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Ugo [173]
3 years ago
12

Define initial set and final set. Briefly discuss one method used to determine them. The following laboratory tests are performe

d:
a. Setting time test of cement paste samples.
b. Compressive strength of mortar cubes.

What are the significance and use of each of these tests?
Engineering
1 answer:
vagabundo [1.1K]3 years ago
3 0

Answer:

Answer in explanation

Explanation:

Initial setting time is the time when the cement starts losing plasticity or begins hardening after addition of water to it.

Final setting time is the time when the cement has completly lost it's plasticity and has set and became hard after addition of water.

The test carried out is vicat's penetration test

The initial setting time can be defined as “time taken by paste to stiffen to such an extent that the Vicatt’s needle is not permitted to move down through the paste through 25 mm.”.

The final setting time can be defined as “ it is the time after which the paste becomes so hard that Vicatt’s 5mm needle doesn’t sinks visibly and leave no impression”.

The importance and use of setting time test of cement paste samples

The importance of setting time test is that it plays a very vital role in the fresh and used(hardened) state of concrete.

It plays a very important role in mixing, transportation, shrinkage, compacting, finishing and removal of formwork

The importance of compressive strength of mortar cubes is that it is the compressive strength that tells us about the performance characteristics of the binder of the concrete which is the cement in this case. The performance characteristics deals with the overall quality of the finished product. Hence, a cement or mortar with good compressive strength is good to ensure the delivery of quality work in the final concrete.

It is important to note that there are applicable standards in form of numbers for compressive strength of a cement or mortar

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One kilogram of water fills a 0.140 m^3 rigid container at an initial pressure of 1.8 MPa. The container is then cooled to 40°C.
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Answer:

T1 = 299.18 °C

P2 = 0.00738443 MPa

Explanation:

From the data, we can get two properties for the initial condition. These are pressure and specific volume.

The pressure is 1.8 MPa and the specific volume, we can get it with the mass and volume of the container, since it’s filled this is also the volume of the water in it.

v=\frac{vol (m^{3})}{mass (kg)} = \frac{0.140 m^{3}}{1 kg} = 0.140 \frac{ m^{3}}{kg}

When we check in the thermodynamic tables, the conditions for saturation at 1.8 MPa we found the following:

P^{sat} = 1.8 MPa

T^{sat} = 207.12 C

v_{g} = 0.1103\frac {m^{3}}{kg} specific volume for the saturated vapor  

v_{l} = 0.001167 \frac{m^{3}}{kg} specific volume for the saturated liquid  

Since the specific volume in our condition is higher that the specific volume for the saturated vapor, we have a superheated steam.  

Looking in the thermodynamic tables for superheated steam we found that the temperature where the steam has a specific volume of 0.140 \frac{ m^{3}}{kg} at 1.8 MPa is 299.18 °C. This is the initial temperature in the container.

Since the only information that we have about the final condition is that the container was cooled. We can assume that it was cooled until a condition of saturation. So, the final pressure for the water will be the pressure of saturation for a temperature of 40°C. From thermodynamic tables we get:

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T_{P}=T_{C}+(n).(m).(p).(T_{D})

Where T_{P} is the production rate for the assembly machine.

Where T_{C} is the ideal cycle time

Where n is the number of stations.

Where m is the number stations that get jam when the defect occurs.

Where p is the defect rate at each station.

And where T_{D} is the average downtime per breakdown

We are looking for the hourly production rate ⇒

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6s=\frac{(6s)(1h)}{(3600s)}= \frac{1}{600}h

60min=1h ⇒

1.2min=\frac{(1.2min)(1h)}{(60min)}=0.02h

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