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>
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Answer:
The answer is 1823.9
Explanation:
Solution
Given that:
m = 5.5 kg/s
= m₁ = m₂ = m₃
The work carried out by the energy balance is given as follows:
m₁h₁ = m₂h₂ +m₃h₃ + w
Now,
By applying the steam table we have that<
p₃ = 50 kPa
T₃ = 100°C
Which is
h₃ = 2682.4 kJ/KJ
s₃ = 7.6953 kJ/kgK
Since it is an isentropic process:
Then,
p₂ = 500 kPa
s₂=s₃ = 7.6953 kJ/kgK
which is
h₂ =3207.21 kJ/KgK
p₁ = 3HP0
s₁ = s₂=s₃ = 7.6953 kJ/kgK
h₁ =3854.85 kJ/kg
Thus,
Since 5 % of this flow diverted to p₂ = 500 kPa
Then
w =m (h₁-0.05 h₂ -0.95 )h₃
5.5(3854.85 - 0.05 * 3207.21 - 0.95 * 2682.4)
5.5( 3854.83 * 3207.21 - 0.95 * 2682.4)
5.5 ( 123363249.32 -0.95 * 2682.4)
w=1823.9