Answer:
The correct option is;
D. The vessel has closed living spaces onboard
Explanation:
Type B-1 Fire extinguishers
A fire extinguisher is required by the law to be installed in a boat that hs the following specifications
1) There are closed compartment in the boat that can be used for fuel storage
2) There exist double double bottom that is only partially filled with flotation materials
3) There are closed living spaces in the boat
4) The fuel tank is permanently installed in the boat
5) The engine is inboard.
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.
A. Physical I/O sensors
Safety switches, operator inputs, travel limit switches etc
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑
= mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑ = mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Answer:
Determine the added thrust required during water scooping, as a function of aircraft speed, for a reasonable range of speeds.= 132.26∪
Explanation:
check attached files for explanation
