy-entropy diagram for intercooling (left) and reheating (right).
Thanks to the intercooler, the specific work needed by the compressor has been reduced to
Ws, 2 − 3 = Ws, 2 − 2. 3 + Ws, 2. 5 − 3 = cp (T 2. 3 − T 2 ) + cp (T 3 − T 2. 5 ). (1.3.3)
The specific gas power Ws,gg has increased by the same amount. The increase in specific gas power strongly depends on the pressure ratios ε 1 and ε 2. It can be shown that, to maximize Ws,gg , we should have ε 1 = ε 2 =
εtot. (1.3.4)
Intercooling may look promising, but it does have one big disadvantage. The combustor needs to do more work. It also needs to provide the heat that was taken by the intercooler. (This is because T 3 has decreased.) So, while Ws,gg increases by a bit, Qs, 3 − 4 increases by quite a bit more. Intercooling therefore increases the work output, at the cost of efficiency.
1.3.3 Reheating
An idea similar to intercooling is reheating. However, reheating is applied in the turbine. Reheating increases the work done by the turbine. A well-known example of applying reheating is the afterburner in an aircraft.
From point 4, we start by expanding the gas in the turbine. We soon reach point 4.3. (Again, we define ε 1 = p 4 /p 4. 3 .) The gas is then reheated, until point 4.5 is reached. (This is done isobarically, so p 4. 3 = p 4. 5 .) From this point, the gas is again expanded in a turbine, until points g and 5 are reached. We now have ε 2 = p 4. 5 /p 5 and εtot = ε 1 ε 2 = p 4 /p 5. The entire process of reheating has been visualized in figure 1.2 (right).
As was said before, reheating increases the work done by the turbine. Let’s suppose that T 4 = T 4. 5. In this case, the increase in turbine work is at a maximum if ε 1 = ε 2 =
εtot, just like in the case of the intercooler. Reheating also has the same downside as intercooling. Although the work increases, more heat needs to be added. So the efficiency decreases.
2.2.1 Isentropic efficiency
Let’s examine the compressor and the turbine. In reality, they don’t perform their work isentropically. To see what does happen, we examine the compressor. In the compressor, the gas is compressed. In an ideal (isentropic) case, the enthalpy would rise from h 02 to h 03 s. However, in reality, it rises from h 02 to h 03 , which is a bigger increase. Similarly, in an ideal (isentropic) turbine, the enthalpy would decrease from h 04 to h 0 gs. However, in reality, it decreases from h 04 to h 0 g , which is a smaller decrease. Both these changes are visualized in figure 2.1.
Figure 2.1: The enthalpy-entropy diagram for a non-ideal cycle.
This effect can be expressed in the isentropic efficiency. The efficiencies for compression and expansion are, respectively, given by
ηis,c =
h 03 s − h 02 h 03 − h 02
T 03 s − T 02 T 03 − T 02
and ηis,t =
h 04 − h 0 g h 04 − h 0 gs
T 04 s − T 0 g T 04 − T 0 gs
You may have trouble remembering which difference goes on top of the fraction, and which one goes below. In that case, just remember that we always have ηis ≤ 1.
By using the isentropic relations, we can rewrite the above equations. We then find that
ηis,c =
p 03 p 02
) kair k^ −^1 air − 1
T 03 T 02 − 1
and ηis,t =
T 0 g T 04 − 1 ( p 0 g p 04
) kgas kgas−^1 − 1
The specific work received by the compressor, and delivered by the turbine, is now given by
W˙s,c = cpair^ T^02 ηis,c
p 03 p 02
) kair k −^1 air − 1
(^) and W˙s,t = cpgas T 04 ηis,t
p 0 g p 04
) kgas kgas−^1 − 1
Note that, through the definition of p 0 g , these two quantities must be equal to each other. (That is, as long as the mass flow doesn’t change, and there are no additional losses when transmitting the work.)
2.2.2 Polytropic efficiency
Let’s examine a compressor with a varying pressure ratio. In this case, it turns out that also the isentropic efficiency varies. That makes it difficult to work with.
To solve this problem, we divide the compression into an infinite number of small steps. All these infinitely small steps have the same isentropic efficiency. This efficiency is known as the polytropic efficiency.
