Steady flow energy equation is Very useful in thermodynamics it helps in finding the compressor & Turbine work and also velocity of Nozzle , etc .

We discussed about Steady flow energy equation in previous post

## The Applications of steady flow energy equation are :

## 1) Compressor :

Compressor is a work or power consumption device.

With the help of steady flow energy equation , we can get the magnitude of work or power consumed in a compressor.

To find the work consumption , assume

1) Insulated compressor

2) Kinetic energy and potential energy are negligible

From steady flow energy equation

m{ h_{1} + v_{1}^{2}/2 + gZ_{1 }} + ∆Q/∆t = m{ h_{2} + v_{2 2}/2 + gZ_{2 }} + ∆W/∆t

As we assumed ,

∆Q = 0

And

v & z ~ 0

So ,

m{ h_{1} + ~~v~~= m{ h_{1}^{2}/2 + gZ_{1 }} + ∆Q/∆t _{2} + ~~v~~} + ∆W/∆t_{2 2}/2 + gZ_{2 }

Therefore ,

**∆W/∆t = m ( h _{1 } – h_{2 } ) , where h_{1 } > h_{2 }**

this is the magnitude of work or power consumed by a compressor.

## 2) Turbine :

Turbine is a work or power producing device.

By the help of steady flow energy equation , we can get the magnitude of work or power produced in a Turbine.

In order to find the work production , assume

(a) Insulated Turbine

(b)Kinetic energy and potential energy are negligible

### As the steady flow energy equation is

m{ h_{1} + v_{1}^{2}/2 + gZ_{1 }} + ∆Q/∆t = m{ h_{2} + v_{2 2}/2 + gZ_{2 }} + ∆W/∆t

From the assumptions

∆Q = 0

And

v_{1 }= v_{2 }& Z_{1 } = Z_{2 }

so ,

m{ h_{1} + ~~v~~= m{ h_{1}^{2}/2 + gZ_{1 }} + ∆Q/∆t _{2} + ~~v~~} + ∆W/∆t_{2 2}/2 + gZ_{2 }

Hence ,

**∆W/∆t = m ( h _{1} – h_{2} ) , where h_{1} < h_{2}**

This is the magnitude of work or power produced by a turbine.

## 3) Nozzle :

It is neither work producing or work consuming device .

**The function of Nozzle is to increase the velocity of the fluid at inlet or outlet.**

To find the magnitude of increase in velocity , assume

(a) No heat transfer i.e , ∆Q = 0

(b) No work transfer i.e , ∆W = 0

(c) No change in Potential energy i.e , ∆Z = 0

From steady flow energy equation :

m{ h_{1} + v_{1}^{2}/2 + ~~gZ~~= m{ h_{1 }} + ∆Q/∆t _{2} + v_{2 2}/2 + ~~gZ~~_{2 }} + ∆W/∆t

Now,

h_{1} + v_{1}^{2}/2 = h_{2} + v_{2 2}/2

If , v_{2 } >>>>>> v_{1}

So ,

h_{1} = h_{2} + v_{2 2}/2

Therefore

**v _{2} = √{2(h_{1} – h_{2})}**

This is the magnitude of increase in velocity by nozzle.

## 4) Throttling Device :

Throttling is the flow of gas from very high pressure to very low pressure through partially open valve .

And

Throttling is

- Very fast process
- Highly irreversible process
- Adiabatic process

A device which is used for throttling is throttling device and it is neither a work producing or consuming device .

Assumption –

Kinetic energy and potential energy are neglected & heat transfer zero

From SFEE ,

m{ h_{1} + ~~v~~ + _{1}^{2}/2~~gZ~~= m{ h_{1 }} + ∆Q/∆t _{2} + ~~v~~ + _{2 2}/2~~gZ~~_{2 }} + ∆W/∆t

Therefore ,

**h _{1 } = h_{2 }**

Conclusion :

During throttling , enthalpy remains constant.

Hence **Throttling is Isoenthalpic process.**

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