Why do engineers need to be particularly aware of the impact of hubris?

The gas stream from a sulfur burner is composed of 15-mol% SO2, 20-mol% O2, and 65-mol% N2. This gas stream at 1 bar and 480?C e

galben [10]

Answer: the equilibrium conversion is 0.15 or 15mole %

K=1.33

Explanation:

Given the mole percent of the gases to be as follows

SO2(g) =15% = 0.15mol

O2(g) =20% = 0.20mol

N2(g) = 65% = 0.65mol

The equilibrium conversion reaction is given by

2SO2(g) + O2(g) <------> 2SO3(g)

2 1. 2

According to the reaction 2 moles of SO2 is converted to get 2 mole of SO3

Therefore 0.15 mole will produce 0.15 mole of SO3

So at equilibrium:

2SO2(g) + O2(g) ------> 2SO3(g)

0.15. 0.075. 0.15

Now the equilibrium constant K is given by:

K = [SO3]²/[SO2]²[O2]

Since the container is constant for all the reactants we can neglect the container capacity and use the mole as percent concentration.

K = [0.15]²/[0.15]²[0.075]

K = 1/0.075

K= 1.33

6 0

3 years ago

You find a publication from a research laboratory that identifies a new catalyst for ammonia synthesis. The article contains the

Sidana [21]

Answer:

Answer: RStoic

Explanation:

The reactor i will select to model this reactor in Aspenplus is RStoic

with the following reason;

Reason: Rstoic is used in Aspen software when the stoichiometry of reaction is known but kinetics isn't available. It can have one or more feed streams attached to it.

Reasons why other reactors are not preferred:

RYield: RYield performs the calculations based on the yield and we are not provided with the yield.

REquil: REquil is used when we are provided with stoichiometry of the reaction and information about equilibrium constant so that to perform chemical and phase equilibrium reactions & we are not provided with the equilibrium information.

RGibbs: Since reactor and products are in the gas phase, so it is not required to use it as it's used to minimise Gibbs free energy.

RCSTR: Reaction kinetics needed but we are not provided with it.

RPlug: Reaction kinetics needed but we are not provided with it.

RBatch: Reaction kinetics needed but we are not provided with it.

6 0

4 years ago

Consider water at 27°C in parallel flow over an isothermal, 1‐m‐long flat plate with a velocity of 2 m/s. a) Plot the variation

yulyashka [42]

Answer:

i) h-bar-L = 4110 W/m^2K

ii ) h-bar-L = 4490 W/m^2K

iii) h-bar-L = 5072 W/m^2K

Explanation:

Given:-

- The temperature of water, T = 27°C

- The velocity of fluid flow, U∞ = 2m/s

- The length of the flat place, L = 1 m

Solution:-

- Using table A-6, to determine the properties of water:

Density ρ = 997 kg/m^3

Dynamic viscosity ν = 0.858*10^-6 m^2/s

Pr = 583 , k = 0.613 W/m.K

- The reynold's number for full length (L = 1m):

Re = U∞*L / ν

Re = (2)*(1) / (0.858*10^-6)

Re = 2.33*10^6

- The boundary layer is mixed with Rex,c = 5*10^5. Evaluate the critical length (xc):

xc = L* ( Rex,c / Re )

= (5*10^5 / 2.33*10^6 )

= 0.215 m

a) Using "IHT correlation tool, External Flow, Local coefficients for laminar or Turbulent flows", h (x) was evaluated and plotted with critical Reynolds number for all 3 cases: (i) 5 × 10^5, (ii) 3 × 10^5, and (iii) 0 (the flow is fully turbulent). - (See attachment 1)

b) Using "IHT correlation tool, External Flow, Average coefficients for laminar or Mixed flows", h - bar- (x) was evaluated and plotted with critical Reynolds number for all 3 cases: (i) 5 × 10^5, (ii) 3 × 10^5, and (iii) 0 (the flow is fully turbulent). - (See attachment 2)

c) The average convection coefficient for the plate can be determined from the graphs presented in (Attachments 1 and 2). Since,

h-bar-L = h-bar-x(L)

The values for the flow conditions are:

( i) h-bar-L = 4110, ii ) h-bar-L = 4490 , iii) h-bar-L = 5072 ) W/m^2K

6 0

4 years ago

Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,

Lyrx [107]

Answer:

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$

Explanation:

$v_0$ = Initial velocity of ball A

$v_A=v_0\cos45^{\circ}$

$v_B$ = Initial velocity of ball B = 0

$(v_A)_n'$ = Final velocity of ball A

$v_B'$ = Final velocity of ball B

$e$ = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

$mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

Coefficient of restitution is given by

$e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

$(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

$v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

Adding the above two equations we get

$2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0$

$\boldsymbol{\therefore v_B'=0.6364v_0}$

$(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0$

From the conservation of momentum along the plane of contact we have

$(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}$

$v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}$

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$ .

5 0

3 years ago

If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam

Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

$\zeta =\dfrac{C}{C_c}$

Where C is the damping coefficient and Cc is the critical damping coefficient.

$C_c=2\sqrt{mK}$

So from above we can say that

$\zeta =\dfrac{C}{2\sqrt{mK}}$

$\zeta \alpha \dfrac{1}{\sqrt K}$

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0

3 years ago