Air is compressed by a compressor from v1 = 1.0 m3/kg to v2 = 0.71 m3/kg in a reversible, isothermal process. The air temperature is maintained constant at 25 oC during this process as a result of heat transfer to the surroundings. Air moves through the compressor at a rate of m = 1 kg/s. a)- Determine the entropy change of the air per kg of air. b)- What is the power required by the compressor? c)- What is the rate at which entropy leave the compressor?

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Netta00 Netta00 · Answer 1 · 2019-07-26T08:14:21+02:00

Answer:

(a)[tex]s_2-s_1[/tex]= -0.098 KJ/kg-K

(b)P= 29.8 KW

(c) [tex]S_{gen}[/tex]= -0.098 KW/K

Explanation:

[tex]V_1=1m^3/kg,V_2=0.71m^3/kg,[/tex] mass flow rate= 1 kg/s.

T=25°C

Air treating as ideal gas

(a)

We know that entropy change for ideal gas between two states

[tex]s_2-s_1=mC_v\ln \frac{T_2}{T_1}+mR\ln \frac{V_2}{V_1}[/tex]

Given that this is isothermal process so

[tex]s_2-s_1=mR\ln \frac{V_2}{V_1}[/tex]

[tex]s_2-s_1=1\times 0.287\ln \frac{0.71}{1}[/tex]

[tex]s_2-s_1[/tex]= -0.098 KJ/kg-K

(b)

Power required

[tex]P=\dot{m}T\Delta S[/tex]

[tex]P=1\times (273+25)(s_2-s_1)[/tex]

[tex]P=1\times (273+25)(-0.098)[/tex]

P= -29.8 KW (Negative sign means it is compression process.)

(c)

Rate of entropy generation [tex]S_{gen}[/tex]

[tex]S_{gen}=\dot{m}T\Delta S[/tex]

[tex]S_{gen}[/tex]=1(-0.098)

[tex]S_{gen}[/tex]= -0.098 KW/K