Chimney draught at maximum discharge — compact formula (mm of water) For a chimney of height H (metres) with outside air absolute temperature T1 (K), the draught required (in mm of water) for maximum discharge of flue gases is given by which relation?

Introduction / Context:Natural draught chimneys rely on buoyancy: lighter hot gases rise and are replaced by cooler outside air. The mass flow through a chimney depends on the draught head and gas density. There is a particular exit-gas temperature that maximizes discharge for a given height and ambient temperature, leading to a compact formula for the required draught in mm of water.Given Data / Assumptions:

• Chimney height H in metres.
• Outside air absolute temperature T1 in kelvin.
• Ideal-gas behavior; standard empirical constant 353 used for mm of water.
• Neglecting frictional and local losses in the maximizing argument (theoretical condition).

Concept / Approach:The general draught head is h = 353 * H * (1/T1 − 1/Tg) in mm of water, where Tg is the absolute temperature of the chimney gases at the outlet. Maximizing discharge with respect to Tg (using continuity and momentum with density proportional to 1/T) yields the condition Tg = 2 * T1. Substituting Tg = 2T1 into the general expression gives the maximum-discharge draught.Step-by-Step Solution:

Verification / Alternative check:If T1 = 300 K and H = 60 m, h_max ≈ 176.5 * 60 / 300 ≈ 35.3 mm of water, which is in the reasonable range for natural draught chimneys.

Why Other Options Are Wrong:

• 353 * H * (1/T1 − 1/Tg): general formula; does not enforce the maximum-discharge condition.
• 353 * H / T1 or 88.25 * H / T1: factors correspond to Tg = ∞ or Tg = (4/3) T1, not the optimum.
• 29.43 * H * (Tg − T1): wrong dependence and units.

Common Pitfalls:Confusing absolute and Celsius temperatures; omitting the constant 353 that converts to mm of water; applying the result when strong frictional losses or dampers alter the optimum.

Final Answer: