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Homework answers / question archive / Starting from the definition of Specific Enthalpy, ? = ? + ??, show that: ? ?? = ?−1 ? Explain the meaning of the word “Adiabatic” and give an example of nonadiabatic flow in the engine
?
??
= ?−1
?
End of Part A Continue to Part B
Part B
Answer any two questions from Part B
iii Calculate the temperature and pressure at the throat for the flow generated in part ii above.
Continued…
End of Part B. End of Question Paper
Formula Sheet – Aeroengines & rocket Science
ISA Table
H 
T 
P 
r 
a 
n 
m 
km 
K 
N.m^{}^{2} 
kg.m^{}^{3} 
m.s^{}^{1} 
m^{2}.s1 
N.s.m^{}^{2} 
0 
288.2 
1.0133E+05 
1.2252 
340.3 
1.4634E05 
1.7930E05 
1 
281.7 
8.9866E+04 
1.1117 
336.4 
1.5837E05 
1.7607E05 
2 
275.2 
7.9480E+04 
1.0065 
332.5 
1.7169E05 
1.7280E05 
3 
268.7 
7.0088E+04 
0.9090 
328.5 
1.8647E05 
1.6950E05 
4 
262.2 
6.1616E+04 
0.8190 
324.5 
2.0290E05 
1.6617E05 
5 
255.7 
5.3993E+04 
0.7359 
320.5 
2.2122E05 
1.6280E05 
6 
249.2 
4.7153E+04 
0.6594 
316.4 
2.4170E05 
1.5939E05 
7 
242.7 
4.1032E+04 
0.5892 
312.2 
2.6466E05 
1.5593E05 
8 
236.2 
3.5571E+04 
0.5248 
308.0 
2.9046E05 
1.5244E05 
9 
229.7 
3.0714E+04 
0.4660 
303.8 
3.1955E05 
1.4891E05 
10 
223.2 
2.6408E+04 
0.4123 
299.4 
3.5246E05 
1.4534E05 
11 
216.7 
2.2605E+04 
0.3636 
295.0 
3.8980E05 
1.4172E05 
12 
216.7 
1.9308E+04 
0.3105 
295.0 
4.5638E05 
1.4172E05 
13 
216.7 
1.6491E+04 
0.2652 
295.0 
5.3434E05 
1.4172E05 
14 
216.7 
1.4085E+04 
0.2265 
295.0 
6.2560E05 
1.4172E05 
15 
216.7 
1.2030E+04 
0.1935 
295.0 
7.3246E05 
1.4172E05 
16 
216.7 
1.0275E+04 
0.1653 
295.0 
8.5756E05 
1.4172E05 
17 
216.7 
8.7763E+03 
0.1411 
295.0 
1.0040E04 
1.4172E05 
18 
216.7 
7.4959E+03 
0.1206 
295.0 
1.1755E04 
1.4172E05 
19 
216.7 
6.4024E+03 
0.1030 
295.0 
1.3763E04 
1.4172E05 
20 
216.7 
5.4684E+03 
0.0879 
295.0 
1.6114E04 
1.4172E05 
1bar = 10^{5} N/m^{2}
Patm = 1.01325x10^{5} Pa
Air density = 1.2252 kg/m3
? (??????) = ?° (???????) + 273 ????????? = ?????? + ??????
γ = 1.4, R= 287 J/kg.K, CP= 1004 J/kg K
γ = 1.3, CP= 1140 J/Kg K, R=283 J/Kg.K
The Lower Calorific Value of hydrocarbon aviation fuel: hPR= 45,000KJ/kg
Thermodynamics:
1 ??? = 100,000 ?? 1 ??? = 101,325 ??
Ideal Gas
?? = ??? ?? = ?? Polytropic process: ??^{?} = ????????
Adiabatic process: ??^{?} = ????????
Work Done
? = ∫?2 ???
?1
Isothermal:
Polytropic: ? = ?1?1−?2?2
?−1
?2
? = ?_{1}?_{1}?? ( )
?1
Specific Heat Capacities
??
= ??
??
??
= ??
??
? = ??
??
??
− ??
= ?
Non Flow Energy Equation
? − ? = ?? ? − ? = ??
Steady Flow Energy Equation
?? − ?? = ?? + ??? + ???
?? − ?? = ?? (? + ?2 + ??)
2
???
− ?? (? + ?2 + ??)
2
??
Where ? = ????????
?? = ??? ? = ? + ?? or ? = ? + ??
Second Law of Thermodynamics
^{2} ??
?? = ?_{2} − ?_{1} = ∫
1 ?
Gibbs Equation:
Isentropic Relationships:
??? = ?? − ??? = ?? + ???
?2 =
?1
?2 ( )
?−1


?2 − ?1 = ???? (
?1
) − ? ?? (
?2
?1
) (Ideal Gas Only)
Thrust Equation:
? = (??0 + ??? )? − ??0?0 + (?? − ?0)??
Speed of Sound: ? = √???
Mach number: ? = ?
?
Total Temperature: ??
Isentropic Area Ratio:
Mass Flow Parameter:
= (1 + ?−1 ?^{2})
2
Subscript 1 refers to conditions before the shock and 2 to conditions after the shock.
Propulsive efficiency:
Thermal efficiency:
Overall efficiency:
?? = ?? × ??
Effective Exhaust velocity:
??
???
=
?? ????
(?? − ??)??
? = ?? +
???
Specific Impulse: ???
= ?
?? ?

Orbital speed: ? = √?
??+?
Velocity increment:
Massey, B.S., Mechanics of fluids, Spon press, 9^{th} Edition, 2012
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