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Aircraft propolsion questions and solutions, Quizzes of Thermodynamics

Questions and solutions. 1-Brayton cycle. 2-combustion 3_Rocket Thrust

Typology: Quizzes

2021/2022

Available from 01/13/2022

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ISTANBUL TECHNICAL UNIVERSITY
FACULTY OF AERONAUTICS AND
ASTRONAUTICS
Jet Propulsion Principles (Quiz 3)
2021-2022 GÜZ
UCK 421E
(CRN: 12133)
Prof. Dr. Ali Kodal
İbrahim Asar
110170043
Deadline: 16.12.2021
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ISTANBUL TECHNICAL UNIVERSITY

FACULTY OF AERONAUTICS AND

ASTRONAUTICS

Jet Propulsion Principles (Quiz – 3 )

2021 - 2022 GÜZ

UCK 421E

(CRN: 12133)

Prof. Dr. Ali Kodal

İbrahim Asar

Deadline: 16.12.

QUESTION 1

Consider air standard ideal Brayton cycle. The cycle parameters are given as θ=2 and α=6.

Compare the increase in the net work output and in the thermal efficiency when we double θ as

for the same . Interpret your result.

SOLUTION 1

Compare net work output and thermal efficiency

Assumptions

Standard air,   1.4 ,   6,  2, T 1  298 Kc p 1.005 kJ / kg K.

Figure 1: Ideal Brayton Cycle

1 - 2 Isentropic compression (in a compressor)

2 - 3 Constant-pressure heat addition

3 - 4 Isentropic expansion (in a turbine)

4 - 1 Constant-pressure heat rejection

The T-s and P-v diagrams of an ideal Brayton cycle are shown in Fig. 1. Notice that all four

processes of the Brayton cycle are executed in steady-flow devices; thus, they should be

analyzed as steady-flow processes. When the changes in kinetic and potential energies are

neglected, the energy balance for a steady-flow process can be expressed, on a unit-mass basis,

as

 q in^ ^ qout^  ^  win^ ^ wout^ ^ hexit^ hinlet Eq.1.

Therefore, heat transfers to and from the working fluid are

qin  h 3  h 2  c (^) P T 3 T 2  Eq.1.

And

qout  h 4  h 1  c (^) P T 4 T 1  Eq.1.

First case

  1.4 ,   6,  2, T 1  298 Kcp 1.005 kJ / kg K.

wnet  1.005 (kJ / kg K. ) 298( K )(6 2)(1 0.5)

wnet 598.98( kJ / kg)

 th  

Second case

  1.4 ,   6,  4, T 1  298 Kcp 1.005 kJ / kg K.

wnet  1.005 (kJ / kg K. ) 298( K)(6  4)(1 0.25)

wnet 449.235( kJ / kg)

 th  

General case

Figure 2-3: wnet and thermal efficiency changing versus

Discussion

Figure 3 shows that while increasing, thermal efficiency is increases. For given =6 there

will be value of for which has maximum value. For    becomes wnetmaximum. For

the same value an increase in  means an increase in temperature T 2. This means that the

compressor is doing more work. So there is an optimum level for wnet. This is the case where

QUESTION 2

Ethane (C 2 H 6 ) at 25°C is burned in a steady-flow combustion chamber at a rate of 5 kg/h with

the stoichiometric amount of air, which is preheated to 500 K before entering the combustion

chamber. An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to

H 2 O but only 95 percent of the carbon burns to CO 2 , the remaining 5 percent forming CO. If

the products leave the combustion chamber at 800 K, determine the rate of heat transfer from

the combustion chamber.

SOLUTION 2

Ethane gas is burned with stoichiometric amount of air during a steady-flow combustion

process. The rate of heat transfer is to be determined.

Assumptions

Steady operating conditions, Ideal gases, Kinetic and potential energies are negligible,

Combustion is complete

Solving the problem

2 6 (25 )

800 500 1

out

C H C Combustion products Air chamber K K atm

Q

The molar mass of C H 2 6 is 30 kg /kmoland theoretical combustion equation of C H 2 6 is

C H 2 6  k O( 2  3.76 N 2 )  2 CO 2  3 H O 2 3.76k N 2

Where k is the stoichiometric coefficient and is determined from the O 2 balance,

k  2  1.5 3.

Then equation becomes,

C H 2 6  3.5(O 2  3.76 N 2 )  1.9CO 2  0.1CO  3 H O 2  0.05 O 2 13.16N 2

The heat transfer for this process is determined from the energy balance

out P (^ f f )^ R (^ f f) P R

 Q  (^)  N h  h  h  (^) N h  h h Eq.2.

Assuming the air and the combustion products to be ideal gases. From the above table

we can determine the enthalpies of formation of gases at different temperatures.

QUESTION 3

A rocket motor burns propellant at a rate of 50 kg/s. The exhaust speed is 3500 m/s and the

nozzle is perfectly expanded. Calculate

(a) the rocket thrust in kN

(b) the rocket motor specific impulse Isp (s).

SOLUTION 3

a) Rocket Thrust

The rocket produces thrust, T, by expelling propellant mass from a thrust chamber with a

nozzle. The propellant (fuel + oxidizer) mass flow rate is mand the ambient pressure of the

surrounding air is P 0.

Figure 4: Rocket thrust schematic.

Ae = nozzle exit area

Pe = area averaged exit gas pressure

e = area averaged exit gas density

Ve = area averaged x − component of velocity at the nozzle exit

Force balance of the rocket is;

2

0  T  P A 0 e  ( eV e Ae P Ae e)

Then our rocket thrust formula is;

2

T  e Ve Ae  ( Pe P 0 )Ae

The propellant mass flow is;

e e e

m  V A

and the rocket thrust formula is often written

T  mVe  ( Pe P 0 )Ae

If the nozzle exit pressure Pe< P 0 the nozzle is over expanded.

Pe =P 0  Perfectly Expanded

Pe >P 0 Under Expanded

In this question the nozzle perfectly expanded so the rocket thrust equation becomes;

T mV

T  50 kg / s (3500 m / s)

T  175, 000 N  175 kN

b) Specific Impulse

The specific impulse defined as the thrust per unit weight flow of propellant and so the

gravitational acceleration at the surface of the Earth is always inserted. The specific impulse is

defined as

(Rocket Thrust Eq.) T  mVe  ( Pe P 0 )Ae

Equivalent Velocity

e e e

P P A F

C V

m m

In this question C Ve because the nozzle is perfectly expanded

I  Total ımpulse F   t F dt   mC dt mC

Total Impulse Specific Impulse: Weight

0 0

sp

T C

I

mg g

In this question, the rocket is assumed at sea level. Therefore, the gravitational acceleration is

taken as 9.81.

2

sp

m s I s m s