This document describes the method of calculating ventilation air or conditioned air pressure loss across a rectangular duct with the help of the Darcy Weisbach equation and Colebrook equation or Moody’s chart.

Let us assume a straight duct of 12” Height x 18” Width x 100” Length, through which 2000 cfm of air is passing. Refer to the below image for details.

We can define the below values from the details given in the above image

Description | Symbol | Formula | Value | Unit |
---|---|---|---|---|

Corss sectionArea of duct | a | $W * H$ | 216 | in2 |

Perimeter of duct | P | $2 * (W+H)$ | 60 | in |

Length of the duct | l | – | 100 | in |

Airflow rate | q | – | 2000 | cfm |

Find below assumed air constants

Description | Symbol | Formula | Value | Unit |
---|---|---|---|---|

Absolute Viscoity of air @ 80° F | µ | – | 0.04462 | lb/ft-hr |

Density of air | ρ | – | 0.00237 | slug/ft3 |

Density of air | ρ | – | 0.0751 | lb/ft3 |

Find below assumed material constants

Description | Symbol | Formula | Value | Unit |
---|---|---|---|---|

Absolute roughness of Galvanized steel | ε | – | 0.0036 | in |

Find below calculated values

Description | Symbol | Formula | Value | Unit |
---|---|---|---|---|

Duct air Velocity | v | $= \dfrac {q} {a}$ | 1333 | fpm |

Wetted perimeter | $p_w$ | = Duct perimeter | 60 | in |

Hydraulic diameter | $d_h$ | $= \dfrac{4 * a} {p_w}$ | 14.40 | in |

$R_e = \dfrac{d_h * v * \rho} {\mu}$

$=\dfrac{(14.4/12) * (1333*60) * 0.0751} {0.04462}$

$= 161578$

If,

$R_e$ < 2300 then the flow is Laminar

$R_e$ > 2300 and $R_e$ < 4000 then the flow is transient

$R_e$ > 4000 then the flow is Turbulent

For Laminar flow

Friction loss co-efficient

$\lambda = \dfrac {64} {R_e}$

For Turbulent flow

We have to use Moody’s chart or Colebrooke equation.

$\lambda$ is a function of $R_e$ and $\dfrac {\epsilon} {d_h}$

$\lambda = f (R_e, \dfrac {\epsilon} {d_h})$

$\dfrac {\epsilon} {d_h} = \dfrac{0.0036} {14.4} = 0.00025$

From Moody’s chart you can find the below value

$\lambda = 0.0144$ @ {$R_e = 161578$ and $\dfrac {\epsilon} {d_h} = 0.00025$}

$\dfrac{1} {\sqrt{\lambda}}= -2 * \log(\dfrac{2.51} {R_e * \sqrt{\lambda}} + \dfrac{k}{dh * 3.72})$

Slove the above equation by trial and error method (iterate) using different values for λ.

Resolved λ = 0.0179

$\Delta P = \lambda * \dfrac{l}{d_h} * \dfrac{\rho*v^2}{2}$

$= 0.0179 * \dfrac{100}{14.4} * \dfrac{0.00237*1333^2}{2}$

$= 0.0727 \dfrac {lb_f}{ft^2}$

$= 0.014$ $inch$ $of$ $H_2 O$

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