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Gelneren [198K]

3 years ago

11

Water steam enters a turbine at a temperature of 400 o C and a pressure of 3 MPa. Water saturated vapor exhausts from the turbin

e at a pressure of 125 kPa. At steady state, the work output of the turbine is 530 kJ/kg. The surrounding air is at 20 o C. Neglect the changes in kinetic energy and potential energy. Determine (20 points) (a) the heat transfer from the turbine to the surroundings per unit mass flow rate, (b) the entropy generation during this process.

1 answer:

yulyashka [42]3 years ago

6 0

Answer:

a) -505.229 kJ/Kg

b) -1.724 kJ/kg

Explanation:

T1 = 400°C

P1 = 3 MPa

P2 = 125 kPa

work output = 530 kJ/kg

surrounding temperature = 20°C = 293 k

<u>A) Calculate heat transfer from Turbine to surroundings </u>

Q = h2 + w - h1

h ( enthalpy )

h1 = 3231.229 kj/kg

enthalpy at P2

h2 = hg = 2676 kj/kg

back to equation 1

Q = 2676 + 50 - 3231.229 = -505.229 kJ/Kg ( i.e. heat is lost )

<u>b) Entropy generation </u>

entropy generation = Δs ( surrounding ) + Δs(system)

= - 505.229 / 293 + 0

= -1.724 kJ/kg

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Answer:

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Explanation:

A) The differential equation of this problem is;

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Where;

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3 years ago

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Answer:

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4 0

3 years ago

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denpristay [2]

Answer:

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Part b: The strain is $8.621 \times 10^{-4} in/in$

Part c: The plastic moment is 600 ksi.

Explanation:

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Here

M is the moment which is to be calculated
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Here

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