Answer:
The PFR is more efficient in the removal of the reactive compound as it has the higher conversion ratio.
Xₚբᵣ = 0.632
X꜀ₘբᵣ = 0.5
Xₚբᵣ > X꜀ₘբᵣ
Explanation:
From the reaction rate coefficient, it is evident the reaction is a first order reaction
Performance equation for a CMFR for a first order reaction is
kτ = (X)/(1 - X)
k = reaction rate constant = 0.05 /day
τ = Time constant or holding time = V/F₀
V = volume of reactor = 280 m³
F₀ = Flowrate into the reactor = 14 m³/day
X = conversion
k(V/F₀) = (X)/(1 - X)
0.05 × (280/14) = X/(1 - X)
1 = X/(1 - X)
X = 1 - X
2X = 1
X = 1/2 = 0.5
For the PFR
Performance equation for a first order reaction is given by
kτ = In [1/(1 - X)]
The parameters are the same as above,
0.05 × (280/14) = In (1/(1-X)
1 = In (1/(1-X))
e = 1/(1 - X)
2.718 = 1/(1 - X)
1 - X = 1/2.718
1 - X = 0.3679
X = 1 - 0.3679
X = 0.632
The PFR is evidently more efficient in the removal of the reactive compound as it has the higher conversion ratio.
Answer:
awnsers should be added to know to show additional
Answer:
Option D
Explanation:
A post development hydrograph will have lower concentration time and lower infiltration losses and hence sooner peak and higher peak and more runoff or higher area under graph. Therefore, all the answers are correct hence option D
Answer:
The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.
Explanation:
From Fluid Mechanics, we remember that absolute pressure (
), measured in pounds per square inch, is the sum of the atmospheric pressure and the working pressure (gauge pressure). That is:
(1)
Where:
- Atmospheric pressure, measured in pounds per square inch.
- Working pressured of the boiler (gauge pressure), measured in pounds per square inch.
If we suppose that 
and 
, then the absolute pressure is:
The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.
Answer:
q1q1 ⇒ 01
Explanation:
The outputs of a positive edge triggered register will match the inputs after a rising clock edge.
q1q1 ⇒ 01 . . . . matching d1d0 = 01
