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Posted: 10/4/2004 11:34:27 AM EST
Question 1:

A nozzle is designed to accelerate the fluid from V1 to V2 in a linear fashion. That is V=ax+b, where a and b are constants. If the flow is constant with V1= 10 m/s at x1 = 0 and V2= 25 m/s at x2= 1 m, determine the local acceleration, convective acceleration, and acceleration of the fluid at points (1) and (2).



Question 2:

The Hoover Dam backs up the Colorado River and creates Lake Meade, which is approximately 115 miles long and has a surface area of appoximately 225 square miles. If during flood conditions the Colorado River flows into the lake at a rate of 45,000 cfs and the outflow from the dam is 8000 cfs, how many feet per 24-hr day will the lake level rise?



thanks
Scott
Link Posted: 10/4/2004 11:43:21 AM EST
[Last Edit: 10/4/2004 11:46:27 AM EST by DriftPunch]

Originally Posted By scottryan:
Question 1:

A nozzle is designed to accelerate the fluid from V1 to V2 in a linear fashion. That is V=ax+b, where a and b are constants. If the flow is constant with V1= 10 m/s at x1 = 0 and V2= 25 m/s at x2= 1 m, determine the local acceleration, convective acceleration, and acceleration of the fluid at points (1) and (2).

You said nozzle...




Question 2:

The Hoover Dam backs up the Colorado River and creates Lake Meade, which is approximately 115 miles long and has a surface area of appoximately 225 square miles. If during flood conditions the Colorado River flows into the lake at a rate of 45,000 cfs and the outflow from the dam is 8000 cfs, how many feet per 24-hr day will the lake level rise?

You can only do this if the lake had vertical shorlines or you know the topography. If you don't know the topo, you're SOL. For example, if the shorline above the current level was inclined at 3 degrees, it would take more water to achieve a 1 foot rise than if it was 90 degrees.
Link Posted: 10/4/2004 11:45:01 AM EST
delete *.*
Link Posted: 10/4/2004 11:46:54 AM EST
#2:

Assuming the water in the lake will rise vertical and not sloped (which it would) the lake is approx 6,272,640,000 sq/feet (5280*5280*225). It is rising at a rate of 37,000 cfs or a total of
3,196,800,00 over 24 hours. 6,272,640,000/3,196,800,000 = Lake rising .5096 ft.

Back to work.
Link Posted: 10/4/2004 11:50:20 AM EST
[Last Edit: 10/4/2004 11:51:23 AM EST by Zaphod]

Originally Posted By scottryan:
Question 1:

A nozzle is designed to accelerate the fluid from V1 to V2 in a linear fashion. That is V=ax+b, where a and b are constants. If the flow is constant with V1= 10 m/s at x1 = 0 and V2= 25 m/s at x2= 1 m, determine the local acceleration, convective acceleration, and acceleration of the fluid at points (1) and (2).



To begin with, A=15 and B=10

As for the rest, I think you need to know mass flow rate, or at least the time it took to go from X1 to X2 in order to answer this....


Link Posted: 10/4/2004 11:50:41 AM EST
the only problem i see is that one is in cubic feet and one is in square feet.......
Link Posted: 10/4/2004 11:51:09 AM EST

Originally Posted By dirk53890:
#2:

Assuming the water in the lake will rise vertical and not sloped (which it would) the lake is approx 6,272,640,000 sq/feet (5280*5280*225). It is rising at a rate of 37,000 cfs or a total of
3,196,800,00 over 24 hours. 6,272,640,000/3,196,800,000 = Lake rising .5096 ft.

Back to work.





RTFQ!
Link Posted: 10/4/2004 12:01:13 PM EST
[Last Edit: 10/5/2004 2:32:17 AM EST by danno-in-michigan]

Originally Posted By scottryan:
Question 1:

A nozzle is designed to accelerate the fluid from V1 to V2 in a linear fashion. That is V=ax+b, where a and b are constants. If the flow is constant with V1= 10 m/s at x1 = 0 and V2= 25 m/s at x2= 1 m, determine the local acceleration, convective acceleration, and acceleration of the fluid at points (1) and (2).




The equation describing the flow field is: v=15x+10. (this gives 10 m/s at x=0 and 25 m/s at x=1).
It's been a few years, but I believe that local acceleration is dv/dt. In this case, it's zero because velocity of the flow field is dependant solely upon position, not time. (The flow field is steady). I believe that convective acceleration is v*dv/dx, which in this case is:
(10m/s)*15/s=150 m/s^2 at point 1
(25m/s)*15/s=375 m/s^2 at point 2.

total acceleration is local acceleration + convective acceleration

It may seem odd that dv/dt is zero, but remember - you're looking at a fluid field, not an individual particle. How does velocity change with time in the field? A: It doesn't because velocity is a function of position only.
Link Posted: 10/4/2004 12:05:29 PM EST

Originally Posted By danno-in-michigan:

It's been a few years, but I believe that local acceleration is dv/dt. In this case, it's zero because velocity of the flow field is dependant solely upon position, not time. (The flow field is steady).



Sorry, ut you are incorrect.

1) If the fluid is going faster at point 1 than point 0, then it's accelerating.

2) Acceleration of a fluid is what a nozzle does.

3) dV/dt is defined as the change in velocity per change in TIME (t).


I don't know about the rest of your answer, though. Been a long time here, too!
Link Posted: 10/4/2004 12:09:20 PM EST
[Last Edit: 10/4/2004 12:25:57 PM EST by danno-in-michigan]

Originally Posted By Zaphod:

Originally Posted By danno-in-michigan:

It's been a few years, but I believe that local acceleration is dv/dt. In this case, it's zero because velocity of the flow field is dependant solely upon position, not time. (The flow field is steady).



Sorry, ut you are incorrect.

1) If the fluid is going faster at point 1 than point 0, then it's accelerating.

2) Acceleration of a fluid is what a nozzle does.

3) dV/dt is defined as the change in velocity per change in TIME (t).


I don't know about the rest of your answer, though. Been a long time here, too!



Sorry, the dv/dt I used were meant to be partial derivitives. Take a look at my edited answer above and it will explain what I meant. Here's the equation for calculating acceleration of a fluid field that is a function of both time and position:
dv/dt = dv/dt + Vxdv/dx (italics = partial derivative)

The first term is local acceleration - it's zero because the partial derivative of velocity wrt time is zero. The second term is convective acceleration, which is acceleration due to position.
It's helpful to remember the physical meaning of a partial derivative - holding all other variables constant, how does the function change with respect to one variable? If you hold position still and measure the velocity at position x=0 (or x=1), how does it change with time? A: it doesn't. On the other hand, if you hold time still, and measure the velocity at various points you find that it does change with position in the nozzle.

Link Posted: 10/4/2004 6:16:24 PM EST
This is an example of why this is the greatest place on the internet. Where else would a thread on fluid dynamics be located on the same page as "Chubby Chick BOTD"?

I rest my case....

Doc H.
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