Q & A ARCHIVE
ME 303 FLUID DYNAMICS/ WESTPHAL/ WSU/ FALL 2000
Here is where you can find e-mail and sometimes telephone transcripts of question and
answer exchanges between your instructor and classmates. The most recent are listed first.
Date: Mon, 6 Nov 2000 12:27:49 -0800 (PST)
From: Russell Westphal
To: "Colgan, Tatyana Y"
Subject: Re: Homework Questions
Tanya:
For number 5, IF I correctly understand your question (which wasn't part
of the problem)... you could find a "transit time" by knowing that all
distances are related by the model/prototype scale, and velocities are
also in a known ratio (the square root of the model scale). Time would
then be distance divided by velocity.
For number 9, the answer in the text is incorrect... the correct answer is
about 30 kW, a bit more or less depending on how you read the Moody chart
and what you choose for an inlet loss factor. But I don't see any way
that one could get as low as 26.7 kW.
Russ
On Mon, 6 Nov 2000, Colgan, Tatyana Y wrote:
> Russ,
>
> I have several questions about the homework:
>
> Problem # 5
> Finding velocity of the prototype is pretty straightforward. To find the height
> of the wave for the prototype, I multiplied each side of the equation by the
> ratio model wave height / prototype wave height. This seems to be the correct
> procedure. What I was wondering, though, is if I had to relate some other
> quantity--such as time--for the model and the prototype, how would I do that?
> In this case, if I multiplied both sides of the equation by the ratio of
> time / prototype time, I am not sure where to go from there.
>
> Problem # 9
> The answer that the book gives for this problem is 26.7 kW. I got $30.4 kW. I
> am assuming this difference is due to the fact that several factors (k/d, f) had
> to be looked up on the Moody chart, and the numbers that I used must be
> different from those used in the book. Is such deviation acceptable?
>
> Thanks for your help.
>
> Tanya
>
Date: Mon, 30 Oct 2000 10:15:49 -0800 (PST)
From: Russell Westphal
To: steve reiman
Subject: Re: HW question
Steve:
Take the vapor pressure (ABSOLUTE!) at the operating temperature and
divide by specific weight. There is a table in the back of the text that
has vapor pressure of water as a function of temperature.
Russ
On Mon, 30 Oct 2000, steve reiman wrote:
> On problem 7.32 what value do I use for the vapor pressure head?
>
Date: Mon, 16 Oct 2000 12:06:22 -0700 (PDT)
From: Russell Westphal
To: Daniel Carpenter
Subject: Re: homework problems 8 and 12
Daniel:
You can't use Bernoulli *through* a device like an engine (that is,
between points at inlet and exhaust), because we assumed steady flow, no
work or heat transfer to/from the fluid mass, as it moved along a known
streamline. Through an engine, a fluid mass would experience both work
and heat transfer.
8 and 12 require different approaches. For 8, just look at the reverser
vane... that is, find the GROSS (reverse) thrust for fluid being turned by
the vane. The velocity INTO your cv (rel) will be 750 m/s, as will the
velocity (rel) out... the in and out will be in different directions,
however. Pressures will be zero gauge everywhere, since the jets are
exposed to atmosphere.
For 8, if you would prefer to figure the NET reverse thrust (not just the
reverse thrust on the vane), you need to know the aircraft speed... at
landing, this would be, say, 50 m/s or less. Your cv's inlet for such an
analysis could extend far upstream of the engine, so that zero gauge
pressure and 50 m/s Vrelin prevailed there.
For 12, your cv could extend well upstream of the boat... in this case,
the velocity in (rel) would always be equal to the boat's speed, at least
if it was moving in non-windy conditions. Downstream, you have a
"slipstream" Vrel given. For both inlet and exit regions of this cv, the
pressures would be zero gauge.
Does this help?
