SOLUTIONS MANUAL for Modern Compressible Flow With Historical Perspective 4th Edition By John Anderson

, CHAPTER 1



p (5.6)(2116)
 = = = 0.00812 slug/ft3
RT (1716)(850)

1
v= = 123 ft3/slug



p (10)(1.01 x 105 )
1.2 = = = 11.0 kg/m3
RT ( 287)( 320)

p . x 10 5 )
(10)(101
n= = −23
= 2.87 x 1026/m3
RT (138
. x 10 )(320)

pv p . x 10 5 )
(10)(101 kg − mole
= = = = 0.0345
RT RT (110. )(8314)(320) kg


1.3 From the definition of enthalpy,

h = e + p v = e + RT (A1)

For a calorically perfect gas, this becomes

cp T = cv T + RT, or cp – cv = R

For a thermally perfect gas, Eq. (A1) is first differentiated

dh = de + Rdt

or,

cp dT = cv dT = Rdt

or,

cp – cv = R

T2 p
1.4 s2 – s1 = cp n - R n 2
T1 p1

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, (a) R = 1716 ft-lb/slugR

R (1.4)(1716)
cp = = = 6006 ft-lb/slugR
 −1 0.4

s2 – s1 = (6006) n (1.687) – (1716) n 4.5

s2 – s1 = 559.9 ft-lb/slugR

(b) R = 287 joule/kgK

R (14
. )( 287)
cp = = = 1004.5 joule/kgK
 −1 0.4

s2 – s1 = (1004.5) n (1.687) – 287 n (4.5)

s2 – s1 = 93.6 joule/kgK



p 2  T2   −1
1.5 = 
p 1  T1 


T   −1 1.4
p2 = p1  2  = 1800 (400/500) 0.4
 T1 

p2 = 824.3 lb/ft2

p2 824.3
2 = = = 0.0012 slug/ft3
RT2 (1716)( 400)


1.6 Volume of room = (20)(15)(8) = 2400 ft3

Standard sea level density = 0.002377 slug/ft3

Mass of air = (0.002377)(2400) = 5.70 slug

Weight = Mass x acceleration of gravity = (5.7)(32.2) = 184 lb



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, 1.7
(a) dp = -VdV

d
and d =   dp, or dp =


Combining:

d
= - VdV


d = -  p2 VdV

d
= -   VdV


d dV
= -   V2
 V

1 1
(b) s = = = 7.07 x 10-6 m2/N
p (1.4)(1.01 x 105 )

d dV
= s  V2 = - (7.07 x 10-6)(1.23)(10)2(0.01)
 V

d
= - 8.7 x 10-6


2
d  1000
(c) Here, will be larger by the ratio   .
  10 

2
d  1000
-6
= (- 8.7 x 10 )   = - 8.7 x 10-2
  10 

Comment: By increasing the velocity of a factor of 100, the fractional change in density is

increased by factor of 104. This is just another indication of why high-speed flows must be

treated as compressible.



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