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docker41 [41]
2 years ago
13

An isothermal CSTR with a first order irreversible reaction A â> B (and rA[mol/(ft3*min)] = - 0.5 CA) has a constant flow rat

e of 29.0 ft^3/min. The reactor volume is 100 ft3. If the inlet concentration CAi is 7.3 moles/ft^3 at t = 0 and the initial outlet concentration CA is 2.3 moles/ft^3.
Required:
Determine the outlet concentration CA after t = 2.9 min
Engineering
1 answer:
worty [1.4K]2 years ago
7 0
Can you help me by solving this question

You might be interested in
The diffusion coefficients for species A in metal B are given at two temperatures:
Kruka [31]

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

D = D_0e^{\frac{-Q_d}{RT} }

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

R = 8.314\ J/mol*K

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}

and

ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}

You might notice that these equations have the form of  

d=y-ax

You can solve this equation system easily using calculator, and you will eventually get

D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol

After you got those 2 parameters, the rest is easy, you can just plug them all   including the given temperature of 1180°C into the Arrhenius equation

6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}

And you should get D = 2.76*10^-16 m^/s as an answer for c)

5 0
3 years ago
An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be
cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, \zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )

Creep rate at 800⁰C, \zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )

\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\

0.01 = C \exp(\frac{-Q}{1073R} ).........................(1)

Creep rate at 700⁰C

\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )

\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2}  \% per hour =5.5 * 10^{-4}

5.5 * 10^{-4}  = C \exp(\frac{-Q}{1073R} ).................(2)

Divide equation (1) by equation (2)

\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\

Take the natural log of both sides

ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol

3 0
3 years ago
Select a research proposal topic that relates to electrical and electronics engineering and write a proposal report taking into
fiasKO [112]

Expand your technical knowledge, form global networks and balance life & work commitments. Our advanced diplomas remain current with technological and industry developments.

8 0
3 years ago
Compute the theoretical density of ZnS given that the Zn-S distance and bond angle are 0.234 nm and 109.5o, respectively. The at
andriy [413]

Answer: the theoretical density is 4.1109 g/cm³

Explanation:  

first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ

θ + ∅ + 90° = 180°

θ = 90° - ∅

θ = 90° - ( 109.5° / 2 )

θ = 35.25°

next we calculate the value of x from the geometry

given that;  distance angle d = 0.234

x = dsinθ

= 0.234 × sin35.25°)

= 0.135 nm = 0.135 × 10⁻⁷ cm

next we calculate the length of the unit cell

a = 4x

a = 4(0.135)

a = 0.54 nm = 0.54 × 10⁻⁷ cm

next we calculate number of formula units

n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)

n' = 8 × 1/8) + ( 6 × 1/2)

= 1 + 3

= 4

next we calculate the theoretical density using  this equation

P = [n'∑(Ac + AA)] / [Vc.NA]

= [n'∑(Ac + AA)] / [(a)³NA]

where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)

∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)

Na is the Avogadro’s number( 6.023 × 10²³ units/mole)

so we substitute

P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]

= 389.88 / 94.84

= 4.1109 g/cm³

therefore the theoretical density is 4.1109 g/cm³

5 0
2 years ago
What's the relationship between energy and time<br>​
boyakko [2]

Answer:

The relationship between power, energy, and time can be described by the following equation : P = Δ E s y s Δ t. P is the average power output, measured in watts (W) ΔEsys is the net change in energy of the system in joules (J) - also known as work. Δt is the duration - how long the energy use takes - measured in seconds (s).

Explanation:

8 0
3 years ago
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