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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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An insulated, vertical piston-cylinder device initially contains 10kg of water, 6kg of which is in the vapor phase. The mass of
Alexeev081 [22]

Answer:

a)120C

b)29kg

Explanation:

Hello!

To solve this exercise follow the steps below

1. we will call 1 the initial state, 2 the steam that enters and 3 the final state

2. We find the quality of the initial state, dividing the mass of steam by the total mass.

q1=\frac{6kg}{10kg} =0.6

3 Find the internal energy in the three states using thermodynamic tables

note:Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

u1=IntEnergy(Water;x=0,6(quality);P=200kPa) =1719KJ/kg

u2=IntEnergy(Water;t=350;P=5000kPa) =2808KJ/kg

u3=IntEnergy(Water;x=1;P=200kPa) =2529KJ/kg

4. use the internal energy and pressure to find the temperature in state 3, using thermodynamic tables

T3=Temperature(Water;P=200kPa;u=u3=2529KJ/kg)=120C

5. Use the first law of thermodynamics in the system, it states that the initial energy in a system must be equal to the final

m1u1+m2u2=(m1+m2)u3

where

m1=inital mass=10kg

m2=the mass of the steam that has entered.

solve for m2

(m1)(u1-u3)=(m2)(u3)-(m2)(u2)

m2=m1\frac{u1-u3}{u3-u2} =10\frac{1719-2529}{2529-2808} =29kg

7 0
3 years ago
**Please Help. ASAP**
natima [27]

Answer:

The answer is below

Explanation:

1)

\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at

2)

\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)

3)

x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}

4)

x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}

5)

x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}

6)

E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\  \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}

7)

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\  \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}

8)

ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\

4 0
3 years ago
How much does 1 gallon of water weigh in pound given that the density of water is 1gram/ cm3
MAXImum [283]

Explanation:

There are 8.35 pounds in a gallon of water. Water weighs 1 gram per cubic centimeter or 1 000 kilogram per cubic meter, i.e. density of water is equal to 1 000 kg/m³; at 25°C (77°F or 298.15K) at standard atmospheric pressure.

6 0
3 years ago
Consider the following ways of handling deadlock: (1) banker’s algorithm, (2) detect
Andrew [12]

Answer:

b

Explanation:

7 0
3 years ago
A cylindrical bar of metal having a diameter of 20.2 mm and a length of 209 mm is deformed elastically in tension with a force o
Rus_ich [418]

Answer:

A) ΔL = 0.503 mm

B) Δd = -0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π(d_o)²)/4

σ = 4F/(π(d_o)²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F/(π(d_o)²) = E(ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o)/(E•π(d_o)²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 × 10^(9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)

ΔL = 0.503 × 10^(-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = -(ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [(Δd/d_o)/(ΔL/L_o)]

v = - [(Δd•L_o)/(ΔL•d_o)]

Making Δd the subject, we have;

Δd = -[(v•ΔL•d_o)/L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = -[(0.33 × 0.503 × 20.2)/209]

Δd = -0.016 mm

8 0
3 years ago
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