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san4es73 [151]
3 years ago
10

Consider a process in which a carbon-based fuel is combusted in the presence of 70% excess oxygen (assume that all of the oxygen

present come from an air feel stream). Based on the information given above, which of the following statements are true? Select all that apply.
a. Exactly 70% of the oxygen presence in the process is not consumed.

b. 70% of the fuel in the process undergoes complete combustion.

c. At least 41% of the oxygen present in the grocers is not consumed

d. There is 70% more air present in the process that would be necessary for the complete combustion of all the fuel

e. There is 70% more oxygen present in the process that would be necessary for the complete combustion of all of the fuel
Engineering
1 answer:
yuradex [85]3 years ago
7 0

Answer:

<u>Option-(A):</u>

  • When a carbon-based fuel is burned(combustion) or we can say burned in the presence of 70% excess oxygen (assuming that all of the oxygen present come from an air feel stream), we can say that there is 70% more oxygen present in the process that would be necessary for the complete combustion of all of the fuel.
  • Oxygen,O₂ is the precursor for the process of combustion as there is the requirement of an optimum level of oxygen for the burning up of the materials.While, there is about enough level of oxygen is present inside the free air or we can say inside the atmosphere for the purpose of combustion or burning of the fuel or consumption of the resources present in the environment.

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Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a
Ann [662]

Answer:

(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K

Explanation:

Solution

Given that:

The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm =0.01 m

Electrical heat rate, q =76 W/m

Wall Temperature, Ts = 370 K

Now,

From the properties table of engine oil we can deduce as follows:

thermal conductivity, k =0.139 W/m .K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg.K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

=76/π *0.01

= 2419.16 W/m²

Now

The oil mean temperature is given as follows:

Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)

Tb =370 - 11/24 * (2419.16 * 0.005/0.139)

Tb = 330.12 K

(b) The center line temperature is given below:

Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)

Tc =304.73 K

(c) The flow velocity is given as follows:

V = m/ρ (πR²)

Now,

The The axial gradient of the mean temperature is given below:

dTb/dx = 2 *qs/ρ *V*cp * R

=2 *qs/ρ*[m/ρ (πR²) *cp * R

=2 *qs/[m/(πR)*cp

dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h =48/11 (k/D)

=48/11 (0.139/0.01)

h =60.65 W/m². K

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3 years ago
The efficiency of a transformer is mainly dependent on: a)- Core losses b)- Copper losses c)- Stray losses d)- Dielectric losses
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The efficiency of a transformer is mainly dependent on a)- Core losses

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3 years ago
What is the difference between science and engineering?
jek_recluse [69]

Answer:

The following table is among some of the discrepancies between some of the different topics.

Explanation:

  • Science seems to be the application of structured organized facts which could be interpreted scientifically because when engineering would be a branch of science which deals with either the profession of acquiring including using technological, computational, economical and severity of adverse effects to design and manufacture equipment and devices that seem to be beneficial to mankind.
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4 0
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Which of the following reduces friction in an engine A)wear B)drag C)motor oil D)defractionation
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It is motor oil, as oil is used to reduce friction

3 0
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Air in a large tank at 300C and 400kPa, flows through a converging diverging nozzle with throat diameter 2cm. It exits smoothly
-Dominant- [34]

Answer:

The answer is "3.74 \ cm\ \ and \ \ 0.186 \frac{kg}{s}"

Explanation:

Given data:  

Initial temperature of tank T_1 = 300^{\circ}\ C= 573 K

Initial pressure of tank P_1= 400 \ kPa

Diameter of throat d* = 2 \ cm

Mach number at exit M = 2.8

In point a:

calculating the throat area:

A*=\frac{\pi}{4} \times d^2

      =\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2

Since, the Mach number at throat is approximately half the Mach number at exit.  

Calculate the Mach number at throat.  

M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4

Calculate the exit area using isentropic flow equation.

\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}}  (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}

Here: \gamma is the specific heat ratio. Substitute the values in above equation.

\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}}  (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm

exit diameter is 3.74 cm

In point b:

Calculate the temperature at throat.

\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K

Calculate the velocity at exit.  

V*=M*\sqrt{ \gamma R T*}

Here: R is the gas constant.  

V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}

Calculate the density of air at inlet

\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\  \frac{kg}{m^3}

Calculate the density of air at throat using isentropic flow equation.  

\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}

Calculate the mass flow rate.  

m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}

5 0
2 years ago
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