Answer:
(a) [tex]Q_1=0.283m^3/s[/tex]
(b) [tex]W=68.8kW[/tex]
Explanation:
Hello,
(a) Based on the inlet conditions, we see that such carbon dioxide has an ideal behavior, so the inlet specific volume is:
[tex]v_1=\frac{RT}{PM}=\frac{8.314\frac{Pa*m^3}{mol*K}*300K}{100000Pa*44g/mol} =0.000567m^3/kg[/tex]
Thus, the volumetric flow rate turns out:
[tex]Q_1=0.000567m^3/g*0.5kg/s*1000=0.283m^3/s[/tex]
(b) Now, by writing and energy balance we have:
[tex]mh_1+W=mh_2[/tex]
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:
[tex]W=0.5kg/s*(351.89-214.34)kJ/kg=68.8kW[/tex]
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