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GarryVolchara [31]
1 year ago
11

Are the stack temperature and oxygen reasonable for these operating conditions? if not, what oxygen and stack temperature would

you recommend if not? what is the efficiency with your chosen values?
Physics
1 answer:
Anton [14]1 year ago
3 0

Stack temperatures typically range from 350 to 450 degrees Fahrenheit. A 2.5% efficiency loss occurs for every 100 degrees over that temperature. The majority of buildings schedule annual boiler cleanings at regular intervals, but if you see those figures rise, it's time for a cleaning.

Excess air is required to completely burn the fuel since the air and fuel cannot combine exactly in a burner. Additionally, any leaks in the heater will draw air into the firebox that doesn't pass through the burners since the furnace or boiler firebox operates at a little negative gauge pressure. Fuels that are gaseous, like natural gas, burn more readily than fuels that are liquid or solid. Depending on the fuel type, different surplus air requirements will apply.

Learn more about temperature here-

brainly.com/question/15267055

#SPJ4

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<span>the same amount of work being done over a longer period of time.</span>
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Kim throws a beach ball up in the air. It reaches its maximum height 0.50s later. We can ignore air resistance. What was the bea
notka56 [123]

Answer:

The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>

Explanation:

Given:

Time taken by the ball to reach maximum height is, t=0.50\ s

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, v=0\ m/s

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, a=g=-9.8\ m/s^2

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

v=u+at

Where, 'u' is the initial velocity.

Plug in the given values and solve for 'u'. This gives,

0=u-9.8(0.5)\\u=9.8\times 0.5\\u=4.9\ m/s

Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s

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3 years ago
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2. How have human activities and natural processes influenced the gradual or sudden change in global temperature? (You need a mi
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Answer:

Human activities and natural processes have influenced the change in the global temperature by the following processes

1) Green house gas such as carbon dioxide, methane, ozone, nitrous oxide and fluorinated gases produced by the combustion of fossil fuels the use of  industrial chemicals, the production of coal, and natural gas

2) Deforestation which reduces the natural process of conversion of carbon dioxide to oxygen, thereby, increasing the greenhouse gases in the atmosphere

3) The accumulation of the greenhouse gases in the atmosphere results in the trapping of heat in the atmosphere, causing the atmospheric temperature to rise

4) Changes in the amount of energy produced by the Sun can result in an increase or decrease in the atmospheric temperature

5) Volcanic activity that occurs at a sufficiently large scale can produce sulfur dioxide that blocks the rays of the Sun from reaching the Earth, resulting in a change of atmospheric temperature.

Explanation:

7 0
3 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s&#10;&#10;0 = 2500 - 186544.54s&#10;&#10;Isolate S next.&#10;&#10;185644.54s= 2500&#10;&#10;s =  2500/(185644.54)&#10;&#10;s=0.0134&#10;

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0
3 years ago
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