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Vilka [71]
3 years ago
7

6.1-2. Diffusion of CO, in a Binary Gas Mixture. The gas CO2 is diffusing at stcady state through a tube 0.20 m long having a di

ameter of 0.01 m and containing N2 at 298 K. The total pressure is constant at 101.32 kPa. The partial pressure of CO2 at one end is 456 mm Hg and 76 mm Hg at the other end. The diffusivity Dan is 1.67 x 10-5 m2/s at 298 K. Calculate the flux of CO2 in cgs and SI units for equimolar counterdiffusion

Engineering
1 answer:
zzz [600]3 years ago
4 0

Answer:

Heat flux of CO₂ in cgs

                 = 170.86 x 10⁻⁹ mol / cm²s

SI units

       170.86 x 10⁻⁸ kmol/m²s  

Explanation:

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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

<u>a) calculate the conversion exiting the first reactor </u>

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 )

<em>where : K = 0.311 ,  Im = 6.97   ( input values into equation 2 )</em>

Xai = 0.684

<u>B) calculate the conversion exiting the second reactor</u>

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 )

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

XA2 = 0.90

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4 0
3 years ago
REVIEW QUESTIONS
babunello [35]
Snap rings, and bearings can be used to keep a gear on a shaft, hope this helps!!
5 0
3 years ago
A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k
goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = \frac{ln\frac{r2}{r1}}{2 \pi *k * L}   .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

so

heat loss is change in temperature divide thermal resistance

Q = \frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}

Q = \frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

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One of the disadvantages of the test of the hypothesis is that the final decision cannot be said to be completely correct eviden
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Answer:

A. True

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Input signal to a controller is​
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were the cord plugs in

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