The resulting process is also known as a polytropic process. This means that there is a polytropic exponent n, satisfying
T 0 T (^0) initial
p 0 p (^0) initial
) (^) nnair air −^1
. (2.2.4)
The polytropic efficiencies for compression η∞c and expansion η∞t are now given by, respectively,
η∞c =
kair − 1 kair
nair nair − 1
ln
p 03 p 02
) kair k^ −^1 air
ln
T 03 T 02
) (^) and η∞t = kgas kgas − 1
ngas − 1 ngas
ln
T 0 g T 04
ln
p 0 g p 04
) kgas kgas−^1.
The full compression/expansion process also has an isentropic efficiency. It is different from the polytropic efficiency. In fact, the relation between the two is given by
ηis,c =
p 03 p 02
) kair k −^1 air − 1
( p 03 p 02
) (^) ηk∞airck −^1 air − 1
and ηis,t =
p 0 g p 04
)η∞tkgas kgas−^1 − 1 ( p 0 g p 04
) kgas kgas−^1 − 1
There are a few important rules to remember. For compression, the polytropic efficiency is higher than the isentropic efficiency. (So, η∞c > ηis,c.) For expansion, the polytropic efficiency is lower than the isentropic efficiency. (So, η∞t < ηis,t.) Finally, if the pressure ratio increases, then the difference between the two efficiencies increases.
2.3 Losses occurring in the gas turbine
In a non-ideal world, losses occur at several places in the gas turbine. There are also several types of losses. We will examine a few.
2.3.1 Pressure losses
Previously, we have assumed that no pressure losses occurred. This is, of course, not true. Pressure losses occur at several places. First of all, in the combustor. The combustion chamber pressure loss is given by ∆pcc = p 03 − p 04. The combustor pressure loss factor is now defined as
εcc =
p 04 p 03
p 03 − ∆pcc p 03
Pressure losses also occur at the inlet and at the exhaust duct. For industrial gas turbines, these pressure losses are defined as
∆p (^0) inlet = pamb − p 01 and ∆p (^0) exhaust = p 05 − pamb. (2.3.2)
We will examine these pressure differences for jet engines in a later chapter.
2.3.2 Mechanical losses
Losses also occur due to internal friction in the system. These mechanical losses are joined together in one term, being the transmission efficiency ηm. It is given by
ηm =
turbine power − mechanical losses turbine power
- Gas turbine types
Gas turbines can be used for many purposes. They can be used to deliver power, heat or thrust. Therefore, different gas turbine types exist. In this chapter, we will examine a few. First, we examine shaft power gas turbines. Then, we examine jet engine gas turbines.
3.1 Shaft power gas turbines
A shaft power gas turbine is a gas turbine whose goal is mainly to deliver shaft power. They are often also referred to as turboshaft engines. These gas turbines are often used in industrial applications. Gas turbines used for electricity production are also of this type.
3.1.1 The shaft power
For shaft power gas turbines, the shaft power is very important. It mainly depends on the temperature drop in the power turbine (PT). This drop is given by
T 0 g − T 05 = T 0 g ηis,P T
p 05 p 0 g
) kgas kgas−^1
(^) = T 0 g
p 05 p 0 g
)η∞,P Tkgas kgas−^1
The efficiencies ηis,P T and η∞,P T are the isentropic and polytropic efficiencies, respectively, of the power turbine. To apply the above equation, we do need to know the properties of point 5. These properties can be derived from the exhaust properties, according to
∆pexh = p 05 − pexh and T 05 = T 0 ,exh = Texh +
c^2 exh 2 cp,gas
where cexh is the velocity of the exhaust gas. (By the way, the exhaust point is often also denoted by point 9.) Once this data is known, the actual shaft power can be derived. For this, we use the equation
Pshaf t = ˙m cp,gas (T 0 g − T 05 ) ηm,P T , (3.1.3)
with ηm,P T being the mechanical efficiency of the gas turbine.
3.1.2 Other performance parameters
There are also some other parameters that are important for shaft power gas turbines. Of course, the thermal efficiency ηthermal is important. (The thermal efficiency is not the thermodynamic efficiency ηth, introduced in chapter 1. In fact, it is lower. This is because the thermal efficiency also takes into account various types of losses.) This efficiency is given by
ηthermal = Pshaf t m ˙f uel Hf uel
where Hf uel is the heating value of the fuel. Other important parameters are the specific fuel con- sumption sf c and the heat rate. The sf c is given by
sf c = m˙f uel Pshaf t
Hf uel ηthermal
The heat rate is given by
heat rate =
m˙f uel Hf uel Pshaf t
ηthermal
Finally, there is the equivalence ratio λ, also known as the percentage excess air. It is defined as
λ = m˙air − m˙st m ˙st
where m˙st is the air mass flow required for a complete combustion of the fuel. If there is just enough air to burn the fuel, then we have a stoichiometric combustion. In this case, λ = 1. However, usually λ is bigger than 1.
