بسم هللا الرحمن الرحيم

Sudan University of Science &Technology

College Of Post Graduates Studies

PhD Program in Mechanical Engineering by Courses and Dissertation

GE 721 - Advanced Combustion

Homework No. (1)

Prepared by student: Sabir Abushousha Ahmed Abushousha Supervisor:

Dr. Mohammed Hassan Mohammed Abuuznien

February 2015

Question 2

A rocket has a thrust of 8896 N and propellant consumption of 3.867 kg/sec. The

vehicle flies at a velocity of 400 m/sec and the propellant specific energy content

(heat of combustion) is 16.911 megajoule/kg (from Sutton). Find the following:

a. Effective exhaust velocity

b. Kinetic energy of the jet for 1 kg of fuel

c. Internal efficiency (

d. Propulsive efficiency

e. Overall efficiency

f. Specific impulse )

g. Specific propellant consumption

Given the following from the problem statement:

F = 8896 N �� = 3.867 kg

sec 𝑢 = 400

m

sec QR =6.911⋅106 J/ kg

a. Effective exhaust velocity:

𝐶 =F

m=

8896

3.867= 2300.5

m

sec

b. Specific kinetic energy of the jet:

𝐾𝐸𝑗𝑒𝑡 = 0.5 C2 = 0.5 ∗ (2300.5 )2 = 2.646130 ×106 J /kg

c. Internal efficiency

𝜂𝑖𝑛𝑡 =𝐾𝐸𝑗𝑒𝑡

QR 𝜂𝑖𝑛𝑡 =

2.646130 × 106

6.911∗106 = 0.38288 = 38.3

d. Propulsive efficiency

The speed ratio

𝑣 =u

c=

400

2300.5= 0.1739

𝜂𝑝 =2 . u

1 + u2=

2 ∗ 400

1 + (2300.5)2= 0.3375 = 33.75%

e. Overall efficiency:

𝜂𝑝 =F . u

m . QR=

8896 ∗ 400

3.867 ∗ 6.911 ∗ 106= 0.1331 = 13.3%

f. Specific impulse:

𝐼𝑠𝑝 =c

g=

2300.5

9.81= 234.6 s

g. Specific propellant consumption:

TSFCW =1

𝐼𝑠𝑝=

1

234.5848= 0.0042629

1

S

Question 4

For the rocket in Problem 2, calculate the specific power, assuming a propulsion system dry

mass of 80 kg and a duration of 3 min.

m =80

3 ∗ 60= 0.4444

𝑃𝑗𝑒𝑡 = 1

2𝐹𝑔0𝐼𝑠 =

1

2𝐹𝑔0

𝐶

𝑔0

=1

2𝐹 𝐶

𝑃𝑗𝑒𝑡 =1

2∗ 𝐹𝑣2 =

1

2∗ 8896 ∗ 2300.5 = 10232624 𝑤

∝=𝑃𝑗𝑒𝑡

𝑚0 =

10232624

80=127907.8 w

Question 7

Plot the variation of the thrust and specific impulse against altitude,

using the atmospheric pressure information given in Appendix 2, and the

data for the Minuteman first-stage rocket thrust chamber in Table 11-3.

Assume that P2 = 8.66 psia.

Solution

CONVERT THE UNITS 8.66 psi = 59708.598158924 pascal

Assuming a ratio of specific heats to be 1.3 and gas constant to be 345.7 kJ/kg K,

FROM TABLE 11

EXIT VELOCITY

𝑣2 = √2𝑘

𝑘 − 1𝑅𝑇 [1 − (

𝑝2

𝑝1)

(𝑘−1)/𝑘

]

Throat area (in 2 ) =164.2 =0.105935272 m2

• Expansion Area Ratio:

10 =𝐴2

𝐴𝑡

𝐴2 = 0.105935272 m2 ∗ 10 = 10.6𝑚2

Mass flow rate =A2*v2/V2

THURST 𝐹 = ��𝑣2 + (𝑝2 − 𝑝3)𝐴2

Specific Impulse 𝐼𝑆 =F

m g0

All data are tabulated in excel sheet attached to the home work

0.1013 MPa

atmospheric pressure which has the value

*A

A

A

A e

throat

exit