Russ
On Mon, 16 Oct 2000, Daniel Carpenter wrote:
> Professor Westphal,
> I'm having some trouble with problems 8 and 12 in the homework. In both
> problems, we have an engine that is sucking air in and exhausting it out to
> give thrust. They give us the exit velocity but no inlet velocity,
> pressures, or areas. I know when using Bernoulli that the exit pressure for
> both problems will be atmospheric, but the inlet pressure cannot be or the
> velocites would be the same, giving us zero thrust. Without knowing the
> inlet pressures or areas in both problems, I don't see how to solve for the
> inlet velocity. What am I missing?
> Thanks for your time,
> Daniel Carpenter
> _________________________________________________________________________
Date: Mon, 16 Oct 2000 10:19:37 -0700 (PDT)
From: Russell Westphal
To: steve reiman
Subject: Re: Help with Me303 HW
Stephen:
6.20: You could interpret the problem statement reasonably in one of two
ways... find the force on the vane itself (*gross* reverse thrust), or to
find the force on the entire engine (*net* reverse thrust). If we look at
the vane alone, your cv INLET would be the flow leaving the main engine
pod, and the cv EXIT would be the deflected flow (different than what you
used). The cv should be fixed to the engine so that the flow is steady.
Your coordinate system can ride with the engine at a fixed speed too,
or be stationary with repsect to "ground". In the work you scanned and
sent to me, you have double-counted velocities, and have not considered
the problem to be steady... not sure where you went astray on these two
points. Take another look, using the cv I suggest?
6.81 The propulsor (an unducted fan, in this case) produces a 3 ft
diameter air stream with Vrel=80 ft/sec... this flow exists as a free jet
within the atmosphere and therefore you may consider its pressure to be
atmospheric. Try putting your cv inlet upstream "far", so that all
entering mass flow can be considered to have a Vrel of Vboat.
6.90 One may assume that the given 300 m/s inflow velocity is
"rel". Further, you don't know if the engine is stationary or moving, so
best to fix the coordinate system (AND the cv!) to the engine and assume
conditions are steady. Take all pressures to be zero gauge.
As to the issue you raise concerning how thrust fits into the momentum
equation: it shows up in these problems as the reaction force required to
restrain the propulsor--so, it is a reaction isolated when your cs cuts
through a "pylon" (just the thing that connects the propulsor to
the rest of the vehicle). There are still only 4 forces on your "force
checklist", we just call this particular type of reaction "thrust". The
need for the thrust (restraint) reaction is created by the large change in
fluid momentum.
Russ
On Mon, 16 Oct 2000, steve reiman wrote:
> Hi. My name is Stephen Reiman from the Wsu campus. I have been able to do
> most of the homework except for the 6.81, 6.20, and 6.90. These problems all
> involve taking a coordinate system not on the Cv I think. On 6.20 I have
> solved out the general equation on the right side as far as I can go, and
> assuming I am correct, I am not sure how to get certain terms or if they go
> away. I have attached a scanned copy of my work in Adobe Photoshop format.
> On6.81 I was wondering if on the first part the pressure right after the
> fan needs to be used in order to solve the equation for the force of the
> fan.
> I think I should be able to solve 6.90 after I get some help on 6.20
> because I don't totally understand how thrust fits into the equation of
> conservation of momentum. I have the right side of the force equation done
> already, but I am not sure about the left. I took my point of reference to
> be located on the Cv, moving with it.
> Thanks for any help. I have to work during Art's office hours
> _________________________________________________________________________
>
Date: Thu, 12 Oct 2000 09:32:58 -0700 (PDT)
From: Russell Westphal
To: Ken_K_Kawabata
Subject: Re: Problem 4 of assignment 4
Ken:
Here's some food for thought...
Suppose that the small end was CLOSED... then, the inlet would be a
STAGNATION POINT, right? Your view of the problem would still tell you
that the inlet plane velocity was equal to the aircraft's; the error in
reasoning is because the flow approaching the inlet may be speeding up or
slowing down, depending on what is going on downstream of the inlet.
