TSFD Qualifying Exam- Fall 2005
Choose 4/6 problems and clearly mark which problems you want graded.
Each problem is worth 25 points.
Fluids1:Problem 1 (estimated time 45 minutes)
Given:
- A two-dimensional, inviscid thermal plume rises from thermal source (e.g., campfire) into a quiescent ambient as sketched below in Figure A).
- Assume a constant ambient pressure and temperature everywhere above the thermal source.
- The flow and ambient fluid possess a common density .
- Note that is the axial direction and is the transverse direction with the plume centerline being . Consider the depth of the plume into the page to be equal to .
- A small distance from the thermal source, the plume has a uniform axial velocity and a half-width . Assume that the only purpose of the thermal source is to provide the initial velocity of at . After that point, the plume velocity is opposed by gravity which points in the negative direction. To first order, assume that the flux of axial momentum remains constant.
- At a distance from the plume exit, the plume has spread to a half width and has an axial velocity profile , where andare unknown.
Find:
a)Determine at a distance .(6 points)
b)Determine at a distance .(6 points)
c)Determine the net mass flux of ambient fluid entrained by the plume between and . (7 points)
d)Qualitatively explain the “smoke in the camper’s face” phenomena shown in figure B). Specifically, provide physical insight as to why the smoke plume from a camp fire always seems to blows in the camper’s face regardless of which side the he stands on. Assume, zero transverse wind. (6 points)
Solution:
Basic Eqns:Momentum across a control surface:
Integral form of the continuity equation:
Bernoulli’s equation:
Schematic: control volume…
Assumptions:a) Steady flow.
b) Incompressible flow.
c) Constant pressure everywhere outside nozzle exit.
Solution:
See the above schematic for the proper control volume sketch.
Fluids 1/Fluids 2: Problem 2 (estimated time 45 minutes)
Given:
- An axisymmetric laminar jet exits into a quiescent ambient as sketched below.
- Consider a cross-section sufficiently down stream of the exit, such that the jet is fully developed and exhibits a self-similar profile.
- Here are the continuity and z-direction Navier-Stokes momentum equations, s
Find:
a)Write down the cylindrical coordinate form of the incompressible continuity equation and the z-momentum equation using the coordinate and velocity system. Transfer this equations to axisymmetric boundary layer coordinates , making the following assumptions: steady flow, axisymmetric flow, constant pressure everywhere, neglect gravity, and thin boundary layer . Your results should be . (4 points)
b)Define a stream function that identically satisfies the continuity equation, . (6 points)
c)Assume a similarity solution given by . Show that this assumption yields .
(6 points)
d)Determine the governing ODE in terms of similarity variables.Specify and list the appropriate number of boundary conditions. (9 points)
Solution:
a)Applying the above assumptions to the continuity and momentum equations yields . Transforming the coordinates yields the axisymmetric boundary layer equations.
b)Choosing a stream function that satisfiesor in boundary layer coordinates,. Plugging these expressions into the identity gives, which when evaluated yieldsand this expression is always true. Thus the chosen form of the stream function is valid.
c)Since this giveswhich yields. Similarly for ,
or.
d)Now taking the expressions from part (c) and substituting into the momentum equation gives the following for each of the terms:
Now combining all of these terms yields
Simplifying this expression results in
The appropriate boundary conditions are as follows:
When these conditions are transformed into similiarity coordinates, the result is
Fluids 2: Problem 3 (estimated time 45 minutes)
Given:
- When a glass channel is filled with an appropriate solution, a phenomenon called electroosmosis (EO) occurs when an electric field, E, is applied to the ends. As a result, a body force, Fx, in the x-direction (Fig. 3) is generated and thus is able to pump liquid along the channel.
- For the 2-D channel, the body force may be expressed as:Fx = , where k-1 is the Debye length (which is typically on the order of 10-9 m; and you can assume k=109 m-1 in this problem), is a dimensional constant that is related to the strength of the electric field and the permittivity of the liquid.
