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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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A fuel cell vehicle draws 50 kW of power at 70 mph and is 40% efficient at rated power. You are asked to size the fuel cell syst
svetoff [14.1K]

Answer:

  a) fuel cell weight and volume: 50 kg, 33.3 L

  b) fuel tank weight and volume: 321 kg, 643 L

Explanation:

<u>Fuel Cell</u>

Delivery of 50 kW from a source with a power density of 1.5 kW/L requires a source that has a volume of ...

  (50 kW)/(1.5 kW/L) = 33.3 L

The weight of the power source is 1 kW/kg, so will be ...

  (50 kW)/(1 kW/kg) = 50 kg

__

<u>Fuel Tank</u>

400 miles at 70 mph will take (400/70) h ≈ 5.71429 h. In that time, the energy used by the vehicle power plant is ...

  (50 kW)(5.71429 h)(3600 s/h) = 1028.57 MJ

Since the power plant is 40% efficient, it must be supplied with 2.5 times that amount of energy, or 2571.43 MJ.

The tank volume and mass will then be ...

  volume = 2571.43 MJ/(4 MJ/L) = 642.9 L

  mass = 2571.43 MJ/(8 MJ/L) = 321.4 kg

_____

<em>Comment on fuel tank volume</em>

It appears the fuel tank would need to be equivalent in size to a sphere about 1.1 m in diameter.

4 0
3 years ago
When it comes to making a good impression in a work setting, it does not apply to an initial contact, since both people are meet
Art [367]

Answer:

you have only seconds in which a person will accept or reject an employee or firm

Explanation:

First impression matters that's why when looking for employment with an organisation, lack of a tie for men may lead to automatic rejection. You have to be smart both intellectually and physically. Therefore, it means that you have only seconds in which a person will accept or reject an employee or firm.

4 0
3 years ago
A piston-cylinder apparatus has a piston of mass 2kg and diameterof
iragen [17]

Answer:

M =2.33 kg

Explanation:

given data:

mass of piston - 2kg

diameter of piston is 10 cm

height of water 30 cm

atmospheric pressure 101 kPa

water temperature = 50°C

Density of water at 50 degree celcius is 988kg/m^3

volume of cylinder is  V = A \times h

                                       = \pi r^2 \times h

                                       = \pi 0.05^2\times 0.3

mass of available in the given container is

M = V\times d

  = volume \times density

= \pi 0.05^2\times 0.3 \times 988

M =2.33 kg

6 0
4 years ago
Water is stored in a tank which has vent open to the atmosphere. The water level is 1.0 m below the top the tank and the water i
sp2606 [1]

Answer:

6.99 x 10⁻³ m³ / s

Explanation:

Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.

h = difference in the height of water column at two ends of delivery pipe

6 - 1 =  5 m

Velocity of flow of water

v = √2gh

= √ (2 x 9.8 x 5)

=  9.9 m /s

Volume of water flowing per unit time

velocity x cross sectional area

= 9.9 x 3.14 x .015²

= 6.99 x 10⁻³ m³ / s

7 0
3 years ago
Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0
3 years ago
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