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Old Jan 21st 2015, 05:37 AM   #1
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Fluid mechanics question

Guys,I am a new member.I am a bit confused about this problem
A simplified schematic of the rain drainage system for a barn is shown in the figure (attached). Rain falling on the slanted roof runs off into gutters around the roof edge; it then drains through several downspouts (only one is shown) into a main drainage pipe M below the basement, which carries the water to an even larger pipe. A floor drain in the basement is also connected to drainage pipe M. The question arises, if such design can cope with heavy storm events such as recently experienced in Australia.
Suppose the following apply:

1. the downspouts have height h1 = 11 m
2. the floor drain has height h2 = 1.2 m
3. pipe M has radius 3.0 cm,
4. the barn has a width w = 30 m and a length L = 60 m,
5. all the water striking the roof goes through pipe M ,
6. the initial water speed in a downspout is negligible,
7. the wind speed is negligible (the rain falls vertically).

(a) By applying Bernoulli’s principle, determine the speed of water in pipe M, when the basement floor is on the verge of being flooded.

(b) Calculate using the continuity equation at what rainfall rate, in centimeters per hour, such flooding will occur?
An attempt at the solution:-

I have made too many assumptions


My assumptions:-
1) Imagine a rectangle and imagine pulling it from its center,raising a pyramid.I assume the roof to be of this pyramidal shape so that,the exposed surface area of this pyramid is same as that of the base.(This is not the case with all pyramids)
2) Pressure inside the house is the same as the atmospheric pressure.
3) Water collected in the pipe joining the floor drainage and the pipe M is stagnant(at rest)
Solution:
Let the point in M directly below the floor drainage be called point 1),the point where a downspout begins as point 3) and the point at the floor drainage as point 2).
Consider a y axis such that y1=0,y2=h2,y3=h1
Between 1) and 2)
P1=P2 + Rg(y2-y1) (water is at rest)
P1= Pa + Rgh2
Since 3) is exposed to the atmosphere,P3=Pa
Therefore,P1-P3=Rgh2
Applying Bernoulli's equation between 1) and 3):-
P1+0.5R(v1^2)+Rgy1=P3+0.5R(v3^2)+Rgy3
Therefore,
P1-P3=Rgh1-0.5R(v1^2)
P1-P3=Rgh2
Therefore,
Rgh2=Rgh1-0.5R(v1^2)
This gives v1=13.85929291m/s
Volume flow rate=A1v1=0.039186227
Continuity:-
Volume flow rate=Rainfall volume rate rate
Now,a point on the roof receives a volume (vdt).da in a time dt where v=speed of rain and da=area of point.Therefore,volume rate=vda
Integrating,flow rate=lbv where l and b anre length and breadth.
lbv=A1v1
Solving the above would give v=2.177 x (10^-5) m/s=7.84 cm/h
Please tell me if I am correct or not
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Old Jan 21st 2015, 09:55 AM   #2
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Barn Flood Prevention...

Hello SOP,

I took a look at this. I'm not sure the pipe diameters are the same.
I'll attach what I have you can see what you think.

Also might I suggest that you spend no more of your precious life
taking time to write science numbers such as -> A1v1 = 0.0391867227 cc/s.
No device exists in any laboratory (here or Russia) that can measure a volume
rate to the nearest 0.0000000001 cc/s. A1v1 = 0.039 cc/s
is enough already.


Also, my hydrostatics work is here:
http://www.thermospokenhere.com/tsh/...d=Ideal Fluids

Good luck with your studies.

TSH for PHF
Attached Thumbnails
Fluid mechanics question-barn_flood_60.png   Fluid mechanics question-bf_01.gif   Fluid mechanics question-bf_02.gif  
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Old Jan 21st 2015, 10:04 AM   #3
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Hi Thermo
thanks for the quick reply. I didn't know the rules so I typed out the full calculator display lol. Shall keep it in mind.
The radius of M is given as 3 cm
What do you think about my assumption about the area?
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Old Jan 21st 2015, 10:07 AM   #4
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I have assumed that water is at rest between 1 and 2.
We don't need to consider the radius of the downspouts because we have to consider the volume received by the roof.
The other thing is that we don't know the number of downspouts so no point talking about its cross sectional area

Last edited by StudentOfPhysics; Jan 21st 2015 at 10:11 AM.
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Old Jan 21st 2015, 12:57 PM   #5
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Hi SOP...

Hello again,

I don't know what the configuration is. I hate to fool around guessing.

It must be there are "many" (/) downpipes. Each leads to its own "M."

The fact of the standpipe (1) - (2) must mean the water downward (through a single one) enters with very small velocity. This way Vm can be calculated. (see my energy equation with V3 = 0 ~ very small but not 0).

Now about the area. Knowing the flow through "M" is not enough to determine the rate of rain.

Good Luck TSH for PHF
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Old Jan 21st 2015, 09:53 PM   #6
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Hi thermo
I don't think that you can apply Bernoulli's equation between 1 and 2 because BE can only be applied along a streamline and there is no streamline that goes from M to the floor drainage and back to M because a streamline cannot turn within a pipe as we have assumed that all the fluid elements within a cross section have the same velocity which is not possible if there 2 opposite streamlines within a cross section. Now, each fluid element travelling in M comes from the roof,assuming no drainage from the house itself. Therefore the rate of volume of rainwater being received on the roof is the same as the volume flow rate through M. Therefore Bernoulli's equation can be applied for a fluid element's journey from the roof to the pipe M. Steady flow is assumed.
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Old Jan 22nd 2015, 09:29 AM   #7
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Momentum Equation

Properties (1) are related to (2) by the momentum equation
(hydrostatic equation, as some would say).

http://www.thermospokenhere.com/wp/0...hydro_eqn.html

I didn't use Bernouilli. In fact I don't use Bernouilli. I prefer the energy
equation for a flow (Bernouilli is part of the energy equation)..

The problem is poorly posed. There must be many more than 1 "M-type"
pathway. All that water through on D=6cm pipe? What text is this?

Good Luck... SOP

TSH for PHF
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