Respuesta :
Answer:
Power consume by compressor=113,726.87 KW
Explanation:
Given:[tex]P_{1}=100KPa ,V_{1}=200 m/s,T_{1}=283 K, A_{1} =2m^2[/tex]
[tex]P_{2}=2000KPa ,T_{2}=513 K,A_{2}=0.5m^2[/tex]
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
[tex]h_{1}+\frac{V_{1} ^{2} }{2}+Q=h_{2}+\frac{V_{2} ^{2} }{2}+W[/tex]
We know that[tex]C_{p}=1.005\frac{Kj}{KgK}[/tex]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= [tex]density\times area\times velocity[/tex]
So mass flow rate =[tex]\rho _{1}V_{1}A_{1}[/tex] ,[tex]\rho =\frac{P}{RT}[/tex]
=1.23×200×2 Kg/s
=541.17 Kg/s
[tex]\rho _{1}V_{1}A_{1}=\rho _{2}V_{2}A_{2}[/tex]
[tex]\rho _{2}=13.58\frac{Kg}{{m}^3}[/tex] ,[tex]\rho =\frac{P}{RT}[/tex]
[tex]V_{2}[/tex]=80.07 m/s
Enthalpy of ideal gas h=[tex]C_{p}\times T[/tex]
So[tex] h_{1}=1.005\times283=284.41\frac{Kj}{Kg}[/tex]
[tex]h_{2}=1.005\times513=515.56\frac{Kj}{Kg}[/tex]
Now by putting the values
[tex]284.41+\frac{220 ^{2} }{2000}+Q=515.56+\frac{80.07 ^{2} }{2000}+W[/tex]
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
