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1 answer:
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Answer:
Explanation:
Given that:
<u>At state 1:</u>
Pressure P₁ = 20 bar
Volume V₁ = 0.03
From the tables at saturated vapour;
Temperature T₁ = 212.4⁰ C ; = 0.0996
/ kg
The mass inside the cylinder is m = 0.3 kg, which is constant.
The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg
<u>At state 2:</u>
Temperature T₂ = 200⁰ C
Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 / kg
From temperature T₂ = 200⁰ C
Since , the saturated pressure at state 2 i.e. P₂ = 15.5 bar
Mixture quality
At temperature T₂, the specific internal energy , also
Thus,
<u>At state 3:</u>
Temperature
Specific volume
Thus; ,
SInce , therefore, the phase is in a superheated vapour state.
From the tables of superheated vapour tables; at and T₃ = 200⁰ C
The pressure = 10 bar and v =0.206
The specific internal energy at the pressure of 10 bar = 2622.3 kJ/kg
The changes in the specific internal energy is:
= (2210.686 - 2599.2) kJ/kg
= -388.514 kJ/kg
≅ - 389 kJ/kg
= (2622.3 - 2210.686) kJ/kg
= 411.614 kJ/kg
≅ 410 kJ/kg
We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.
Answer:
Step 1 of 3
Case A:
AISI 1018 CD steel,
Fillet radius at wall=0.1 in,
Diameter of bar
From table deterministic ASTM minimum tensile and yield strengths for some hot rolled and cold drawn steels for 1018 CD steel
Tensile strength
Yield strength
The cross section at A experiences maximum bending moment at wall and constant torsion throughout the length. Due to reasonably high length to diameter ratio transverse shear will be very small compared to bending and torsion.
At the critical stress elements on the top and bottom surfaces transverse shear is zero
Explanation:
See the next steps in the attached image
Answer:
a)
b)
c) Impulse = 0 kg m/s²
d) percent decrease in kinetic energy = 47.85%
Explanation:
Let be the initial velocity of rod A
Let be the initial velocity of rod B
Let be the final velocity of rod A
Let be the final velocity of rod B
Using the principle of conservation of momentum:
............(1)
Coefficient of restitution,
........................(2)
Substitute equation (2) into equation (1)
..............(3)
Solving for in equation (3) above:
....................(4)
From equation (2):
......(5)
Substitute equation (5) into (1)
..........(6)
Solving for in equation (6) above:
.........(7)
b)
Rod A is said to be at rest after the impact,
Substitute these parameters into equation (7)
To calculate the final velocity, , substitute the given parameters into (4):
c) Impulse,
I = 0
d) %
%
%