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