Air expands through an ideal turbine from 1 MPa, 900 K to 0.1 MPa, 500K. The inlet velocity is small compared to the exit veloci
ty of 100 m/s. The turbine operates at steady state and develops a power output of 3,200 kW. Heat transfer between the turbine and its surroundings and any potential energy effects are negligible.
a. Calculate the mass flow rate of air, in kg/s, and the exit area, in m^2.
a. Calculate the mass flow rate of air, in kg/s, and the exit area, in m^2.
1 answer:
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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