Problem of a tank with trapezoidal section.

[Queren Suriano]

Homework Statement

A deposit of 20 feet long and 10 feet high, has a width of 8 feet at the bottom and 18 feet at the top. In the bottom is an orifice an area of 24 in2 and discharge coefficient of 0.60. If the tank is full calculate the time required for the lower level 6 feet. Consider full contraction of the jet.[/B]

Homework Equations

a= area of the orifice
c= discharge coefficient

The Attempt at a Solution

I have found a relationship between the variable X (width) and the variable H (height). I prolongate the sides of the trapeze until have a triangle. 18 / 8 = Y / (Y-10), where Y is the height from the vertix of the triangle to the bottom of the triangle. So I have that Y=18 ft. After that I did another relation of triangles: x/18 = H/(Y=18)-------x=H

I applied the diferential equation dt= 24.922 H ^(1/2) dH; with height limits from 18 feet to 14 feet.
But I don't know if it's right or not

 

[Chestermiller]

Homework Statement

A deposit of 20 feet long and 10 feet high, has a width of 8 feet at the bottom and 18 feet at the top. In the bottom is an orifice an area of 24 in2 and discharge coefficient of 0.60. If the tank is full calculate the time required for the lower level 6 feet. Consider full contraction of the jet.[/B]

Homework Equations

View attachment 87699

a= area of the orifice
c= discharge coefficient

The Attempt at a Solution

I have found a relationship between the variable X (width) and the variable H (height). I prolongate the sides of the trapeze until have a triangle. 18 / 8 = Y / (Y-10), where Y is the height from the vertix of the triangle to the bottom of the triangle. So I have that Y=18 ft. After that I did another relation of triangles: x/18 = H/(Y=18)-------x=H

I applied the diferential equation dt= 24.922 H ^(1/2) dH; with height limits from 18 feet to 14 feet.
But I don't know if it's right or not

Show us your equation for A(h).

Chet

 

[Queren Suriano]

Show us your equation for A(h).

Chet

I have surface area =20 H, because the length of 20 is constant. And from the relation between triangles I know that X=H

 

[haruspex]

I have surface area =20 H, because the length of 20 is constant. And from the relation between triangles I know that X=H

The width increases by 10 ft as the height increases by 10 ft, but you can't set h=0 at 8 feet below the base of the container. That would make your $ac\sqrt{2gh}$ wrong.

 

[Chestermiller]

I get 20(8+h).

 

[Queren Suriano]

The width increases by 10 ft as the height increases by 10 ft, but you can't set h=0 at 8 feet below the base of the container. That would make your $ac\sqrt{2gh}$ wrong.

The width increases by 10 ft as the height increases by 10 ft, but you can't set h=0 at 8 feet below the base of the container. That would make your $ac\sqrt{2gh}$ wrong.

I think I understand, but there are some exercises in which is valid to take the zero down or away from the element considered. For example in this exercise (strain energy) and I only integrate in the real part of the element. So when will know whether it is valid or not take away the element origin?

 

[Queren Suriano]

I get 20(8+h).

Thank you, I got this too, when I put the origin in the base of the tank

 

[Chestermiller]

Thank you, I got this too, when I put the origin in the base of the tank

I got this by just fitting a straight line between between (0,8) and (10,18).

Chet

 

[haruspex]

I think I understand, but there are some exercises in which is valid to take the zero down or away from the element considered. For example in this exercise (strain energy) and I only integrate in the real part of the element. So when will know whether it is valid or not take away the element origin?

It's valid as long as you are consistent. Your problem was that in different equations you were measuring h from different origins.

 

[Queren Suriano]

It's valid as long as you are consistent. Your problem was that in different equations you were measuring h from different origins.

Thank you