Question

Air enters a compressor at 100 kPa, 10°C, and 220 m/s through an inlet area of 2 m2. The air exits at 2 MPa and 240°C through an area of 0.5 m2. Including the change in kinetic energy, determine the power consumed by this compressor, in kW.

Answer 1

Power consume by compressor=113,726.87 KW

Step-by-step explanation:

Given:
`P_(1)=100KPa ,V_(1)=200 m/s,T_(1)=283 K, A_(1) =2m^2`

`P_(2)=2000KPa ,T_(2)=513 K,A_(2)=0.5m^2`

Actually compressor is an open system, so here we will use first law of thermodynamics for open system .

We know that first law of thermodynamics for steady flow

`h_(1)+(V_(1) ^(2) )/(2)+Q=h_(2)+(V_(2) ^(2) )/(2)+W`

We know that
`C_(p)=1.005(Kj)/(KgK)`and we take the air as ideal gas.

System is in steady state then mass flow rate in =mass flow rate out

Mass flow rate=
`density* area* velocity`

So mass flow rate =
`\rho _(1)V_(1)A_(1)` ,
`\rho =(P)/(RT)`

=1.23×200×2 Kg/s

=541.17 Kg/s

`\rho _(1)V_(1)A_(1)=\rho _(2)V_(2)A_(2)`

`\rho _(2)=13.58\frac{Kg}{{m}^3}` ,
`\rho =(P)/(RT)`

`V_(2)`=80.07 m/s

Enthalpy of ideal gas h=
`C_(p)* T`

So
`h_(1)=1.005*283=284.41(Kj)/(Kg)`

`h_(2)=1.005*513=515.56(Kj)/(Kg)`

Now by putting the values

`284.41+(220 ^(2) )/(2000)+Q=515.56+(80.07 ^(2) )/(2000)+W`

Here Q=0 because heat transfer is zero here.

W= -210.15 KJ/kg

So power consume by compressor=541.17×210.15

=113,726.87 KW