3.1.3 Exhaust gas of the shaft power turbine
Shaft power gas turbines usually produce quite some heat. This heat can be used. When applying cogeneration, we use the heat of the exhaust gas itself. (For example, to produce hot water or steam.) In a combined cycle the heat is used to create additional power. This can be done by expanding the heat in a steam turbine.
The process of creating steam deserves some attention. In this process, heat is exchanged from the exhaust gas to the water/vapor. This is caused by the temperature difference between the exhaust gas and the water/vapor. An important parameter is the temperature difference ∆Tpinch at the so-called pinch point. This is the point where the water just starts to boil. The entire process, including the pinch point, is also visualized in figure 3.1.
Figure 3.1: Exchanged heat versus temperature diagram for creating steam, using exhaust gas.
We want to minimize the exhaust losses. One way to do that, is to increase the initial exhaust gas temperature. This makes the heat exchange a lot easier. Another option is to split up the process into multiple steps, where each step has a different pressure. In this case, there will also be multiple pinch points.
3.2 Jet engine gas turbines
A jet engine gas turbine is a turbine whose goal is mainly to deliver thrust. This can be done in two ways. We can let the gas turbine accelerate air. We can also let the gas turbine shaft power a propeller. Often, both methods are used. Jet engine gas turbines are mainly used on aircraft.
3.2.1 Finding the thrust
The goal of a jet engine is to produce thrust. The net thrust FN can generally be found using
FN = ˙m (cj − c 0 ). (3.2.1)
This is a measure of how well the energy from the power turbine has been used to accelerate air. And now, we can finally define the total efficiency. It is given by
ηtotal = Pthrust m ˙f uel Hf uel
= ηpropηthermal. (3.2.11)
3.2.4 Other important parameters
Next to the efficiencies, there are also quite some other parameters that say something about jet engines. First, there is the specific thrust Fs. It is defined as the ratio between the net thrust and the air intake. So,
Fs =
FN
m ˙
Second, we have the thrust specific fuel consumption T SF C, which is the ratio between the fuel flow and the thrust. We thus have
T SF C =
m˙f uel FN
c 0 ηtotal Hf uel
3.2.5 Improving the jet engine
We have previously noted that it’s best to give a small velocity increment to a large amount of air. However, jet engines theirselves usually give quite a big acceleration to the air. For this reason, most commercial aircraft engines have big propellers, called turbofans. These fans are driven by the shaft of the jet engine. And they give, as iss required, a small velocity increment to a large amount of air. This air then bypasses (flows around) the jet engine itself.
The relation for the thrust of a turbofan engine is similar to that of a normal jet engine. The only difference, is that we need to add things up. So,
FN = ˙mjet (c 8 − c 0 ) + Ajet (p 8 − p 0 ) + ˙mf an (c 8 − c 0 ) + Af an (p 8 − p 0 ). (3.2.14)
The rest of the equations also change in a similar way.
- Combustion
The combustion chamber is the part where energy is inserted into the gas turbine. In this chapter, we’re going to examine it in detail.
4.1 The combustion process
First, we will look at the combustion process. What reaction is taking place? And what parameters influence this reaction?
4.1.1 The combustion process
The combustor provides the energy input for the gas turbine cycle. It receives air, inserts fuel, mixes the two components and then it lets the mixture combust. This process is known as internal combustion. In a gas turbine, it is generally done at constant pressure. (Although small pressure losses are generally present.)
An important parameter during combustion is the T 04 temperature. It effects the power output and the thermal efficiency. T 04 is generally limited by material properties. The materials must be able to withstand large temperatures and temperature gradients. If not, the gas turbine might fail.
