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Lina20 [59]
3 years ago
5

The gradual and uniform cooling of glass that has been heated in order to relieve random internal stresses is... Group of answer

choices annealing tempering quenching heat treating
Engineering
1 answer:
kirill115 [55]3 years ago
7 0

Answer: Annealing

Explanation:

In the manufacturing of glass, Processes such as expansion when glass heats and cooling when it contracts are involved. These build up random internal stresses which can cause cracks and breakage when finally cooled at room temperature. To relieve these stresses, Annealing is carried out on glass during the manufacturing process.

Annealing is the process whereby the temperature during the process of the manufacturing of glass is reduced and maintained at a uniform rate so that it undergoes a gradual cooling process. so as to make the glass tough and reduce internal stress.

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Asolid rectangular rodhas a length of 90mm, made of steel material (E =207,000 MPa, Syield= 300 MPa), the cross section of the r
erastova [34]

Answer:

13.6mm

Explanation:

We consider diameter to be a chord that runs through the center point of the circle. It is considered as the longest possible chord of any circle. The center of a circle is the midpoint of its diameter. That is, it divides it into two equal parts, each of which is a radius of the circle. The radius is half the diameter.

See attachment for the step by step solution of the problem

3 0
3 years ago
A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas
ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × 10^{5} Pa

P (gauge ) = 2 × 10^{5} Pa

so from idea gas equation

\frac{P1*V1}{T1} = \frac{P2*V2}{T2}   ................1

so {P2} = \frac{P1*T2}{T1}

{P2} = \frac{2*10^5*350}{300}

P2 = 2.33 × 10^{5} Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × 10^{5}  - 1.0 × 10^{5}

gauge pressure = 1.33 × 10^{5} Pa

so gauge pressure is 133 kPa

4 0
3 years ago
Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the
alex41 [277]

Answer:

The mass flow rate of cooling water required to cool the refrigerant is 123.788\,\frac{kg}{min}.

Explanation:

A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:

\dot Q_{ref} - \dot Q_{w} = 0

\dot Q_{ref} = \dot Q_{w}

\dot m_{ref}\cdot (h_{ref, in} - h_{ref,out}) = \dot m_{w}\cdot (h_{w, out} - h_{w,in})

The mass flow rate of the cooling water is now cleared:

\dot m_{w} = \dot m_{ref }\cdot \frac{h_{ref,in}-h_{ref,out}}{h_{w,out}-h_{w,in}}

Given that h_{ref,in} = 808.34\,\frac{kJ}{kg}, h_{ref, out} = 88.82\,\frac{kJ}{kg}, h_{w,out} = 104.83\,\frac{kJ}{kg} and h_{w,in} = 62.98\,\frac{kJ}{kg}, the mass flow of the cooling water is:

\dot m_{w} = \left(7.2\,\frac{kg}{min} \right)\cdot \left(\frac{808.34\,\frac{kJ}{kg}-88.82\,\frac{kJ}{kg} }{104.83\,\frac{kJ}{kg}-62.98\,\frac{kJ}{kg} } \right)

\dot m_{w} = 123.788\,\frac{kg}{min}

The mass flow rate of cooling water required to cool the refrigerant is 123.788\,\frac{kg}{min}.

4 0
3 years ago
Astronauts who landed on the moon during the Apollo 15, 16, and 17 missions brought back a large collection of rocks to the eart
stiks02 [169]

Answer: a) W(earth) = 935.62 lbs

b) Mass of rocks in slugs = 29.06 slugs

Explanation:

a) From Newton's law, W = mg. Whether on the moon or on earth. Although, the mass of the rocks everywhere is the same, that is, mass of rocks on the moon = mass of rocks on earth.

W(moon) = mg(moon)

W(moon) = 154 lbs

g(moon) = 5.30 ft/s2

m = W(moon)/g(moon) = 154/5.3 = 29.06 lb.s2/ft

W(earth) = m g(earth)

g(earth) = 32.2 ft/s2

W(earth) = 29.06 × 32.2 = 935.62 lbs.

b) A slug = 1 lb.s2/ft, therefore the mass of the rocks in slugs is 29.06 slugs.

QED!

8 0
3 years ago
an inclined manometer is connected to a pitor tube to measure the velocity at the center of a circular duct. If the inclined man
jekas [21]
57.5 m/s
I did 2.3/0.04
I’m not sure if it’s correct though
8 0
3 years ago
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