Answer
given,
P₁ = 8 bar T₁ = 500 K V₁ = 150 m/s
P₂ = 1 bar T₂ = 320 K V₂ = 10 m/s
writing energy equation
h₁ + (KE)₁ + (PE)₁ + Q m = h₂ + (KE)₂ + (PE)₂ + W
ideal gas property of air
T₁ = 500 K h₁ = 503.02 KJ/kg S₁ = 2.21952 kJ/kgK
T₂ = 320 K h₂ = 320.29 KJ/kg S₂ = 1.7679 kJ/kgK
W = 193.93 KJ/Kg
calculation of energy destruction
=
=
=
=
=43.54 KJ/Kg
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer:
a) The proportional limit is 2.99MPa.
b) The modulus of elasticity is 0.427GPa.
C) The poisson´s ratio is 0.021
Explanation:
a) The proportional limit is the maximum stress for wich a tension bar stops acting as a linear material in a stress-strain curve. So this stress can be obtained as:
b) The modulus of elasticity E is the proportion between the strain and the stress in the linear section of the stress-strain curve.
The strain in for the proportional limit is:
Therefore:
c) The Poisson´s ratio is the negative proportion between the transverse strain and axial strain.
In this case, the transverse strain is:
So the poisson´s ratio is:
ANSWER:
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EXPLANATION:
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Answer:
second-law efficiency = 62.42 %
Explanation:
given data
temperature T1 = 1200°C = 1473 K
temperature T2 = 20°C = 293 K
thermal efficiency η = 50 percent
solution
as we know that thermal efficiency of reversible heat engine between same temp reservoir
so here
efficiency ( reversible ) η1 = 1 - ............1
efficiency ( reversible ) η1 = 1 -
so efficiency ( reversible ) η1 = 0.801
so here second-law efficiency of this power plant is
second-law efficiency =
second-law efficiency =
second-law efficiency = 62.42 %