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/m ² ))

ρ = density (kg/m ³ )

v = flow velocity (m/s)

g = acceleration of gravity (9.81 m/s ² )

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 (ft ² /s ² = ft lb/slug or m ² /s ² = 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/m ³ )

The dimensions of equation (5) are

• energy per unit volume (ft lb/ft ³ = lb/ft ² or Nm/m ³ = N/m ² )

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  (m ³ /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 ft ³ /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/ft ³ )(2 ft ³ /s)

= 17.6 ft

• specific weight of water - 62.4 lb/ft ³
• 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