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iVinArrow [24]
3 years ago
6

The organic acid, ACOOH, reacts reversibly with the alcohol BOH, to form the ester ACOOB according to the stoichiometric equatio

n: ACOOH BOH ACOOB H20 The reaction will be carried out in an ideal batch reactor and the water will be removed rapidly by stripping with an inert gas as the reaction proceeds. Therefore, the reverse reaction can be neglected. The rate equation for the forward reaction is: The value of the rate constant s k = 0.16 M-2h-1 at 373 K. The reaction is carried out in solution in a 2000 L reactor. The initial concentration of ACOOH is ACOOHJo = 2 M and the initial concentration of BOH is [BOHo 3 M. The reactor is operated isothermally at 373 K. How long must the liquid phase reaction run in an ideal, isothermal batch reactor to obtain a fractional conversion of A of 0.9?

Engineering
1 answer:
kompoz [17]3 years ago
4 0

Answer:

The time taken to achieve the 0.9 fractional conversion, t = 28.125 hours

Explanation:

The first image attached is the picture showing the question clearly.

The 2nd attached image is page 1 of the solution and the 3rd attached image is the page 3 of the solution.

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What are the common approximations made in the analysis of heat exchangers?
Oksanka [162]

Answer: making sure that they are up to date

Explanation:

7 0
3 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / 0.02^{0.22

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

l_i = l_0e^{ET

given that l_0 = 480 mm

we substitute

l_i =480mm × e^{0.04481

l_i =  501.998 mm

Now we find the elongation;

Elongation = l_i - l_0

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

6 0
3 years ago
Which statements describe how the Fed responds to high inflation? Check all that apply.
Sveta_85 [38]
Answer:
• it charges banks more interest
• it sells more securities
• it decreases the money supply

In response to high inflation, the Fed charges banks more interests and pays the banks less interests. It also sells not securities.
8 0
3 years ago
"Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is
Katarina [22]

Answer:

Total wight =640.7927 KN

Explanation:

Given that

do= 61 cm

L =120

t= 0.9 cm

That is why inner diameter of the pipe

di= 61 - 2 x 0.9 cm

di=59.2 cm

Water density ,ρ = 1 kg/L = 1000 kg/m³

Weight of the pipe ,wt = 2500 N/m

wt = 2500 x 120 N = 300,000 N

The wight of the water

wt ' = ρ V g

wt'=1000\times \dfrac{\pi}{4}\times (0.61^2-0.0592^2)\times 9.81\times 120 N

wt'=340792.47 N

That is why total wight

Total wight = wt + wt'

Total wight =300,000+ 340792.47 N

Total wight =640,792.47 N

Total wight =640.7927 KN

7 0
4 years ago
A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u
Darina [25.2K]

Answer:

The pressure upstream and downstream of a shock wave are related as

\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}

where,

\gamma= Specific Heat ratio of air

M = Mach number upstream

We know that \gamma _{air}=1.4

Applying values we get

\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)

Similarly the temperature downstream is obtained by the relation

\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}

Applying values we get

\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C

The Mach number downstream is obtained by the relation

M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616

3 0
4 years ago
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