SOLUTIONS MANUAL FOR
Aircraft Propulsion
and
Gas Turbine Engines
by
Ahmed El-Sayed
86898.indd 1 8/14/08 10:25:48
, CHAPTER 3
PULSEJET and RAMJET ENGINES
3.1 A pulsejet similar to the German engine, has the following data:
The net thrust 1400 lb
Flight altitude 10,000 ft
Speed 400 mph
Inlet diameter 1.34 ft
Total pressure drop in the inlet 20%
Thrust specific fuel consumption 10 lb / lb f hr
Fuel heating value 14,000 Btu / lbm
Assume γ = 1.4 and Cp = 0.24 Btu / lbm 0 R
Calculate:
a- Fuel-to-air ratio
b- Maximum temperature in the combustion chamber
c- Exhaust velocity
d- Propulsive efficiency
Thermal efficiency
solution
1) Data: A pulse jet generates a thrust force of 1400 lbf
At Altitude h=10000 ft, then the ambient conditions are
( Pa = 10108 lb f / ft 2 , Ta = 483.03 R , ρ a = 1.756 × 10-3 slug / ft 3 )
U = 400 mph = 400 ×1.467 = 586.8 ft / s , Di = 1.34 ft
lbm 10 lbm
TSFC = 10 = = 2.778 × 10−3
lbf .hr 3600 lbf .s
Btu
QR = 14000 , γ = 1.4
Ibm
Btu
C p = 0.24
Ibm.R
Solution:
a) The thrust specific fuel consumption is expresses by the relation
m&
TSFC = f
T
, m& f = TSFC × T = 2.778 × 10-3 × 1400 = 3.889 lbm / s = 0.12089 slug / s
π
m& a = ρ a AU
i = 1.756 × 10-3 × × (1.34) 2 × 586.8 = 1.4531 slug / s
4
m& f
f = = 0.0831
m& a
b) The fuel to air ratio is expressed by the relation
C × T − C pc × To 2
f = ph o 3
QR − C ph × To3
For ideal cycle C ph = C pc = C p
U2 586.8
To 2 = Ta + = 483.03 + ( ) = 511.68 R
2C p 2 × 6009.07
f × QR + C p × To 2 0.0831×14, 000 + 0.24 × 511.68
To 3 = = = 4,948 R
C p (1 + f ) 0.24 × (0.0831 + 1)
Maximum temperature in the combustion chamber ( T03 = 4,948 R)
c) T = m& a [(1 + f )U e - U ] + Ae ( Pe - Pa )
Since Pe = Pa for unchoked nozzle
Then T = m& a [(1 + f )U e - U ]
T 1400
+U + 586.6
m& a 1.4531
Thus Ue = = = 1431.13 ft / s
(1 + f ) 1 + 0.0831
Exhaust velocity is U e = 1431.13 ft / s
d) The propulsive efficiency ( η p ) is expressed by the relation
T ×U
ηp =
T × U + 0.5 × ma × (1 + f ) × (U e − U ) 2
&
1400 × 586.8
ηp = = 59.42%
1400 × 586.8 + 0.5 ×1.4531× (1 + 0.0831) × (1413.13 − 586.8) 2
e) Thermal efficiency (ηth )
T × U + 0.5 × m& a × (1 + f ) × (Ue − U ) 2
ηth =
m& f × QR
=
1400 × 586.8 + 0.5 × 1.4531× (1 + 0.0831) × (1431.13 − 586.8) 2
ηth = = 32.6%
0.12082 × 35044660
3.2 Why is not mechanical efficiency an issue with ramjets?
Aircraft Propulsion
and
Gas Turbine Engines
by
Ahmed El-Sayed
86898.indd 1 8/14/08 10:25:48
, CHAPTER 3
PULSEJET and RAMJET ENGINES
3.1 A pulsejet similar to the German engine, has the following data:
The net thrust 1400 lb
Flight altitude 10,000 ft
Speed 400 mph
Inlet diameter 1.34 ft
Total pressure drop in the inlet 20%
Thrust specific fuel consumption 10 lb / lb f hr
Fuel heating value 14,000 Btu / lbm
Assume γ = 1.4 and Cp = 0.24 Btu / lbm 0 R
Calculate:
a- Fuel-to-air ratio
b- Maximum temperature in the combustion chamber
c- Exhaust velocity
d- Propulsive efficiency
Thermal efficiency
solution
1) Data: A pulse jet generates a thrust force of 1400 lbf
At Altitude h=10000 ft, then the ambient conditions are
( Pa = 10108 lb f / ft 2 , Ta = 483.03 R , ρ a = 1.756 × 10-3 slug / ft 3 )
U = 400 mph = 400 ×1.467 = 586.8 ft / s , Di = 1.34 ft
lbm 10 lbm
TSFC = 10 = = 2.778 × 10−3
lbf .hr 3600 lbf .s
Btu
QR = 14000 , γ = 1.4
Ibm
Btu
C p = 0.24
Ibm.R
Solution:
a) The thrust specific fuel consumption is expresses by the relation
m&
TSFC = f
T
, m& f = TSFC × T = 2.778 × 10-3 × 1400 = 3.889 lbm / s = 0.12089 slug / s
π
m& a = ρ a AU
i = 1.756 × 10-3 × × (1.34) 2 × 586.8 = 1.4531 slug / s
4
m& f
f = = 0.0831
m& a
b) The fuel to air ratio is expressed by the relation
C × T − C pc × To 2
f = ph o 3
QR − C ph × To3
For ideal cycle C ph = C pc = C p
U2 586.8
To 2 = Ta + = 483.03 + ( ) = 511.68 R
2C p 2 × 6009.07
f × QR + C p × To 2 0.0831×14, 000 + 0.24 × 511.68
To 3 = = = 4,948 R
C p (1 + f ) 0.24 × (0.0831 + 1)
Maximum temperature in the combustion chamber ( T03 = 4,948 R)
c) T = m& a [(1 + f )U e - U ] + Ae ( Pe - Pa )
Since Pe = Pa for unchoked nozzle
Then T = m& a [(1 + f )U e - U ]
T 1400
+U + 586.6
m& a 1.4531
Thus Ue = = = 1431.13 ft / s
(1 + f ) 1 + 0.0831
Exhaust velocity is U e = 1431.13 ft / s
d) The propulsive efficiency ( η p ) is expressed by the relation
T ×U
ηp =
T × U + 0.5 × ma × (1 + f ) × (U e − U ) 2
&
1400 × 586.8
ηp = = 59.42%
1400 × 586.8 + 0.5 ×1.4531× (1 + 0.0831) × (1413.13 − 586.8) 2
e) Thermal efficiency (ηth )
T × U + 0.5 × m& a × (1 + f ) × (Ue − U ) 2
ηth =
m& f × QR
=
1400 × 586.8 + 0.5 × 1.4531× (1 + 0.0831) × (1431.13 − 586.8) 2
ηth = = 32.6%
0.12082 × 35044660
3.2 Why is not mechanical efficiency an issue with ramjets?
