Attached.

Part 1
Question # 1:

Sol:
Given data
𝑃𝑜3
= 25
𝑃𝑜2
𝑄𝑟 = 18400
𝑃𝑜13
=6
𝑃𝑜2
𝛾 = 1.4
𝑇𝑜4 = 3250 𝑅
𝑀 = 1.6
𝐶𝑝 = 0.24
𝐵 = 0.5
From this attitude 40000ft we have
𝑇𝑎 = 389.97 𝑅
𝑃𝑎 = 2.730 𝑝𝑠𝑖
Then
𝑇𝑜 2 = 𝑇𝑜𝑎 = 𝑇𝑎 (1 +

𝛾−1 2
𝑀 )
2

= 389.97 ∗ (1 +

1.4 − 1 2
1.6 )
2

= 589.63 𝑅
𝛾
𝛾−1
𝑇𝑜 2
𝑃𝑜 2 = 𝑃𝑎 (1 + (
− 1))
𝑇𝑎
1.4
1.4−1
589.63
= 2.730 (1 + (
− 1))
389.97

= 11.55 𝑝𝑠𝑖
𝑃𝑜 3 =

𝑃𝑜3
∗ 𝑃𝑜2
𝑃𝑜2

= 25 ∗ 11.55
= 288.75 𝑝𝑠𝑖
𝛾−1

𝑃𝑜3 𝛾
)
𝑇𝑜 3 = 𝑇𝑜 2 ∗ (
𝑃𝑜2

1.4−1

= 589.63 ∗ (25) 1.4
= 1479.08 𝑅
𝑃𝑜 13 =

𝑃𝑜13
∗ 𝑃𝑜2
𝑃𝑜2

= 6 ∗ 11.55
= 69.30 𝑝𝑠𝑖
𝛾−1

𝑃𝑜13 𝛾
)
𝑇𝑜 13 = 𝑇𝑜 2 ∗ (
𝑃𝑜2
1.4−1

= 589.63 ∗ (6) 1.4
= 983.80 𝑅

𝑇𝑜 4
)
𝑇𝑜 3 − 1
𝑓=
𝑇𝑜
𝑄𝑟
( 4)
−
𝐶𝑝 ∗ 𝑇𝑜 3
𝑇𝑜 3
(

=

3250
(1479.08) − 1
18400
3250
−
(
. 24 ∗ 1479.08
1479.08)
= 0.0241

𝑃𝑜4 = 𝑃𝑜3 ∗ (1 + (

𝛾
𝑇𝑜4
− 1))𝛾−1
𝑇𝑜3

1.4
3250
= 288.75(1 + (
− 1))1.4−1
1479.08

= 4540.91
𝑇𝑜5 = 𝑇𝑜4 − (𝑇𝑜3 − 𝑇𝑜2 ) − 𝐵(𝑇𝑜13 − 𝑇𝑜𝑎 )
= 3250 − (1479.08 − 589.63) − .5(983.80 − 589.63)
= 2163.465 𝑅
𝛾

𝛾−1
𝑇𝑜5
𝑃𝑜7 = 𝑃𝑜5 = 𝑃𝑜4 ∗ (1 + (
− 1)
𝑇04
1.4
2163.465
= 4540.91 ∗ (1 + (
− 1))1.4−1
3250

= 1092.88 𝑝𝑠𝑖
𝑈 = 𝑀 ∗ √𝛾𝑅𝑇𝑎
= 1.6 ∗ √1.4 ∗ 1716 ∗ 389.97
= 1548.6 𝑓𝑡/𝑠
𝛾−1

𝛾
𝑃𝑎 𝛾
) 𝑅𝑇𝑜13 ∗ (1 − (
) )
𝑈𝑒𝑓 = √2 (
𝛾−1
𝑃𝑜13

1.4−1

1.4
2.730 1.4
) 1716 ∗ 983.80 ∗ (1 − (
)
)
= √2 (
1.4 − 1
69.30
= 2669.6 𝑓𝑡/𝑠
𝛾−1

𝛾
𝑃𝑎 𝛾
) 𝑅𝑇𝑜5 ∗ (1 − ( ) )
𝑈𝑒 = √2 (
𝛾−1
𝑃𝑜7

1.4−1

1.4
1.4
2.730
) 1716 ∗ 2163.465 ∗ (1 − (
)
)
= √2 (
1.4 − 1
1092.88

= 4614.87 𝑓𝑡/𝑠
𝑇
= (1 + 𝑓)𝑈𝑒 − 𝑈 = (1 + .0241)4614.87 − 1548.67
𝑚̇𝑎
= 3177.42 𝑙𝑏𝑓𝑠/𝑙𝑏𝑚
𝑇
= (1 + 𝑓)𝑈𝑒 + 𝐵𝑈𝑒𝑓 − (1 + 𝐵)𝑈
𝑚̇ 𝑜
= (1 + .0241) ∗ 4614.87 + .5 ∗ 2669.63 − (1 + .5) ∗ 1548.67
= 3737.90 𝑙𝑏𝑓𝑠/𝑙𝑏𝑚
𝑇𝑆𝐹𝐶 =

𝑓
. 0241
=
𝑇
3737.90
𝑚̇ 𝑜

= 6.45 ∗ 10−6 ∗ 3600
= 0.0240 𝑙𝑏𝑚/𝑙𝑏𝑓ℎ𝑟
Question:

