In a steady-flow air compressor, air enters at a speed of 5 m/s with a pressure of 1 bar and leaves at a speed of 7.5 m/s with a pressure of 7 bar. If the inlet specific volume is 0.5 m³/kg and the outlet specific volume is 0.15 m³/kg, what is the ratio of the inlet pipe diameter to the outlet pipe diameter?

Concept:

Continuity equation:

• The continuity equation represents the law of conservation of mass.
• In general, for unsteady flow the continuity equation is: ρ1A1V1 = ρ2A2V2  
• For steady flow density is constant.
• The mass flow rate into the system - Mass flow rate out of the system = 0 ⇒ A1V1 = A2V2, Where A = Area and V = Velocity

Calculation:

Given:

V₁ = 5 m/s, V₂ = 7.5 m/s, inlet specific volume (\(\nu _1\)) = 0.5 m3/kg ⇒ ρ₁ = \(\frac{1}{\nu _1}~=~\frac{1}{0.5}~=~2~kg/m^3\), outlet specific volume (\(\nu _2\)) = 0.15 m3/kg ⇒ ρ₂ = \(\frac{1}{\nu _2}~=~\frac{1}{0.15}~=~6.67~kg/m^3\)

∵ For steady flow and with non-uniform density, ρ₁A1V1 = ρ₂A2V2

⇒ \(ρ_1~\times~{D_1}^2 ~\times~V_1~=~ρ_2~\times~{D_2}^2 ~\times~V_2\)

⇒ \(2~\times~{D_1}^2 ~\times~5~=~6.67~\times~{D_2}^2 ~\times~7.5\)

⇒ \(\frac{{D_1}^2}{{D_2}^2}~=~\frac{7.5~\times~6.67}{5~\times~2}\)

⇒ \(\frac{{D_1}}{{D_2}}~=~\sqrt{\frac{7.5~\times~6.67}{5~\times~2}}\)

⇒ \(\frac{{D_1}}{{D_2}}~=~2.236\)