1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Yuliya22 [10]
3 years ago
6

Benzene gas (C6H6) at 25° C and 1 atm, enters a combustion chamber operating at steady state and burns with 95% theoretical air

entering at 25° C and 1 atm. The combustion products exit at 1000 K and include only CO2 , CO , H2O , and N2.
Determine:
The mass flow rate of the fuel, in kg/s, to provide heat transfer at a rate of 1000 kW
Engineering
2 answers:
Natalka [10]3 years ago
8 0

Answer:

The mass flow rate of the fuel, in kg/s = 0.037 kgs^{-1}

Explanation:

The actual balanced equation for the combustion reaction can be written as :

C_6H_6 +(7.5)(O_2+3.76N_2) ------> 6CO_2 +3H_2O+28.2N_2

At 95% theoretical air entering at 25° C and 1 atm, the equation of the reaction can be represented as :

C_6H_6 + 7.125(O_2+3.76N_2) ------> 5.25 CO_2+ 0.75CO+3H_2O+26.79N_2

The Energy rate formula can be use to determine the mass flow rate of the fuel and which is given as:

\frac{Q_{Cv}}{n_{fuel}}} = 5.25( \bar h_f^0 + \delta \bar h)_{co_2} + 0.75 (\bar h^0_f +\delta \bar h )_{co} +3( \bar h_f^0 + \delta \bar h)_{H_2O} + 26.79( \bar h_f^0 + \delta \bar h)_{N_2}- (\bar h_f^0)_{fuel}

from ideal gas table at respective amount for each compound; we have:

\frac{Q_{Cv}}{n_{fuel}}} = 5.25 (-393520 + 42769-9364) + 0.75 (-110530+30355-8664 ) +3(-241820+35882-9904) + 26.79( 30129-8669)-82930

\frac{Q_{Cv}}{n_{fuel}}} = -2112780 kJ/mol.k

n_{fuel} = \frac{Q_{cv}}{-2112780kJ/kmol}

Given that;

The combustion products exit at  1000 K = Q_{cv}

n_{fuel}= \frac{-1000}{-2112780}

n_{fuel}= 4.73*10^{-3} kmol/s

From the ideal gas tables M = 78.11 kg/kmol

∴

n_{fuel}= 4.73*10^{-3} kmol/s * 78.11 kg/kmol

n_{fuel} = 0.037 kgs^{-1}

∴ The mass flow rate of the fuel, in kg/s = 0.037 kgs^{-1}

tiny-mole [99]3 years ago
5 0

Answer:

Explanation:

Benzene gas is burned with 95 percentage theoretical air during a steady â flow combustion process. The mole fraction of the CO in the products and the heat transfer from the combustion chamber are to be determined

Assumption

steady operating conditions exit .

Air and combustion gases are gases

Kinetic and potential energies are negligible

The fuel is burned with insufficient amount of air and thus the products will contain some CO as well as CO2, H2O and H2

Combustion equation is

C6H6 + ath (O2 + 3.76N2) â 6CO2 +3H2O +3.76ath N2

Where ath is the stoichiometric coefficient and is determined from the O2 balance

ath = 6+1.5 = 7.5

then the actual combustion equation can be written as

C6H6 + 0.95 * 7.5(O2 + 3.76N2) â xCO2 + (6-x) CO + 3H2O + 26.79N2

O2 balance: 0.95 * 7.5 = x +(6-x)/2 +1.5 â x=5.25

Thus C6H6 + 7.125(O2 +3.76 N2) â 5.25 CO2 + 0.75CO + 3H2O + 26.79N2

The mole fraction of CO in the products is

yCO = N CO/ N total = 0.75/5.25 + 0.75+3+26.79

=0.021 or 2.1% mole fraction of the CO in the products

You might be interested in
A double-pane insulated window consists of two 1 cm thick pieces of glass separated by a 1.8 cm layer of air. The window measure
Elanso [62]

Answer:

(b). T = 22.55 ⁰C

(c). q = 557.8 W

Explanation:

we take follow a step by step process to solving this problem.

from the question, we have that

The two glass pieces is separated by a 1.8 cm distance layer of air.

the thickness of glass piece is 1 cm

width = 4 m

the height = 3 m

(a). the sketch of the thermal circuit is uploaded in the picture below.

