# 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*= 0 - equation

_{ in }*(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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## Related Documents

### 1st Law of Thermodynamics

The First Law of Thermodynamics simply states that energy can be neither created nor destroyed (conservation of energy). Thus power generation processes and energy sources actually involve conversion of energy from one form to another, rather than creation of energy from nothing.

### Bernoulli Equation

Conservation of energy in a non-viscous, incompressible fluid at steady flow.

### Energy

Energy is the capacity to do work.

### Energy Equation - Pressure Loss vs. Head Loss

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### Fluid Flow - Equation of Continuity

The Equation of Continuity is a statement of mass conservation.

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### Potential Energy - Hydropower

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### Pumps - NPSH (Net Positive Suction Head)

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