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.
Answer:
Efficiency based on Otto cycle.
Effotto = 47.47%
Explanation:
Efficiency based on Otto cycle.
effotto = 1 – (V2 / V1)^γ-1
effotto = 1 – (1 / 5)^1.4 - 1
effotto = 47.47%
Answer:
See the attached picture for answer.
Explanation:
See the attached picture for explanation.
Answer:
BOD concentration at the outflow = 17.83 mg/L
Explanation:
given data
flow rate of Q = 4,000 m³/day
BOD1 concentration of Cin = 25 mg/L
volume of the pond = 20,000 m³
first-order rate constant equal = 0.25/day
to find out
What is the BOD concentration at the outflow of the pond
solution
first we find the detention time that is
detention time t = 
detention time t = 
detention time = 5 days
so
BOD concentration at the outflow of pond is express as
BOD concentration at the outflow = 
here k is first-order rate constant and t is detention time and Cin is BOD1 concentration
so
BOD concentration at the outflow = 
BOD concentration at the outflow = 17.83 mg/L
Answer:
While calculating the stresses in a body since we we assume a constant distribution of stress across a cross section if the body is loaded along the centroid of the cross section , this assumption of uniformity is assumed only on the basis of Saint Venant's Principle.
Saint venant principle states that the non uniformity in the stress at the point of application of load is only significant at small distances below the load and depths greater than the width of the loaded material this non uniformity is negligible and hence a uniform stress distribution is a reasonable and correct assumption while solving the body for stresses thus greatly simplifying the analysis.