Find:
a)Starting from the mass conservation equation and momentum equation, obtain a differential equation governing the x-component velocity (u) for a 2-D channel of constant height h. Make appropriate assumptions as you think necessary and state them clearly. Provide details of simplifications. State the boundary conditions for u(y). (7 points)
b)Solve the velocity profile u(y), and obtain the shear stress profile. What is the maximum velocity? (6 points)
c)If h=100 m (=10-4m), how thick is the Debye length comparing with the channel height 2h? What is implication of this value in terms of the velocity profile? How do you describe the velocity profile in the bulk of the channel away from the wall? (6 points)
d)Evaluate the wall shear stress. In your best judgment, why is the wall shear stress independent of the viscosity? Explain it in terms of physics.(6 points)
Solution:
a)continuity:(1)
momentum:(2)
driving force: EO force (assumed to be constant in time) => steady flow;
fluid = liquid => incompressible and =constant
constant channel height h => =>fully developed flow
channel flow => flow is 2-D; u = (u, v)
=> =>
Since v on the wall is 0 due to no penetration condition =>
(3)
v-momentum equation =>(4)
; =>
V=0; 2-D & => =ex
=>(5)
Since each side is a respect function of x and y, both must equal to a same const.
However, since there is NO applied pressure gradient, it is clear that must be 0. Hence (6)
Boundary conditions for u(y) are:
u(y=h/2) = u(y=-h/2) = 0(no-slip) (7)
b)Integrating Eq. (6) twice w.r.t. y =>
Symmetry of the BC and geometry requires that .
No-slip condition =>.
Hence(8)
Obviously, the maximum velocity is attained at y=0 (centerline) so that
(9)
c)For k-1=10-9m, h=10-4m, k-1/h=10-5 « 1.
Consider the wall near y=-h/2.
Because cosh(kh/2) ~ exp(kh/2) exponentially, the leading order of as kh is exp[k(-h/2-y)] for –h/2y0.
Hence ~ 0 for most of the region in –h/2y0 => u~.
When y is near –h/2, exp[k(-h/2-y)] ~O(1) so that ~O(1) => u~O(1).
That is, in the region, say, k|-h/2-y|<3 we have u~O(1) because exp[k(-h/2-y)]~O(1).
Outside this region, exp[k(-h/2-y)]~0 so that u~.
Hence, the velocity profile is pretty much flat with u~ in the channel except in a very small region about 3 time of Debye length where velocity profile transits from u~ to 0 at the wall.
- Since => =>
which is independent of the viscosity.
This flow is caused by EO force. The wall shear stress is generated in response to the applied force. From force balance point of view, the wall shear stress should balance the driving force applied to the fluid element across the channel. Thus
Fluid 1 Problem 4
Given:
- Consider an unsteady, incompressible viscous flow over an airfoil as shown on the right.
- The prototype has a length of Lp=2 m and is designed to fly at a cruise speed of U=70 m/s in standard atmospheric condition.
- Due to the wind gust, the airfoil effectively experiences an upstream velocity ofU(1+sint) where U is the amplitude of the fluctuating velocity and is the angular frequency. You can view U as the time-average upstream velocity.
Find:
a)Write down the governing equations and necessary boundary conditions for the unsteady, incompressible, viscous flow that is relevant to the present flow condition. (5 points)
b)Non-dimensionalize the governing equations and boundary conditions the using:u* = u/U, x* = x/L, t* = t, p* = p/U2, What are the important dimensionless parameters? (5 points)
c)For air viscosity of =0.000014 m2/s, calculate the Reynolds number and Mach number. You may use a=340 m/s for the speed of sound. Estimate the percent change of the density from the upstream to the front stagnation point by assuming U=70 m/s. (5 points)
d)Experimental studies are to be conducted in a wind tunnel using a model of Lm/Lp=1/5. For Reynolds numbers to match (so that Rem=Rep), what should the average upstream velocity Um be in the model? For the unsteady effects to be modeled correctly, what frequency m should be used if p= 5 rad/s under the condition that Rem=Rep? (5 points)
e)If you use the value of the average upstream velocity Um based on the above calculations using Rem=Rep, what additional serious problem will you encounter in achieving dynamic similarity? How do you solve this problem? Why? (5 points)
Solution:
a.For incompressible flow, we have
mass conservation:(1)
momentum:(2)
At U=70 m/s, it is basically a low speed aerodynamic problem. The energy equation does not need to be considered as the temperature change due to the change of the kinetic energy is relatively small (see part c.).
