Overview

 

Topic 1 - CC Engines

CC is an abbreviation used for continuous combustion engines.  These are engines in which the combustion of the fuel-air mixture occurs continuously, rather than intermittently during the engine cycle.  Some examples are:  

• Rocket Engines
  • Solid fuel rocket engines have a sold mass which contains both the fuel and an oxidizer. When ignited, this mass of fuel continues to burn until it is exhausted.  Thrust rate can be controlled by the shape of the mass of fuel/oxidiser.
  • Liquid Fuel rocket engines contain a tank of fuel and a tank of oxidizer and both are combined in a combustion chamber to create a combustible fuel-air mixture. When ignited, this fuel air mixture will continue to burn long as both the fuel and the oxidizer are supplied to the combustion chamber.

• Steam Engines - Steam engines can take many forms, but the key aspect that makes them CC engines is that the combustion for the fuel and air occurs continuously. The working fluid (typically water) moves through the system and is heated by this combustion, creating steam, which is then used to power a piston system and drive the output of the engine.  The working fluid moves through the heating chamber where the heat generation, from combustion, is on-going rather than intermittent.

• Stirling Engines – To demonstrate this, strike a match, let it burn, and then blow it out. You will notice that most of the smoke occurs during the ignition and the extinguishing – not during the steady burn.  A Stirling engine makes use of this.  One side of the engine is continuously heated (thus CC) and the other side is continuously cooled.  The working fluid is always contained (again adding to clean operation as it is never released to atmosphere) and is frequently air, although other gases can be used.  This working air is moved to the hot side, causing it to expand, which pushes a piston up.  Then the air is moved through a regenerator to the cold side, where it cools and contracts, pulling the piston down.  The temperature change of the operating fluid produces the pressure change required to move the piston.  Note that as the hot air flows through the regenerator, it heats the cool air on its way to be heated, thus improving the efficiency of the engine by warming the cool air on the way to being heated and reducing the temperature of the hot air on its way to the cooling chamber.

Wikimedia / CC BY-SA 3.0

• Gas Turbines – These are rotating CC engines in which ambient air is passed through a compressor which increases its pressure above ambient. The high pressure air is mixed with fuel for combustion, and the hot exhaust gasses are allowed to expand through a turbine, which extracts energy from them.  Some of the energy extracted by the turbine is used to drive the compressor, and the remainder is available through an output shaft.  These will be examined in more detail in ongoing sections.

• Jet Engines – Jet engines are actually gas turbines, in which the excess energy of the hot gasses is not extracted via a turbine to drive an external output shaft, but rather, is passed through a nozzle to create a jet stream of hot gases which produce a thrust that propels the engine forward.

• Wave Rotor Engines – These replace the combustion chamber of a gas turbine with a wave rotor. The device uses gas-dynamic waves to transfer energy directly to and from the working fluid through which the waves travel. It consists of a series of constant area passages that rotate about an axis. Through rotation, the ends of the passages are periodically exposed to various circumferentially arranged ports which initiate the travelling waves within the passages. Because each passage of the wave rotor is periodically exposed to both hot and cold flow, the mean temperature of the rotor material remains considerably below the peak cycle temperature.

Topic 2 - Types of gas turbines

There are a number of different types, or configurations, of gas turbines.  They will be discussed as follows:

• Turbo Prop
  • Single Spool
  • Multi Spool
• Turbo Fan
  • Low Bypass
  • High Bypass
• Turbo Jet
• Turbo Shaft
• Turbo Shaft w/ Free Turbine

Turbo Prop engines are used to power propellers which pull prop-driven aircraft through the sky.  In this configuration, air enters the engine inlet, and passes through the compressor.  The high pressure air is then mixed with fuel and ignited in the combustor, before passing through the turbine, which extracts energy from the expanding hot gasses.  If the engine is a single spool (i.e. containing only one rotor) configuration, then the turbine drives a shaft which drives the compressor.  The shaft also extends outward from the front of an engine, to drive a step-down gearbox which turns the propeller which pulls the aircraft through the air. 

Wikimedia / CC BY 2.5

A dual spool (i.e. containing two separate rotors running at different speeds) version of the turbo-prop engine has two separate turbines.  The first (sometimes referred to as the high-pressure, or HP, turbine) drives the compressor via a shaft connecting the two.  The second turbine (sometimes referred to as the low-pressure, or LP, turbine because the exhaust gasses passing through it are at a lower pressure, having already passed through the LP turbine) is often referred to as the power turbine, or PT, because the output power delivered external to the engine comes from this turbine and is delivered by a separate shaft (often referred to as the Power Take-Off, or PTO, shaft).  An example is shown.

