Air is compressed by an adiabatic compressor from 100 kPa and 20°C to 1.8 MPa and 400°C. Air enters the compressor through a 0.15-m2 opening with a velocity of 30 m/s. It exits through a 0.078-m2 opening. Calculate the mass flow rate of air and the required power input. The constant pressure s

A) Flow rate = 2.783 kg/s

B) Power input = 1088.65 Kw

Explanation:

A) P1 = 100KPa; v = 30m/s; T1 = 20°C = 273 + 20k = 293K; A = 0.078 m²; R= 0.287 kPa.m³/kg

The mass flow rate is given as;

m' = ρ1V1' = (P1A1v1)/RT1

Thus, m' = (100 x 0.078 x 30)/(0.287 x 293) = 234/84.091 = 2.783 kg/s

B) The power input is going to be derived from the energy balance equation which is;

m'(h1 + v1²/2) + W' = m'h2

Factorizing out, we have;

W' = m'(h2 - h1 - v1²/2)

Now we have to find the specific enthalpies at temperature of 293K and 400 + 273 = 673K

Thus, looking at the table i attached for ideal gas properties of air,

At T =293K,when we interpolate, we have h1 approximately = 293. 166 Kj/kg and at T=673K,when we interpolate, we have h2 approximately = 684.344 Kj/Kg

Thus, since W' = m'(h2 - h1 - v1²/2)

W' = 2.783(684.344 - 293. 166) = 1088.65 Kw

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