Now consider the problem at hand: the small end open. The flow *escaping*
at the small end exhausts to the surroundings as a jet. Hence, the *exit*
plane pressure must be atmospheric, right?
Hope this helps.
Russ
On Thu, 12 Oct 2000 Ken_K_Kawabata wrote:
> Hi Dr. Westphal,
>
> I'm having difficulty arriving at the inlet velocity to the funnel. The
> problem hint stated that the velocity in to the funnel would be different
> from the velocity of the airplane. I looked at a streamline arriving at the
> edge of the funnel and with no change in elevation and no pressure
> difference the velocity would be the same. I must be missing something can
> you provide an additional hint.
>
> Thanks,
>
> Ken Kawabata
Date: Mon, 9 Oct 2000 11:12:56 -0700 (PDT)
From: Russell Westphal
To: Eric David Lindsley
Subject: Re: Homework#4
Eric:
On problem 4-1 (text, 5.52), your error was right at the start, in
applying conservation of mass: with a diameter ratio of 2, the area ratio
would be 4, so that the velocity at the exit would be FOUR times the
piston velocity.
On problem 4-4, yes, the small (downstream-facing) end is open to
atmosphere. If that end was CLOSED (which it is NOT), then the inlet face
(large end of the funnel) would be a stagnation point. HINT: The exit
(small end) pressure must be equal to that of the surrounding atmosphere
(WHY?).
Hope this helps,
Russ
On Mon, 9 Oct 2000, Eric David Lindsley wrote:
> Dr. Westphal
>
> On problem #5.52 I calculated the velocity at exit to be 2 times the
> piston velocity. Then, setting the pressure at exit(A jet issuing to the
> atmosphere) equal to 0 in Bernoulli's Eq., I find the pressure at the
> piston to be .505 psig. I used a rho of 1.94 slugs/ft^3. My pressure at
> piston is of by EXACTLY a factor of 5 from the pressure that would give
> the force the back of the book has.
> I was hoping you might be able to find my error.
>
> Also on problem #4 is the small end open to the atmosphere?
> Would the entrance to the big end be a stagnation point?
>
> Thanks for any assistance you can provide.
> Eric
>
Date: Wed, 4 Oct 2000 13:40:49 -0700 (PDT)
From: Russell Westphal
To: Ward Trythall
Subject: Re: Fluids for Lunch
Ward:
Thanks for the suggestion. If I can get it together for tomorrow, I'll
try it then. If not, next week!
Russ
On Wed, 4 Oct 2000, Ward Trythall wrote:
> Hi Russ,
>
> Here's an idea I have for improving the test scores. It means a little more
work for you however. Sorry about that:)
>
> What about including 2 or 3 short questions from you to the students in the
Fluids for lunch period. These could be similar to test questions and everyone
would have say five minutes to do them each followed by you explaining the
solution. We could do 2 or 3 per class, not necessarily all at once and no
actual scoring would take place besides our own internal "I screwed this one up
didn't I".
>
> Cheers
>
> Ward
>
Date: Wed, 4 Oct 2000 13:38:24 -0700 (PDT)
From: Russell Westphal
To: Ben Spence
Subject: Re: ME 303 fluids homework
Ben:
We can imagine a stagnation process anytime--even if there isn't one in a
particular situation. Hence, we can always talk about the "stagnation
pressure" at any point as being the pressure which would occur if the flow
at the point in question were brought to rest according to Bernoulli's
"S" equation.
Russ
On Wed, 4 Oct 2000, Spence wrote:
> Professor Westphal:
> I have a question about the third part of question 10 in the homework for ME
303.
> It asks for the stagnation pressure of the exiting jet of water from the
nozzle. What/where would the
> stagnation point be? Seems like it should be somewhere where the velocity of
the water is zero.
>
> Any information you could impart on this would be appreciated.
