Aircraft Propulsion 2E Solution Manual 1.1
Chapter 1 Solutions
Problem 1.1
The Carnot cycle sets the limit on p2 p3
thermal efficiency of a heat engine
T2 2 3
operating between two temperature
limits. Show that ideal Carnot efficiency is:
p1 p4
T1
ηth = 1 − T1
T2 1 4
What is the thermal efficiency if T 1 =288 K
and T 2 =2000 K? s
Solution:
Following conservation of energy, the amount of work done by the system per unit mass is:
w = ∫ dq
For a reversible heat engine operating between two reservoirs at temperatures TH and TL ,
∫ dq = ∫ Tds = (T H − TL )∆s
Therefore, net work, per unit mass is w
w = (TH − TL )∆s
The heat input in the cycle takes place between stations 2 and 3, i.e.,
3
qin = ∫ Tds = TH ∆s
2
Therefore thermal efficiency of the cycle is:
w (TH − TL )∆s T
η th = = = 1− L
qin TH ∆s TH
For the cycle defined in the diagram, the thermal efficiency is
T1
η th = 1 −
T2
T1 288 K
Thermal efficiency: ηt =1 − =1 − ⇒ ηt =0.856 or 85.6%
T2 2000 K
,Aircraft Propulsion 2E Solution Manual 1.2
Problem 1.2 The ideal Brayton cycle operates between
two pressure limits as shown. It is
the model of an airbreathing jet engine,
such as a turbojet or ramjet engine. p2 = p 3
T2 2
Show that ideal Brayton cycle efficiency
is: p1 = p4 4
T1
T 1
ηth = 1 − 1
T2
s
What is the thermal efficiency of the Brayton
That has T 1 =288 K and T 2 =864 K? Note that maximum
cycle temperature T 3 has no effect on cycle thermal efficiency.
Solution: Net cycle heat exchange is:
∫ δq = ∫ Tds
Gibbs equation is:
Tds = dh - vdp
Therefore for a constant pressure process, Tds = dh,
3 3
∫ Tds = ∫ dh = h
2 2
3 − h2 = c p (T3 − T2 )
In a cycle, the net work output is equal to the net heat input (according to the 1st law of thermo)
∫ δw = ∫ δq = c p (T3 − T2 ) − c p (T4 − T1 )
By definition, cycle thermal efficiency is:
Wnet qin − qout
ηB = =
qin qin
Thus,
T
T1 4 − 1
c p (T4 − T1 )
= 1− 1
q 4−1 T
ηB = 1− = 1−
q 2 −3 c p (T3 − T2 ) T3
T2 − 1
T2
Since processes 1-2 and 3-4 are isentropic and p 3 = p 2 and p 4 = p 1 , we can write:
γ −1 γ −1
T4 T4 T3 T2 p 4 γ T3 p 2 γ T3
= = =
T1 T3 T2 T1 p3 T2 p1 T2
Therefore, the ideal Brayton cycle efficiency is simplified to:
T1
ηB = 1 −
T2
T1 288 K
Thermal efficiency: ηt =1 − =1 − ⇒ ηt ≈ 0.667 or 66.7%
T2 864 K
,Aircraft Propulsion 2E Solution Manual 1.3
The Brayton cycle operates between two isobars (constant pressure lines), therefore, it is the
pressure ratio that sets the thermal efficiency of the ideal Brayton cycle. The maximum cycle
temperature changes the amount of heat input and the work output in the same proportion such
that the ratio remains constant.
Problem 1.3
Humphrey cycle operates a constant-volume
combustor instead of a constant-pressure T3=Tmax 3
cycle like Brayton. Show that:
1
T v2 = v3
3 − 1 / T3 − 1
T T γ
ηth = 1 − γ 1
T2 T2
2
T2
p1 = p4
4
is the thermal efficiency of an ideal
Humphrey cycle. 1
Let us use the same T 1 as in Problems 1.1 s1=s2 s3=s4
and 1.2, i.e., T 1 =288 K. Let use the same
temperature T 2 as in Problem 1.2, i.e., T 2 =864 K.