4.1.2 The reaction
Gas turbines mainly use kerosene-type fuels. These fuels consist of hydrocarbons with the chemical composition CxHy. They are mixed with air and then combusted. The ideal reaction is then given by
CxHy + ε (XO 2 O 2 + XN 2 N 2 + XCO 2 CO 2 + XAr Ar ) → nCO 2 CO 2 + nH 2 O H 2 O + nN 2 N 2 + nAr Ar. (4.1.1)
In this equation, the X’s indicate the composition of air. We thus have
XO 2 = 0. 2095 , XN 2 = 0. 7808 , XCO 2 = 0. 0003 and XAr = 0. 0094. (4.1.2)
The term ε is the number of moles of air necessary for every mole of fuel. It is given by
ε =
x + 14 y XO 2
Finally, the n′s denote the amount of reaction products. We find that we have
nCO 2 = x + εXCO 2 , nH 2 O =
y, nN 2 = εXN 2 and nAr = εXAr. (4.1.4)
In reality, we don’t have this ideal reaction. In the real world, not all fuel gets combusted. Not all carbon atoms form carbon dioxide CO 2. (We will also have carbon monoxide CO.) And there will be various other reaction products as well. We won’t examine all those reaction products though.
4.1.3 Reaction parameters
When combusting the mixture, the fuel-to-air ratio F AR = m˙f uel/ m˙air is an important parameter. We know that, to combust 1 mole of fuel, we need ε moles of air. In stoichiometric conditions, there is precisely enough air to combust all fuel. The stoichiometric fuel-to-air ratio is therefore given by
F ARst =
m˙f uel m ˙air
st
ε
MCxHy Mair
XO 2
x + 14 y
MCxHy Mair
4.2 The layout of the combustion chamber
Designing a combustion chamber is not an easy thing. There are several complicated parts in it. So let’s examine how a combustion chamber is build up.
4.2.1 The general layout
There are several types of combustion chambers. But virtually all combustion chambers have a diffuser, a casing, a liner, a fuel injector and a cooling arrangement. The entire general layout is visualized in figure 4.1.
Figure 4.1: The layout of the combustion chamber.
4.2.2 The diffuser
The gas entering the combustion chamber usually has quite a high velocity. This velocity will be respon- sible for a pressure drop. (This pressure drop is called the cold loss.) Also, the flame in the combustion chamber can not survive if the air has a high velocity. Therefore, the airflow needs to be slowed down. And this is exactly the task of the diffuser.
Diffusers can be distinguished by how quickly they decrease the velocity. In the aerodynamic diffuser, the flow is slowed down gradually, while in the dump diffuser, it is slowed down quickly. The dump diffuser has more losses, but is also smaller. It is therefore mainly used in aircraft jet engines.
4.2.3 The casing and the liner
After the airflow has passed the diffuser, it is split up by the liner. One part of the airflow goes through the region between the liner and the casing. This region is called the annulus. Another part of the airflow enters the mixing chamber, where fuel is injected.
There are several reasons for splitting up the flow. First, the air-to-fuel should have the right value. If it is too high, the mixture will not ignite. Also, the velocity of the flow leaving the diffuser is still too high. The part of the flow that will be ignited has to be slowed down even further.
The liner is divided into three sections. There is a primary zone (PZ), a secondary/intermediate zone (SZ/IZ) and a tertiary/dilution zone (TZ/DZ). The main function of the PZ is to provide enough time for the fuel to mix and combust. The goal of the SZ is to provide enough time to achieve full combustion. This significantly reduces bad reaction products like carbon monoxide CO and unburned
hydrocarbons (UHC). Finally, the goal of the DZ is to reduce the temperature of the outlet stream, such that it is acceptable for the turbine.
4.2.4 The fuel injector
The fuel injector injects the fuel into the flow. It is important that the fuel is vaporized before it enters the flame zone. Otherwise, it might not combust properly.
To promote vaporization, the fuel should be atomized. This means that the fuel is converted into small drops. This increases vaporization rates. To accomplish this, an atomizer is used. To atomize fuel, it has to be given a high relative velocity, with respect to the airflow. So-called pressure-assist atomizers give the fuel a high velocity. On the other hand, air blast atomizers inject slow-moving fuel into a high-velocity air stream.
4.2.5 Flame stabilization
After the fuel has been injected into the flow, the flow will enter the flame region. It does this with quite a high velocity. To make sure the flame isn’t blown away, flow reversal can be applied in the PZ. This causes the flow to reverse direction. The best way to reverse the flow, is to swirl it. This is done using swirlers. The two most important types of swirlers are axial swirlers and radial swirlers.
