Applications of steady flow energy equation or SFEE

By
Aman Rai
-
Throttling valve

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

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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

Insulated compressor

From steady flow energy equation

m{ h1 + v12/2 + gZ1 } + ∆Q/∆t = m{ h2 + v2 2/2 + gZ2 } + ∆W/∆t

As we assumed ,

∆Q = 0

And

v & z ~ 0

So ,

m{ h1 + v12/2 + gZ1 } + ∆Q/∆t = m{ h2 + v2 2/2 + gZ2 } + ∆W/∆t

Therefore ,

∆W/∆t = m ( h1 – h2 ) , where h1 > h2

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

Turbine

As the steady flow energy equation is

m{ h1 + v12/2 + gZ1 } + ∆Q/∆t = m{ h2 + v2 2/2 + gZ2 } + ∆W/∆t

From the assumptions

∆Q = 0

And

v1 = v2 & Z1 = Z2

so ,

m{ h1 + v12/2 + gZ1 } + ∆Q/∆t = m{ h2 + v2 2/2 + gZ2 } + ∆W/∆t

Hence ,

∆W/∆t = m ( h1 – h2 ) , where h1 < h2

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

Nozzle

From steady flow energy equation :

m{ h1 + v12/2 + gZ1 } + ∆Q/∆t = m{ h2 + v2 2/2 + gZ2 } + ∆W/∆t

Now,

h1 + v12/2 = h2 + v2 2/2

If , v2 >>>>>> v1

So ,

h1 = h2 + v2 2/2

Therefore

v2 = √{2(h1 – h2)}

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

Throttling Device

From SFEE ,

m{ h1 + v12/2 + gZ1 } + ∆Q/∆t = m{ h2 + v2 2/2 + gZ2 } + ∆W/∆t

Therefore ,

h1 = h2

Conclusion :

During throttling , enthalpy remains constant.

Hence Throttling is Isoenthalpic process.

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