Finally, let us use the same maximum cycle temperature as in Carnot
(Problem 1.1), i.e., T max =2000 K. With the ratio of specific heats γ=1.4,
calculate the thermal efficiency of Humphrey cycle. Compare the answer with
Brayton cycle efficiency.
3
3’
v=const.
T
p=const.
2
4
p=const.
1
s
Net cycle heat exchange is:
3 1
∫ δq = ∫ Tds = ∫ Tds + ∫ Tds
2 4
Since Gibbs equation is
,Aircraft Propulsion 2E Solution Manual 1.4
Tds = de + pdv
And the process from 2 to 3 is constant volume heating, Tds = de for a constant volume process,
3 3
∫ Tds = ∫ de = e
2 2
3 − e2 = cv (T3 − T2 )
Another form of Gibbs equation is
Tds = dh - vdp
Therefore for a constant pressure process, Tds = dh, therefore
1 1
∫ Tds = ∫ dh = h
4 4
1 − h4 = c p (T1 − T4 )
In a cycle, the net work output is equal to the net heat input (according to the 1st law of thermo)
∫ δw = ∫ δq = c (T v 3 − T2 ) − c p (T4 − T1 )
Thermal investment in the cycle is the integral of δq from 2 to 3.
3 3 3
∫ δq =∫ Tds = ∫ de = e3 − e2 = cv (T3 − T2 )
2 2 2
Therefore thermal efficiency of the ideal Humphrey cycle is:
T4
−1
cv (T3 − T2 ) − c p (T4 − T1 ) T4 − T1 T1 T1
η th = = 1− γ = 1 − γ
cv (T3 − T2 ) T3 − T2 T2 T3 − 1
T2
Now, we show that
1/ γ
T4 T3
=
T1 T2
Using chain rule, we may write:
( γ −1) / γ ( γ −1) / γ
T4 T4 T3' T2 p4 T p T3'
= = . 3' . 2 =
T1 T3' T2 T1 p3' T2 p1 T2
Note that p 3’ =p 2 and p 4 =p 1 .
γ −1 γ −1 γ −1
1− 1/ γ
T3' T3' T3 p2 γ T T v γ T T γ T
= = . 3 = 2 3 . 3 = 3 = 3 note that v 3 =v 2
T2 T3 T2 p3 T2 v2 T3 T2 T2 T2
Therefore, we show that the thermal efficiency of the ideal Humphrey cycle is:
1
T1 T3 γ T
ηth = 1 − γ − 1 / 3 − 1 QED
T2 T2
2
T
,Aircraft Propulsion 2E Solution Manual 1.5
Substituting numbers for T 1 , T 2 and T 3 in cycle thermal efficiencies, we get
ηth − Brayton ≈ 66.67%
ηth − Humphrey ≈ 70.85%
Problem 1.4 The rotor of a millimeter-scale gas turbine engine has a radius of 1 mm. It has to
reach a tip, or rim speed of near speed of sound for an effective compression.
Assuming that the speed of sound is 340 m/s, calculate the rotor rotational speed
in revolutions per minute (rpm).
R
Vtip
Rotor radius, R rotor = 1 mm = 0.001 m
Tip or rim speed, Vtip ≈ 340 m / s
Rotor rotational speed, N is given by:
1 rev 60 s
⇒ N = (340,000 rad / s )⋅
Vtip 340 m / s
N= = ⋅
Rrotor 0.001 m 2π rad 1 min
The final product yields:
N ≈ 3,250,000 rpm
Problem 1.5
Specific fuel consumption (sfc) projects the fuel economy of an engine, i.e. it
measures the fuel flow rate (say in pound-mass per hour or g/s) that leads to a
production of a unit thrust (say 1 pound-force or 1 Newton). Two sets of
numbers are copied from Table 1.1 (from EJ200 specification), which are:
Sfc (max. power) 0.81 lbm/hr/lbf (with AB-off)
Sfc w. AB 1.75 lbm/hr/lbf
Thrust (SL) 13,500 lbf
Thrust w. AB 20,250-22,250 lbf
,Aircraft Propulsion 2E Solution Manual 1.6
First note that afterburner (AB) use more than doubles the fuel consumption
while boosting the thrust by only ~50%. This explains the sparse use of an
afterburner in aircraft mission. Now to quantify, calculate the amount of
additional fuel burned in 30 minutes of afterburner use (producing 21,000 lbf
thrust) as compared to 30 minutes of no afterburner use (producing 13,500 lbf
thrust).
