TEST BANK FOR Gas Dynamics 4TH Edition By Ethirajan Rathakrishnan

1 SomePreliminaryThoughts 1
2 BasicEquations ofCompressibleFlow 3
3 Wave Propagation 23
4 One-DimensionalFlow 25
5 NormalShockWaves 79
6 ObliqueShock andExpansionWaves 119
7 PotentialEquationforCompressibleFlow 157
8 SimilarityRules 161
9 TwoDimensionalCompressibleFlows 165
10 Prandtl-Meyer Flow 169
11 Flow with Friction and Heat Transfer 173
12 MOC 205
13 Measurements in Compressible Flow 207
iii
Chapter 2
Basic Equations of
Compressible Flow
2.1 In the reservoir, the air is at stagnation state. So, the entropy relation
would be
s2 − s1 = cp ln
!
T02
T01
"
− Rln
!
p02
p01
"
But, T01 = T02 for adiabatic process. Therefore,
"s = Rln
!
p01
p02
"
= Rln
!
p01
12
p01
"
= Rln 2
= 198.933 J/(kg K)
Note: It should be noted that, for entropy only subscripts 2 and 1 are used;
since entropy is not defined like static or stagnation entropy.
2.2 Let the initial state be denoted by subscript 1 and expanded state by subscript
2.
(a) Since the cylinder is insulated, preventing any heat transfer what-so-ever,
the process is adiabatic. The governing equation for this process is given by
p1V!
1 = p2V!
2 = constant (1)
Also, from ideal gas state equation
p1V1
T1
=
p2V2
T2
= R (2)
3
4 Basic Equations of Compressible Flow
From Eqs. (1) and (2), we get
p1
p2
=
!
V2
V1
"!
=
!
T1
T2
"!/(!−1)
Therefore,
T2 = T1
#
10(!−1) = 557.35K
"T = −842.65K
(b)
Work =
$
pdv =
$
dh −
$
du −
$
vdp (3)
Also,
pv! = constant from equation (1)
Differentiating equation (1), we have,
p!v!−1dv + v!dp = 0
Dividing throughout by v!−1 and integrating, we get
$
p!dv +
$
vdp = 0
$
vdp = −!w (4)
Substituting equation (4) in equation (3) and simplifying, we get
(1 − !)w = R"T
w =
R"T
1 − !
=
287 × (−842.65)
(−0.4)
= 6.04 × 105 J/kg
Note: Since the process undergone is expansion from a high pressure, the work
removed is positive, i.e, work is done by the gas.
(c) Also, from equation (1)
p1
p2
=
!
V2
V1
"!
= 101.4 = 25.1189
Therefore, The pressure ratio = 25.1189
5
2.3 p1v!
1 = p2v!
2, where v is specific volume, i.e. volume per unit mass = V/m.
Therefore,
p1
!
V1
m1
"!
= p2
!
V2
m2
"!
Also, V1 = V2 = V = volume of the tank.
p2 = p1
!
m2
m1
"!
= 5× 105 ×
!
1
2
"1.4
= 1.8946 × 105 Pa
From equation of state for a calorically perfect gas,
p1
p2
=
"1
"2
T1
T2
T2 =
!
p2
p1
"!
m1
m2
"
T1
=
!
1.8946
5
"
× 2 × 500
= 378.92K
2.4
p1
p2
=
!
T1
T2
"!/(!−1)
(a) Therefore,
T2 =
!
p2
p1
"(!−1)/!
T1
= 61/3.5 × 290 = 483.868K
The change in the temperature is
"T = T2 − T1 = 483.868 − 290
= 193.868K
(b) By first law of thermodynamics, we have
du + d(pe) + d(ke) = dq + dw
6 Basic Equations of Compressible Flow
Here, velocity changes are neglected. Therefore,
d(ke) = 0
Also, assuming
d(pe) = 0
The first law of thermodynamics reduces to
du = dq + dw
But the process is isentropic, thus dq = 0. Therefore,
du = dw = cv"T = 717.5 × 193.868
= 1.39 × 105 J/kg
(c) The work done is negative, i.e. work is done on the gas. It has been
computed in (b) above.
2.5 Work done by the weight on the piston goes towards increasing the internal
energy of the gas. From the first law of thermodynamics
E2 − E1 = Q +W
where, E, Q, and W are respectively the internal energy, heat transfered, and
work done. Since no heat is transfered, Q = 0. Therefore,
E2 − E1 = W =
$
F .ds
where, F is force and ds is distance. At the new equilibrium position, the force
acting on the piston face is F = p2Ap, Ap is the area of the piston face. The
distance traveled by the piston is ds = (V1 − V2)/Ap, V1 and V2 are the initial
and final volumes. Thus we have,
E2 − E1 = p2 .Ap(V1 − V2)/Ap
= −p2(V2 − V1)
For unit mass,
e2 − e1 = −p2(V2 − V1)
For calorically perfect gas, e = cvT. Therefore
cv(T2 − T1) = −p2
!
RT2
p2 −
RT1
p1
"
cv
R
!
T2
T1 − 1
"
= −
T2
T1
+
p2
p1
T2
T1
%
1 +
cv
R
&
=
cv
R
+ # (where # = p2/p1