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forsale [732]
3 years ago
5

Fuel Combustion and CO2 Sequestration [2016 Midterm Problem] Long-term storage of carbon dioxide in underground aquifers or old

oil fields is one method to prevent release of CO2 to the atmosphere (to help mitigate climate change). This is referred to as carbon sequestration. A power plant burns fuel oil containing 39.0 mol% C, 60.3 mol% H, and 0.70 mol% S at a rate of 290 kmol/hr. An air stream flowing at 945 kmol/hr provides oxygen for the combustion process. Conversion of the fuel is 95%. Of the C that burns, 90% goes to CO2. The gases are then separated and the carbon dioxide is sequestered underground. Assume 100% separation of the CO2 from the other gases. (a) Draw a flowchart (reactor and separation unit) and label all streams. Incorporate all of the information given above (b) Calculate the percent excess air fed. (c) Determine the molar flowrate of the O2 in the stack gas leaving the plant (d) What is the mass flowrate of CO2 into the underground aquifer?

Engineering
1 answer:
Sholpan [36]3 years ago
7 0

Answer:

The answers are on the attachment.

Explanation:

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In a tensile test on a steel specimen, true strain is 0.171 at a stress of 263.8 MPa. When true stress is 346.2 MPa, true strain
swat32

Answer:n=0.973

Explanation:

Given

When True strain\left ( \epsilon _T_1\right )=0.171

at \sigma _1=263.8 MPa

When True stress\left ( \sigma _2\right )=346.2 MPa

true strain \left ( \epsilon _T_2\right )=0.226

We know

\sigma =k\epsilon ^n

where \sigma=True stress

\epsilon=true strain

n=strain hardening exponent

k=constant

Substituting value

263.8=k\left ( 0.171\right )^n------1

346.2=k\left ( 0.226\right )^n-----2

Divide 1 & 2 to get

\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n

1.312=\left ( 1.3216\right )^n

Taking Log both side

ln\left ( 1.312\right )=nln\left ( 1.3216\right )

n=0.973

6 0
3 years ago
Refrigerant-134a enters the expansion valve of a refrigeration system at 120 psia as a saturated liquid and leaves at 20 psia. D
Shkiper50 [21]

Solution :

$P_1 = 120 \ psia$

$P_2 = 20 \ psia$

Using the data table for refrigerant-134a at P = 120 psia

$h_1=h_f=40.8365 \ Btu/lbm$

$u_1=u_f=40.5485 \ Btu/lbm$

$T_{sat}=87.745^\circ  F$

∴ $h_2=h_1=40.8365 \ Btu/lbm$

For pressure, P = 20 psia

$h_{2f} = 11.445 \ Btu/lbm$

$h_{2g} = 102.73 \ Btu/lbm$

$u_{2f} = 11.401 \ Btu/lbm$

$u_{2g} = 94.3 \ Btu/lbm$

$T_2=T_{sat}=-2.43^\circ  F$

Change in temperature, $\Delta T = T_2-T_1$

                                         $\Delta T = -2.43-87.745$

                                           $\Delta T=-90.175^\circ  F$

Now we find the quality,

$h_2=h_f+x_2(h_g-h_f)$

$40.8365=11.445+x_2(91.282)$

$x_2=0.32198$

The final energy,

$u_2=u_f+x_2.u_{fg}$

   $=11.401+0.32198(82.898)$

   $=38.09297 \ Btu/lbm$

Change in internal energy  

$\Delta u= u_2-u_1$

   = 38.09297-40.5485

  = -2.4556        

5 0
2 years ago
Today I meant to look for my missing watch, but I could never find the time.
xxMikexx [17]

Answer:

haha i get it

Explanation:

5 0
2 years ago
Read 2 more answers
A box-shaped aquarium has horizontal dimensions 0.5 m by 1 m, and depth 0.5 m, and is filled two-thirds of the way to the surfac
antoniya [11.8K]

Answer:

3270 N/m^2

Explanation:

we can calculate the pressure difference between the bottom and surface of the tank by applying the equation for the net vertical pressure

Py = - Ph ( g ± a )

for a downward movement

Py = - Ph ( g - a )  ------ ( 1 )

From the above data given will  be

p = 1000 kg/m^3, h = 2/3 * 0.5 = 0.33 m , a =2g , g = 9.81

input values into equation 1  becomes

Py =  -Ph ( g - 2g ) = Phg ------ ( 3 )

Py = 1000 * 0.33 * 9.81

    = 3270 N/m^2

6 0
3 years ago
A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the mome
Vikki [24]

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

                           mag(BC) = sqrt ( 48^2 + 36^2) = 60 in

                           BE / BC = 45 / 60 = 0.75

Hence,                E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in

- Find unit vector EF:

                           mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

                           vec(EF) = < -15 , -110 , 30 >

                           unit(EF) = (1/115) * < -15 , -110 , 30 >

- Tension            T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb

- Find unit vector AD:

                           mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

                           vec(AD) = < 48 , -12 , 36 >

                           unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >

                           unit (AD) = <0.7845 , -0.19612 , 0.58835 >

Next:

                           M_AD = unit(AD) . ( E x T_EF)

                           M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right]

                            M_AD = 1835.73 + 116.49528 - 593.0568

                            M_AD = 1359.17 lb-in

3 0
3 years ago
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