We were discussing the basic principle of operations, functions of various parts and performance characteristics of different types of fluid machines such as hydraulic turbinescentrifugal pumps and reciprocating pumps in our previous posts. We have seen there that working fluid for above mentioned fluid machines was liquid such as water.

Now it’s time to discuss few other types of fluid machines such as centrifugal compressors, axial flow compressors, fans and blowers where working fluid will be steam, air or gases.

We have already seen the working principle of centrifugal compressor in our last post, where we have also seen the parts and their functions of a centrifugal compressor. Today we will be interested here to discuss the velocity diagram of centrifugal compressor. We will also find out here the work done on air or energy transferred to the fluid in the centrifugal compressor.

### Velocity diagram of centrifugal compressor

Following figure shown here indicates the velocity diagram of a centrifugal compressor. Inlet velocity triangle and outlet velocity triangle are drawn here in following figure.

Let us first understand here the various nomenclatures that we are using here.

u1 = Mean blade velocity at inlet = πD1N/60
u2 = Mean blade velocity at outlet = πD2N/60
V1 = Absolute velocity of air at inlet to rotor or impeller
V2 = Absolute velocity of air at outlet to rotor or impeller
Vr1 = Relative velocity of air at inlet
Vr2 = Relative velocity of air at outlet
Vω1 = Velocity of whirl at inlet or tangential component of absolute velocity of air at inlet
Vω2 = Velocity of whirl at outlet or tangential component of absolute velocity of air at outlet
Vf1 = Velocity of flow at inlet
Vf2 = Velocity of flow at outlet
α1=Absolute angle at inlet or outlet angle
α2=Inlet angle to the diffuser
β1=Inlet angle to the rotor blade
β2= Outlet angle to the rotor blade
m = Mass flow rate of air in Kg/sec

Centrifugal compressor parts are designed in such way that air will enter and leave the compressor without any shock. Let us assume that condition is ideal and there is no slip and no pre whirl.

We will see here two cases and velocity diagram will be different for these two cases. So let us understand here the first case.

### Energy transferred to the fluid

#### Case -1: Air enters the impeller eye in an axial direction (α1 = 900) and leaves the impeller radially

Following figure indicates the velocity triangle at inlet and outlet of the impeller blade.

Air enters at the inlet axially and hence α1i.e. absolute angle at inlet will be 900.
α1= 900,
Vω1 = 0, Vf1 =V

As compressed air will leave the rotor radially and hence we can write following equation as mentioned here.
β2= 900,
Vr2 =Vf2
Vω2 =u

Now we will determine the work done by the impeller on air or energy transferred to the fluid and it could be written as mentioned here
E = m [Vω2u2- Vω1u1

E = m [Vω2u2- Vω1u1

E = m [u2x u2

Energy transferred per unit mass of air could be written by following equation

### E/m = u2 2

#### Case -2: Air enters the impeller eye in an axial direction (α1 = 900) but not leaving the impeller radially i.e. β2 < 900

This is a case, where we are going to consider the phenomenon of slip in centrifugal compressor. We will see the phenomenon of slip in centrifugal compressor in our next post, but we must understand here that due to this phenomenon of slip in centrifugal compressor, velocity triangle at the outlet of the impeller blade will be changed.

Work done by the impeller on air or energy transferred to the fluid will be determined by following equation as mentioned here.

E = m [Vω2u2- Vω1u1]

E = m Vω2u

Energy transferred to the fluid per unit mas, E/m = Vω2u

Let us see here the outlet velocity triangle and we can write the following equation
Tan β2=Vf2/ (u2 - Vω2)
Cotβ2= (u2 - Vω2) / Vf2
Vω2= u2 -Vf2Cot β

Therefore, we can say here that with the consideration of phenomenon of slip in centrifugal compressor, velocity of whirl at outlet i.e. Vωwill be less than the Mean blade velocity at outlet i.e. u2
Vωu2

Here, we will consider one important factor i.e. slip factor in deriving the expression for energy transferred to the fluid. Slip factor will be discussed in detail in our next post.

Slip factor, σ = Velocity of whirl at outlet i.e. VωMean blade velocity at outlet i.e. u2
σ = Vωu2
Vω2 = σ u2

Now we will use this value of Vω2 in above mentioned energy equation and we will have following equation for energy transferred to the fluid per unit mas

Energy transferred to the fluid per unit mas, E/m = σ u22

### Energy transferred to the fluid per unit mas, E/m = σ u22

We must note it here that the value of slip factor i.e. σ will be less than 1 and hence we can easily conclude that less amount of energy will be transferred to the fluid with consideration of slip factor as compared to the amount of energy transferred to the fluid in no slip.

Here, if we considered one more factor i.e. power input factor (ψ), we will have following equation of energy transferred to the fluid per unit mass as mentioned below.

### Energy transferred to the fluid per unit mas, E/m = ψ σ u22

So, we have seen here the velocity triangles at inlet and outlet of impeller blades. We have also secured here the expression for the energy transferred to the fluid per unit mas or work done on air in a centrifugal compressor.

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Further we will find out, in our next post, slip phenomenon and slip factor for centrifugal compressor

### Reference:

Fluid mechanics, By R. K. Bansal