The heat loss from a boiler is to be held at a maximum of 900Btu/h ft2 of wall area. What thickness of asbestos (k= 0.10 Btu/h ft ℉) is required if the inner and outer surfaces of the insulation are to be 1600 and 500℉, respectively? Now if a 3-in.-thick layer of kaolin brick (k= 0.07 Btu/h ft ℉) is added to the outside of the asbestos, what heat flux will be result if the outside surface of the kaolin is 250℉? What will be the temperature at the interface between the asbestos and kaolin for this condition?

Answers

Answer 1

Answer:

a. 0.122 ft b. -70 Btu/h ft² c. 633.33 °F

Explanation:

a. Since the rate of heat loss dQ/dt = kAΔT/d where k = thermal conductivity, A = area, ΔT = temperature gradient and d = thickness of insulation.

Now [dQ/dt]/A = kΔT/d

Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), ΔT = T₂ - T₁ = 500 °F - 1600 °F = -1100 °F. We need to find the thickness of asbestos, d. So,

d = kΔT/[dQ/dt]/A

d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²

d = 0.122 ft

b. If the 3 in thick Kaolin is added to the outside of the asbestos, and the outside temperature of the asbestos is 250℉, the heat loss due to the Kaolin is thus

[dQ/dt]/A = k'ΔT'/d'

k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft

[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft

[dQ/dt]/A  = -70 Btu/h ft²

c. To find the temperature at the interface, the total heat flux equals the individual heat loss from the asbestos and kaolin. So

[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d' where  [dQ/dt]/A = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), k' = 0.07 Btu/h ft ℉(for Kaolin), T₁ = 1600 °F, T₂ = unknown and T₃ = 250℉.

Substituting these values into the equation, we have

-900Btu/h ft² = 0.10 Btu/h ft ℉(T₂ - 1600 °F)/0.122 ft + 0.07 Btu/h ft ℉(250℉ - T₂)/0.25 ft

-900Btu/h ft² = 0.82 Btu/h ft ℉(T₂ - 1600 °F) + 0.28Btu/h ft ℉(250℉ - T₂)

-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82T₂  - 1312°F + 70 °F - 0.28T₂

collecting like terms, we have

-900 °F + 1312°F - 70 °F = 0.82T₂   - 0.28T₂

342 °F = 0.54T₂

Dividing both sides by 0.54, we have

T₂ = 342 °F/0.54

T₂ = 633.33 °F

Answer 2

The thickness of asbestos required is 0.122 ft.

The heat flux will be -70 Btu/h ft²

And the temperature of the interface is 633.33 °F.  

(i) the rate of heat loss :

dQ/dt = kAΔT/d

where k = thermal conductivity, A = area, ΔT = temperature gradient, and

d = thickness of insulation.

[dQ/dt]/A = kΔT/d

Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft²,

k = 0.10 Btu/h ft ℉,

ΔT = 500 °F - 1600 °F = -1100 °F

We have to find the thickness of asbestos that is d.

d = kΔT/[dQ/dt]/A

d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²

d = 0.122 ft is the thickness required.

(ii) a 3-in thick Kaolin is added to the outside of the asbestos

outside temperature of the asbestos is 250℉,

the heat loss due to the Kaolin is:

[dQ/dt]/A = k'ΔT'/d'

k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft

[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft

[dQ/dt]/A  = -70 Btu/h ft²

(iii) temperature at the interface

the total heat flux :

[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d'

where  [dQ/dt]/A = -900 Btu/h ft²,

k = 0.10 Btu/h ft ℉  (for asbestos),

k' = 0.07 Btu/h ft ℉  (for Kaolin),

T₁ = 1600 °F and T₃ = 250℉.

-900 = 0.10(T₂ - 1600 °F)/0.122 + 0.07(250℉ - T₂)/0.25

-900 = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82T₂  - 1312°F + 70 °F - 0.28T₂

-900 °F + 1312°F - 70 °F = 0.82T₂   - 0.28T₂

342 °F = 0.54T₂

Dividing both sides by 0.54, we have

T₂ = 342 °F/0.54

T₂ = 633.33 °F

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Answers

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edge2020

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Answers

Answer:

A) 12.08 m/s

B) 19.39 m/s

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g(sinθ) – μ_k(g cosθ) = a

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v = √(v_o² + 2ad)

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velocity is 1 m/s

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A water-skier is pulled behind a boat by a rope that is at an angle of 13° and has a tension of 490 N. The water-skier has a mass of 49 kg.

What are the magnitudes of the x- and y- components of the tension?

What is the normal force acting on the skier?

Answers

Answer:

normal force= 480.2 x component=477.4  y component= 110.4

Explanation:

I'm sorry I just did this today in class but i think normal force has to be equal to mg or the mass * gravity of the person and that would be 480.2 newtons. I multiplied 9.8 (gravity) by the mass of 49g

I think for the x and y components you can make a triangle with the angled string and then use sohcahtoa to solve for the tension. I used cos(13) = x/490 to solve for x component and then I used the pythagoran theroem to get the remaining side which would be : a^2 + 477.4^2 = 490^2

hope this can help and that it is correct. good luck

The magnitudes of the x- and y- components of the tension is 477.44N and 110.23N

The normal force acting on the skier is 480.2N

The magnitude of the x- and y- components of the tension is expressed as:

x = Tcos 13°y = T sin  13°

Given the following

Tension T = 490N

x = 490 cos13° = 477.44N

y = 490sin13° =  110.23N

Hence the magnitudes of the x- and y- components of the tension is 477.44N and 110.23N

The normal force acting on the skier is equal to the weight.

N = W = mg

N = W = 49 * 9.8

N = W = 480.2N

Hence the normal force acting on the skier is 480.2N

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Answers

Answer:

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Group of answer choices

A.Because atoms have the same number of protons and electrons

B. Because atoms want 8 valence electrons

C. To make our lives difficult

D. Because atoms gained neutrons

Answers

Answer:

B. Because atoms want 8 valence electrons

Explanation:

Ions are formed because atoms want 8 valence electrons. Most atoms attain stability when they 8 electrons in their outermost shell.

This is the reason why they form ions.

To form ions, atoms will gain or lose electrons or share them. This loss, gain, or sharing of electrons often lead to an atom become stable. One group of atoms, the noble gases, do not form ions readilyThey have complete numbers of electrons in their valence shells. So they do not gain, lose or share their electrons to form ions unlike the others.

what is the acceleration of an object that entered of a force 200 N if the object mass was 100 kg?

Answers

Answer:

The answer is 2 m/s²

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

where

f is the force

m is the mass

From the question we have

[tex]a = \frac{200}{100} \\ [/tex]

We have the final answer as

2 m/s²

Hope this helps you

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