Question

1. For airflow through a fan, the mechanical energy balance with fan work along with conservation of mass are:

-W(dot)fan=sum((m(dot)_exit)*(e_exit)) - sum((m(dot)_inlet)*(e_inlet))

so...

0=sum(m(dot)_exit) - sum(m(dot)_inlet)

Using the mechanical energy balance and conservation of mass, show that the theoretical fan power added to the air is flow rate multiplied by the pressure rise:

W(dot)fan=?*?*P

Answer 1

solution :

The power gained by the fluid from a pump or fan can be expressed as:

P = m w         (1)

where

P = power (W)

m = mass flow rate (kg/s)

w = specific work (Nm/kg, J/kg)

Specific work of a pump or fan working with an incompressible fluid can be expressed as:

w = (p₂ - p₁) / ?         (2)

where

w = specific work (Nm/kg, J/kg, m²/s²)

p = pressure (N/m²)

? = density (kg/m³)

put the value of equation 2 in equation 1

P = m (p₂ - p₁) / ?

P = Q (p₂ - p₁)

( Q = m / ? )

P = Q ?P (3)

so from above equation we can say that theoretical fan power added to the air is the flow rate times the pressure rise.