Pa
P
02
101200
=
P
02
P
atm
:=
and hence
Pa
P
atm
101200
:=
P
01
P
atm
1
2
ρ
⋅
0
2
⋅
+
=
P
atm
=
- therefore, the total pressure in the test section is the same as the total pressure in the room
- but for 1 well away from the bellmouth, V
1
= 0 and P
1
= P
atm
- thus,
P
01
P
02
=
or
P
1
1
2
ρ
⋅
V
1
2
⋅
+
P
2
1
2
ρ
⋅
V
2
2
⋅
+
=
- thus, (1) reduces to
(since the working fluid is a gas, the potential energy change
would be small even if elevation changed between 1 and 2)
y
1
y
2
=
taking
m/s
2
g
9.81
:=
(1)
P
1
1
2
ρ
⋅
V
1
2
⋅
+
ρ
g
⋅
y
1
⋅
+
P
2
1
2
ρ
⋅
V
2
2
⋅
+
ρ
g
⋅
y
2
⋅
+
=
- consider the streamline from 1 to 2
- Bernoulli is applicable (steady, incompressible flow with negligible friction)
(a)
Total pressure in the test section:
P
atm
= 101,200 Pa
T
atm
= 18C
Water
15 cm
P = 95,000 Pa
1
2
4.1
Flow in open circuit wind tunnel:
MAAE2300/4.1

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