What are the reasons why fine grained of alkali igneous rocks can not be used in cement

1 answer:

8 0

Fine grained of alkali igneous rocks cannot be used in cement because of the volume expansion caused by the Alkali-silica reaction, fine-grained igneous rocks cannot be used as aggregates in cement.

<h3>What does fine grained mean in an igneous rock?</h3>

Extrusive igneous rocks have a fine-grained or aphanitic texture, with grains that are too small to see without a magnifying glass. The fine-grained texture suggests that the rapidly cooling lava did not have enough time to form large crystals. A petrographic microscope can be used to examine these tiny crystals.

The texture of an igneous rock (fine-grained vs coarse-grained) is determined by the rate at which the melt cools: slow cooling produces large crystals, while fast cooling produces small crystals.

The chemical reaction that occurs in both alkali cations and hydroxyl ions in the pore solution of hydrated cement paste and certain reactive silica phases present in concrete aggregates is known as the alkali-silica reaction (ASR).

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Vesnalui [34]

<h3><u>CSMA/CD Protocol: </u></h3>

Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.

But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.

There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.

<u>Example: </u>

$T_{P}=1 H r$ , at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.

12:00 AM A will see collisions

Pocket Size to detect the collision.

$\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}$

CSMA/CD is widely used in Ethernet.

<u>Efficiency of CSMA/CD:</u>

In the previous example we have seen that in worst case $2 T_{P}$ time require to detect a collision.

There could be many collisions may happen before a successful completion of transmission of a packet.

We are given number of collisions (contentions slots)=4.

$\text { Propagation day }=\frac{\text {distance}}{\text {speed}}$

Distance = 1km = 1000m

$\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}$