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Sidana [21]
3 years ago
15

The measured engine fuel flow rate is 0.4g/s, airflow rate is 5.6 g/s and exhaust gas composition (measured dry) is CO2 =13.0% ,

CO = 2.8% with O2 essentially zero.Unburned hydrocarbon emissions can be neglected. Compare the equivalence ratio calculated from the fuel and airflow with the equivalence ratio calculated from the exhaust gas composition. The fuel is Gasoline with an H/C ratio of 1.87. Assume an H2 concentration equal to one third the CO concentration.
Engineering
1 answer:
Verizon [17]3 years ago
7 0

Answer:

b1 = 0.8228, b2 = 1.078, b3 = 0.1772, b5 = 5.472, Φ2 = 1.012 rad, Φ1 = 1.043 rad,

EF= 14 KJ, EFs = 14.6 K,

Explanation:

We are given that the measured engine fuel flow rate is 0.4g/s = 0.004 kh, airflow rate is 5.6 g/s = 0.0056 kg . Therefore the ratio of airflow rate to the engine fuel flow rate = EF = 5.6/0.4 = 14.

We also know that the "fuel is Gasoline with an H/C ratio of 1.87" that is x = 1.87.

The equation for the reaction is given below;

CHx + 1/Φ × (1 + x/ 4) (O2 + 3. 773 N2) ----------------> b1 CO2 + b2 H2O + b3 CO + b4 H2 + b5 N2.

So, if n1 + n2 = 1 ; x = 2 × b2 + 2 × b4.

Note that our EFs = 34.56 [ 4 + x / 12.01 + 1.08 + x ].

Note that Φ1 = EFs/EF = 1.043 radian.

(Where EFs= 14.6).

Then, 1/ Φ2 × [ 1 + x/4 ] = 2 × b1 + b2 + b3. -------------------------------(1).

1/ Φ2 × [1 + x/4] ×2 × 3.773 = 2 × b5. ----------------------(1).

If we solve the above equations we will have b1 = 0.8228, b2 = 1.078, b3 = 0.1772, b5 = 5.472, Φ2 = 1.012 radian.

Pressure of the dry composition = b1 + b3 + b4+ b5 = 6.329 kPa.

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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

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How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor
zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

4 0
3 years ago
According to the
zysi [14]

Answer:

The part of the system that is considered the resistance force is;

B

Explanation:

The simple machine is a system of pulley  that has two pulleys

The effort, which is the input force at A gives the value of the tension at C and  D which are used to lift the load B

Therefore, we have;

A = C = D

B = C + D = C + C = 2·C

∴ C = B/2

We have;

C = B/2 = A

Therefore, with the pulley only a force, A equivalent to half the weight, B, of the load is required to lift the load, B

The resistance force is the constant force in the system that that requires an input force to overcome in order for work to be done

It is the force acting to oppose the sum of the other forces system, such as a force acting in opposition to an input force

Therefore, the resistance force is the load force, B, for which the input force, A, is required in order for the load to be lifted.

3 0
3 years ago
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