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

Now itâ€™s time to discuss few other types of fluidÂ machinesÂ such as centrifugal compressors, axial flow compressors, fans and blowers where working fluid will be steam, air or gases.Â

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.Â

### Velocity diagram of centrifugal compressorÂ

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.Â

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

u1 = Mean blade velocity at inlet = Ï€D1N/60
u2 = Mean blade velocity at outlet = Ï€D2N/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
VÏ‰1 = Velocity of whirl at inlet or tangential component of absolute velocity of air at inlet
VÏ‰2 = 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
Î±1=Absolute angle at inlet or outlet angle
Î±2=Inlet angle to the diffuser
Î²1=Inlet angle to the rotor blade
Î²2= Outlet angle to the rotor blade
m = Mass flow rate of air in Kg/secÂ

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.Â

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

### Energy transferred to the fluid

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

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

Air enters at the inlet axially and hence Î±1i.e. absolute angle at inlet will be 900.
Î±1= 900,
VÏ‰1 = 0, Vf1 =V1Â

As compressed air will leave the rotor radially and hence we can write following equation as mentioned here.
Î²2= 900,
Vr2 =Vf2
VÏ‰2 =u2Â

Now we will determine the work done by the impeller on air orÂ energy transferred to the fluid and it could be written as mentioned hereÂ
E = m [VÏ‰2u2- VÏ‰1u1]Â

E = m [VÏ‰2u2- VÏ‰1u1]Â

E = m [u2x u2]Â

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

### E/m = u2 2Â

#### Case -2: Air enters the impeller eye in an axial direction (Î±1 = 900) but not leaving the impeller radially i.e. Î²2 < 900

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

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

E = m [VÏ‰2u2- VÏ‰1u1]

E = m VÏ‰2u2Â

Energy transferred to the fluid per unit mas, E/m = VÏ‰2u2Â

Let us see here the outlet velocity triangle and we can write the following equation
Tan Î²2=Vf2/ (u2 - VÏ‰2)
CotÎ²2= (u2 - VÏ‰2) / Vf2
VÏ‰2= u2 -Vf2Cot Î²2Â

Therefore, we can say here that with the consideration ofÂ phenomenon of slip in centrifugal compressor, velocity of whirl at outlet i.e.Â VÏ‰2Â will be less than theÂ Mean blade velocity at outletÂ i.e.Â u2.Â
VÏ‰2Â <Â u2

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.Â

Slip factor,Â Ïƒ = Velocity of whirl at outlet i.e.Â VÏ‰2Â /Â Mean blade velocity at outletÂ i.e.Â u2
Ïƒ = VÏ‰2Â /Â u2
VÏ‰2Â =Â ÏƒÂ u2

Now we will use this value of VÏ‰2Â in above mentioned energy equation and we will have following equation for energy transferred to the fluid per unit mas.Â

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

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

We must note it here that the value of slip factor i.e.Â Ïƒ 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 theÂ amount of energy transferred to the fluid in no slip.Â

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

### Energy transferred to the fluid per unit mas,Â E/m =Â Ïˆ ÏƒÂ 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 orÂ work done on air in a centrifugal compressor.Â

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

### Reference:Â Â

Fluid mechanics, By R. K. BansalÂ