# Math Help - open ended pipe long question.

1. ## open ended pipe long question.

an open ended pipe has an internal diameter of 0.35m an dis 3.0m long.
determine:
1) the cross sectional area
2)its circumference
3)the volume of the pipe
4)the internal surface area

the pipe is flowing part full of water. the water surface subtends an angle of 110 degrees at the centre.
determine:
1) the cross sectioanl area of flow.
2)the wetted perimeter

2. Given:
$d = 0.35 m$ (has an internal diameter of 0.35m)

$r = \frac{0.35}{2} = 0.175 m$ ( $r = d/2$)

$h = 3 m$(and is 3.0m long.)

The pipe's shape is cylinderlike.

Formulas involving cylinder-shaped (closed at both ends) objects:
1. Cross-sectional area (the area of the base only): A = $\pi r^2$

2. Circumference (perimeter of the circle): $C = 2\pi r$

3. Volume: $V = \pi r^2 h$

4. Surface area: $S = 2\pi r h + 2\pi r^2$

Use what I gave you and try to solve. If you face problems, come back.

EDIT: Please remember that this is an OPEN-ended pipe. Don't count the surface area of the base/top twice.

3. i worked out the 1st part i think.

a= 0.10m
c=1.09m
v=0.29m
s=3.46m.

how do i go about working out the wetted perimeter and the cross sectional area of flow? thanks by the way.

4. A circle is 360 degrees, and the water surface is only 110 degrees (this is, btw, is called the central angle). Now, the water cross-sectional area is NOT the cross-sectional area of the pipe, but rather:

$\frac{110}{360} \cdot A_{cross-sectional pipe}$ = ?

I'm sure you can do number 2 now.

5. ok im not to sure if this is correct but i think i have calculate the cross sectional area of flow and the wetted perimeter.

1) 0.0843m
2)0.058m

the figures seem small to me tho.

6. I will leave the computations for you. And by the way, if you don't know how to find the wetted perimeter, just compute this:

$\frac{110}{360} \cdot C_{cross-sectional area}$

7. thanks i think i get it. but im not to sure what the sectional area is?

8. Okay, first of all, I want to make sure that I made a mistake earlier, and wrote cone-like instead of cylinder-like. The equations I wrote were for cylinders, so don't worry about any calculation mistake.

Cross-sectional area is simply the area of the circle. Suppose you have a cyliner. Slice it perfectly into halves. Now, if you look at its insides (i'm talking about solid, not hollow, cylinder ), you see a cross-section, which is the circle. This is the area that the question wants you to find. O

9. Originally Posted by Chop Suey
A circle is 360 degrees, and the water surface is only 110 degrees (this is, btw, is called the central angle). Now, the water cross-sectional area is NOT the cross-sectional area of the pipe, but rather:

$\frac{110}{360} \cdot A_{cross-sectional pipe}$ = ?

I'm sure you can do number 2 now.
I don't want to pick at you but as far as I understand the situation the cross section of the flow consist of the sector minus a isosceles triangle.

See attachment.

So the value of the dark blue area is calculated by:
$
A_{blue} = \frac{110}{360} \cdot \pi R^2 - R \cdot \sin\left(\frac{110^\circ}2 \right) \cdot R \cdot \cos\left(\frac{110^\circ}2 \right) =$
$R^2 \left(\frac{11 \pi}{36} - \sin\left(\frac{110^\circ}2 \right) \cdot \cos\left(\frac{110^\circ}2 \right) \right)$

By the way: The wording of the problem allows a second solution when the level of the water is higher than the radius of the pipe.

10. You're absolutely right. I misunderstood the problem. Thanks.