# Mechanical Energy and Bernoulli Equation

## The mechanical energy equation related to energy per unit mass, energy per unit volume and energy per unit weight involving head

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**:

p_{in}/ ρ + v_{in}^{2}/ 2 + g h_{in}+ E_{shaft}= p_{out}/ ρ + v_{out}^{2}/ 2 + g h_{out}+ E_{loss}(1)

where

p= static pressure(Pa)

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}= loss due to friction (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** 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 a turbine or similar (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)* a larger amount of loss - *w _{loss}* - result in more shaft work required for the same rise of output energy. The efficiency of a

**pump or fan process**can be expressed as:

η= (E_{shaft}- E_{loss}) / Ew_{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 a 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}+ρE_{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= lossheaddue 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*and

_{in}= p_{out}= 0*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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