Carbon dioxide (CO2) at 1 bar, 300 K enters a compressor operating at steady state and is compressed adiabatically to an exit st
ate of 10 bar, 520 K. The CO2 is modeled as an ideal gas, and kinetic and potential energy effects are negligible. For the compressor, determine (a) the work input, in kJ per kg of CO2 flowing, (b) the rate of entropy production, in kJ/K per kg of CO2 flowing, and (c) the isentropic compressor efficiency.
1 answer:
Answer:
A.) 0.08 kJ/kg.K
B.) 207.8 KJ/Kg
C.) 0.808
Explanation:
From the question, the use of fluids mechanic table will be required. In order to get the compressor processes, the kinetic energy and the potential energy will be negligible while applying the ideal gas model.
Since the steam is a closed system, the carbon dioxide will be compressed adiabatically.
Please find the attached files for the solution and the remaining explanation.
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Answer:
Explanation:
We can solve Von Karman momentum integral equation as seen below using following in the attached file
Answer:
(a) The heat transferred is 2552.64 kJ
(b) The entropy change of the mixture is 1066.0279 J/K
Explanation:
Here we have
Molar mass of H₂ = 2.01588 g/mol
Molar mass of N₂ = 28.0134 g/mol
Number of moles of H₂ = 500/2.01588 = 248 moles
Number of moles of N₂ = 1200/28.0134 = 42.8 moles
P·V = n·R·T
V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³
Since the volume is doubled then
V₂ = 2 × 7.25 = 14.51 m³
At constant pressure, the temperature is doubled, therefore
T₂ = 600 K
If we assume constant specific heat at the average temperature, we have
Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂
cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K
cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K
Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ
b)
Where:
and
are the mole fractions of Hydrogen and nitrogen respectively.
Therefore, = 248 /(248 + 42.8) = 0.83
= 42.8/(248 + 42.8) = 0.1472
∴ = 1066.0279 J/K