HVAC duct air pressure loss

by Dasa . 07 Jul 2021
Enter real values to get proper results:
Cross Section Area [a] (in2):
Perimeter [p] (in):
Air Velocity [v] (fpm):
Wetted Perimeter [p​w​​] (in):
Hydraulic Diameter [dh] (in):
Reynold Number [Re]:
Friction loss co-efficient [λ]:
Pressure Loss [ΔP] (in of H2O):

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 WHW * H 216 in2
Perimeter of duct P 2(W+H)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 =qa= \dfrac {q} {a} 1333 fpm
Wetted perimeter pwp_w = Duct perimeter 60 in
Hydraulic diameter dhd_h =4apw= \dfrac{4 * a} {p_w} 14.40 in

Renoyld number (Re):

Re=dhvρμR_e = \dfrac{d_h * v * \rho} {\mu}

=(14.4/12)(133360)0.07510.04462=\dfrac{(14.4/12) * (1333*60) * 0.0751} {0.04462}

=161578= 161578

Friction loss co-efficient (Lambda):


ReR_e < 2300 then the flow is Laminar
ReR_e > 2300 and ReR_e < 4000 then the flow is transient
ReR_e > 4000 then the flow is Turbulent

For Laminar flow

Friction loss co-efficient
λ=64Re\lambda = \dfrac {64} {R_e}

For Turbulent flow

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

Moody’s chart:

λ\lambda is a function of ReR_e and ϵdh\dfrac {\epsilon} {d_h}

λ=f(Re,ϵdh)\lambda = f (R_e, \dfrac {\epsilon} {d_h})

ϵdh=0.003614.4=0.00025\dfrac {\epsilon} {d_h} = \dfrac{0.0036} {14.4} = 0.00025

From Moody’s chart you can find the below value

λ=0.0144\lambda = 0.0144 @ {Re=161578R_e = 161578 and ϵdh=0.00025\dfrac {\epsilon} {d_h} = 0.00025}

Colebrooke equation:

1λ=2log(2.51Reλ+kdh3.72)\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

Pressure loss using Darcy Weisbach equation:

ΔP=λldhρv22\Delta P = \lambda * \dfrac{l}{d_h} * \dfrac{\rho*v^2}{2}

=0.017910014.40.00237133322= 0.0179 * \dfrac{100}{14.4} * \dfrac{0.00237*1333^2}{2}

=0.0727lbfft2= 0.0727 \dfrac {lb_f}{ft^2}

=0.014= 0.014 inchinch ofof H2OH_2 O

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