A thermodynamic cycle analysis of the Goble jet engine.
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Tecnología Aerospacial 2008/2009 Sistemas de propulsión Goblin II Thermodynamic Cycle Analysis Jorge García Tíscar October 29, 2009 All calculations and results given in SI if not indicated otherwise. The gas turbine engines that you see attached to the wing of a jet air plane consists of a diffuser, compressor, combustor, turbine, and nozzle that are arranged in series. (Figure 1) Figure 1: Schematic of a gas turbine jet engine. Figure 2 shows the Goblin II turbojet engine (used as the primary engine on the F80 airplane) sectioned so that the compressor, combustion chambers, and turbine are visible. This homework problem analyzes the jet engine shown in Figure 1 at the conditions that are approximately consistent with the Goblin II jet engine; the analysis will be carried out on a component-by-component basis. Model the air using like an ideal gas, assume that it has constant specific heat capacities. 1
Transcript
Tecnología Aerospacial 2008/2009Sistemas de propulsión
Goblin II Thermodynamic Cycle AnalysisJorge García Tíscar
October 29, 2009
All calculations and results given in SI if not indicated otherwise.
The gas turbine engines that you see attached to the wing of a jet air plane consists of a diffuser, compressor,combustor, turbine, and nozzle that are arranged in series. (Figure 1)
Figure 1: Schematic of a gas turbine jet engine.
Figure 2 shows the Goblin II turbojet engine (used as the primary engine on the F80 airplane) sectioned sothat the compressor, combustion chambers, and turbine are visible.
This homework problem analyzes the jet engine shown in Figure 1 at the conditions that are approximatelyconsistent with the Goblin II jet engine; the analysis will be carried out on a component-by-component basis.Model the air using like an ideal gas, assume that it has constant specific heat capacities.
1.1 DifusserAir entering the engine first encounters the diffuser which has the purpose of reducing the velocity of the air(with respect the engine) and increasing its pressure; note that a diffuser is a lot like a nozzle - it is simply aduct with a varying cross-sectional area. The air enters the diffuser with a velocity that is equal to the velocityof the jet, V0 = 300 mile/hr. The jet is flying at an altitude where the inlet air pressure is P0 = 70 kPa andthe temperature is T0 = -10°C. Assume that the diffuser is reversible (i.e., it generates no entropy). Assumethat the exit velocity of the diffuser is small enough that the kinetic energy of the air leaving the diffuser isnegligible. Further, assume that the diffuser operates at steady state and is adiabatic. You may ignore thepotential energy of the flow entering and leaving the diffuser.
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a. What is the pressure and temperature of the air leaving the diffuser, P2t and T2t?
T2t = T0 + V02
2cp= 272.105 (16)
P2t = P0
(T2t
T0
)γ
γ−1 = 78697.6 (17)
1.2 CompressorThe compressor for the Goblin II engine is a single stage centrifugal compressor which has a pressure ratioPRc = 5.0; that is, the pressure at the compressor exit (P3r) is 5.0 times larger than the pressure at thecompressor inlet (P2r). The compressor efficiency is η23 = 0.85. You may assume that the compressor operatesat steady state and is adiabatic.
b. What is the power required by the compressor, W23?
c. What is the temperature of the air leaving the compressor, T3t?
η23 = (π23)γ−1γ −1
τ23cpT2t
⇒ τ23 = 187698 (18)
Wc =G0τ23 = 4.4109×106 (19)
P3t =π23P2t = 393488 (20)
τ23 = cp (T3t −T2t)⇒ T3t = 458.999 (21)
(22)
1.3 CombustorAs the air ,passes through the combustor it is mixed with fuel which is ignited, releasing heat. Here, we willmodel this combustion as a heat addition, qeb, which bring the temperature of the air leaving the combustorto temperature T4t = 950°C. You may assume that there is no pressure loss in the combustor so that P4t = P3t.The combustor operates at steady state and you can ignore the kinetic and potential energy of the flow enteringand leaving the combustor.
e. What is the rate of heat transfer that must be provided to the combustor, qeb?
P4t = P3t = 393488 (23)
qcb =G0cp (T4t −T3t)= 1.80348×107 (24)
1.4 TurbineThe turbine provides sufficient mechanical power to drive the compressor (sometime additional power is ex-tracted to run auxiliary systems like electronics, etc. but we’ll neglect that here); therefore, Wt = We. Theturbine efficiency is ηt = 0.85. You may assume that the turbine operates at steady state and is adiabatic.
Hint: tackle the analysis of the turbine using the following steps: (1) carry out an energy balance on the actualturbine to get h5t and T5t, (2) use the definition of the turbine efficiency to get Wt,s, the power that would beproduced by a reversible turbine, (3) carry out an entropy and enthalpy balance on the reversible turbine toget h5t,s, (4) determine the exit pressure P5t from h5t,s and s5,s.
f. What is the temperature of the air leaving the turbine, T5t?
g. What is the pressure of the air leaving the turbine, P5t?
