# Air Ducts - Major Friction Loss vs. Temperature and Pressure

## The influence of temperature and air pressure on major 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