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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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What are the three reasons why nebula contribute more to Stellar formation than any other region in the universe?
monitta
I think it is a, b, and e. Hope it helps! :)
7 0
3 years ago
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A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.48° apart. What is the angular
larisa [96]

Answer:

0.38°

Explanation:

\theta = Angle

m = Number

d = Distance

n = Refractive index of liquid = 1.25

a denotes air

l denotes liquid

In the case of double split interferance we have the relation

m\lambda=dsin\theta

For air

m\lambda_a=dsin\theta_a

For liquid

m\lambda_l=dsin\theta_l

Dividing the two equations

\frac{m\lambda_a}{m\lambda_l}=\frac{dsin\theta_a}{dsin\theta_l}\\\Rightarrow \frac{\lambda_a}{\lambda_l}=\frac{sin\theta_a}{sin\theta_l}

Wavelength ratio = n

n=\frac{sin\theta_a}{sin\theta_l}\\\Rightarrow \frac{sin0.48}{1.25}=sin\theta_l\\\Rightarrow \theta_l=sin^{-1}\frac{sin0.48}{1.25}\\\Rightarrow \theta_l=0.38^{\circ}

The angular separation is 0.38°

5 0
3 years ago
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8 0
2 years ago
A 0.250 kg block on a vertical spring with spring constant of 4.45 ✕ 103 N/m is pushed downward, compressing the spring 0.080 m.
Elza [17]

Answer:h=5.81 m

Explanation:

Given

Mass of block(m)=0.250 kg

Spring Constant k=4.45\times 10^3 N/m

Initial elongation =0.080 m=8 cm

Thus Initial Potential Energy stored =Final Potential Energy stored in Block

P_i=\frac{kx^2}{2}

P_i=\frac{4.45\times 10^3\times 64\times 10^{-4}}{2}=14.24 J

P_f=mgh=0.25\times 9.8\times h

P_i=P_f

14.24 =0.25\times 9.8\times h

h=\frac{14.24}{0.25\times 9.8}=5.81 m

6 0
2 years ago
A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid
Galina-37 [17]

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm=\frac{6}{100}=0.06 m

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

dF=IBdr

dF=IBdr

\tau=rdF=IBrdr

\tau=\int_{0}^{R}IBr dr

\tau=IB(\frac{R^2}{2}

Torque due to friction

\tau=R\times F

Where friction force=F

R\times F=\frac{IBR^2}{2}

F=\frac{IBR}{2}

Substitute the values

F=\frac{3\times 1\times 0.06}{2}

F=0.09 N

7 0
3 years ago
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