Example:
The tank shown rolls along a level track. Water received from a jet is retained in the tank. The tank is to
accelerate from rest toward the right with constant acceleration, a. Neglect wind and rolling resistance.
Find an algebraic expression for
Continuous-Duty Wind Tunnels
Continuous duty wind tunnels utilize a compressor to produce the driving pressure gradient for the flow. In
order to minimize the required compressor power, the wind tunnel should operate as efficiently as possible,
i.e. as cl
Image from: http:/history.nasa.gov/SP-440/ch5-6.htm
Notes:
1.
There is a fixed amount of time for which the device will operate at the design test section Mach
number, MaTS, since the tank mass will decrease with time. To extend the duration of the test,
Again, in order to minimize the compressor power requirements, the losses in the system should be
minimized. The ideal case (shown below) is to have an isentropic deceleration from supersonic to subsonic
speeds.
Mat2 = 1
Mat1 = 1
MaTS > 1
p01
Ma < 1
Ma <
Example:
A blowdown wind tunnel exhausting to atmospheric pressure (14.7 psia) is to be designed. The test section
cross-sectional area is specified to be 1 ft2, and the desired test section Mach number is 2.0. The supply
tank can be pressurized to 150 ps
Supersonic Wind Tunnel Design
There are three common designs for supersonic wind tunnels:
1. high-pressure gas storage tanks (and/or vacuum tanks) for blowdown wind tunnels,
2. a compressor and diffuser for continuous-duty wind tunnels, and
3. shock tubes
Example:
A converging-diverging nozzle, with an exit to throat area ratio, Ae/At, of 1.633, is designed to operate with
atmospheric pressure at the exit plane, pe = patm.
a. Determine the range(s) of stagnation pressures for which the nozzle will be free
A converging-diverging nozzle with pressure taps along the
length of the device. The flow is from left to right.
The pressure ratio as a function of the axial distance in the CD nozzle for various back pressures.
Note the gradual
pressure rise due to the
4.
The location of a shock wave for a back pressure in the range corresponding to case 3 and case 5 can
be determined through iteration.
a. Assume a location for the shock wave (e.g. pick a value for A/At since the geometry is known).
b. Determine the Mac
Now consider what happens if we make the 2nd throat just a little bit larger than the 1st throat. As we
decrease the back pressure we will reach a case where the flow in the 1st throat becomes choked and a
shock wave forms in the diverging section of the
and we decrease the back pressure further, then the shock will be swallowed by the 2nd throat and the flow
within the test section will, at last, be supersonic (shown below).
Mat2 > 1
Mat1 = 1
p01
MaTS > 1
Ma < 1
Ma < 1
pback
At1 = A1*
p/p01
1
p*/p01
1st
3.
Over-speeding the diffuser is often impractical. For example, consider a diffuser designed to operate
at a Mach number of 1.7 (Ai/A* = Ai/At = 1.338). The critical Mach number for swallowing the shock
will be:
Ma,crit > 1
Note: A2* = At
1 2
1
1
Ma 2
2
As the upstream Mach number increases, the sonic area approaches the throat area, i.e. A* At, and the
shock moves closer to the inlet (the shock gets weaker and less flow needs to be diverted around the
diffuser). Eventually well reach design conditions b
Supersonic Diffuser Design
Another application where the efficient deceleration of a supersonic flow is of interest is a supersonic
diffuser at the inlet of aircraft jet engines. The flow entering a jet engine typically needs to be subsonic in
order to av
Example:
Consider a supersonic wind tunnel starting as shown in the figure below. The upstream nozzle throat area
is 1.25 ft2, and the test section design Mach number is 2.50. As the tunnel starts, a normal shock stands in
the divergence of the nozzle whe
5.
In real nozzles flows, the flow will typically separate from the nozzle walls as a result of the large
adverse pressure gradient occurring across a shock wave. Interaction of the shock with the separated
boundary layer results in a more gradual pressur
Once the wind tunnel is running and weve decreased the 2nd throat area, we should try to minimize the
stagnation pressure loss through the shock wave in the 2nd diverging section (and, hence, increase the tunnel
efficiency). To do this we increase the bac
Example:
A converging-diverging nozzle, with Ae/At = 1.633, is designed to operate with atmospheric pressure at the
exit plane. Determine the range(s) of stagnation pressures for which the nozzle will be free from normal
shocks.
C. Wassgren
Chapter 12: Ga
Notes:
1. The critical back pressure ratio corresponding to case 3 can be found from the isentropic relations (the
flow throughout the entire device is isentropic). Assume that the geometry, and hence the exit-tothroat area ratio, Ae/At, is given. Since f
Example:
A jet of water is deflected by a vane mounted on a cart. The water jet has an area, A, everywhere and is
turned an angle with respect to the horizontal. The pressure everywhere within the jet is atmospheric.
The incoming jet velocity with respect
Example:
A variable mesh screen produces a linear and axi-symmetric velocity profile as shown in the figure. The
static pressure upstream and downstream of the screen are p1 and p2 respectively (and are uniformly
distributed). If the flow upstream of the
The LME for Non-Inertial Frames of Reference
Recall that Newtons 2nd law holds strictly for inertial (non-accelerating) frames of reference. Now lets
consider frames of reference that are non-inertial (accelerating). First lets examine how we can describe
Example:
Incompressible fluid of negligible viscosity is pumped, at total volume flow rate Q, through a porous
surface into the small gap between closely spaced parallel plates as shown. The fluid has only horizontal
motion in the gap. Assume uniform flow
Now apply the linear momentum equation in the X-direction to the same control volume.
d
u X dV + u X ( u rel dA ) = FB, X + FS , X
dt
where
d
dt
CV
CV
u
X
(3.82)
dV = 0
CV
r=R
u X ( u rel dA ) = V12 R 2 +
CV
left side
= dA
2
3 r
2 V1 ( 2 rdr )
R
r =0
r
Substitute.
Fx = 2 V 2 RH
(3.75)
To determine V, apply conservation of mass to the same control volume.
d
dV + u rel dA = 0
dt
where
d
dt
CV
CS
dV = 0
(3.76)
(steady flow)
CV
= dA
=
u rel dA = Q +
inlet
CS
V Rd H = Q + V RH
=0
outlet
Substitute and s
Now apply the linear momentum equation in the X-direction to the same control volume.
d
u X dV + CS u X ( urel dA ) = FBX + FSX
dt CV
where
d
u X dV = 0 (steady flow)
dt CV
d
d
dA ) = V 2 hw + ( V 2 hw ) ( 1 dx ) + V 2 hw + ( V 2 hw ) ( 1 dx )
2
2
d
Example:
Water is sprayed radially outward through 180 as shown in the figure. The jet sheet is in the horizontal
plane and has thickness, H. If the jet volumetric flow rate is Q, determine the resultant horizontal
anchoring force required to hold the noz