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loris [4]
3 years ago
12

Write short notes on: (any four) a) Suspended ground floor b) Soil exploration c) Baulking of sand d) Bearing capacity of soil e

) Pointing DI INC CON​
Engineering
1 answer:
vredina [299]3 years ago
6 0

Answer:

a) A suspended floor is a ground floor with a void underneath the structure. The floor can be formed in various ways, using timber joists, precast concrete panels, block and beam system or cast in-situ with reinforced concrete. However, the floor structure is supported by external and internal walls.

b) Soil exploration consists of determining the profile of the natural soil deposits at the site, taking the soil samples and determining the engineering properties of soils using laboratory tests as well as in-situ testing methods

c) Bulking in sand Occurs When dry sand interacts with the atmospheric moisture. Presence of moisture content forms a thin layer around sand particles. This layer generates the force which makes particles to move aside to each other. This results in the increase of the volume of sand.

d) In a nutshell, bearing capacity is the capacity of soil to support the loads that are applied to the ground above. It depends primarily on the type of soil, its shear strength and its density. It also depends on the depth of embedment of the load – the deeper it is founded, the greater the bearing capacity.

Explanation:

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A furnace wall is to be built of 20-cm firebrick and building (structural) brick of same thickness. The thermal conductivities o
Norma-Jean [14]

Answer:

q=2313.04W/m^2

T=690.86°C

Explanation:

Given that

Thickness t= 20 cm

Thermal conductivity of firebrick= 1.6 W/m.K

Thermal conductivity of structural brick= 0.7 W/m.K

Inner temperature of firebrick=980°C

Outer temperature of structural brick =30°C

We know that thermal resistance

R=\dfrac{t}{KA}

These are connect in series

R=\left(\dfrac{t}{KA}\right)_{fire}+\left(\dfrac{t}{KA}\right)_{struc}

R=\dfrac{0.2}{1.6A}+\dfrac{0.2}{0.7A}\ K/W

R=\dfrac{23}{56A}\ K/W

Heat transfer

Q=\dfrac{\Delta T}{R}

Q=56A\times \dfrac{980-30}{23}\ W

So heat flux

q=2313.04W/m^2

Lets temperature between interface is T

Now by equating heat in both bricks

\dfrac{980-T}{\dfrac{0.2}{1.6A}}=\dfrac{T-30}{\dfrac{0.2}{0.7A}}

So T=690.86°C

6 0
3 years ago
Xyxyydfufggivivihogcufuf​
Genrish500 [490]

Answer:

ummm why is you doing this

Explanation:

It doesnt make sense.

7 0
4 years ago
What do we need to do to get CO2 emissions all the way to zero?
Svetlanka [38]

Answer:

A key element is powering economies with clean energy, replacing polluting coal - and gas and oil-fired power stations - with renewable energy sources, such as wind or solar farms. This would dramatically reduce carbon emissions. Plus, renewable energy is now not only cleaner, but often cheaper than fossil fuels

Explanation:

here is your answer if you like my answer please follow

3 0
3 years ago
Calculate the rate at which body heat is conducted through the clothing of a skier in a steady- state process, given the followi
olga2289 [7]

Answer:

230.4W

Explanation:

Heat transfer by conduction consists of the transport of energy in the form of heat through solids, in this case a jacket.

the equation is as follows

Q=\frac{KA(T2-T1)}{L} \\

Where

Q=heat

k=conductivity=0.04

A=Area=1.8m^2

T2=33C

T1=1C

L=thickness=1cm=0.01mQ=\frac{(0.04)(1.8m^2)(33-1)}{0.01m}

Q=230.4W

the skier loses heat at the rate of 230.4W

4 0
4 years ago
. En la facultad de Ingeniería Industrial se realizó una encuesta a 200 personas para saber que lenguaje de programación preferí
GarryVolchara [31]

Answer:

lalalalapumpe

Explanation:

4 0
3 years ago
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