Who here likes to play project gotham racing? will mark brainlyest

Close to 16 billion pounds of ethylene glycol (EG) were produced in 2013. It previously ranked as the twenty-sixth most produced

bekas [8.4K]

Answer:

a) 0.684

b) 0.90

Explanation:

Catalyst

EO + W → EG

a) calculate the conversion exiting the first reactor

CAo = 16.1 / 2 mol/dm^3

Given that there are two stream one contains 16.1 mol/dm^3 while the other contains 0.9 wt% catalyst

Vo = 7.24 dm^3/s

Vm = 800 gal = 3028 dm^3

hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins

next determine the value of conversion exiting the reactor ( Xai ) using the relation below

KIm = $\frac{Xai}{1-Xai}$ ------ ( 1 )

make Xai subject of the relation

Xai = KIm / 1 + KIm --- ( 2 )

where : K = 0.311 , Im = 6.97 ( input values into equation 2 )

Xai = 0.684

B) calculate the conversion exiting the second reactor

CA1 = CA0 ( 1 - Xai )

therefore CA1 = 2.5438 mol/dm^3

Vo = 7.24 dm^3/s

To determine the value of the conversion exiting the second reactor ( Xa2 ) we will use the relation below

XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )

where : Xai = 0.684 , Im = 6.97, and K = 0.311 ( input values into equation 3 )

XA2 = 0.90

4 0

2 years ago

a coil consists of 200 turns of copper wire and has a cross-sectional area of 0.8mm square . The mean length per turn is 80 cm a

Amanda [17]

Answer:

The picture below with the answer. Hope it helps, have a great day/night and stay safe! Length of the coil,

8 0

2 years ago

The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod

Vsevolod [243]

Answer:

Explanation:

From the information given:

$E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN$

The total load is distributed across both the rod and tube:

$P = P_1+P_2 --- (1)$

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

$\delta_1=\delta_2$

$\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}$

$\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}$

$P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})$

$P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}$

$P_2 = 6.6212 \ P_1$

Replace $P_2$ into equation (1)

$P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN$

Finally, to determine the normal stress in aluminum rod:

$\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}$

$\sigma_1 = - 23.523 \ MPa}$

Thus, the normal stress = 23.523 MPa in compression.

8 0

3 years ago

3. Briefly explain the conduction mechanism in metals?

Bingel [31]

Answer:

conduction occurs when a substance is heated

Explanation:

3 0

2 years ago

On highways, the far left lane is usually the _____. A. emergency lane B. merge lane C. slowest D. fastest

slamgirl [31]

On highways, the far left lane is usually the fastest moving traffic.

Answer: Option D.

Explanation:

For the most part, the right lane of a freeway is for entering and leaving the traffic stream. It is an arranging path, for use toward the start and end of your interstate run. The center paths are for through traffic, and the left path is for passing. On the off chance that you are not passing somebody, try not to be driving in the left path.

Regular practice and most law on United States expressways is that the left path is saved for passing and quicker moving traffic, and that traffic utilizing the left path must respect traffic wishing to surpass.

7 0

3 years ago

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