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aleksklad [387]
4 years ago
12

Explain the advantages of continuous insulation in envelope assemblies and why it is better to have the continuous insulation on

the exterior face of the envelope. Where is inside placement of continuous insulation in the envelope desirable
Engineering
1 answer:
Luden [163]4 years ago
6 0

Answer:

Continuous insulation helps in eliminating the  necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.

Continuous insulation helps building last much longer over a period of time,without the need to upgrade and repair.

Inside the placement of continuous insulation in the envelope needed includes the following; the foundation wall or slab insulation, balcony interface, bond joist insulation, insulation against sub floor, on the interior of masonry wall.

Explanation:

Solution:

The analytical path to success contains a well planned and designed building exterior, which avoids the loss of energy, control cost and the maximization of technology advancement in materials.

Property installed continuous insulation on the exterior can also execute as an air barrier. the flashing of wall penetrations can form into a drainage plane. this plane can stop potentially damaging moisture from entering into the wall assembly.

By making use of continuous insulation it helps in removing necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.

Continuous insulation helps building to withstand the test of time without the need to upgrade and repair

Now inside the placement of continuous insulation in the envelope is needed as follows:

  • Foundation wall or slab insulation
  • Insulation against sub floor
  • Balcony interfaces
  • On the interior of masonry wall
  • Bond  joist insulation
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A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur
Amanda [17]

Question in order:

A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?

Mirror Radius (mm) Bending Failure Stress (MPa)

0.603                         225

0.203                         368

0.162                         442

Answer:

191 MPa

Explanation:

Failure stress of bending is Inversely proportional to the mirror radius

Bending Stress = \frac{1}{(Mirror Radius)^{n}}

At mirror radius 1 = 0.603 mm   Bending stress 1 = 225 Mpa..............(1)

At mirror radius 2 = 0.203 mm  Bending stress 2 = 368 Mpa...............(2)

At mirror radius 3 = 0.162 mm   Bending stress 3 = 442 Mpa...............(3)

comparing case 1 and 2 using the above equation

\frac{Stress 1}{Stress 2} = ({\frac{Radius 2}{Radius 1}})^{n_1}

\frac{225}{368} = ({\frac{0.203}{0.603}})^{n_1}

0.6114 = (0.3366)^{n_1}

Taking the natural logarithm of both side

ln(0.6114) = n ln(0.3366)

n₁ = ln(0.6114)/ln(0.3366)

n₁ = 0.452

comparing case 2 and 3 using the above equation

\frac{Stress 2}{Stress 3} = ({\frac{Radius 3}{Radius 2}})^{n_2}

\frac{368}{442} = ({\frac{0.162}{0.203}})^{n_2}

0.8326 = (0.7980)^{n_2}

Taking the natural logarithm of both side

ln(0.8326) = n ln(0.7980)

n₂ = ln(0.8326)/ln(0.7980)

n₂ = 0.821

comparing case 1 and 3 using the above equation

\frac{Stress 1}{Stress 3} = ({\frac{Radius 3}{Radius 1}})^{n_3}

\frac{225}{442} = ({\frac{0.162}{0.603}})^{n_3}

0.5090 = (0.2687)^{n_3}

Taking the natural logarithm of both side

ln(0.5090) = n ln(0.2687)

n₃ = ln(0.5090)/ln(0.2687)

n₃ = 0.514

average for n

n = \frac{n_1 + n_2 + n_3}{3}

n = \frac{0.452 +0.821 + 0.514}{3}

n = 0.596

Hence to get bending stress x at mirror radius 0.796

\frac{Stress x}{Stress 3} = ({\frac{Radius 3}{Radius x}})^{0.596}

\frac{Stress x}{225} = ({\frac{0.603}{0.796}})^{0.596}

\frac{Stress x}{225} = 0.8475

stress x = 191 MPa

3 0
4 years ago
One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter
lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150×10^{-3} m^{3}

m = 1 Kg

P_{1} = 2 MPa

T_{2}  = 40°C

The waters specific volume is calculated:

v_{1} = V/m

Here, the waters specific volume at initial condition is v_{1}, the containers volume is V, waters mass is m.

v_{1} = 150×10^{-3} m^{3}/1

v_{1} = 0.15 m^{3}/ Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 m^{3}/ Kg and 0.13 m^{3}/ Kg.

T_{1}= 350+(400-350) \frac{0.15-0.13}{0.1522-0.1386}

T_{1} = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

v_{2} = v_{1}= 0.15 m^{3}/ Kg

In saturated water labels for T_{2}  = 40°C.

v_{f} = 0.001008 m^{3}/ Kg

v_{g} = 19.515 m^{3}/ Kg

The final state is two phase region v_{f} < v_{2} < v_{g}.

In saturated water labels for T_{2}  = 40°C.

P_{2} = P_{Sat} = 7.3851 kPa

P_{2} = 7.3851 kPa

7 0
3 years ago
Question #4
mihalych1998 [28]

Answer:

C. they need on the job experience I think

7 0
3 years ago
Will give brainliest to whoever answers correctly!!!!
frosja888 [35]

Answer:

B) m<1 + m<2 < m<2 + m<3 + m<4

5 0
3 years ago
Read 2 more answers
A sinusoidal voltage source produces the waveform, v t = 1 + cos 2πft. Design a system with v t as its input such that an LED wi
DerKrebs [107]

Answer:

See explaination

Explanation:

LM358 is the useful IC which works as buffer. It enables circuit to remove overloading effect on each other. Image is in attachment.

We can define a light-emitting diode (LED) as a semiconductor light source that emits light when current flows through it. Electrons in the semiconductor recombine with electron holes, releasing energy in the form of photons

See attached file for detailed solution of the given problem.

3 0
3 years ago
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