Solution: As stated in the problem, the afterburner use produces disproportionate thrust for
the extra fuel consumption.
The amount of fuel burned in 30 minutes with afterburner-off is:
1 hr
Fuel AB − Off = sfc ∗ Thrust = (0.81 lbm / hr / lbf ) * (13,500 lbf ) * 30 min×
60 min
Fuel AB − Off = 5,467.5 lbm
The amount of fuel burned in 30 minutes with afterburner-on is:
1 hr
Fuel AB − On = sfc ∗ Thrust = (1.75 lbm / hr / lbf ) * (21,000 lbf ) * 30 min×
60 min
Fuel AB − On = 18,375 lbm
Amount of additional fuel burned with the afterburner operating is ~13,000 lbm, which is more
than doubled the fuel consumed without the afterburner.
, Aircraft Propulsion 2E Solution Manual
Chapter 2 Solutions
Problem 2.1
Given that:
1 2
at=400 m/s
a1=300 m/s
γ=1.4
N.S.
Solution:
a) Using Equation 2.102-a,
a2 u2 a2
+ = t = const.
γ −1 2 γ −1
Rearranging and solving for u1,
⇒ u1 = 591.6 m / s
In order to get M1:
u1
M1 = ⇒ M 1 = 1.972
a1
b) First, we need to find M2 of the flow
Refer to Appendix C, Normal Shock Table (γ=1.4),
M 2 ≈ 0.58 for M1=1.972
Then, with the help of Equation 2.104,
1
Chapter 1 Solutions
Problem 1.1
The Carnot cycle sets the limit on p2 p3
thermal efficiency of a heat engine
T2 2 3
operating between two temperature
limits. Show that ideal Carnot efficiency is:
p1 p4
T1
ηth = 1 − T1
T2 1 4
What is the thermal efficiency if T 1 =288 K
and T 2 =2000 K? s
Solution:
Following conservation of energy, the amount of work done by the system per unit mass is:
w = ∫ dq
For a reversible heat engine operating between two reservoirs at temperatures TH and TL ,
∫ dq = ∫ Tds = (T H − TL )∆s
Therefore, net work, per unit mass is w
w = (TH − TL )∆s
The heat input in the cycle takes place between stations 2 and 3, i.e.,
3
qin = ∫ Tds = TH ∆s
2
Therefore thermal efficiency of the cycle is:
w (TH − TL )∆s T
η th = = = 1− L
qin TH ∆s TH
For the cycle defined in the diagram, the thermal efficiency is
T1
η th = 1 −
T2
T1 288 K
Thermal efficiency: ηt =1 − =1 − ⇒ ηt =0.856 or 85.6%
T2 2000 K
,Aircraft Propulsion 2E Solution Manual 1.2
Problem 1.2 The ideal Brayton cycle operates between
two pressure limits as shown. It is
the model of an airbreathing jet engine,
such as a turbojet or ramjet engine. p2 = p 3
T2 2
Show that ideal Brayton cycle efficiency
is: p1 = p4 4
T1
T 1
ηth = 1 − 1
T2
s
What is the thermal efficiency of the Brayton
That has T 1 =288 K and T 2 =864 K? Note that maximum
cycle temperature T 3 has no effect on cycle thermal efficiency.
