Answer:
Explanation:
Products of oil in our everyday life:
(1) Petro-Chemical Feedstock: These are by product of Refining of Oil which it is used extensively to make PET bottles, Paints, Polyester Shirts, Pocket combs e.t.c
(2) Asphalt : Used extensively to make Motor Road, highways
(3) Plastics : we use plastics in our everyday life, this is also a product of Refining of crude oil e.g PVC, Telephone casing, Tapes e.t.c
(4) Lubricating Oil/Grease : This is another product from crude oil Fractional Distillation.
(5) Propane/ Cooking Gas: This is also a product from oil which is used in our everyday life for cooking, grilling etc.
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:
We would need background context,
Explanation:
Then I would be happy to help!
The maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing is 28.98 MPa.
<h3>What is power?</h3>
Power is the energy transferred per unit time.
Torque is find out by
P = 2πNT/60
10000 = 2π x 2000 x T / 60
T =47.74 N.m
The gear ratio Ne / Ns =4/1
Ns =2000/4 = 500
Ts =Ps x 60/(2π x 500)
Ts =190.96 N.m
Maximum shear stress τ = 16/π x (T / (d₀⁴ - d₁⁴))
τ max =T/J x D/2
where d₁ = 30mm = 0.03 m
d₀ = 30 +(2x 4) = 38mm =0.038 m
Substitute the values into the equation, we get
τ max = 16 x 190.96 x 0.038 /π x (0.038⁴ - 0.03⁴)
τ max = 28.98 MPa.
Thus, the maximum shear stress in the tube is 28.98 MPa.
Learn more about power.
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Answer: its an Ignition coil