Turbo Fan engines are also designed to pull aircraft through the air, but rather than using a propeller, they incorporate a very large diameter compressor rotor, referred to as a “fan.”  This fan serves a dual purpose.  Some of the air passing through it (the portion at a low diameter) enters the compressor, but has thus already seen a slight pressure boost, which increases the overall pressure that the air has when it reaches the combustor.  The remainder of the air (the portion at a higher diameter) bypasses the compressor and is merely released back to the atmosphere.  This is the portion of air that essentially pulls the engine through the air under the effect of the many fan blades, in similar fashion to what the propeller did with its 4 or 5 blades.  Two types of fan engine exist:  those with a low bypass ratio (smaller percentage of the air passing through the fan is bypassed) or high bypass ratio (larger percentage of the air passing through the fan is bypassed. 

Wikimedia / CC BY-SA 4.0

Turbo Shaft engines are non-aircraft engines, which are used to power ground-based installations, such as generators for emergency (back-up) power at hospitals, or every-day power on things like large ships or off-shore oil rigs, or pumping stations for oil or natural gas in petroleum rich drilling fields.  They come in two basic configurations.  The first looks very much like the Turbo Prop applications previously discussed for aircraft, except that they are ground-based and the output shaft drives (through a step-down gearbox) a pump or generator, rather than a propeller. 

The second configuration, known as a Free Turbine configuration, is a two-spool configuration which has the second, or LP turbine mechanically disconnected from the rest of the engine.  Its only connection to the main unit is the air flowing out of the HP turbine, and into the LP turbine, which then drives an output shaft to deliver power to an external system.

Wikimedia / CC BY 2.5

Topic 3 - Gas turbine cycles

The cycle used to represent the operation of a gas turbine is the Brayton Cycle.  It is shown here for a simple gas turbine configuration.

The idealized Brayton cycle where P = pressure, V = volume, T = temperature, S = entropy, and Q = the heat added to or rejected by the system.

Wikimedia / CC BY-SA 3.0

The air pulled into the turbine is the working fluid.  Work is done to the air (W_1-2) as it is compressed between points 1 and 2 in the cycle.  Then heat is added to the system (Q_2-3) as the fuel is mixed with the air and combustion occurs between points 2 and 3.  The turbine then extracts work (W_3-4) from the expanding hot gasses as they flow out of the engine between points 3 and 4.

${\text{W}}_{12}=-{\text{c}}_{\text{p}}({\text{T}}_2-{\text{T}}_1)$ 

${\text{Q}}_{23}={\text{c}}_{\text{p}}({\text{T}}_3-{\text{T}}_2)$ 

${\text{W}}_{34}={\text{c}}_{\text{p}}({\text{T}}_3-{\text{T}}_4)$ 

$\text{where}$ 

$\text{Q = heat transfer}$
$\text{W = work transfer}$
$\text{T = temperature}$
$\text{C = heat capacity}$
${\text{C}}_{\text{p}}=\text{specific heat of fluid}$ 

Since the turbine drives the compressor, the net available work out, W_net, is the work extracted from the hot air by the turbine, W_3-4, minus the work the turbine has to supply to drive the compressor, W_1-2.

Therefore, cycle efficiency is

$\text{η}=\frac{{\text{c}}_{\text{p}}({\text{T}}_3-{\text{T}}_4)-{\text{c}}_{\text{p}}({\text{T}}_2-{\text{T}}_1)}{{\text{c}}_{\text{p}}({\text{T}}_3-{\text{T}}_2)}$ 

If we assume an ideal Brayton cycle to be an isotropic process (constant entropy in steps 1-2 and 3-4).

Then ${\text{T}}_2/{\text{T}}_1={\text{T}}_3/{\text{T}}_4={\text{r}}^{\left(\text{γ - 1}\right)/\text{γ}}$ 

$\text{where r is the pressure ratio, }\text{r}={\text{p}}_2/{\text{p}}_1={\text{p}}_3/{\text{p}}_4$   $\text{and }\text{γ}={\text{c}}_{\text{p}}/{\text{c}}_{\text{v}}$ 

c_p = amount of heat to raise one kg of substance 1 degree at constant pressure

c_v = amount of heat to raise one kg of substance 1 degree at constant volume

Note:  g = 1.4 for air

Rearranging the efficiency equation, we see

$\text{η}=\frac{{\text{c}}_{\text{p}}({\text{T}}_3-{\text{T}}_4)-{\text{c}}_{\text{p}}({\text{T}}_2-{\text{T}}_1)}{{\text{c}}_{\text{p}}({\text{T}}_3-{\text{T}}_2)}=1-{(1/\text{r})}^{\left(\text{γ - 1}\right)/\text{γ}}$ 

If we think about this, it tells us that for a given gas and for an ideal Brayton cycle, the efficiency is a function purely of the pressure ratio.