Solution: Net cycle heat exchange is:
∫ δq = ∫ Tds
Gibbs equation is:
Tds = dh - vdp
Therefore for a constant pressure process, Tds = dh,
3 3
∫ Tds = ∫ dh = h
2 2
3 − h2 = c p (T3 − T2 )
In a cycle, the net work output is equal to the net heat input (according to the 1st law of thermo)
∫ δw = ∫ δq = c p (T3 − T2 ) − c p (T4 − T1 )
By definition, cycle thermal efficiency is:
Wnet qin − qout
ηB = =
qin qin
Thus,
T
T1 4 − 1
c p (T4 − T1 )
= 1− 1
q 4−1 T
ηB = 1− = 1−
q 2 −3 c p (T3 − T2 ) T3
T2 − 1
T2
Since processes 1-2 and 3-4 are isentropic and p 3 = p 2 and p 4 = p 1 , we can write:
γ −1 γ −1
T4 T4 T3 T2 p 4 γ T3 p 2 γ T3
= = =
T1 T3 T2 T1 p3 T2 p1 T2
Therefore, the ideal Brayton cycle efficiency is simplified to:
T1
ηB = 1 −
T2
T1 288 K
Thermal efficiency: ηt =1 − =1 − ⇒ ηt ≈ 0.667 or 66.7%
T2 864 K
,Aircraft Propulsion 2E Solution Manual 1.3
The Brayton cycle operates between two isobars (constant pressure lines), therefore, it is the
pressure ratio that sets the thermal efficiency of the ideal Brayton cycle. The maximum cycle
temperature changes the amount of heat input and the work output in the same proportion such
that the ratio remains constant.
Problem 1.3
Humphrey cycle operates a constant-volume
combustor instead of a constant-pressure T3=Tmax 3
cycle like Brayton. Show that:
1
T v2 = v3
3 − 1 / T3 − 1
T T γ
ηth = 1 − γ 1
T2 T2
2
T2
p1 = p4
4
is the thermal efficiency of an ideal
Humphrey cycle. 1
Let us use the same T 1 as in Problems 1.1 s1=s2 s3=s4
and 1.2, i.e., T 1 =288 K. Let use the same
temperature T 2 as in Problem 1.2, i.e., T 2 =864 K.
Finally, let us use the same maximum cycle temperature as in Carnot
(Problem 1.1), i.e., T max =2000 K. With the ratio of specific heats γ=1.4,
calculate the thermal efficiency of Humphrey cycle. Compare the answer with
Brayton cycle efficiency.
3
3’
v=const.
T
p=const.
2
4
p=const.
1
s
Net cycle heat exchange is:
3 1
∫ δq = ∫ Tds = ∫ Tds + ∫ Tds
2 4
Since Gibbs equation is
,Aircraft Propulsion 2E Solution Manual 1.4
Tds = de + pdv
And the process from 2 to 3 is constant volume heating, Tds = de for a constant volume process,
3 3
∫ Tds = ∫ de = e
2 2
3 − e2 = cv (T3 − T2 )
Another form of Gibbs equation is
Tds = dh - vdp
Therefore for a constant pressure process, Tds = dh, therefore
1 1
∫ Tds = ∫ dh = h
4 4
1 − h4 = c p (T1 − T4 )
In a cycle, the net work output is equal to the net heat input (according to the 1st law of thermo)
∫ δw = ∫ δq = c (T v 3 − T2 ) − c p (T4 − T1 )
Thermal investment in the cycle is the integral of δq from 2 to 3.
3 3 3
∫ δq =∫ Tds = ∫ de = e3 − e2 = cv (T3 − T2 )
2 2 2
Therefore thermal efficiency of the ideal Humphrey cycle is:
T4
−1
cv (T3 − T2 ) − c p (T4 − T1 ) T4 − T1 T1 T1
η th = = 1− γ = 1 − γ
cv (T3 − T2 ) T3 − T2 T2 T3 − 1
T2
Now, we show that
1/ γ
T4 T3
=
T1 T2
Using chain rule, we may write:
( γ −1) / γ ( γ −1) / γ
T4 T4 T3' T2 p4 T p T3'
= = . 3' . 2 =
T1 T3' T2 T1 p3' T2 p1 T2
Note that p 3’ =p 2 and p 4 =p 1 .
