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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

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Example 3 :
kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × \frac{0. 06}{0. 12* 1* 0. 9}

friction factor = 0. 52 × \frac{0. 06}{0. 108}

friction factor = 0. 52 × 0. 55

friction factor = 0. 289

b. When V = 3mls

Friction factor = 0. 52 × \frac{0. 06}{0. 12 * 3* 0. 9}

Friction factor = 0. 52 × \frac{0. 06}{0. 324}

Friction factor = 0. 52 × 0. 185

Friction factor = 0.096

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × \frac{1}{2*6. 6743 * 10^-11} × \frac{1^2}{0. 120} × \frac{1}{100}

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = 0. 096 × \frac{1}{1. 334 *10^-10} × \frac{3^2}{0. 120} × \frac{1}{100}

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

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2 years ago
the two rotating systems shown in the figure differ only in that the two identical movable masses are positioned at different di
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Answer: the block at the right lands first

Explanation:

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