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3 years ago

Why excess air is required to burn a fuel completely

Engineering

2 answers:

netineya [11]3 years ago

7 0

Answer:

Excess air is supplied to the combustion process to ensure each fuel molecule is completely surrounded by sufficient combustion air. As a burner tune-up improves the rate at which mixing occurs, the amount of excess air required can be reduced.

Explanation:

8_murik_8 [283]3 years ago

5 0

Explanation:

excess air is required to ensure adequate mixing of fuel and air, avoid smoke, minimize sg in Coal burning, and to ensure maximum steam output.

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HELP!

olya-2409 [2.1K]

The thickness is thick

5 0

3 years ago

Tech A says that thrust angle refers to the direction the front wheels are pointing. Tech B says that scrub radius refers to the

kenny6666 [7]

Both of the technicians are correct.

<u>Explanation:</u>

The thrust angle is defined as the angle formed by the imaginary line drawn perpendicular to the rear axis center line. It is used in alignment of the four wheels.

It is also used to determine the direction the front wheels are pointing. While the scrub radius is the intersecting point of the vertical center line of front tires with the imaginary line drawn from the steering knuckles.

Thus both the technicians are saying correct. The thrust angle and scrub radius are used to determine the alignment of wheels, if there is any misalignment in wheels then it needs to make it correct to prevent accidents.

4 0

3 years ago

One kilogram of air, initially at 5 bar, 350 K, and 3 kg of carbon dioxide (CO2), initially at 2 bar, 450 K, are confined to opp

pentagon [3]

Answer:

Check the explanation

Explanation:

Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

$m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}$

1x0.723x $(T_{eq} -350)$ =3x0.780x $(450-T_{eq} )$ ⇒ $T_{eq}$ = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

$V_{a,i} = \frac{mRT_{a,i} }{P_{a,i} }$ = $\frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar}$ = 0.201 $m^{3}$

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

$P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }$

$P_{eq}$ = 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

(0.201+1.275)

= 246.67 KPa = 2.47 bar

6 0

3 years ago

Water vapor at 10 MPa, 600°C enters a turbine operating at steady state with a volumetric flow rate of 0.36 m3/s and exits at 0.

elena-s [515]

Answer:

Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point

8 0

3 years ago

Here, we want to become proficient at changing units so that we can perform calculations as needed. The basic heat transfer equa

netineya [11]

Answer:

9500 kJ; 9000 Btu

Explanation:

Data:

m = 100 lb

T₁ = 25 °C

T₂ = 75 °C

Calculations:

1. Energy in kilojoules

ΔT = 75 °C - 25 °C = 50 °C = 50 K

$m = \text{100 lb} \times \dfrac{\text{1 kg}}{\text{2.205 lb}} \times \dfrac{\text{1000 g}}{\text{1 kg}}= 4.54 \times 10^{4}\text{ g}\\\\\begin{array}{rcl}q & = & mC_{\text{p}}\Delta T\\& = & 4.54 \times 10^{4}\text{ g} \times 4.18 \text{ J$\cdot$K$^{-1}$g$^{-1}$} \times 50 \text{ K}\\ & = & 9.5 \times 10^{6}\text{ J}\\ & = & \textbf{9500 kJ}\\\end{array}$

2. Energy in British thermal units

$\text{Energy} = \text{9500 kJ} \times \dfrac{\text{1 Btu}}{\text{1.055 kJ}} = \text{9000 Btu}$

7 0

3 years ago