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

Luden [163]

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

6 0

4 years ago

The head difference between the inlet and outlet of a 1km long pipe discharging 0.1 m^3/s of water is 0.53 m. If the diameter is

Reptile [31]

Answer:

The correct option is 'e': f = 0.05.

Explanation:

The head loss as given by Darcy Weisbach Equation is as

$h_{l}=\frac{flv^{2}}{2gD}$

where

$h_{l}$ is head loss in the pipe

'f' is the friction factor

'l' is the length of pile

'v' is the velocity of flow in pipe

'D' is diameter of pipe

From equation of contuinity we have $v=\frac{Q}{A}$

Thus using this in darcy's equation we get

$h_{l}=\frac{flQ^{2}}{2gDA}$

where

'Q' is discharge in the pipe

'A' is area of the pipe $A=\frac{\piD^2}{4}$

Applying the given values we get

$h_{l}=\frac{8flQ^{2}}{\pi ^{2}gD^{5}}$

Solving for 'f' we get

$f=\frac{0.53\times \pi ^{2}\times 9.81\times 0.6^{5}}{1000\times 0.1^{2}\times 8}\\\\f=0.05$

7 0

3 years ago

Lawn maintenance is an alternative energy source<br> -true<br> -false

Reika [66]

Answer:

false

Explanation:

4 0

3 years ago

Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia

Mrrafil [7]

Answer:

The fin temperature in °C at a distance of 10 cm from the base = 33.78°C

Explanation:

The following assumptions will be made to solve this problem

- The heat transfer coefficient does not change with the time or distance.

- The temperature of the fins varies just in only one direction.

The temperature of the fin at x = 10 cm = 0.10 m from the base can be calculated from the temperature variation with distance formula for a very long fin.

(T - T∞) = (T₀ - T∞)e⁻ᵐˣ

T = T(x) = temperature at any point along the fin

T∞ = temperature at the tip of the fin = ambient temperature = 25°C

T₀ = temperature at the base of thw fin = 50°C

x = any distance along the length of the fin from the base of the fin = 0.1 m

m = √(hP/KA)

h = Heat transfer coefficient = 123 W/m².K

P = perimeter in contact with the base = πD = π × 0.03 = 0.0943 m

K = thermal conductivity = 150 W/m.K

A = surface area in contact with the base = πD²/4 = π(0.03)²/4 = 0.0007071 m²

m = √(123 × 0.0943)/(150 × 0.0007071)

m = 10.46

mx = 10.46 × 0.1 = 1.046

(T - 25) = (50 - 25) e⁻¹•⁰⁴⁶

T = 25 + 25 e⁻¹•⁰⁴⁶ = 25 + 8.78 = 33.78°C

8 0

3 years ago

Technician A says that latent heat is hidden heat and cannot be measured on a thermometer. Technician B says that latent heat is

AysviL [449]

Answer: C

Both A and B are correct

Explanation:

Latent heat is the hidden heat.

Latent heat is the heat energy required to change one state of matter to another state of matter without change in temperature. For example, solid state to liquid state, or liquid state to gaseous state.

Thermometer can not detect the latent heat. That is why it is called hidden heat.

If Technician A says that latent heat is hidden heat and cannot be measured on a thermometer. And Technician B says that latent heat is hidden heat that is required for a change of state of matter, then we can therefore conclude that both Technician A and Technician B are correct.

5 0

3 years ago