# Air Ducts - Temperature, Pressure and Friction Loss

## Temperature and air pressure influence on air ducts friction loss

Air ducts friction loss and pressure drop diagrams are normally made for air at Normal Temperature and Pressure (NTP) condition - with a temperature of *20 ^{ o}C *and a atmospheric pressure of

*1 bar*.

### Influence of Temperature

The friction coefficient in the D'Arcy-Weisbach Equation increases with higher temperature but the coefficient will also be outweighed by the reduced pressure loss due to reduced density of the air.

- the pressure loss in an air duct is reduced with higher temperature

With air temperature other than *20 ^{ o}C* - a

*"Temperature Compensating Coefficient"*can be used to modify the pressure loss:

Δp = k_{t}Δp_{NTP}(1)

where

Δp = actual friction loss (pressure or head)

k_{t}= Temperature Compensating Coefficient

Δp_{NTP }= friction loss at NTP conditions (pressure or head)

"Temperature Compensating Coefficient" for various air temperatures:

### Example - Air with Temperature *100*^{o}C

^{o}C

For air with temperature *100 ^{o}C* the "Temperature Compensating Coefficient" is approximately

*0.83*. So,

- the pressure loss in air with temperature
*100*is approximately^{o}C*83 %*of the pressure loss in air at standard NTP conditions*20*^{o}C

### Influence of Pressure

If the actual pressure differ from atmospheric pressure (*1 bar abs*) - friction loss in air ducts should be compensated with a "Pressure Compensating Coefficient":

Δp = k_{p}Δp_{NTP }(2)

where

Δp = actual friction loss (pressure or head)

k_{p}= Pressure Compensating Coefficient

Δp_{NTP }= friction loss at NTP conditions (pressure or head)

The "Pressure Compensating Coefficient" can be estimated for various pressures from the diagram below:

*1000 mbar = 1 bar = 10*^{5}Pa (N/m^{2}) = 0.1 N/mm^{2}= 10,197 kp/m^{2}= 10.20 m H_{2}O = 0.9869 atm = 14.50 psi (lb_{f}/in^{2}) = 10^{6}dyn/cm^{2}= 750 mmHg