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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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How will you maintain the orderliness of your storage area 2pts?​
fiasKO [112]

Explanation:

Label and group products. One would think that a general cleanup would be the first step, but no, it's not. ...Clean up the area. ...Put up demarcation lines. ...Stack properly. ...Keep the aisles, paths and ramps clear. ...Have all the safety signs in place.

4 0
2 years ago
One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catal
7nadin3 [17]

Answer: maximum length of the nanowire is 510 nm

Explanation:

 

From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K

Thermal conductivity of silicon carbide k = 30 W/m.K

Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m  

lets consider the equation for the value of m

m = ( (hP/kAc)^1/2 )  = ( (4h/kD)^1/2 )  

m =  ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04    

now lets find the value of h/mk    

h/mk = 10⁵ / ( 942809.04 × 30) =  0.00353

lets consider the value θ/θb by using the equation

θ/θb = (T - T∞) / (T - T∞)

θ/θb =  (3000 - 8000) / (2400 - 8000)

= 0.893

the temperature distribution at steady-state is expressed as;

θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)]   / [cosh mL+  (h/mk) sinh mL]

θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)]   / [cosh mL+  (h/mk) sinh mL]

θ/θb = [ 1 ]  / [cosh mL+  (h/mk) sinh mL]

so we substitute

0.893 =  [ 1 ]  / [cosh (942809.04 × L) +  (0.00353) sinh (942809.04 × L)]

L = 510 × 10⁻⁹m

L = 510 nm

therefore maximum length of the nanowire is 510 nm

4 0
3 years ago
Given the following phasors and the information related to the frequency of that phasor, provide the corresponding time-domain r
Evgesh-ka [11]

Answer:

Explanation:

In Engineering and Physics a Phasor That is a portmanteau of phase vector, is a complex number that represents a sinusoidal function whose Amplitude (A), Angular Frequency (ω), and Initial Phase (θ) are Time-invariant.

For the step by step solution to the question you asked, go through the attached documents.

4 0
3 years ago
In a 5V system if you were asked to take one input HIGH and another LOW what would you do (i.e. where would you connect them)?
dangina [55]

Answer:

HIGH from the supply voltage

LOW from ground

Explanation:

The answer depends on the kind of system and the purpose of the signal. But for practical reasons, in a DIGITAL system where 5V is HIGH and 0 V is LOW, 5 volts can be taken from the supply voltage (usually the same as high,  BUT must be verified), and the LOW signal from ground.

If the user has a multimeter, it must be set to continuous voltage on 0 to 20 V range.  Then place the probe in the ground of the circuit (must be a big copper area). Finally  leave one probe in the circuit ground and place the other probe in some test points to identify 5 v.

4 0
3 years ago
Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:
Xelga [282]

Answer:

(a) Current density at P is J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\.

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\

where \textbf{a}_{\rho} and \textbf{a}_{\phi} unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation.  Therefore

J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\

(b) Total current flowing outward can be calculated by using the relation,

I=\int {\textbf{J} \, \textbf{ds}

where integral is calculated through the circular band given in the question. We can write the integral as below,

I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\

due to unit vector multiplication. Then,

I=10\int\(\rho^3z.dz.d\phi

where \rho=3,\ 0. Therefore

I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A

4 0
3 years ago
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