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777dan777 [17]
3 years ago
13

Describe the role of C-S-H in providing strength for cement. Discuss which compounds produce C-S-H and why balancing the amounts

of those compounds is important.
Engineering
1 answer:
wariber [46]3 years ago
8 0

Answer:

Following are the solution to this question:

Explanation:

Whenever a chemical reaction occurs between water and cement the heat is released, and a CaOSiO_2H_2O (C-S-H gel) gel constructs gel is also recognized as "tobermorite gel."

This one gel acts like a pack of gum and also has a cement quality, that holds its particles intact and therefore contributes to the overall compression mix. An increase in supply explicitly causes the movement in the outcome of power. C3S and C2S are both the compounds of Bouge that produce hydration C-S-H gel.

It mixture must be balanced as Ca(OH)_2 with C-S-H gel also is given as a byproduct. It Ca(OH)_2 , that cause sudden with sulphate and form CaSO_4 , is an unacceptable substance. Sulfate attack or later deterioration of its cement is caused by this CaSO_4 .

All C3S and C2S generate various amounts of C-S-H gel so, the required strength can be maintained without compromising on real term durability.

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If you were to plot the voltage versus the current for a given circuit, what would you expect the slope of the line to be? If no
Brut [27]

Answer:

Part 1: It would be a straight line, current will be directly proportional to the voltage.

Part 2: The current would taper off and will have negligible increase after the voltage  reaches a certain  value. Graph attached.

Explanation:

For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.

V=I*R

where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.

In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.

7 0
3 years ago
A battery is connected to a resistor. Increasing the resistance of the resistor will __________. A battery is connected to a res
belka [17]

Answer: the increase in the external resistor will affect and decrease the current in the circuit.

Explanation: A battery has it own internal resistance, r, and given an external resistor of resistance, R, the equation of typical of Ohm's law giving the flow of current is

E = IR + Ir = I(R + r)........(1)

Where IR is the potential difference flowing in the external circuit and Or is the lost voltage due to internal resistance of battery. From (1)

I = E/(R + r)

As R increases, and E, r remain constant, the value (R + r) increases, hence the value of current, I, in the external circuit decreases.

8 0
3 years ago
Write the heat equation for each of the following cases:
jok3333 [9.3K]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

\dfrac{\partial^2T}{\partial x^2}=  \ 0  \  ;  \ if \  T = f(x)  \\ \\ \dfrac{\partial^2T}{\partial y^2}=  \ 0  \  ;  \ if \  T = f(y)  \\ \\ \dfrac{\partial^2T}{\partial z^2}=  \ 0  \  ;  \ if \  T = f(z)

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}

where;

Q_g = heat generated per unit volume

\alpha = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0

where;

The radial directional term = \dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) and the axial directional term is \dfrac{\partial^2 T}{\partial z^2 }

d) The heat equation for a wire going through a furnace is:

\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]

since;

the steady-state is zero, Then:

\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]'

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}

4 0
3 years ago
The common type of defects found when soldering on a printed circuit board
sammy [17]

Explanation:

Solder Bridges

Plating Voids

Non-wetting or dewetting.

5 0
3 years ago
Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclin
Alborosie

Answer:

a) Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

b) principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Explanation:

Combined normal stress and shear stress  sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

бy =  tensile stress in y direction

Txy = y component of shear stress acting on the perpendicular plane to x axis

бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

A) Normal and shear stresses on inclined plane

Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

B) principal and maximum shear stresses

principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

6 0
2 years ago
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