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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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Nitrogen at an initial state of 300 K, 150 kPa and 0.2 m3 is compressed slowly in an isothermal process to a final pressure of 8

s344n2d4d5 [400]

Answer:

Work = - 4175.23 J

Heat Transfer = -4175.23 J

Explanation:

The work done in an isothermal process, is given by the following formula:

W = RT ln (P1/P2)

where,

W = Work done

R = Universal Gas Constant = 8.314 J/mol.k

T = Constant Temperature = 300 K

P1 = initial pressure = 150 KPa

P2 = final pressure = 800 KPa

using these values, we get:

W = (8.314 J/mol.K)(300 k) ln (150/800)

Here, negative sign shows that work is done on the system.

In isothermal process, from 1st law of thermodynamics:

Heat Transfer = Q = W (Since change in internal energy is zero for isothermal processes)

Here, negative sign shows that heat is transferred from the system to surrounding.

6 0

3 years ago

Read 2 more answers

C&A's potato chip filling process has a lower specification limit of 9.5 oz. and an upper specification limit of 10.5 oz. Th

Law Incorporation [45]

Answer:

Process capability index = (USL - LSL)/6Sigma = 1/6(0.3)= 0.56

Answer (a) 0.56

Explanation:

3 0

4 years ago

Before turning in a test, it would be best to

Blababa [14]

Answer: D mate

Explanation:

4 0

4 years ago

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a horizontal jet of water (at 100c) that is 6 cm in diameter and has a velocity of 20 m/s is deflected by the vane as shown. if

Reptile [31]

The net resultant direct force and angle on the vane is created when the water jet exits the vane at position 2 with 92% of its initial velocity.

<h3>What is mean by velocity?</h3>

The speed at which a body or object is moving determines its direction of motion. A scalar quantity, speed is primarily. As a matter of fact, velocity is a vector quantity.

The rate at which distance changes is what it is. It measures the displacement's rate of change. A body's velocity is defined as its speed in a particular direction.

Velocity is a measure of how quickly a distance changes in relation to time. Having both magnitude and direction, velocity is a vector quantity.

The rate of change in a body's displacement with respect to time is referred to as velocity. In the SI, m/s is its unit.

Given,

External angle of Curved Vane = 158°

mean velocity at 1 = 12 m/s

Volumetric flow rate = $55 \mathrm{~m}^3 / \mathrm{h}=\frac{55}{3600} \mathrm{~m}^3 / \mathrm{s}$.$

mean velocity at $2=12 \times 0.92=11.04 \mathrm{~m} / \mathrm{s}$

i) Force exerted in x - friction A C 1 = Volume

$F_{S_x} &=\rho A C_1\left[C_2 \cos \theta-C_1\right] \\$

$&=1000 \times \frac{55}{3600}\left[\left(11.04 \cos 158^{\circ}\right)-12\right]$

i $\rangle F_{\text {sc }}=\supseteq A c_1\left[C_2 \sin \theta\right] \\$

$&=1000 \times \frac{55}{3600} \times \text { TI. 04 } \sin (1589 \\$

$&F_{\text {syn }}=63.18 \mathrm{~N} \\$

$&\text { Angle } \Rightarrow \frac{F_{s y}}{F_{3 x}}-\tan \theta \\$

$&\tan \theta=\frac{63.18}{339.18}, \theta=160-10-5.3 \\$

$&\theta=\tan ^{-}\left(\frac{-63 \cdot 18}{339728}\right) \\$

$&\theta=-10.540^{\circ} \\$

The complete question is:

A horizontal jet of water strikes a curved vane as shown in Figure C.1. The external angle of the curved vane is 158°.The mean velocity and volumetric flow rate of the water jet at position 1 are 12 m/s and 55 m³/h respectively. Due to friction, the water jet leaves the vane at position 2 with 92 % its original velocity.

(i) Direct force exerted by the water jet on the vane in the x - direction.

(ii) Direct force exerted by the water jet on the vane in the y - direction.

(ii) Net resultant direct force and angle on the vane.

To learn more about velocity, refer to:

brainly.com/question/24681896

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4 0

2 years ago

What is the resultant force on one side of a 25cm diameter circular plate standing at the bottom of 3m of pool water?

Tom [10]

Answer:

F=1.47 KN

Explanation:

Given that

Diameter of plate = 25 cm

Height of pool h = 3 m

We know that force can be given as

F= P x A

P=ρ x g x h

Now by putting the values

P=1000 x 10 x 3

P= 30 KPa

$A=\dfrac{\pi}{4}\times 0.25^2\ m^2$

$A=0.049\ m^2$

F= 30 x 0.049 KN

F=1.47 KN

So the force on the plate will be 1.47 KN.

4 0

3 years ago