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.
STP stands for standard temperature pressure and NTP stands for normal temperature pressure
Answer:
launch- The first stage is ignited at launch and burns through the powered ascent until its propellants are exhausted. The first stage engine is then extinguished, the second stage separates from the first stage, and the second stage engine is ignited. The payload is carried atop the second stage into orbit
powered ascent-The first stage is ignited at launch and burns through the powered ascent until its propellants are exhausted. The first stage engine is then extinguished, the second stage separates from the first stage, and the second stage engine is ignited. The payload is carried atop the second stage into orbit
coasting flight-
When the rocket runs out of fuel, it enters a coasting flight. The vehicle slows down under the action of the weight and drag since there is no longer any thrust present. The rocket eventually reaches some maximum altitude which you can measure using some simple length and angle measurements and trigonometry.
ejection charge-At the end of the delay charge, an ejection charge is ignited which pressurizes the body tube, blows the nose cap off, and deploys the parachute. The rocket then begins a slow descent under parachute to a recovery. The forces at work here are the weight of the vehicle and the drag of the parachute.
slow decent- slow downs (i guess)
recovery-A recovery period is typically characterized by abnormally high levels of growth in real gross domestic product, employment, corporate profits, and other indicators. This is a turning point from contraction to expansion and often results in an increase in consumer confidence
Explanation:
Answer:
a) 42.08 ft/sec
b) 3366.33 ft³/sec
c) 0.235
d) 18.225 ft
e) 3.80 ft
Explanation:
Given:
b = 80ft
y1 = 1 ft
y2 = 10ft
a) Let's take the formula:
1 + 8f² = (20+1)²
= 8f² = 440
f² = 55
f = 7.416
For velocity of the faster moving flow, we have :
V1 = 42.08 ft/sec
b) the flow rate will be calculated as
Q = VA
VA = V1 * b *y1
= 42.08 * 80 * 1
= 3366.66 ft³/sec
c) The Froude number of the sub-critical flow.
V2.A2 = 3366.66
Where A2 = 80ft * 10ft
Solving for V2, we have:
= 4.208 ft/sec
Froude number, F2 =
F2 = 0.235
d) 
= 18.225ft
e) for critical depth, we use :
![y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3](https://tex.z-dn.net/?f=%20y_c%20%3D%20%5B%5Cfrac%7B%28%5Cfrac%7B3366.66%7D%7B80%7D%29%5E2%7D%7B32.2%7D%5D%5E1%5E%2F%5E3%20)
= 3.80 ft