The most important advantage of flow reversal, is that the flow speed varies a lot. So there will be a point at which the airflow velocity matches the flame speed. (The flame speed is the speed, relative to the airflow, at which the flame can move.) This is the point where the flame anchors.
4.2.6 Cooling
The liner is exposed to high temperatures, during combustion. It therefore needs to be cooled, using the airflow in the annulus. There are several ways to do this.
In film cooling, stacks of holes are put in the liner. These holes inject air along the inner surface of the liner, prodiving a protective cooling film. A downside is that the liner is not cooled evenly. The best way to cool the liner evenly, is by using transpiration cooling. Now, the liner wall is constructed from a porous material that allows air to pass through it.
Next to these cooling methods, we can also put metallic tiles onto the inner surface of the liner. This provides some protection. Finally, the airflow passing through the annulus will also automatically provide some convection cooling.
4.2.7 Combustor types
We can distinguish combustors by their shape. Can-type combustors (also known as tubular com- bustors) consist of several cylindrical tubes, placed around the turbine shaft. Each tube has both a liner and a casing. Although these types of turbines are easy to develop, their weight is relatively high.
Another type of combustor is the annular combustor. The annular combustor has a single ring-shaped liner mounted inside a single ring-shaped casing. Due to the low weight, this type is mostly used in aircraft jet engines. A downside is that it’s sensitive to buckling loads.
Finally, we can combine the can-type and the annular combustor. We then get a can-annular combus- tor. We now have several cylindrical liners (like in the tubular combustor) placed in a single ring-shaped casing (like in an annular combustor).
hydrocarbons U HCs. The amount of gaseous pollutants is usually given by the emission index (EI). This is defined as
EI =
mass of the produced pollutant in g mass of fuel used in kg
The second group of remaining combustion products is called smoke. It mainly consists of soot parti- cles, which are particles with a high amount of carbon in them.
- Compressor and turbines
In this chapter, we will look at the compressor and the turbine. They are both turbomachinery: machines that transfer energy from a rotor to a fluid, or the other way around. The working principle of the compressor and the turbine is therefore quite similar.
5.1 Axial compressors
We will mainly look at axial compressors. This is because they are the most used type of compressors. Also, axial compressors work very similar to axial turbines.
5.1.1 Euler’s equation for turbomachinery
Let’s examine a rotor, rotating at a constant angular velocity ω. The initial radius of the rotor is r 1 , while the final radius is r 2. A gas passes through the rotor with a constant velocity c. The rotor causes a moment M on the gas. The power needed by the rotor is thus P = M ω.
It would be nice if we can find an expression for this moment M. For that, we first look at the force F acting on the gas. It is given by
F =
d(mc) dt
= ˙mc, (5.1.1)
where we have used the assumption that c stays constant. Only the tangential component Fu contributes to the moment. Every bit of gas contributes to this tangential force. It does this according to
dFu = ˙mdcu, (5.1.2)
where cu is the tangential velocity of the air. Let’s integrate over the entire rotor. We then find that
M =
1
dM =
1
r dFu = ˙m
1
r dcu = ˙m (cu, 2 r 2 − cu, 1 r 1 ). (5.1.3)
The power is now given by
P = M ω = ˙m (cu, 2 r 2 − cu, 1 r 1 ) ω = ˙m (cu, 2 u 2 − cu, 1 u 1 ). (5.1.4)
In this equation, u denotes the speed of the rotor at a certain radius r. We have also used the fact that ω = u 1 /r 1 = u 2 /r 2. The above equation is known as Euler’s equation for turbomachinery.
5.1.2 The axial compressor power
Let’s examine an axial compressor. This compressor has rotors and stators. The rotors are moving. They increase the kinetic energy of the gas. The stators are not moving. They are used to turn the kinetic energy into an increase in pressure.
Let’s examine the velocities of the gas, as it passes through a rotor and a stator. The situation is shown in figure 5.1. At the point we’re examining, the rotor is moving with a velocity u. The velocity of the gas, relative to the rotor, is denoted by v. The angle between the flow velocity c and the shaft axis is denoted by α. The angle between the rotor blade angle and the shaft axis is denoted by β.
The component of the velocity c in axial direction is denoted by ca. It is assumed to be constant along the compressor. There is a relation between ca and u. It is given by
u = ca (tan α 1 + tan β 1 ) = ca (tan α 2 + tan β 2 ). (5.1.5)