γ −1 γ −1 γ −1
1− 1/ γ
T3' T3' T3 p2 γ T T v γ T T γ T
= = . 3 = 2 3 . 3 = 3 = 3 note that v 3 =v 2
T2 T3 T2 p3 T2 v2 T3 T2 T2 T2
Therefore, we show that the thermal efficiency of the ideal Humphrey cycle is:
1
T1 T3 γ T
ηth = 1 − γ − 1 / 3 − 1 QED
T2 T2
2
T
,Aircraft Propulsion 2E Solution Manual 1.5
Substituting numbers for T 1 , T 2 and T 3 in cycle thermal efficiencies, we get
ηth − Brayton ≈ 66.67%
ηth − Humphrey ≈ 70.85%
Problem 1.4 The rotor of a millimeter-scale gas turbine engine has a radius of 1 mm. It has to
reach a tip, or rim speed of near speed of sound for an effective compression.
Assuming that the speed of sound is 340 m/s, calculate the rotor rotational speed
in revolutions per minute (rpm).
R
Vtip
Rotor radius, R rotor = 1 mm = 0.001 m
Tip or rim speed, Vtip ≈ 340 m / s
Rotor rotational speed, N is given by:
1 rev 60 s
⇒ N = (340,000 rad / s )⋅
Vtip 340 m / s
N= = ⋅
Rrotor 0.001 m 2π rad 1 min
The final product yields:
N ≈ 3,250,000 rpm
Problem 1.5
Specific fuel consumption (sfc) projects the fuel economy of an engine, i.e. it
measures the fuel flow rate (say in pound-mass per hour or g/s) that leads to a
production of a unit thrust (say 1 pound-force or 1 Newton). Two sets of
numbers are copied from Table 1.1 (from EJ200 specification), which are:
Sfc (max. power) 0.81 lbm/hr/lbf (with AB-off)
Sfc w. AB 1.75 lbm/hr/lbf
Thrust (SL) 13,500 lbf
Thrust w. AB 20,250-22,250 lbf
,Aircraft Propulsion 2E Solution Manual 1.6
First note that afterburner (AB) use more than doubles the fuel consumption
while boosting the thrust by only ~50%. This explains the sparse use of an
afterburner in aircraft mission. Now to quantify, calculate the amount of
additional fuel burned in 30 minutes of afterburner use (producing 21,000 lbf
thrust) as compared to 30 minutes of no afterburner use (producing 13,500 lbf
thrust).
Solution: As stated in the problem, the afterburner use produces disproportionate thrust for
the extra fuel consumption.
The amount of fuel burned in 30 minutes with afterburner-off is:
1 hr
Fuel AB − Off = sfc ∗ Thrust = (0.81 lbm / hr / lbf ) * (13,500 lbf ) * 30 min×
60 min
Fuel AB − Off = 5,467.5 lbm
The amount of fuel burned in 30 minutes with afterburner-on is:
1 hr
Fuel AB − On = sfc ∗ Thrust = (1.75 lbm / hr / lbf ) * (21,000 lbf ) * 30 min×
60 min
Fuel AB − On = 18,375 lbm
Amount of additional fuel burned with the afterburner operating is ~13,000 lbm, which is more
than doubled the fuel consumed without the afterburner.
, Aircraft Propulsion 2E Solution Manual
Chapter 2 Solutions
Problem 2.1
Given that:
1 2
at=400 m/s
a1=300 m/s
γ=1.4
N.S.
Solution:
a) Using Equation 2.102-a,
a2 u2 a2
+ = t = const.
γ −1 2 γ −1
Rearranging and solving for u1,
⇒ u1 = 591.6 m / s
In order to get M1:
u1
M1 = ⇒ M 1 = 1.972
a1
b) First, we need to find M2 of the flow
Refer to Appendix C, Normal Shock Table (γ=1.4),
M 2 ≈ 0.58 for M1=1.972
Then, with the help of Equation 2.104,
1
