Mechanical Energy Equation vs. Bernoulli Equation

The Energy Equation is a statement based on the First Law of Thermodynamics involving energy, heat transfer and work. With certain limitations the mechanical energy equation can be compared to the Bernoulli Equation .

The Mechanical Energy Equation in Terms of Energy per Unit Mass

The mechanical energy equation for a pump or a fan can be written in terms of energy per unit mass where the energy into the system equals the energy out of the system.

E _pressure,in + E _velocity,in + E _elevation,in + E _shaft

= E _pressure,out + E _velocity,out + E _{elevation,out} + E _loss (1)

or

p _in / ρ + v _in ²/ 2 + g h _in + E _shaft

= p _out / ρ + v _out ²/ 2 + g h _out + E _loss (1b)

where

p = static pressure (Pa, (N/m²))

ρ = density (kg/m³ )

v = flow velocity (m/s)

g = acceleration of gravity (9.81 m/s²)

h = elevation height  (m)

E _shaft = net shaft energy per unit mass for a pump, fan or similar (J/kg)

E _loss = hydraulic loss through the pump or fan (J/kg)

The energy equation is often used for incompressible flow problems and is called the Mechanical Energy Equation or the Extended Bernoulli Equation .

The mechanical energy equation for a turbine - where power is produced - can be written as:

p _in / ρ + v _in ²/ 2 + g h _in

= p _out / ρ + v _out ²/ 2 + g h _out + E _shaft + E _loss (2)

where

E _shaft = net shaft energy out per unit mass for the turbine (J/kg)

Equation (1) and (2) dimensions are

• energy per unit mass (ft²/s²= ft lb/slug or m²/s²= N m/kg)

Efficiency

According to (1) more loss requires more shaft work to be done for the same rise of output energy. The efficiency of a pump or fan process can be expressed as:

η = (E _shaft - E _loss ) / E _shaft (3)

The efficiency of a turbine process can be expressed as:

η = E _shaft / (E _shaft + E _loss )                                     (4)

The Mechanical Energy Equation in Terms of Energy per Unit Volume

The mechanical energy equation for a pump or fan (1) can also be written in terms of energy per unit volume by multiplying (1) with the fluid density - ρ :

p _in + ρ v _in ²/ 2 + γ h _in + ρ E _shaft

= p _out + ρ v _out ²/ 2 + γ h _out + ρ E _loss (5)

where

γ = ρ g = specific weight (N/m³ )

The dimensions of equation (5) are

• energy per unit volume (ft lb/ft³ = lb/ft²or Nm/m³ = N/m²)

The Mechanical Energy Equation in Terms of Energy per Unit Weight involving Heads

The mechanical energy equation for a pump or a fan (1) can also be written in terms of energy per unit weight by dividing with gravity - g :

p _in / γ + v _in ²/ 2 g + h _in + h _shaft

= p _out / γ + v _out ²/ 2 g + h _out + h _loss (6)

h _shaft = E _shaft / g = net shaft energy head per unit mass for a pump, fan or similar  (m)

h _loss = E _loss / g = loss head due to friction  (m)

The dimensions of equation (6) are

• energy per unit weight (ft lb/lb = ft or Nm/N = m)

Head is the energy per unit weight .

h _shaft can also be expressed as:

h _shaft = E _shaft / g

= E _shaft / m g = E _shaft / γ Q (7)

where

E _shaft = shaft power (W)

m = mass flow rate  (kg/s)

Q = volume flow rate  (m³ /s)

Example - Pumping Water

Water is pumped from an open tank at level zero to an open tank at level 10 ft. The pump adds four horse powers to the water when pumping 2 ft³ /s .

Since v _in = v _out = 0, p _in = p _out = 0 and h _in = 0 - equation (6) can be modified to:

h _shaft = h _out + h _loss

or

h _loss = h _shaft - h _out (8)

Equation (7) gives:

h _shaft = E _shaft / γ Q

= (4 hp)(550 ft lb/s/hp) / (62.4 lb/ft³ )(2 ft³ /s)

= 17.6 ft

• specific weight of water - 62.4 lb/ft³
• 1 hp ( English horse power ) = 550 ft lb/s

Combined with (8) :

h _loss = (17.6 ft ) - (10 ft)

= 7.6 ft

The pump efficiency can be calculated from (3) modified for head:

η = (( 17.6 ft) - ( 7.6 ft) ) / (17.6 ft)

= 0.58