Answer:
Option (d) 2 min/veh
Explanation:
Data provided in the question:
Average time required = 60 seconds
Therefore,
The maximum capacity that can be accommodated on the system, μ = 60 veh/hr
Average Arrival rate, λ = 30 vehicles per hour
Now,
The average time spent by the vehicle is given as
⇒ 
thus,
on substituting the respective values, we get
Average time spent by the vehicle = 
or
Average time spent by the vehicle = 
or
Average time spent by the vehicle = 
or
Average time spent by the vehicle =
hr/veh
or
Average time spent by the vehicle =
min/veh
[ 1 hour = 60 minutes]
thus,
Average time spent by the vehicle = 2 min/veh
Hence,
Option (d) 2 min/veh
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.
Examples of quality assurance activities include process checklists, process standards, process documentation and project audit. Examples of quality control activities include inspection, deliverable peer reviews and the software testing process. You may like to read more about the quality assurance vs quality control.
Answer:
Relative density = 0.545
Degree of saturation = 24.77%
Explanation:
Data provided in the question:
Water content, w = 5%
Bulk unit weight = 18.0 kN/m³
Void ratio in the densest state,
= 0.51
Void ratio in the loosest state,
= 0.87
Now,
Dry density, 

= 17.14 kN/m³
Also,

here, G = Specific gravity = 2.7 for sand

or
e = 0.545
Relative density = 
= 
= 0.902
Also,
Se = wG
here,
S is the degree of saturation
therefore,
S(0.545) = (0.05)()2.7
or
S = 0.2477
or
S = 0.2477 × 100% = 24.77%