Answer:
1) The exergy of destruction is approximately 456.93 kW
2) The reversible power output is approximately 5456.93 kW
Explanation:
1) The given parameters are;
P₁ = 8 MPa
T₁ = 500°C
From which we have;
s₁ = 6.727 kJ/(kg·K)
h₁ = 3399 kJ/kg
P₂ = 2 MPa
T₂ = 350°C
From which we have;
s₂ = 6.958 kJ/(kg·K)
h₂ = 3138 kJ/kg
P₃ = 2 MPa
T₃ = 500°C
From which we have;
s₃ = 7.434 kJ/(kg·K)
h₃ = 3468 kJ/kg
P₄ = 30 KPa
T₄ = 69.09 C (saturation temperature)
From which we have;
h₄ = 
+ x₄×
= 289.229 + 0.97*2335.32 = 2554.49 kJ/kg
s₄ = 
+ x₄×
= 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)
The exergy of destruction, 
, is given as follows;
= T₀ × 
= T₀ × 
× (s₄ + s₂ - s₁ - s₃)
= T₀ × 
×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)
∴ = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW
The exergy of destruction ≈ 456.93 kW
2) The reversible power output, 
= 
+ ≈ 5000 + 456.93 kW = 5456.93 kW
The reversible power output ≈ 5456.93 kW.
Attached is the solution to the above question.
Answer:
0
Explanation:
output =transfer function H(s) ×input U(s)
here H(s)=
U(s)=
for unit step function
output =H(s)×U(s)
=×
=
taking inverse laplace of output
output=t×
at t=0 putting the value of t=0 in output
output =0
True the use many abstract power
Answer:
Velocity of ball B after impact is 
and ball A is 
Explanation:
= Initial velocity of ball A
= Initial velocity of ball B = 0
= Final velocity of ball A
= Final velocity of ball B
= Coefficient of restitution = 0.8
From the conservation of momentum along the normal we have
Coefficient of restitution is given by
Adding the above two equations we get
From the conservation of momentum along the plane of contact we have
Velocity of ball B after impact is and ball A is .
