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Exhaust Outlet

Reference data and engineering information about exhaust outlet for miscellaneous applications.

exhaustoutlet

Overview

Engineering reference data for Exhaust Outlet in miscellaneous.

Key Formulas

Unit Conversion

y=xky = x \cdot k

Multiply by conversion factor.

Linear Interpolation

y=y1+(xx1)(y2y1)x2x1y = y_1 + \frac{(x - x_1)(y_2 - y_1)}{x_2 - x_1}

Estimate between two known points.

Percentage

p=partwhole×100%p = \frac{\text{part}}{\text{whole}} \times 100\%

Part as fraction of whole.

Variables

SymbolDescriptionUnit
xxInput value
yyOutput value
kkConversion factor

Detailed Capture Velocity Formulas

The capture air velocity at a distance x from an exhaust outlet is given by the following relationships, derived from a point source diffusion model.

For smaller exhaust outlets, the formula is: vc=q4πx2=voA12x2=voπd248x2v_c = \frac{q}{4 \pi x^2} = \frac{v_o A}{12 x^2} = \frac{v_o \pi d^2}{48 x^2}

Important Note: According to this point-source approximation, the capture velocity is independent of the duct size. This equation is valid only for exhaust outlets with relatively small diameters.

For larger exhaust outlets, a modified formula accounts for the finite source size: vc=voAA+10x2=qA+10x2v_c = \frac{v_o A}{A + 10 x^2} = \frac{q}{A + 10 x^2}

Example Calculation

Problem: Calculate the capture air velocity at a distance of 250 mm from a 250 mm diameter duct with an internal air velocity of 3 m/s.

Solution: Using the formula for smaller outlets: vc=(3 m/s)π(0.250 m)248(0.250 m)2=3π480.2 m/sv_c = \frac{(3 \text{ m/s}) \cdot \pi \cdot (0.250 \text{ m})^2}{48 \cdot (0.250 \text{ m})^2} = \frac{3 \pi}{48} \approx 0.2 \text{ m/s}

Result: The capture velocity is approximately 0.2 m/s.

Key Observation: The air velocity at a distance of one duct diameter from the outlet is less than 10% of the duct air velocity.

Exhaust Outlets in Walls or with Flanges

If the exhaust outlet is installed in a wall or has a flange, the evacuation efficiency is improved. The capture velocity formula is modified by a factor of 1.33: vc=1.33voAA+10x2=1.33qA+10x2v_c = \frac{1.33 \cdot v_o A}{A + 10 x^2} = \frac{1.33 \cdot q}{A + 10 x^2}

For calculation using the standard calculator for larger outlets, the volume flow (q) should be multiplied by 1.33 before input.

The use of exhaust hoods further improves evacuation efficiency beyond this modification.

References