Volume of Silo bin with Cone top

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SUMMARY

The discussion focuses on calculating the volume of a silo bin with a conical top using specific geometric formulas. When the height of the stored product, h, is less than or equal to the height of the conical section (hcone), the volume is calculated using the formula V=1/3πr²h. If h exceeds hcone, the total volume combines the cone and cylindrical sections, represented by V=1/3πr²hcone + πr²(h-hcone). Given a conical height of 3.0 m, a cylindrical radius of 2.0 m, and a total height of 10.0 m, the maximum volume of material that can be stored is approximately 100 m³.

PREREQUISITES
  • Understanding of geometric volume calculations
  • Familiarity with conical and cylindrical shapes
  • Knowledge of trigonometric functions, specifically tangent
  • Ability to manipulate algebraic formulas
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Engineers, architects, and anyone involved in storage solutions or agricultural engineering will benefit from this discussion on calculating silo volumes.

zeshkani1985
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To calculate the volume of the contents you use the formula for a cone, as long as the height of the product, h, is less than or equal to the height of the conical section, hcone.
V=1/3rh2h if h ≤ hcone
and rh is the radius at height h: rh=tan∅ if rh ≤ R.

If the height of the stored product is greater than the height of the conical section, the equation for a cylinder must be added to the volume of the cone:
V=1/3∏r2hcone+∏r2(h-hcone) if h > hcone.
If the height of the conical section is 3.0 m, the radius of the cylindrical section is 2.0 m, and the total height of the storage bin is 10.0 m, what is the maximum volume of material that can be stored?

here is what i did. I found the volume of the cone since it says we are given a conical height of 3 and the radius of 2 so i get a volume of about 12m3 and I get the cylinidrical volume by just V=∏r2h since were are given the total height of the bin at 10m so the cylinder must be 7m tall since (10-3hcone=7) and the volume of the cylinder is about 87m3 so the total volume is them combined to give me a total volume of around V=100m3

is this right?
 
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zeshkani1985 said:
To calculate the volume of the contents you use the formula for a cone, as long as the height of the product, h, is less than or equal to the height of the conical section, hcone.
V=1/3rh2h if h ≤ hcone
and rh is the radius at height h: rh=tan∅ if rh ≤ R.

If the height of the stored product is greater than the height of the conical section, the equation for a cylinder must be added to the volume of the cone:
V=1/3∏r2hcone+∏r2(h-hcone) if h > hcone.
If the height of the conical section is 3.0 m, the radius of the cylindrical section is 2.0 m, and the total height of the storage bin is 10.0 m, what is the maximum volume of material that can be stored?

here is what i did. I found the volume of the cone since it says we are given a conical height of 3 and the radius of 2 so i get a volume of about 12m3 and I get the cylinidrical volume by just V=∏r2h since were are given the total height of the bin at 10m so the cylinder must be 7m tall since (10-3hcone=7) and the volume of the cylinder is about 87m3 so the total volume is them combined to give me a total volume of around V=100m3

is this right?
The formula V=1/3∏r2hcone+∏r2(h-hcone) is correct. So one is looking for the total volume subject to the constraint of the given dimensions.
 

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