Answer:
Explanation:
The pressures given are relative
p1 = 2000 psi
P1 = 2014 psi = 13.9 MPa
p2 = 4 psi
P2 = 18.6 psi = 128 kPa
Values are taken from the steam pressure-enthalpy diagram
h2 = 2500 kJ/kg
If the output of the turbine has a quality of 85%:
t2 = 106 C
I consider the expansion in the turbine to adiabatic and reversible, therefore, isentropic
s1 = s2 = 6.4 kJ/(kg K)
h1 = 3500 kJ/kg
t2 = 550 C
The work in the turbine is of
w = h1 - h2 = 3500 - 2500 = 1000 kJ/kg
The thermal efficiency of the cycle depends on the input heat.
η = w/q1
q1 is not a given, so it cannot be calculated.
Answer:
the maximum length of specimen before deformation is found to be 235.6 mm
Explanation:
First, we need to find the stress on the cylinder.
Stress = σ = P/A
where,
P = Load = 2000 N
A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4
A = 1.0752 x 10^-5 m²
σ = 2000 N/1.0752 x 10^-5 m²
σ = 186 MPa
Now, we find the strain (∈):
Elastic Modulus = Stress / Strain
E = σ / ∈
∈ = σ / E
∈ = 186 x 10^6 Pa/107 x 10^9 Pa
∈ = 1.74 x 10^-3 mm/mm
Now, we find the original length.
∈ = Elongation/Original Length
Original Length = Elongation/∈
Original Length = 0.41 mm/1.74 x 10^-3
<u>Original Length = 235.6 mm</u>
Explanation:
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