SOLUTIONS MANUAL FOR
Aircraft Propulsion
and
Gas Turbine Engines
by
Ahmed El-Sayed
86898.indd 1 8/14/08 10:25:48
, Chapter 2
PERFORMANCE PARAMETERS OF JET ENGINES
2.1 A turbojet engine powers an aircraft which is flying at a speed of 240 m/s, has
an exhaust speed of 560 m/s and a specific thrust of 525 N ⋅ s / kg . Using the
three different formulae, calculate the propulsive efficiency. What are your
comments?
(Neglect the fuel-to-air ratio)
Solution
Given data:
Engine: Turbojet
U =240 m/s, U e =560 m/s, T / m& a =525 N.s/kg , f =0
Required: Calculate η P using three different formulae
T ×U
a) η P =
T × U + 0.5 × m& e × (U e − U ) 2
m& e = m& a (1 + f ) = m& a
2(T / m& a )U 2(525)(240)
ηp = = = 71.106% =
2(T / m& a )U + (U e − U ) 2
2(525)(240) + (560 − 240) 2
TU
b) ηP =
0.5 × m& e × U − 0.5 × m& a × U 2
2
e
2(T / m& a )U 2(525) × (240)
ηP = = = 98.4375%
Ue −U
2 2
(560) 2 − (240) 2
2U (2)(240)
c) ηP = = = 60%
U + U e 240 + 560
Summary
Expression 2(T / m& a )U TU 2U
2(T / m& a )U + (U e − U ) 2 0.5 × m& e × U − 0.5 × m& a × U 2
2
e U + Ue
ηP 71.106% 98.4375% 60%
, ___________________________________________________________
2.2 Starting from the thrust force equation for an unchoked nozzle engine,
Prove that:
(1 + f ) 2 u e2
ηo =
max
4 fQ R
If λ is the ratio between the maximum overall efficiency and the overall
efficiency ( λ = η omax / η o ), then prove that the flight speed u may be
expressed by the relation:
(1 + f )u e ⎡ + λ −1⎤
u= ⎢1 ⎥
2 ⎣ λ ⎦
Now, consider an aircraft fitted with a single engine and has the
following data:
Air mass flow rate m& a = 100kg / s , Exhaust speed u e = 770m / s
Overall efficiency η o = 0.16 , Maximum overall efficiency η omax = 0.18
Fuel-to-air ratio f = 0.02 , Lift/ Drag ratio L / D = 10
Aircraft mass ratio at take off and landing m1 / m 2 = 1.2
Calculate:
2.1.1 Possible value(s) for flight speed
2.1.2 Heating value ( QR )
2.1.3 Possible range(s) of aircraft
2.1.4 Trip time(s)
2.1.5 The aircraft masses m1 and m 2
2.1.6 Thrust force(s)
Solution:
Given data: starting from the thrust force equation for an unchoked nozzle
engine
(1 + f ) 2 u e2
a) Prove that η omax =
4 fQ R
Tu
Since η0 = η P ×ηth =
m& f × Qr
and T = m& a [(1 + f )ue - u ] + ( Pe - Pa ) Ae
For unchoked nozzle; Pe = Pa , thus:
Aircraft Propulsion
and
Gas Turbine Engines
by
Ahmed El-Sayed
86898.indd 1 8/14/08 10:25:48
, Chapter 2
PERFORMANCE PARAMETERS OF JET ENGINES
2.1 A turbojet engine powers an aircraft which is flying at a speed of 240 m/s, has
an exhaust speed of 560 m/s and a specific thrust of 525 N ⋅ s / kg . Using the
three different formulae, calculate the propulsive efficiency. What are your
comments?
(Neglect the fuel-to-air ratio)
Solution
Given data:
Engine: Turbojet
U =240 m/s, U e =560 m/s, T / m& a =525 N.s/kg , f =0
Required: Calculate η P using three different formulae
T ×U
a) η P =
T × U + 0.5 × m& e × (U e − U ) 2
m& e = m& a (1 + f ) = m& a
2(T / m& a )U 2(525)(240)
ηp = = = 71.106% =
2(T / m& a )U + (U e − U ) 2
2(525)(240) + (560 − 240) 2
TU
b) ηP =
0.5 × m& e × U − 0.5 × m& a × U 2
2
e
2(T / m& a )U 2(525) × (240)
ηP = = = 98.4375%
Ue −U
2 2
(560) 2 − (240) 2
2U (2)(240)
c) ηP = = = 60%
U + U e 240 + 560
Summary
Expression 2(T / m& a )U TU 2U
2(T / m& a )U + (U e − U ) 2 0.5 × m& e × U − 0.5 × m& a × U 2
2
e U + Ue
ηP 71.106% 98.4375% 60%
, ___________________________________________________________
2.2 Starting from the thrust force equation for an unchoked nozzle engine,
Prove that:
(1 + f ) 2 u e2
ηo =
max
4 fQ R
If λ is the ratio between the maximum overall efficiency and the overall
efficiency ( λ = η omax / η o ), then prove that the flight speed u may be
expressed by the relation:
(1 + f )u e ⎡ + λ −1⎤
u= ⎢1 ⎥
2 ⎣ λ ⎦
Now, consider an aircraft fitted with a single engine and has the
following data:
Air mass flow rate m& a = 100kg / s , Exhaust speed u e = 770m / s
Overall efficiency η o = 0.16 , Maximum overall efficiency η omax = 0.18
Fuel-to-air ratio f = 0.02 , Lift/ Drag ratio L / D = 10
Aircraft mass ratio at take off and landing m1 / m 2 = 1.2
Calculate:
2.1.1 Possible value(s) for flight speed
2.1.2 Heating value ( QR )
2.1.3 Possible range(s) of aircraft
2.1.4 Trip time(s)
2.1.5 The aircraft masses m1 and m 2
2.1.6 Thrust force(s)
Solution:
Given data: starting from the thrust force equation for an unchoked nozzle
engine
(1 + f ) 2 u e2
a) Prove that η omax =
4 fQ R
Tu
Since η0 = η P ×ηth =
m& f × Qr
and T = m& a [(1 + f )ue - u ] + ( Pe - Pa ) Ae
For unchoked nozzle; Pe = Pa , thus:
