Question

A two stage centrifugal compressor operating at 3000 rpm is to compress refrigerant R134a from an evaporator temperature of 0°C (h_i = 398.6 kJ/kg) to a condensing temperature of 32°C (h_e = 419.8 kJ/kg). If the impeller diameter of both the stages have to be same, what is the diameter of the impeller?

Assume suction condition to be dry saturated, compression process isentropic, impeller blades radial and refrigerant enters impeller radially
1. 0.65 m
2. 0.8 m
3. 1.0 m
4. 1.5 m

Answer 1

Correct Answer - Option 1 : 0.65 m

Concept:

Work done in the centrifugal compressor, W = ϕ_s ϕ_w u²

where, velocity of impeller, \(u = \frac{π DN}{60} \), ϕ_s = slip factor,  ϕw = Power work input factor

From SFEE equation compressor workdone, W = he - hi 

Therefore work done, W = he - hi = ϕs ϕw u2

Calculation:

Given:

Evaporator temperature = 0^∘C, Condensing temperature = 32^∘C, Number of stages = 2, rotational speed N = 3000 RPM,

Enthalpy of refrigerant at compressor inlet, h_i = 398.6 kJ/kg, Enthalpy of refrigerant at compressor exit, h_e = 419.8 kJ/kg

entalpy drop in stage 1 is equal to stage 2, Δh1 = Δh2 , Diameter of impeller is same for both stages.

Work done by compressor w = Δhstage  = u²

Since the blade are radial with no tangential velocity component at inlet, the enthalpy rise in each stage is equal

Δh₁ = Δh₂ = Δh_stage 

Enthalpy rise across the compressor = Δh1 + Δh2 = 2 Δhstage = h_e - h_i 

 \( Δh_{stage} =\frac {h_e - h_i}{2} =\frac {419.8 - 398.6}{2} = 10.6\ kJ/kg\)

∴ u2  = 10.6 x 1000 ⇒ u =103 m/s

\(u = \frac{π DN}{60} = \frac{π\ \times\ D\ \times\ 3000}{60}\) 

103 = π x D x 50 

D = 0.655 m

Note: If in problem ϕs and ϕw are not mentioned then consider them equal to 1.