Assignment #1: 09/29/2004 Due on 10/06/2004
1. Consider the isentropic flow through a supersonic wind-tunnel nozzle. The
reservoir properties are T0=500K and p0=10atm. If p =1atm at the nozzle exit,
calculate the exit temperature and density. (10points)
Answer:
From isentropic relation:
T
p
=( )
p
T
0
γ
y −1
0
p
1
T = T ( ) = 500( )
p
10
γ −1
γ
0
1.4 −1
1.4
= 259 K
0
p
1.01× 10
ρ=
=
= 1.359kg / m
RT (287)(259)
5
3
2. An airfoil is in a freestream where p∞ =0.61atm, ρ∞ =0.819 kg/m3, and V∞=300
m/s. At a point on the airfoil surface, the pressure is 0.5atm. (40 points)
a. Assuming isentropic flow, calculate the velocity at that point.
b. Calculate the percentage error obtained if you use the incompressible
Bernoulli equation for the same problem as (2.a).
c. For the same freestream condition, calculate the velocity at a point where
the pressure is 0.3atm.
d. Calculate the percentage error obtained if you use the incompressible
Bernoulli equation for the same problem as (2.c).
Answer:
a.
(0.61)(1.01× 10 )
T =
=
= 262.1K
(0.819)(287)
ρR
p
5
∞
∞
∞
because
p
T
=( )
p
T
∞
γ
y −1
∞
p
0.5
T = T ( ) = 262.1(
)
p
0.61
γ −1
γ
∞
1.4 −1
1.4
= 247.6 K
∞
Since the flow is isentropic, it is also adiabatic. Hence,
h = const.
0
h +
∞
V
V
= h+
2
2
∞
V = 2(h − h) + V = 2c (T − T ) + V
2
∞
∞
p
∞
2
∞
= 2(1004.5)(262.1 − 247.6) + (300) = 345m / s
2
b.
From Bernoulli equation
V
V
p +ρ
= p+ρ
2
2
2
2
∞
∞
V=
2( p − p )
∞
ρ
= 342.2m / s
2(1.01× 10 )(0.61 − 0.5)
+V =
+ (300)
0.819
5
2
∞
2
% Error = (
345 − 342.2
) × 100 = 0.81
345
C.
p
0.3
T = T ( ) = 262.1(
)
p
0.61
γ −1
γ
∞
1.4 −1
1.4
= 214 K
∞
V = 2(h − h) + V = 2c (T − T ) + V
2
∞
∞
p
∞
2
∞
= 2(1004.5)(262.1 − 214) + (300) = 432m / s
2
D.
V=
2( p − p )
∞
ρ
= 408m / s
% Error = (
2(1.01 × 10 )(0.61 − 0.3)
+V =
+ (300)
0.819
5
2
∞
432 − 408
) × 100 = 5.55
432
2