Hence we neglected the variation of viscosity in Eq. (2).
For a coordinate system attached to the airfoil, the B.C.’s are:
u = 0 on the surface of the airfoil(3)
u = U(1+sint)exas |x| (4)
b.(5)
momentum:(6)
whereSt = L/U = Strouhal number(7a)
Re=UL/ = Reynolds number(7b)
These two are important dimensionless numbers for low speed unsteady aerodynamic problems.
c.Re = UL/ =70*2/0.000014 = 107.
Mach number: M=U/a=70/340=0.20588 which can be considered to be small.
Temperature variation from upstream to the stagnation point is:
T0/T = 1+(-1)/2
based on inviscid energy equation following the streamline.
Hence the density variation for an isentropic process (from upstream to stagnation point) is
0 / = (T0/T)1/(= [1+(-1)/2]1/(~ 1+/2 for small /2
For = 0.20588 => /2= 0.0212
Thus the density variation will be around 2%. (Hence it is acceptable to assume that the flow is nearly incompressible.)
d.For Lm/Lp=1/5 and Rem=Rep, and using the same air (so that m=p), we have
Um= 5Up =5*70= 350 m/s
For the Strouhal number to match, we require
Stm = mLm/Um = Stp = pLp/Up= 5*2/70 = 0.142857
Hence m = pLpUm /(LmUp) = 5 * 5 * 5 = 125 rad/s
e.Problem: at Um = 350 m/s, the Mach number of the model is
Mam = 350/340 = 1.0294
Thus the model flow will be transonic while the prototype flow is compressible.
To address this issue, we recognize that Rep = 107 is very high and the flow is fully turbulent. Since the drag on the airfoil in the incompressible flows is controlled mainly by turbulence, even if Rem =106, the characteristics of the drag coefficient will not change. Hence, we can still perform the experiment at, say, Um= Up =70 m/s so that Rem= 2 x 106, this will keep Ma number the same so that the effects of compressibility remain the same between the prototype and the model. For the behavior of lift, it is largely controlled by the pressure which in turn depends on the separation point. Since the separation point is fairly insensitive to Re, the behavior of the lift coefficient therefore will not be affected by Re.
Thus, we use Lm/Lp=1/5 and Um= Up =70 m/s, the frequency of the fluctuation should be
m = pLpUm /(LmUp) = 5 * 5 = 25 rad/s
Compressible Flow: Problem 1 (estimated time 50 minutes)
Given:
- A two-dimensional duct of height H=30 cm passing through a confined region (shaded areas), as shown in the sketch below.
- Originally designed as a constant height duct delivering a uniform flow of air at p=p1=1 atm and M=M1=3 to the ambient atmosphere where p=pa=1 atm.
- Embedding instrumentation in the duct where it passes through the confined region resulted in inserting a symmetrical section with half-diamond walls (the section between the dotted lines) to replace the original constant area section
- The contractor claims that this insert produces zero axial force on the welds connecting the two sections, a net loss in total pressureof less than 5% (assuming that the flow is inviscid throughout), and minimal loss in flow uniformity at the exit.
Find:
(a) Sketch the expected flow field in the duct insert showing the centerline streamline and an intermediate streamline; include shocks and expansions as necessary (4 points)
(b) The pressure and Mach number in regions 2, 3, 4, and 5. (12 points)
(c) The net axial force on the inserted section (3 points)
(d) The percent change in total pressure across the inserted section (3 points)
(e) The extent of flow uniformity, i.e., constant M, at the exit plane (3 points)
Solution (for air with =1.4):
(a) The expected flow field is shown in the sketch below. The shocks are shown as bold lines and the expansion fans are shown as light lines. The actual flow will be calculated subsequently.
(b) Region 1: The total pressure in section 1 is pt,1=p1/(p1/pt,1)= (1 atm)/0.02722=36.74, where the ratio of static to stagnation is given by
p/pt= [1+0.2M2]-3.5(1)
Alternatively the ratio may be obtained from the isentropic flow tables.
