Question

A compressor is used to steadily compress Air which enters the compressor at a pressure of 220 kPa, a temperature of 320 K, and a velocity of 8 m/s through an inlet with area of 0.13 m². The air exits the compressor at a pressure of 770 kPa, temperature is 490 K, and a of velocity is 4 m/s. The compressor consumes 430 kW of power. Assuming variable specific heats, find: A) The heat transfer during this process. B) The area of the compressor outlet.

Answer 1

Let suffix 1 denote entry point and 2 denotes exit of compressor.

For air, Cp = 1.005 kJ/kgk, density = 1.2 kg/m3

Properties at entry point,

P1 = 220 kPa

T1 = 320 K

V1 = 8 m/s

A1 = 0.13 m2

So, h1 = Cp*T1 = 1.005*320 = 321.6 kJ/kg

Properties at exit of compressor,

P2 = 770 kPa

T2 = 490 K

V2 =4 m/s

A2 =?

h2 = 1.005*490 = 492.45 kJ/kg

W = - 430 kW (negative sign indicate that work is consumed by condeser that is work input)

Q =?

um the to

(A) Apply steady flow energy equation

​​​​​h1 + kE1 + PE1 + Q = h2 + kE2 + PE2 +W

Assume potential energy is constant l.

So,

h1 + (V12/2000) + Q = h2 + (V22 / 2000) + W

Or, Q = h2 - h1 + {(V22 - V12) / 2000} +W

Put all values,

Or, Q = 492.45 - 321.6 +{ (16 - 64)/2000} - 430

​​​​Here we have W in kW and all other in kJ/kg so we need to multiply by mass flow rate ,then it also became kW

Mass flow rate=m = density* Velocity*area = 1.2*8*0.13 = 1.248 kg/s

Now,

Q = 1.248*(170.85) + 1.248*(-48/2000) - 430

Q = 213.2208 - 0.029952 - 430 = - 216.809152

So, Q = - 216.809 kW (negative sign indicate, heat rejection from compressor)

(B) apply mass conversation between entry and exit,

A1V1 = A2V2

OR, 0.13*8 = A2* 4

Or, A2 = 0.13*2 = 0.26 m2

So area at exit is 0.26 m2.