Answer:
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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
Answer:
The digitization of performance management not only provides more precise data but also positively influences management processes and strategic development. Technology-enabled performance management tools simplify the manager's evaluation process and turn employees into active participants in their review sessions
Answer:
So the exit velocity of water is 4.5 m/s
Explanation:
Given that
Water entering pressure = 1.5 MPa
Temperature = 150°C
Velocity = 4.5 m/s
From first law of thermodynamics for open system
Here given that valve is adiabatic so Q= 0
In valve W= 0
Wen also also know that throttling process is an constant enthaply process so
So from above equation we can say that
So the exit velocity of water is 4.5 m/s
Answer:
with a square cross section and length L that can support an end load of F without yielding. You also wish to minimize the amount the beam deflects under load. What is the free variable(s) (other than the material) for this design problem?
a. End load, F.
b. Length, L.
c. Beam thickness, b
d. Deflection, δ
e. Answers b and c.
f. All of the above.