Region 2: The flow deflection in region 2 is =arctan(7.25/51.61)=8o and the corresponding shock angle is =25.61o. This may be obtained from the vs. graph for M1=3, and/or using the appropriate equation, for example:
tan = 2cot (M2sin2 – 1)/[M2(+ cos2) + 2](2)
The Mach number normal to the shock, M1,n=M1sin2=3(sin 25.61)= 1.30 and the normal component downstream of the shock is M2,n=0.79, as may be determined from the normal shock tables or from
M2,n= [(M1 sin 2)2 + 5)/(7M1sin 2)2 – 1)]1/2 (3)
.
The corresponding Mach number may be found from
M2= M2,n/sin(2 – )= 2.61 (4)
The static-to-total pressure ratio may be found from the isentropic flow tables or from Eq. 1, which yields the result p2/pt,2=0.049. The total pressure ratio across the shock is given by
pt,2/pt,1=[6M1,n2/(M1,n2+5)]3.5[6/(7M1,n2 – 1)]2.5
Instead, the ratio may be read directly from the normal shock tables with the result that pt,2/pt,1=0.98 so that pt,2=0.98(36.74)=36 atm, and p2=0.049(36)=1.76 atm.
Region 3: In region 3 the flow must turn back through 3=8o to become axial again, and for this 3=28.96o. The normal Mach numbers are M2,n=M2sin3=2.61sin(28.96)=1.26 and M3,n=0.80. Then M3= M3,n/[sin(3-3)]=0.8/sin(20.96)= 2.23, for which the Prandtl-Meyer angle =32.50o. The pressure ratios are p3/pt,3=0.089 and pt,3/pt,2=0.98 so that pt,3=0.98(35.99)=35.27 and p3=0.089(35.27)=3.13 atm
Region 4: The flow is turned another 8o into region 4, but this is an expansion and 4=3+8=40.50o. Here the Mach number is M4=2.56 and p4/pt,4=0.053. Since pt,4=pt,3 the pressure p4=0.053(35.27)=1.86 atm .
Region 5: In region 5 the flow turns through again but this time the boundary condition is that the pressure on the slip surface leaving the lip of the duct is equal to the ambient pressure of pa=1 atm. However these expansion waves do not interact with the surface of the duct so they contribute nothing to the forces on the duct. On the other hand, the Mach number in region 5 is given by that appropriate to 5=4+8=48.50o, or M5=2.94 and p5/pt,5=0.029. Then, since pt,5=pt,4=pt,3=35.27 atm, we have p5=0.029(35.27)=1.02 atm.
The net axial force, in the absence of friction and taking the depth of the duct to be equal to D, is given by
F=(p2-p4,avg.)A=(p2 – p4’)2(0.0725 m) D=[1.76(0.155)-1.78(0.155-0.0243)-p’(0.00738)]D
F= (0.273 – 0.233)D – p’(.0243)D= 0.04D - .0243p’D atm-m2
The last characteristic of the expansion fan to hit the surface of the duct would be at an inclination of 23.17o, corresponding to a value of =35.53o and an expansion of 2o. Examination of the expansion from the corner in 2o increments shows that the first increment results in a Mach number of M=2.35 which has =25.18o. Thus the inclination of this line to the horizontal is 25.18o -2o =23.18o, which is considered close enough to the required value of 23.17o. This characteristic represents an additional 2o expansion from n4=40.51 and M4=2.56 to n=42.51 and M=2.65. Then the static-to-total pressure ratio is p/pt =0.046, so the pressure p=0.046(35.27) =1.62 atm. If we assume a linear pressure variation across the last 5.25 cm of the inserted section then p’=(1.78+1.62)/2=1.7 atm and the total axial force is approximately given by
F= 0.04D - .0243p’D atm-m2=[.04-.0243(1.7)]D= -0.00131D atm-m2= -0.133D kPa
(c) The percent change in total pressure across the section is given by
[(pt,5-pt,1)/pt,1]x100=(pt,5/pt,4)(pt,4/pt,3)(pt,3/pt,2)(pt,2/pt,1) – 1 =(1)(1)(0.99)(0.98) -1= -3%
(d) The flow is axial and uniform at the exit plane over an extent of the central portion of that plane bounded by the last two characteristics. From geometrical considerations this vertical extent can be found to be 11.42 cm, as shown in the sketch.
