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3 years ago

Example 12: Write an algorithm and draw a flowchart to calculate

Engineering

1 answer:

Ilya [14]3 years ago

7 0

Answer:

An algotherum is a finite set of sequential instructions to accomplish a task where instructions are written in a simple English language

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The reverse water-gas shift (RWGS) reaction is an equimolar reaction between CO2 and H2 to form CO and H2O. Assume CO2 associati

klasskru [66]

Answer:

a) Check explanation for this

b)Rate law is $Rate = \frac{k_{1}k_{4} }{k_{3}+ 2k_{4} } [H_{2} ]$

c) The rate does not depend on the concentration of CO₂

Explanation:

a) Elementary steps for the RWGS reaction:

Dissociative adsorption of the H₂ Molecule

$H_{2} $\xrightarrow{\text{k1}}$H + H$ (Fast process)

Reversible Reaction between CO₂ and H

$CO_{2} + H\mathrel{\mathop{\rightleftarrows}^{\mathrm{k2}}_{\mathrm{k3}}}COOH$ (Fast Process)

Slow dissociation of COOH into gaseous CO and absorbed OH

$COOH $\xrightarrow{\text{k1}}$ CO + OH$ (Slow process)

Fast hydrogenation of the OH to form H₂O

$OH + H $\xrightarrow{\text{k5}}$H_{2} O$ (Fast process)

b) Derivation of the rate law

We need to determine the rate law for H, OH and COOH because these are the intermediates for this reaction.

The steady state approximation is applied to a consecutive reaction with a slow first step and a fast second step (k1≪k2). If the first step is very slow in comparison to the second step, there is no accumulation of intermediate product.

Rate of consumption = Rate of production

For COOH:

Using steady state approximation

$\frac{d[COOH]}{dt} = 0$

$k_{2} [CO_{2} ][H] = k_{3} [COOH] k_{4} [COOH]\\$

$[COOH] = \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4} } \\$

For H:

$\frac{d[H]}{dt} = 0$

$k_{1}[H_{2}] = k_{2}[CO_{2} [H]+k_{5} [ OH][H]$

$[H]= \frac{k_{1}[H_{2}] }{k_{5}[OH] +k_{2}[CO_{2}]}\\$

For OH:

$\frac{d[OH]}{dt} = 0$

$k_{4} [COOH] = k_{5} [OH][H]\\\k[OH] = \frac{k_{4} [COOH]}{k_{5} H}\\$

The rate of the overall reaction is determined by the slowest step of the reaction. The slowest process is the dissociation of COOH

Therefore the overall rate of reaction is:

$Rate = k_{4} [COOH]\\$

$Rate = k_{4} \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4} }\\Rate = k_{4} \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}] }{k_{5}[OH] +k_{2}[CO_{2}]} }{k_{3}k_{4}}\\Rate = k_{4} \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}] }{k_{5}\frac{k_{4}COOH }{k_{5}H } +k_{2}[CO_{2}]} }{k_{3}k_{4}}$

Simplifying the equation above, the rate law becomes

$Rate = \frac{k_{1}k_{4} }{k_{3}+ 2k_{4} } [H_{2} ]$

c) It is obvious from the rate law written above that the rate of the RWBG reaction does not depend on the concentration of CO₂

7 0

3 years ago

A company specification calls for a steel component to have a minimum tensile strength of 1240 MPa. Tension tests are conducted

Pavlova-9 [17]

Answer:

a) The minimum acceptable value is 387.5 HV using Vickers hardness test.

b) The minimum acceptable value is 39.4 HRC using Rockwell C hardness test.

Explanation:

To get the tensile strength of a material from its hardness, we multiply it by an empirical constant that depends on things like yield strength, work-hardening, Poisson's ratio and geometrical factors. The incidence of cold-work varies this relationship.

According to DIN 50150 (a conversion table for hardness), the constant for Vickers hardness is ≈ 3.2 (an empirical approximate):

$\mbox{Tensile strength}=HV*3.2\\\\HV = \frac{\mbox{Tensile strength}}{3.2} =\frac{1240}{3.2}=387.5$

According to DIN 50150, the constant for Rockwell C hardness test is ≈31.5 around this values of tensile strength:

$\mbox{Tensile strength}=HRC*31.5\\\\HRC = \frac{\mbox{Tensile strength}}{31.5} =\frac{1240}{31.5}=39.4$

8 0

3 years ago

Which of the following is a purpose of sports biomechanics?

Firdavs [7]

I say A because, During a Biomechanical assessment we are able to track even further up the body, to lower back pain caused by the feet. Biomechanics is the tool that understands human movement and has also become hugely popular amongst sports athletes as it is used to enhance athletic performance and prevent injury, too.

7 0

3 years ago

A flow of 12 m/s passes through a 6 m wide, 2 m deep rectangular channel with a bed slope of 0. 001. If the mean velocity of flo

prohojiy [21]

Answer:

manning's coefficient is 0.0357

Explanation:

Given:

Velocity of flow, v = 12 m/s

Width of the channel, b = 6 m

Depth of the channel, d = 2 m

bed slope, s = 0.001