Chimney Draught — Exit Velocity Under a Given Static Draught Head For a chimney providing a static draught head of H' metres (effective head), the theoretical velocity V of flue gases at the chimney exit is given by which expression?

Introduction:
Natural draught chimneys create a small pressure difference that accelerates flue gases upward. When this draught is expressed as an effective static head H' (metres of gas), we can estimate the ideal exit velocity using energy principles. This relationship aids preliminary sizing and performance checks.

Given Data / Assumptions:

• H' denotes effective static draught head expressed in metres of the flue gas.
• Inviscid, one-dimensional flow; friction and thermal mixing neglected for the ideal estimate.
• g is the gravitational acceleration.

Concept / Approach:
The pressure energy per unit weight corresponding to head H' is converted to kinetic energy at the exit. Using Bernoulli-style reasoning for the ideal case, the velocity head V^2/(2 * g) equals H'. Therefore, V = sqrt(2 * g * H'). If H' is instead given indirectly from density difference and chimney height, then H' = H * (rho_a - rho_g) / rho_g and the velocity expression can be written in extended form. Here, H' is already the effective head, so the compact square-root relation applies.

Step-by-Step Solution:
Start with energy balance: velocity head at exit = available head.Set V^2 / (2 * g) = H'.Solve for V: V = sqrt(2 * g * H').

Verification / Alternative check:
If H' is derived from chimney height H via H' = H * (rho_a - rho_g) / rho_g, substitution yields V = sqrt(2 * g * H * (rho_a - rho_g) / rho_g), consistent with the extended formula.

Why Other Options Are Wrong:

• sqrt(2 * g * H' * (rho_a - rho_g) / rho_g): double-counts density ratio when H' already represents effective head.
• 2 * g * H' and H'/(2g): dimensionally incorrect for velocity.

Common Pitfalls:
Mixing “height of chimney” H with “effective head” H'; always check which head the problem states.

Final Answer:
V = sqrt(2 * g * H')