Answer:
a) 0.684
b) 0.90
Explanation:
Catalyst
EO + W → EG
<u>a) calculate the conversion exiting the first reactor </u>
CAo = 16.1 / 2 mol/dm^3
Given that there are two stream one contains 16.1 mol/dm^3 while the other contains 0.9 wt% catalyst
Vo = 7.24 dm^3/s
Vm = 800 gal = 3028 dm^3
hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins
next determine the value of conversion exiting the reactor ( Xai ) using the relation below
KIm =
------ ( 1 )
make Xai subject of the relation
Xai = KIm / 1 + KIm --- ( 2 )
<em>where : K = 0.311 , Im = 6.97 ( input values into equation 2 )</em>
Xai = 0.684
<u>B) calculate the conversion exiting the second reactor</u>
CA1 = CA0 ( 1 - Xai )
therefore CA1 = 2.5438 mol/dm^3
Vo = 7.24 dm^3/s
To determine the value of the conversion exiting the second reactor ( Xa2 ) we will use the relation below
XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )
<em> where : Xai = 0.684 , Im = 6.97, and K = 0.311 ( input values into equation 3 )</em>
XA2 = 0.90
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Answer: Create lessons learned at the end of the project.
Explanation:
Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.
The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.
Answer:
138.9 °C
Explanation:
The datum of quality is saying to us that liquid water is in equilibrium with steam. Saturated water table gives information about this liquid-vapour equilibrium. In figure attached, it can be seen that at 350 kPa of pressure (or 3.5 bar) equilibrium temperature is 138.9 °C
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.