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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Answer:

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Answer:

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Answers

do u have to find z? how do yk what z is?

Answer:

(if for solving for z)

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Explanation:

az-(b+c) [set equal to zero]

az-(b+c)=0

+(b+c) +(b+c)

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distillation is correct

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Answers

Answer:

The x-direction and y-direction refer are used to refer to the location of an object as well as to the components of the motion of the object in a two dimensional space

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In physics, in the analysis of the motion of an object in two dimensional space, the relative location of the object is specified using the measured distances in the x-direction from the y-axes and y-direction from the x-axes

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You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill (Fig. 6-25). You find that the slope of the hill is u " 12.03, that the cars were separated by distance d " 24.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 " 18.0 m/s.With what speed did car A hit car B if the coefficient of kinetic friction was (a) 0.60 (dry road surface) and (b) 0.10 (road surface covered with wet leaves)

Answers

Answer:

A) 12.08 m/s

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mg(sinθ) – F_k = ma

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F_k = (μ_k) × F_n

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F_k = μ_k(mg cosθ)

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A man jumps from a 45 m cliff into a large body of water. How long will he fall before he enters the wate

Answers

Answer:

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Explanation:

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Answers

Answer:

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Explanation:

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What is Electromagnet ?

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Answer:

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Answers

Answer:

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Answers

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edge2020

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Answers

Answer:

Tiny particles in solids, liquids and gases are always in motion. It is the motion of particles that creates a form of energy called thermal (heat) energy that is present in all matter. The particles in solids are tightly packed and can only vibrate. Matter in its solid state has the lowest amount of thermal energy (for that type of matter). Because solids have less thermal energy than liquids or gasses, the atoms in their solid state move very little.

Explanation:

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Which of the following substance used to selectively absorb certain frequencies of light and transmit or reflect the rest; responsible for object's colour?
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Answers

Answer:

[tex]D. \: Tyndall \: scattering.[/tex]

Steve stands on a scale and it reads 65 kilograms. What is his weight in Newtons? *

Need help

Answers

Answer:

It would be 637.43 Newtons

Explanation:

Because

1 Kilogram:

The kilogram is defined as being equal to the mass of the International Prototype Kilogram (IPK), which is almost exactly equal to the mass of one liter of water.

1 Newton:

1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth. This is derived using Newton's second law f=ma and assuming Earth gravity of 9.80665 m/s2. 1 N (Earth) = 0.101971621297793 kg.

PLS ANSWER ASAP!! I PROMISE I'LL GIVE YOU BRAINLIEST!!

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.

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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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What is the net force required to accelerate a 5.3 kg box at a rate of 4.2 m/s^2?

Answers

Answer:

The answer is 22.26 N

Explanation:

The force acting on an object given the mass and acceleration we use the formula

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mass = 5.3 kg

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We have

force = 5.3 × 4.2

We have the final answer as

22.26 N

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a car on a circular track accelerates from rest. the radius of the track is 0.30km and the magnitude of the constant angular acceleration is 4.5x10-3 rads/s squared. find the centripetal acceleration of the car when it has completed half of a lap. ​

Answers

Answer:

The centripetal acceleration of the car is 50.4 m/s²

Explanation:

Given;

radius of the track, r = 0.3km = 300 m

constant angular acceleration, α = 4.5 x 10⁻³ rad/s²

during half of a lap, θ = π

Determine time of motion

θ = ωt + ¹/₂αt²

π = 0 + ¹/₂αt²

t² = 2π / α

t² = ( 2 x 3.142) / (4.5 x 10⁻³)

t² = 1396.44

t = √1396.44

t = 37.369 s

Determine the angular velocity

[tex]\omega_f = \omega_i + \alpha t\\\\\omega_f = 0 \ + \ 37.369*4.5*10^{-3}\\\\\omega_f = 0.168 \ rad/s[/tex]

Determine the centripetal acceleration of the car

[tex]\alpha _c = \omega^2 r\\\\\alpha _c = 0.168*300\\\\\alpha _c =50.4 \ m/s^2[/tex]

Therefore, the centripetal acceleration of the car is 50.4 m/s²

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Answers

The spring constant is the number of Newtons needed to cause an extension of 1metre. So in this case it is 15/0.3 = 50 N/m

Answer:

50 on edge

Explanation:

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Answer:

workdone = force x displacement

= 27.5 x 12.3

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Answers

Answer:

3 m/s east

Explanation:

I took a test a got it right

Answer:

3 m/s east

Explanation:

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Answers

Answer:

See the explanation below.

Explanation:

An ammeter is an instrument for measuring electrical currents in a circuit. To measure the current the ammeter must be connected in series with the circuit so that the current flows through it. At present time there are modern instruments such as the amperimetric clamp, which also serves to measure currents in a more easy way

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Answers

Answer:

The principle is that the slope of the line on a velocity-time graph reveals useful information about the acceleration of the object. If the acceleration is zero, then the slope is zero. If the acceleration is positive, then the slope is positive

Explanation:

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Answer:

velocity is 1 m/s

Explanation:

Answer:

Include: The object starts away from the origin.

The object moves toward the origin at a constant velocity.

The object stops for one second.

The object moves away from the origin at a greater constant velocity.

Explanation:

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