Solutions Manual For Aircraft Propulsion 1st Edition By Saeed Farokhi

Chapter 11.1

Fvacuum  242500 lb Fsealevel  190400 lb

Psl  14.7 lb/ft 2
Guess A: A  1
Given
Fvacuum  Fsealevel ( Psl  0)  A

( Aval )  Find ( A )
Aval
 24.612622826908541194
144

Nozzel Area=24.61 ft 2


24.61
D2  4 


D2  5.598 ft 2

,Chapter 11.2


Dth  0.2355m
D2  1.430m
kg
mdotp  2393
sec
P2  7300Pa

Is  337.sec (vacuum)

Nozzle Area Ratio

AR=A2/Ath=(D2/Dth) 2

2
AR   
D2

 Dth 

AR  36.871

Thrust=F=Is x mdot x g 0


Thrust  Is  g  mdotp
6 -2
Thrust  7.908  10 s  m kg
Thrust
c 
mdotp

3 -1
c  3.305  10 s  m

Thrust Pressure Thrust
PThrust 
2
D2

4

6 -2 -1
PThrust  4.924  10 s  m  kg Pressure Thrsut=4.924 MPa

,Chapter 11.3


kg
mdotp  1000
sec
m
C  3500
sec

Thrust  C  mdotp
6 -2
Thrust  3.5  10 s  m kg
Thrust
Is 
g  mdotp

Is  356.901 s

, Chapter 11.4


lb
pc  1000
2
in
2
Ath  1.5ft


  1.2

lb
p0  14.7
2
in
1    1

 2  2  1   p0   
CF      2   1   
   1     1   pc  

CF  1.597 For Perfect Expansion Case


Thrust  CF  pc  Ath
5
Thrust  1.564  10 kg


   1  
  pc    2 
M2    p0   1   
     1 

M2  3.194

AR=(A2/Ath)

   1
  1 
 1     1 2 
1     2   M2  
( AR )      
 M2    1     1   12  
   2   
AR  8.87