Examining things further,

${\text{W}}_{\text{out}}={\text{c}}_{\text{p}}({\text{T}}_3-{\text{T}}_4)-{\text{c}}_{\text{p}}({\text{T}}_2-{\text{T}}_1)={\text{T}}_3{\text{c}}_{\text{p}}[1-1/{\text{r}}^{\left(\text{γ - 1}\right)/\text{γ}}]-{\text{T}}_1{\text{c}}_{\text{p}}[{\text{r}}^{\left(\text{γ - 1}\right)/\text{γ}}-1]$ 

Thus, the Output Work is a function of only r and the difference between T₃ and  T₁ …..i.e. the pressure ratio and how much we raise the temperature.

If we think about this, it tells us that if efficiency is purely a function of pressure ratio, and Work Output is a function of pressure ratio and max temperature………..then  it means that the more we can compress the air above the ambient and the hotter we make it………. the better off we are. 

So logic would say that we should always use more compressor stages…….. and run hotter. 

But wait, there are other issues.  We are trying to produce output power (which is work/time).  Which means we want the difference between the power extracted by the turbine and the power required to drive the compressor to be maximized.   

And if we make a bigger compressor, it takes more of the turbine’s power to drive it.  Which leaves less power for output.  And, we cannot just keep making the turbine run hotter and hotter, because at some point we will literally melt the turbine blade metal, like this:

© Peter Hylton

Thus, we enter a regime where we have multiple conflicting parameters to optimise.

Perhaps we can find a way to make the turbine blades last longer, which would allow us to run hotter.  One example would be if we could take cooler air and bring it into the centre of the turbine blade, letting it bleed out into the flow path through small holes in the surface, as shown.

However, this cooling air must come from somewhere, and the only practical path is to steal it from the compressor.  Air coming from the later stages of the compressor will be much lower in temperature than the exhaust air and will be at sufficiently high pressure that its delivery through passages in the turbine shafts and turbine wheels to arrive at the turbine blades is possible.  However, by stealing this air from the compressor, we are reducing the overall efficiency.   So, once again, we have to make trade-offs between conflicting criteria.

This is actually just one of a number of losses in efficiency that the compressor will experience.  Here is a more complete list:

• Fluid Kinetic Energy
• Fluid Friction
• Hardware Friction (rub)
• Bearing Friction
• Windage Losses
• Combustion process is not as simple as we have thus far treated it
• Calculations are approximate, as mass flow is not truly constant, since fuel is added at combustion
• Compressor air is often “stolen” as “bleed flow” for cooling or for sealing

Let us consider the individual component efficiencies:     

Compressor Efficiency

Note:  We will assume T₁ and p₁ are ambient atmospheric, i.e. T_a and p_a

${\text{η}}_{\text{c}}=\frac{{\text{T}}_{\text{a}}[{({\text{p}}_2/{\text{p}}_{\text{a}})}^{(\text{γ}-1)/\text{γ}}-1]}{({\text{T}}_2-{\text{T}}_{\text{a}})}$  

Turbine Efficiency

Note:  We will assume p₄ is atmospheric, i.e.  p_a, However, clearly T₄ is NOT atmospheric

${\text{η}}_{\text{t}}=\frac{({\text{T}}_{\text{3}}-{\text{T}}_{\text{4}})}{{\text{T}}_3[1-{({\text{p}}_{\text{a}}/{\text{p}}_3)}^{(\text{γ}-1)/\text{γ}}]}$  

Mechanical Efficiency

Note:  We will assume T₁ is atmospheric, i.e. T_a

${\text{η}}_{\text{m}}=\frac{{\text{c}}_{\text{p}}({\text{T}}_2-{\text{T}}_{\text{a}})}{{\text{W}}_{\text{t}}}$ 

Pressure Loss in Combustion Chamber

Note:  We will assume p₃ = p₂ per the ideal Brayton cycle, minus losses

${\text{p}}_3={\text{p}}_2(1-{\text{Δp}}_{\text{b}}/{\text{p}}_2)$ 

where p_b = pressure loss in combustion chamber

Topic 3 - Application

A turbo shaft engine with a free power turbine is shown.  Determine the specific work available as output to the driven load.

