Question

An activated sludge aeration blower is delivering air to diffusers in an aeration tank. You want to estimate the isentropic efficiency and the work input (by the motor). At the blower inlet, the volumetric flow rate is measured at 85 m3/min, the pressure is 0.98 atm and the temperature is 20⁰C. If the pressure is increased to 1.61 atm and the blower outlet temperature is 350 K: a) what is the blower isentropic efficiency, and b) what is the work input (kW) to the blower (by the motor)? Hints: k for air can be taken (Table A-2b) as 1.40; you can use Equation 8-43 to determine the isentropic process T2 value. You can use the Ideal Gas Law to determine the mass flow rate based on the inlet conditions and volumetric flow rate. Cp can be obtained from Table A-2b (From: Introduction to Thermodynamics and Heat transfer).

Answer 1

a)

For isentropic process, T_{2_is} / T₁ = (P₂ / P₁)^1-1/k

T_{2_is} / (273 + 20) = (1.61 / 0.98)^1-1/1.4

T_{2_is} = 337.7 K

Isentropic efficiency = Isentropic enthalpy rise / Actual enthalpy rise

= (h_{2_is} - h₁) / (h₂ - h₁)

= C_p(T_{2_is} - T₁) / [C_p(T₂ - T₁)]

= (T_{2_is} - T₁) / (T₂ - T₁)

= (337.7 - (273+20)) / (350 - (273+20))

= 0.783 or 78.3 %

b)

At inlet, Density = P / RT

= (0.98*101325) / 287 / (273+20)

= 1.18 kg/m³

Mass flow rate m = Density * Volumetric flow rate

= 1.18 * 85

= 100.4 kg/min

= 100.4 / 60 kg/s

= 1.673 kg/s

For air Specific heat C_p = 1.005 kJ/kg/K

Work input = m * C_p * (T₂ - T₁)

= 1.673 * 1.005 * (350 - (273+20))

= 95.8 kW