Sol:
(a)
𝐾1 = 0.32,
𝐾2 = 0.0,
𝐶𝐷𝑜 = 0.027667
𝑉 = 𝑀 ∗ √𝛾𝑅𝑇𝑎
= 1.6 ∗ √1.4 ∗ 1716 ∗ 389.97
= 1548.67

𝑓𝑡
𝑠

𝐿
𝑆𝑤
𝐶𝐿 =
1 2
2 𝜌𝑉
40000 ∗ .92
720
=
0.5 ∗ 5.87 ∗ 10−4 ∗ (1548.67)2
= 0.0726
𝐶𝐷 = 𝐾1 𝐶𝐿2 + 𝐾2 𝐶𝐿 + 𝐶𝐷𝑜

= 0.32 ∗ 0. 07262 + 0 + 0.027667
= 0.0294
D = ½  x V2 x S x CD
1

=2 ∗ 5.87 ∗ 10−4 ∗ 1548.672 ∗ 720 ∗ 0.0294
= 9.621lbf
n = 100 – 0.05’
= 0.95
Uninstalled thrust per engine = 9.621/0.95
= 10.128lbf
b)
mp = 146.3 lbm/s
Minimum uninstalled specific thrust = Uninstalled thrust per engine / (mp/2)
=

10.128
146.3
2

= 0.138 𝑙𝑏𝑓𝑠/𝑙𝑏𝑚
c)
𝐶𝑙
0.0726
=
𝐶𝐷 0.0294
= 2.469
√T

TSFC = (0.4 + 0.45M) x 𝑇𝑠𝐶𝑑
216

= (0.4 + 0.45 x 1.6) x √288.14
=0.9697

Then

Cycle Analysis of Turbofan Engine
T/Mdot o vs. TSFC

4400.0000

4200.0000

4000.0000

TSFC

3800.0000

3600.0000

3400.0000

3200.0000

3000.0000
0.0200

0.0210

0.0220

0.0230

T/Mdot o

0.0240

0.0250

0.0260

Part 2

Range Factor vs. Mach # At Diff. Altitudes
3000.000Sea Level

10000 ft

30000 ft

36000 ft

40000 ft

0.8
MACH NO.

1

50000 ft

20000 ft

2500.000

RANGE FACTOR M)

2000.000

1500.000

1000.000

500.000

0.000
0

0.2

0.4

0.6

1.2

1.4

1.6

Endurance Factor vs. Mach # At Diff. Altitudes
sea level

10000 ft

30000 ft

36000 ft

40000 ft

50000 ft

20000 ft

12.000

ENDURANCE FACTOR(H)

10.000

8.000

6.000

4.000

2.000

0.000
0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

MACH NO.

d)
1.
The greatest tenacity The Mach number often drops off as you ascend. This is due to the fact that
the air is less thick and there is less drag at higher altitudes, necessitating slower speeds for
maximum endurance.
2.
At relatively low altitudes, where the aircraft can maintain the ideal balance of air density, engine
efficiency, and airspeed for least fuel consumption per unit of time, the optimum fuel consumption
for endurance typically happens.

3.
The ideal voyage (range) Altitude often causes the Mach number to rise. By reducing drag and
enabling greater speeds to be maintained with less fuel used per unit of distance, air density drops
as altitude rises.
4.
At an altitude where the aircraft can maintain an ideal balance between engine efficiency, air
density, and airspeed, fuel consumption for range is at its lowest. This is frequently encountered in
the mid- to high-altitude range because higher altitudes offer less dense air and less drag, enabling
effective cruising speeds with lower fuel consumption per unit of distance.
5.
From a social and worldwide standpoint, it is essential to comprehend EF (engine fuel flow) and
RF (range factor) plots for aviation engine design since it enables engineers to maximize fuel
economy and reduce environmental effect. Engineers can determine the best operating parameters
for aircraft engines by examining these graphs, thereby lowering fuel use, pollutants, and related
expenses.
6.
Optimizing range and endurance variables has a direct impact on fuel consumption efficiency for
a particular amount of flight time or distance to be covered. Reduced fuel usage helps to reduce
greenhouse gas emissions, dependence on fossil fuels, and the overall environmental effect of air
travel, all of which are important when taking into account social and global implications.


To2 (R)
Po2 (lbf/ft2)
Po13 (lbf/ft2)
To13 (R)
Uef (ft/s)
Ua (ft/s)

B

589.63464
11.604
69.622
983.812
2670.810
1548.668

Given Data
P_a (lbf/ft2)
2.73
R(ftlbf/(lb*°F)
1716.000
Ta (R)
389.97
To4 (R)
3250
Po13/Po2
6
Po3/Po2
25
Po4/Po3
0.95
Cp (Btu/lbmR)
0.24
Cp_b
1.235
M
1.6
q_r (Btu/lbm)
18400
B
0.5
y
1.4

0.1
Po3 (lbf/ft2)
185.6580988
208.8653611
232.0726235
255.2798858
278.4871482
290.0907793
324.9016729

B

To3 (R)
1302.019
1346.580
1387.733
1426.042
1461.938
1479.089
1527.766

Po4 (lbf/ft2)
4561.894
4561.894
4561.894
4561.894
4561.894
4561.894
4561.894

To5 (R)
2498.198
2453.637
2412.484
2374.175
2338.279
2321.128
2272.451

Po4 (lbf/ft2)
4561.894
4561.894
4561....