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Assuming steam to be an ideal gas, calculate its specific volume and density at a pressure of 90 lb/in2 and a temperature of 650F Compare your answer using data from the steam tables (appendix of your textbook). Problem 2 (25 points). Find the mass of air in a closed chamber measuring 35ft by 20ft by 10ft, when the pressure is 17 lb/in? and the temperature is 75F, assuming air to be an ideal gas.

Answer :

Answer:

1) Sv = 0.4584 m³/Kg...assuming steam as an ideal gas

% deviation from the values in the steam tables

⇒ % dev = 45 %

2) mass air = 272.617 Kg; assuming air to be an ideal gas

Explanation:

ideal gas:

PV = RTn

molar volume:

⇒ V/n = RT/P

∴ P = 90 psi * ( 0.06895 bar/psi ) = 6.2055 bar

∴ T = 650 F = 343.33 °C = 616.33 K

∴ R = 0.08314 bar.L/mol.K

⇒ V/n = (( 0.08314 )*(616.33 K )) / 6.2055 bar

⇒ V/n = 8.2574 L/mol * ( m³/1000L ) = 8.2574 E-3 m³/mol

specific volume ( Sv ):

∴ Mw = 18.01528 g/mol

⇒ Sv = 8.2574 E-3 m³/mol * ( mol / 18.01528 g ) * ( 1000 g/Kg )

⇒ Sv = 0.4584 m³/Kg

steam table:

∴ P = 6.2055 bar ≅ 6 bar → Sv = 0.3157 m³/Kg

⇒ % deviation = (( 0.4584 - 0.3157 ) / 0.3157) * 100

⇒ % dev = 45.2 %; significant value, assuming  steam to be a ideal gas

2) mass air, assuming ideal gas:

∴ V = 20ft * 35ft * 10ft = 7000ft³ * ( 28.3168 L/ft³ ) = 198217.6 L

∴ P = 17 psi * ( 0.06895 bar/psi ) = 1.172 bar

∴ T = 75 °F = 23.89 °C = 296.89 K

∴ R = 0.08314 bar.L/K.mol

⇒ n air = PV/RT = (( 1.172 )*( 198217.6 )) / (( 0.08314 )*( 296.89 ))

⇒ n air = 9411.616 mol air

∴ Mw air = 28.966 g/mol

⇒ mass air = 9411.616 mol * ( 28.966 g/mol ) = 272616.892 g = 272.617 Kg

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