Wikimedia / CC BY-SA 4.0

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Let us begin with the compressor efficiency equation:

${\text{η}}_{\text{c}}=\frac{{\text{T}}_{\text{a}}[{({\text{p}}_2/{\text{p}}_{\text{a}})}^{(\text{γ}-1)/\text{γ}}-1]}{({\text{T}}_2-{\text{T}}_{\text{a}})}$  

$0.86=\frac{288[{(12/1)}^{(1.4-1)/1.4}-1]}{({\text{T}}_2-{\text{T}}_{\text{a}})}$ 

Solve for (T₂ – T_a) = 346.3 K   

Now use the compressor work equation, taking into account the mechanical efficiency and realizing that T₁ = T_a:

${\text{W}}_{12}=-{\text{c}}_{\text{p}}({\text{T}}_2-{\text{T}}_1)/{\text{η}}_{\text{m}}=\text{1005(346.3)/0.99 = 351.5 kJ/kg}$  

The compressor exit pressure will be 12 bar, since the inlet pressure is 1 bar and the pressure ratio is 12.  We can now find the turbine inlet pressure:

${\text{P}}_3={\text{P}}_2(1-{\text{ΔP}}_{\text{b}}/{\text{P}}_2)=\text{12(1 – 0.06) = 11.28 bar}$  

Now let us find the temperature drop as the air passes through the high pressure turbine.  We have to recognize that the HP turbine supplies the work to turn the compressor, which we found as W₁₂ above:

${\text{W}}_{34}={\text{c}}_{\text{p}}({\text{T}}_3-{\text{T}}_4)$ 

$351.5=1148({\text{T}}_3-{\text{T}}_4)$ 

So (T₃ – T₄) = 306.2 K    and since we know the turbine inlet temperature (TIT which would be T₃) from thermocouple measurement, to be 1350K, then T₄   = 1043.8 K

Recall that at station 4, the pressure P₄ is NOT ambient, because it still has pressure and has to pass through the power turbine and have more power extracted before it is exhausted to ambient.  We can, however, figure out the temperature at T₃ from the turbine efficiency equation and the (T₃ – T₄) value that we just found:

${\text{η}}_{\text{t}}=\frac{({\text{T}}_3-{\text{T}}_4)}{{\text{T}}_3[1-{({\text{p}}_3/{\text{p}}_4)}^{\left(\text{γ - 1}\right)/\text{γ}}]}$ 

$0.89=\frac{\text{(306.2)}}{1350[1-{({\text{p}}_3/{\text{p}}_4)}^{\text{(1.333 - 1) / 1.333}}]}$ 

Which yields (p₃/p₄) = 3.242, which is the pressure ratio for the HP turbine, and from this we can find

${\text{P}}_4=\text{11.28 bar/3.242 = 3.478 bar}$  

P₅ will be ambient pressure, since after exiting the power turbine, the exhaust gasses become atmospheric.  Note that our given data indicated a 3% exhaust pressure loss, so the pressure ratio through the PT will be

${\text{P}}_4/{\text{P}}_5=\text{3.478/(1 + 0.03) = 3.377}$ 

Now we can use the turbine efficiency equation for the power turbine

${\text{η}}_{\text{t}}=\frac{({\text{T}}_4-{\text{T}}_5)}{{\text{T}}_4[1-{({\text{p}}_5/{\text{p}}_4)}^{(\text{γ}-1)/\text{γ}}]}$ 

$0.89=\frac{({\text{T}}_4-{\text{T}}_5)}{1043.8[1-{\left(3.377\right)}^{(1.333-1)/1.333}]}$ 

Solve for (T₄ – T₅) = 243.7K which makes T₅ = 1043.8 – 243.7 = 800 K

The work which can be done by the power turbine then is:

${\text{W}}_{45}={\text{c}}_{\text{p}}({\text{T}}_4-{\text{T}}_5)=1148\left(243.7\right)=279\text{kJ/kg}$ 

But after the loss due to the mechanical efficiency, the work which can be supplied to the outside world is   279(0.99) = 277 kJ/kg

And applying the known flow rate of 1.75 kg/sec, the power output is:

$\text{(277 kJ/kg)(1.75 kg/sec) = 485 kW}$ 

We picked a relatively easy configuration (although not the easiest).  Consider what would have been required had we chosen a two-spool turbojet with afterburner.  The number of stations to consider would have been much greater, as shown here:

Reference and bibliography

Falempin, F. (2008) “Continuous Detonation Wave Engine.” Advances on Propulsion Technology for High Speed Aircraft, Paper #8.

Nguyen, Q. & and Jacqmin, D. (). “A Study of Cavitation-Ignition Bubble Combustion.”  NASA/TM—2005-213599 Retrieved from https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20050215681.pdf on 14 August 2019.

Saravanamutto, H., Rogers, G., Cohen, H. & Strznicky, P. (2009).  Gas Turbine Theory.  Essex, UK: Pearson.

Sforza, P. (2011). Theory of Aerospace Propulsion. New York: Butterworth-Heinemann.

Wilson, D & Korakianitis, T. (2014).  The Design of High-Efficiency Turbomachinery and Gas Turbines.  Cambridge, USA:  MIT Press.