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

Answer:

Power consume by compressor=113,726.87 KW

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

$h_{1}+\frac{V_{1} ^{2} }{2}+Q=h_{2}+\frac{V_{2} ^{2} }{2}+W$

We know that$C_{p}=1.005\frac{Kj}{KgK}$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= $density\times area\times velocity$

So mass flow rate =$\rho _{1}V_{1}A_{1}$     ,$\rho =\frac{P}{RT}$

                                   =1.23×200×2 Kg/s

                                  =541.17 Kg/s

$\rho _{1}V_{1}A_{1}=\rho _{2}V_{2}A_{2}$

$\rho _{2}=13.58\frac{Kg}{{m}^3}$  ,$\rho =\frac{P}{RT}$

$V_{2}$=80.07 m/s

Enthalpy of ideal gas h=$C_{p}\times T$

So$h_{1}=1.005\times283=284.41\frac{Kj}{Kg}$

             $h_{2}=1.005\times513=515.56\frac{Kj}{Kg}$

Now by putting the values

$284.41+\frac{220 ^{2} }{2000}+Q=515.56+\frac{80.07 ^{2} }{2000}+W$

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