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Grain Conical Pile Volume

Calculate the volume of a conical pile of granular material formed by free pouring, V = (π/3)·r³·tan(θ), from the pile base radius r (m) and the material's angle of repose θ (degrees). When granular material is freely poured onto a surface, it naturally forms a CONE whose side slope is the angle of repose — a characteristic property of each material (dry sand ~30-35°, grain ~25-30°, crushed stone ~37-40°) reflecting inter-particle friction. Since the cone height is h = r·tan(θ), the cone volume (1/3·π·r²·h) becomes (π/3)·r³·tan(θ), a function of radius and angle of repose only. This is widely used in practice to estimate, from a survey or base-radius measurement, the volume (and with density, the mass) of open stockpiles — piles of grain, sand, crushed stone, coal, ore, fertilizer in yards and warehouses. It is the basis of bulk-material inventory in piles, a quick alternative to weighing. It also guides stockyard area and height sizing and bulk-warehouse design. For elongated piles (prismatic with conical ends), add the central part. Enter the base radius and the angle of repose.

Resultado

Volume de pilha cônica de grãos

O volume de uma pilha cônica de material granular formada por despejo livre é V = (π/3)·r³·tan(θ), a partir do raio da base r e do ângulo de repouso θ do material. Quando um material granular é despejado livremente sobre uma superfície, ele forma naturalmente um cone cuja inclinação das laterais é o ângulo de repouso — uma propriedade característica de cada material (areia seca ~30-35°, grãos ~25-30°, brita ~37-40°), que reflete o atrito interno entre as partículas. Como a altura do cone é h = r·tan(θ), o volume do cone (⅓·π·r²·h) torna-se (π/3)·r³·tan(θ), função apenas do raio e do ângulo de repouso. Este cálculo é muito usado na prática para estimar volumes de estoque: a partir de um levantamento topográfico ou da simples medição do raio da base, obtém-se o volume (e, com a densidade aparente, a massa) de pilhas a céu aberto — montes de grãos, areia, brita, carvão, minério ou fertilizante em pátios e armazéns. É a base do inventário de materiais a granel estocados em pilhas, uma alternativa rápida e barata à pesagem (hoje muitas vezes feita por drones ou escâneres a laser que medem a pilha). Também orienta o dimensionamento da área e da altura de pátios de estocagem e o projeto de armazéns graneleiros. Para pilhas alongadas (prismáticas, com extremidades cônicas), soma-se a parte central prismática a este volume das pontas. Informe o raio da base e o ângulo de repouso.

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Silo Emptying Time

Calculate the time to empty a silo by gravity discharge, t = M ÷ W, from the stored product mass M (kg) and the mass discharge rate W (kg/s). Since the discharge rate of a granular material through an orifice is practically CONSTANT (independent of the product height above, by the Janssen effect and per the Beverloo equation), the emptying time is simply total mass divided by rate — a direct relation, unlike a liquid's emptying, which slows as the level falls. This time is an important operational parameter in silo, hopper and storage-unit design and operation: it sets the dispatch capacity (how fast a truck, rail car or ship is loaded), sizes the downstream conveying systems (belts, bucket elevators, screws) that must match the discharge rate, and frames shift logistics and vehicle queues at grain terminals. The discharge rate W can be estimated by the Beverloo equation from the outlet diameter, closing the calculation: larger outlets discharge faster (W ∝ D₀^2.5), reducing emptying time. Enter the stored mass and the discharge rate.

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Calculate the theoretical air volume needed for complete combustion of a hydrocarbon C_xH_y, V_air = (x + y/4) × 22.4 ÷ (0.21 × M), from the carbon x, hydrogen y atoms and the molar mass M. The result, in Nm³ of air per kg of fuel (at normal conditions), is the stoichiometric air — the basis for sizing fans, burners and combustion air systems. The factor 22.4 is the ideal gas molar volume (L/mol at STP) and 0.21 the volume fraction of O₂ in air. Methane needs ~13.3 Nm³/kg. Multiplied by the excess air, it gives the actual air supplied. Enter x, y and the molar mass.

The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.