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Mariulka [41]
3 years ago
10

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.

Engineering
1 answer:
3241004551 [841]3 years ago
4 0

Answer:

Mass flow rate= 7.53 kg/sec

Exit Area= 0.108m^2

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

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Answer:

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(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

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Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588  = 248 moles

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Since the volume is doubled then

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At constant pressure, the temperature is doubled, therefore

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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)  \Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)

Where:

x_A and x_B are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, x_A = 248 /(248 + 42.8) = 0.83

x_B = 42.8/(248 + 42.8) = 0.1472

∴ \Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472) =  1066.0279 J/K

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