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Wt =Wc =G0cp (T4t −T5t)⇒ T5t = 1036.26 (25)
ηt =1−
(T5tT4t
)1−π45
γ−1γ
⇒π45 = 0.499793 (26)
P5t =π45P4t = 196663 (27)
1.5 NozzleThe nozzle accelerates the fluid to a high velocity so that the air can be propelled out the back of the turbojetengine at a velocity that is faster than the plane itself is flying; this produces thrust (like a rocket engine). Thenozzle efficiency is η59 = 0.85. The nozzle exit pressure is the same as the diffuser inlet pressure, P9 = P0. Youmay assume that the nozzle operates at steady state and is adiabatic.
i. What is the velocity of the air leaving the nozzle, V9?
η59 =V 2
9
V 29s
⇒V9 =pη59
√2cpT5t
(1−
(P0
P5t
)γ−1γ
)= 672.431 (28)
η59 =1− T9
T5t
1−(
P0P5t
) γ−1γ
⇒ T9 = 811.142 (29)
2 System Analysis
k. Prepare an h-s plot for air and overlay your state points onto the plot. Make sure that the states lookcorrect to you and that you understand why enthalpy and entropy are changing as the air flows througheach component.
First we calculate the enthalpy in each station:
h0 = cpT0 = 264282 (30)
h2 = cpT2t = 273275 (31)
h3 = cpT3t = 460972 (32)
h4 = cpT4t = 1.22841×106 (33)
h5 = cpT5t = 1.04071×106 (34)
h9 = cpT9 = 814630 (35)
The entropy change in each station is given by:
∆S02 = cp ln[
T2t
T0
]−Rs ln
[P2t
P0
]=−0.0134852 (36)
∆S23 = cp ln[
T3t
T2t
]−Rs ln
[P3t
P2t
]= 63.1071 (37)
∆S34 = cp ln[
T4t
T3t
]−Rs ln
[P4t
P3t
]= 984.352 (38)
∆S45 = cp ln[
T5t
T4t
]−Rs ln
[P5t
P4t
]= 32.5638 (39)
∆S59 = cp ln[
T5t
T5t
]−Rs ln
[P0
P5t
]= 296.529 (40)
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So the points of the plot will be:
Ciclo= { (41)
{S0,h0} , (42)
{S0 +∆S02,h2} , (43)
{S0 +∆S02 +∆S23,h3} , (44)
{S0 +∆S02 +∆S23 +∆S34,h4} , (45)
{S0 +∆S02 +∆S23 +∆S34 +∆S45,h5} , (46)
{S0 +∆S02 +∆S23 +∆S34 +∆S45 +∆S59,h9}} (47)
Plotting them with the aid of the following Mathematica script:
The whole point of a turbojet engine is to produce a thrust force that pushes the jet forward. The thrust forceis given by:
E =G0 (V9 −V0) (48)
l. What is the thrust force produced by this engine? (Note that the thrust rating for the Goblin II engine wasabout 13.8 kN so your answer shouldn’t be too different from that).
E =G0 (V9 −V0)= 12650.5N (49)
The heat provided to the combustor is obtained by burning fuel. The amount of fuel required is obtained fromthe fuel’s heating value, HV, which is the ratio of the heat provided to the mass of the fuel burned. The heatingvalue of jet fuel is approximately HV = 4.2× l07 J/kg.
l. Determine the rate at which the engine consumes fuel, c.
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HV= 4.2∗107 (50)
c = qcb
HV= 0.429399 (51)
(52)
There are two important figures of merit for a jet engine; one is the thrust produced and the other is calledthe specific thrust, which is defined as the amount of thrust you get divided by the rate at which the engineconsumes fuel (1/Ce).
m. Determine the specific thrust of the engine.
Ec = Ec= 29460.9 (53)
n. Plot the specific thrust of your engine as a function of the compressor pressure ratio, for pressure ratiosfrom 2 to 50; if you were trying to optimize the specific thrust of your engine then what pressure ratiowould you try to achieve?
Using the following Mathematica code, which has T4t, π23, cp and γ as input variables, we can solve the enginein a symbolic manner, instead of the numerical component-by-component analysis performed above:
1 Clear [ " Global ‘ * " ] ;2 G0=23.5;3 P0=70000;4 T0=273.15−10;5 \[Eta ]23=0.85;6 \[Eta ] c =0.85;7 \[Eta ] t =0.85;8 \[Eta ]59=0.85;9 P9=70000;
10 " Spec i f i c Thrust (N/ ( Subscript [ kg , fue l ] / s ) ) " } ]
We can trace the plot we were searching for:
10 20 30 40 50
20 000
25 000
30 000
35 000
40 000
Π23
Spe
cific
Thr
ustHN
�Hkg fu
el�sL
L
Figure 4: Specific thrust against π23
Is easy to find that the value of π23 that optimizes the specific thrust is around 30.
n. Plot the thrust force produced by your engine as a function of the compressor pressure ratio, for pressureratios from 2 to 50; overlay on your plot curves that correspond to several values of turbine inlet tem-perature including T4t = 950°C, 1000 °C, 1100 °C, and 1200 °C. If you were trying to optimize the thrustprovided by your engine, then what pressure ratio would you try to achieve? Your plots should indicatethat the performance of your engine gets much better as T4t increases; this trend explains the intenseeffort towards developing high temperature materials for gas turbine engines.
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Again, using the first Mathematica code and the following script:
Note (easily seen in the code) that the curves represent T4t temperatures varying from 900 to 1300 ºC in stepsof 50ºC. Also, although the code for it is more obcure, as the curves get red, the temperature (T4t) that theyrepresent increases; so we can see that thrust effectively increases with T4t.