>
> Thanks,
>
> Ben Spence
>
Date: Mon, 18 Sep 2000 10:30:33 -0700 (PDT)
From: Russell Westphal
To: charles_h_mulkey
Subject: A-15
Chuck:
Per your question about the moment of inertia values shown on page A-15 of
the text for the semicircle: I_xx=0.11*r^4 is consistent with the diagram
and correct; the second I value is incorrectly labelled I_xx also. It
should have been labelled I_yy; that is, I_yy=pi*r^4/8 would be correct.
Thanks for bringing this typo to my attention.
Russ
Date: Thu, 14 Sep 2000 09:40:27 -0700 (PDT)
From: Russell Westphal
To: Ken_K_Kawabata
Subject: Re: Me303 Assignment 2 Homework
Ken:
Try looking at just a 2 ft wide (into the page) segment of the form.
Since there is one tie every 2 ft, a single tie provides all of the
necessary reaction to support the bottom of each 2 ft segment of form.
Don't overlook the fact that the upper tie also provides a reaction... to
eliminate it from consideration, how about taking moments about the upper
tie point?
Hope this helps,
Russ
On Thu, 14 Sep 2000 Ken_K_Kawabata wrote:
> Dr. Westphal,
>
> On problem 5 of the homework (text problem 3.68) I'm having difficulty
> describing the particular forces acting on the bottom brace. I know the
> pressure exerted by the liquid (concrete) and I can convert that to force by
> dividing by the area but I'm unsure how the force is distributed among the
> braces.
> ...
>
> Thank you for any help you can send in my direction...
> Ken Kawabata
>
Date: Tue, 5 Sep 2000 11:24:39 -0700 (PDT)
From: Russell Westphal
To: Eric David Lindsley
Subject: Re: Homework ?
Eric:
Your symbolic result for the shear stress is correct.
One reference (Crane TP 410, page A-3) gives a dynamic viscosity of
about 700 Centipoise for SAE 30 lubricant at 10 C. If you look elsewhere,
don't be surprised to find a somewhat different result... the SAE rating
only indicates a range of viscosity-temperature behavior. Hence, SAE 30
oil from different manufacturers can have somewhat different values of
viscosity at a given temperature.
Russ
On Tue, 5 Sep 2000, Eric David Lindsley wrote:
> I'm working on problem number 3 in the homework.
> Thats the one with the cranckshaft main bearing.
> I have derived the equation :
>
> Tau = Mu(omega*d)/2*delta y
>
> We are given that the SAE 30 oil is at 10 C
> The table in the book gives the Dynamic viscosity values for oil
> only at 38 C.
> Since viscosity is a function of temperature is, there a way to find the
> viscosity at a temperature given the viscosity at another temperature?
> Or, should I find an alternative table that lists the viscosities of oil
> at different temperatures and interpolate?
> Thanks for your time.
> Eric
>
Date: Wed, 30 Aug 2000 14:15:06 -0700 (PDT)
From: Russell Westphal
To: Jay Colgan
Subject: Re: Rounding for significant digits
Jay:
Short answer: Report final results with 3 significant digits. Carry
"calculator" precision through all intermediate results, however. Do not
round off input values--use whatever precision is available.
Long answer: The maximum number that are justified depends on the
precision with which one knows the input data, among other factors, and
can then be determined for some result computed from the input data by a
rather complex process known as "uncertainty analysis". However, if fewer
digits than are justified from uncertainties are needed for a particular
application, the fewer number is reported. For example, one cannot buy a
22.32 hp motor, so if you analysis is aimed at specifying a motor, you'd
specify 25 hp.
Russ
On Wed, 30 Aug 2000 Jay Colgan wrote:
> Dr. Westphal,
>
> What is your policy concerning the rounding of digits? For instance if two
> numbers are multiplied (conversion factor and variable) would you prefer
> that we round the product to the least number of significant digits of the
> two numbers being multiplied or just include all the decimal points as far
> as our calculator can go?
>
> Sincerely,
>
> Jay Colgan
>