A hydraulic power unit is sometimes referred to as a ?

For the following conditions determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste

andrew11 [14]

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.

3 0

3 years ago

Biblical studies of john

nevsk [136]

Answer:

<h2>the answer of sols brother is correct</h2><h3>hope it helps you have a good day</h3><h2 />

5 0

3 years ago

g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress

zvonat [6]

Answer:

a) $\mathbf{\sigma _ 1 = 4800 psi}$

$\mathbf{ \sigma _2 = 0}$

b) $\mathbf{\sigma _ 1 = 6000 psi}$

$\mathbf{ \sigma _2 = 3000 psi}$

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = 0$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 4800 psi}$

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = \frac{Pd}{4t}$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 6000 psi}$

$\sigma _2 = \frac{Pd}{4t} \\ \\ \sigma _2 = \frac{250(12)}{4(0.25)}$

$\mathbf{ \sigma _2 = 3000 psi}$

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0

3 years ago

214Bi83 --> 214Po84 + eBismuth-214 undergoes first-order radioactive decay to polonium-214 by the release of a beta particle,

Zolol [24]

Answer:

(C) ln [Bi]

Explanation:

Radioactive materials will usually decay based on their specific half lives. In radioactivity, the plot of the natural logarithm of the original radioactive material against time will give a straight-line curve. This is mostly used to estimate the decay constant that is equivalent to the negative of the slope. Thus, the answer is option C.

3 0

3 years ago

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A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane-strain fracture tough

jeyben [28]

Complete question:

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

Answer:

Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection

so that critical flow is subject to detection

Explanation:

We are given:

Plane strain fracture toughness K $= 98.9 MPa \sqrt{m}$

Yield strength Y = 860 MPa

Flaw detection apparatus = 3.0mm (12in)

y = 1.0

Let's use the expression:

$oc = \frac{K}{Y \sqrt{pi * a}}$

We already know

K= design

a = length of surface creak

Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.

Therefore

$a= \frac{1}{pi} * [\frac{k}{y*a}]^2$

Substituting figures in the expression above, we have:

$= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2$

= 0.0168 m

= 17mm

Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection

3 0

3 years ago

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