Carbon dioxide enters an adiabatic compressor at 100 kPa and 300 K at a rate of 0.5 kg/s and leaves at 600 kPa and 450 K. Neglecting

Carbon dioxide enters an adiabatic compressor at 100 kPa and 300 K at a rate of 0.5 kg/s and leaves at 600 kPa and 450 K. Neglecting kinetic energy changes, determine (a) the volume flow rate of the carbon dioxide at the compressor inlet and (b) the power input to the compressor. Amswers: (a) 0.283 m^3/s, (b) 68.8 kW

Answers

(a) Q_1=0.283m^3/s

(b) W=68.8kW

Explanation:

Hello,

(a) Based on the inlet conditions, we see that such carbon dioxide has an ideal behavior, so the inlet specific volume is:

v_1=frac{RT}{PM}=frac{8.314frac{Pa*m^3}{mol*K}*300K}{100000Pa*44g/mol} =0.000567m^3/kg

Thus, the volumetric flow rate turns out:

Q_1=0.000567m^3/g*0.5kg/s*1000=0.283m^3/s

(b) Now, by writing and energy balance we have:

mh_1+W=mh_2

Thus, since the ideal gas enthalpy does not depend on the pressure, in Cengel's A-20 table, the corresponding enthalpies at 300K and 450 K are 9431kJ/kmol and 15483kJ/kmol whose values in kJ/kg are 214.34 kJ/kg and 351.89 kJ/kg respectively, thus:

W=0.5kg/s*(351.89-214.34)kJ/kg=68.8kW

Best regards.

a)dot V = 0.28335 m3/s

b)dot W_{in} = 259.63 kW

Explanation:

part a)

inlet volume of air V1 is given as = frac{RT_1}{P_1}

putting all value in the above formula

V_1 = = frac{0.1889 kpa - m3/kg .K * 300 K}{100kPa}

V_1 = 0.5667 m3/kg

we know that  volume of flow rate

dot V = dot mV_1

dot V = 0.5 kg/s *0.5667 m3/kg = 0.28335 m3/s

dot V = 0.28335 m3/s

part b

we know that total energy remain constant, so we have

E_in - E_out = Change in energy in system

E_in = E_out

therefore

dot W_{in} + dot m h_1 =dot m h_2

dot W_{in} = dot m h_2- dot m h_1

the power input to the compressor is

dot W_{in} = dot m c_p (h_2- h_1 )

dot W_{in} = 0.5*5.1926*(450-350 )

dot W_{in} = 259.63 kW

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