Mechanical Energy Equation vs. Bernoulli Equation
The Mechanical Energy Equation compared to the Extended 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
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