# Equation of Mechanical Energy

## The equation of mechanical energy in terms of Energy per Unit Mass, Energy per Unit Volume and Energy per Unit Weight involving head

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The Energy Equation is a statement of the first law of thermodynamics. The energy equation involves 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}+ w_{shaft}= p_{out}/ ρ + v_{out}^{2}/ 2 + g h_{out}+ w_{loss}(1)

where

p= static pressure

ρ= density

v= flow velocity

h= elevation height

w_{shaft}= net shaft energy per unit mass for a pump, fan or similar

w_{loss}= loss due to friction

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}+ w_{shaft}+ w_{loss}(2)

where

w_{shaft}= net shaft energy out per unit mass for a turbine or similar

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= (w_{shaft}- w_{loss}) /w_{shaft}(3)

The efficiency of a **turbine process** can be expressed as:

η=w_{shaft }/ (w_{shaft}+ w_{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 fluid density - *ρ*:

p_{in}+ ρ v_{in}^{2}/ 2 +γ h_{in}+ρ w_{shaft}= p_{out}+ρ v_{out}^{2}/ 2 + γ h_{out}+ρw_{loss}(5)

where

γ=ρ g =specific weight

The dimensions of equation (5) are

*energy per unit volume**(ft.lb/ft*^{3}= lb/ft^{2}or N.m/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)

where

γ=ρ g =specific weight

h_{shaft}=w_{shaft}/ g= net shaft energy head per unit mass for a pump, fan or similar

h_{loss}=w_{loss}/ g= loss head due to friction

The dimensions of equation (6) are

*energy per unit**weight**(ft.lb/lb = ft or N.m/N = m)*

Head is the energy per unit weight.

*h _{shaft}* can also be expressed as:

h_{shaft}=w_{shaft}/ g=W_{shaft}/ m g = W_{shaft}/ γ Q(7)

where

W_{shaft}= shaft power

m= mass flow rate

Q= volume flow rate

### 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 horsepowers *to the water when pumping *2 ft ^{3}/s*.

Since *v _{in}*

*=*

*v*

_{out}= 0,*p*and

_{in}= p_{out}= 0*h*= 0 - equation (6) can be modified to:

_{in}

h_{shaft}= h_{out}+ h_{loss}

or

h(8)_{loss}= h_{shaft}- h_{out }

Equation (7) gives:

h_{shaft}= W_{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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