Compressible Flow: Problem 2 (estimated time 50 min.)
Given:
- A long smooth circular pipe 14.6 cm in diameter connects a large high pressure air reservoir (pressure po=13 MPa and temperature To=420 K) having a contoured contraction section with an exit Mach number M1=0.14 to a contoured nozzle supplying a circular pipe 30.04 cm in diameter with supersonic flow.
- Just upstream of the nozzle is a heat exchanger that is used to raise the temperature of the air flow passing through the 14.60 cm diameter pipe. The heated section is designed for an entrance Mach number M2=0.5.
- A schematic diagram of the facility is shown below
Find:
(a) Find the thermodynamic properties (p,T, po, To,) in sections 1, 2, 3, and 4
(16 points)
(b) Find the thermal power required from the heater (5 points)
(c) Find the length L of the smooth pipe section (4 points)
Solution (for air with =1.4):
(a) Conditions at the various stations. Here we use the subscript “o” for stagnation conditions:
Location 1 The Mach number is given as M1=0.14, and this is the inlet to the smooth pipe. Assuming that this is a pipe flow with only friction acting we can use the adiabatic flow with friction (Fanno flow) tables to find po,1/po,1*=4.182, p1/p1*=7.809, T1/T1*=1.195, and 4fL1*/D=32.511. We also know from the isentropic flow tables that at station 1, for M1=0.14, we have p1/po,1=0.9864 and T1/To,1=0.9961. We assume that the flow from the stagnation conditions in the reservoir (po=13 MPa and To=420K) through the contoured contraction is isentropic up to location 1. From these relations we find
p1=12.833 MPa; p1*=1.642 MPa; po,1=13 MPa
T1=418.36 K; T1*=350.10 K; To,1=420 K
Location 2 The Mach number is given as M2=0.5 and this station is the exit of the long smooth pipe and the entrance to the heater section. Assuming that this is a pipe flow with only friction acting we can use the adiabatic flow with friction (Fanno flow) tables to find po,2/po,2*=1.340, p2/p2*=2.138, T2/T2*=1.143, and 4fL2*/D=1.0691. We also know from the isentropic flow tables that at station 2, for M2=0.50, we have p2/po,2=0.8430 and T2/To,2=0.9524. The critical conditions at station 1 are the same throughout the smooth pipe section right up to station 2 and the total temperature is a constant in the pipe. From these relations and the critical conditions determined at station 1 we find
p2=3.511 MPa; p1*= p2*= 1.642 MPa; po,2=4.165 MPa
T2=400.20 K; T1*= T2*= 350.10 K; To,2=To,1= 420 K
Location 3 The Mach number at station 3 is not specified. This station is the exit of the heated section and the entrance to nozzle. Assuming that this is a pipe flow with only heating and no friction acting we can use the one-dimensional flow with heat exchange (Rayleigh flow) tables to find the appropriate flow properties anywhere in the heated section. Although we know the actual flow properties at station 2 we must find the pertinent Rayleigh flow variables there from the table. These are as follows:
M2=0.5; po,2/po,2*=1.1141, p2/p2*=1.7778, T2/T2*=0.7901, and To,2To,2*=0.6914
Then, using the known state at station 2 we can find the appropriate critical properties as follows:
M2=0.5; po,2*=3.738 MPa; p2*=1.975 MPa; T2*=506.27 K; and To,2*=607.46 K
Though the Mach number at station 3 is not specified, we can safely infer that it must be M3=1 in order to provide supersonic flow through the nozzle and into the larger pipe section attached to the nozzle. Thus we have
M3=1.0; po,2/po,2*=1.0, p2/p2*=1.0, T2/T2*=1.0, and To,2To,2*=1.0
We also know from the isentropic flow tables that at station 3, for M3=1.0, we have p3/po,3=0.5283 and T3/To,3=0.8334. The critical conditions at station 2are the same throughout the heated pipe section right up to station 3 and the total temperature is of course, not a constant in this section of the pipe. From these relations and the critical conditions determined at station 2 we find