(b).  the thermal resistance due to the conduction in the first glass plane is given thus;

R₁ = Lg / Kg A ................(1)

given that Kg rep. the thermal conductivity of the glass plane

A = conduction surface area

Lg = Thickness of glass plane4

taking the thermal conductivity of glass plane as Kg = 0.78 w/mk

inputting values into equation (1) we have,

R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]

R₁ = 1.068 ˣ 10 ⁻³ k/w

Being that we have same thermal resistance in the first and second plane,

therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w

⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus

R₂ = La/KaA .....................(2)

given Ka = thermal conductivity of air

A = surface area

La = thickness of air

substituting values into the equation we have

R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]

R₂ = 5.73 ˣ 10⁻² k/w

Given the thermal resistance on the outer surface due to convection, we have

R₄ = 1/hA

inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w

R₄ = 6.94 ˣ 10⁻³k/w

Finally the sum total of thermal resistance = R₁ + R₂ + R₃ + R₄

R-total = 0.0663 kw

From this we can calculate the rate of heat loss

using  q = Ti - To / R-total ..............(3)

given Ti and To is the inside and outside temperature i.e. 27⁰C and -10⁰C

from equation (3),

q = 27- (-10) / 0.0063 = 557.8 W

q = 557.8 W  

⇒ Applying the heat transfer formula for inside surface glass temperature gives;

q = Ti - T₂ / R₃ + R₄

T₂ = Ti - q (R₃ + R₄)

T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C

T₂ = 22.55°C

cheers i hope this helps

8 0
3 years ago
Who can help me with electric systems for cars?
hoa [83]

Answer: i can see if i can what is the problem

Explanation:

7 0
2 years ago
The boy in the wagon begins throwing bricks out of the wagon to simulate rocket propulsion. The wagon begins at rest, and the bo
Digiron [165]

Q:What velocity does the boy attain if he throws the bricks one at a time?

Answer:Linear velocity since it moves back and firth and does not rotate like angular velocity.

5 0
2 years ago
Describe carbonation as it applies to the four-stroke engine.
valentinak56 [21]
Carbonation is more of a healer to the engine
5 0
2 years ago
Read 2 more answers
A 2-m3insulated rigid tank contains 3.2 kg of carbon dioxide at 120 kPa.Paddle-wheel work is done on the system until the pressu
AleksandrR [38]

Answer:

The change in entropy is found to be 0.85244 KJ/k

Explanation:

In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.

P1/T1 = P2/T2

T2/T1 = P2/P1

T2/T1 = 180 KPa/120KPa

T2/T1 = 1.5

Now, the change in entropy is given as:

ΔS = m(s2 - s1)

where,

s2 = Cv ln(T2/T1)

s1 = R ln(V2/V1)

ΔS = change in entropy

m = mass of CO2 = 3.2 kg

Therefore,

ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]

Since, V1 = V2, therefore,

ΔS = mCv ln(T2/T1)

Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K

Therefore,

ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)

<u>ΔS = 0.85244 KJ/k</u>

3 0
2 years ago
Other questions:
  • Is someone an engineer that can help me?plz
    11·1 answer
  • When determining risk, it is necessary to estimate all routes of exposure in order to determine a total dose (or CDI). Recognizi
    8·1 answer
  • A cubic transmission casing whose side length is 25cm receives an input from the engine at a rate of 350 hp. If the vehicle's ve
    5·1 answer
  • Consider the equation y = 10^(4x). Which of the following statements is true?
    9·1 answer
  • How many times has the ITU-R revised the CCIR 601 international standard? A. four B. five C. six D. seven
    8·1 answer
  • The boost converter of Fig. 6-8 has parameter Vs 20 V, D 0.6, R 12.5 , L 10 H, C 40 F, and the switching frequency is 200 kHz. (
    12·1 answer
  • Crank OA rotates with uniform angular velocity 0  4 rad/s along counterclockwise. Take OA= r= 0.5
    11·1 answer
  • How do you explain the application of regulations in locations containing baths, showers and electric floor heating, including t
    10·1 answer
  • How buy airpods in my phone​
    7·2 answers
  • A properly fitted wearable pfd should have which characteristics
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!