Answer:
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A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt = 60 kpsi and Syc = 75 kpsi. Using
the ductile Coulomb-Mohr theory, determine the factor of safety for the states of plane stress.
1 answer:
Answer:
2.135
Explanation:
Lets make use of these variables
Ox 16.5 kpsi, and Oy --14,5 kpsi
To determine the factor of safety for the states of plane stress. We have to first understand the concept of Coulomb-Mohr theory.
Mohr–Coulomb theory is a mathematical model describing the response of brittle materials such as concrete, or rubble piles, to shear stress as well as normal stress.
Please refer to attachment for the step by step solution.
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Answer:
See the explanation below.
Explanation:
First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.
Find h₁:
Using Saturated Water Table and given pressure p₁ = 100 kPa
h₁ = 417.5 kJ/kg
Find h₂:
In order to find h₂, add the to h₁, where
is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.
But first we need to compute To computer
Pressures:
p₁ = 100 kPa
p₂ = 15,000 kPa
and
Using saturated water pressure table, the volume of water = 1.0432
Dividing 1.0432/1000 gives us:
Volume of water = v₁ = 0.001043 m³/kg
Compute the value of h₂:
h₂ = h₁ + v₁ (p₂ - p₁)
= 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)
= 417.5 + 0.001043 (14900)
= 417.5 + 15.5407
= 433.04 kJ/kg
Find h₃
Using steam table:
At pressure p₃ = 15000 kPa
and Temperature = T₃ = 450°C
Then h₃ = 3159 kJ/kg
The entropy s₃ = 6.14 kJ/ kg K
Find h₄
Since entropy s₃ is equal to s₄ So
s₄ = 6.14 kJ/kgK
To compute h₄
s₄ = +
= 6.14 - 2.45 / 3.89
= 0.9497
The enthalpy h₄:
h₄ =
= 908.4 + 0.9497(1889.8)
= 908.4 + 1794.7430
= 2703 kJ/kg
This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa
Find h₅
Using steam table:
At pressure p₅ = 2000 kPa
and Temperature = T₅ = 450°C
Then h₅ = 3358 kJ/kg
Find h₆:
Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to So
s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK
To compute h₆
s₆ = +
= 7.29 - 1.3028 / 6.0562
= 0.988
The enthalpy h₆:
h₆ =
= 417.51 + 0.988 (2257.5)
= 417.51 + 2230.41
h₆ = 2648 kJ/kg
This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa
Compute power used by pump:
is found by using:
mass flow rate = m = 1.74 kg/s
Volume of water = v₁ = 0.001043 m³/kg
p₁ = 100 kPa
p₂ = 15,000 kPa
= ( m ) ( v₁ ) ( p₂ - p₁ )
= (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)
= (1.74 kg/s) (0.001043 m³/kg) (14900)
= 27.04
= 27 kW
Compute heat added and heat rejected
from boiler using computed enthalpies:
= ( h₃ - h₂ ) + ( h₅ - h₄ )
= ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )
= 2726 + 655
= 3381 kJ/kg
= h₆ - h₁
= 2648 kJ/kg - 417.5 kJ/kg
= 2232 kJ/kg
Compute net work
W =
-
= 3381 kJ/kg - 2232 kJ/kg
= 1150 kJ/kg
Compute power produced by the cycle
mass flow rate = m = 1.74 kg/s
W = 1150 kJ/kg
P = m * W
= 1.74 kg/s * 1150 kJ/kg
= 2001 kW
Compute rate of heat transfer in the reheater
Q = m * ( h₅ - h₄ )
= 1.74 kg/s * 655
= 1140 kW
Compute Thermal efficiency of this system
μ = 1 -
/
= 1 - 2232 kJ/kg / 3381 kJ/kg
= 1 - 0.6601
= 0.34
= 34%