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cupoosta [38]
3 years ago
11

When storing used oil, it need to be kept in________ container?

Engineering
1 answer:
lisov135 [29]3 years ago
6 0

Answer: In a clean plastic or metal container with a tightly sealed lid

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What are factor of safety for brittle and ductile material
galben [10]

Explanation:

Step1

Factor of safety is the number that is taken for the safe design of any component. It is the ratio of failure stress to the maximum allowable stress for the material.

Step2

It is an important parameter for design of any component. This factor of safety is taken according to the environment condition, type of material, strength, type of component etc.

Step3

Different material has different failure stress. So, ductile material fails under shear force. Ductile material’s FOS is based on yield stress as failure stress as after yield point ductile material tends to yield. Brittle material’s FOS is based on ultimate stress as failure stress.

The expression for factor of safety for ductile material is given as follows:

FOS=\frac{\sigma_{yp}}{\sigma_{a}}

Here,\sigma_{f} is yield stress and \sigma_{a} is allowable stress.

The expression for factor of safety for brittle material is given as follows:

FOS=\frac{\sigma_{ut}}{\sigma_{a}}

Here,\sigma_{ut} is ultimate stress and \sigma_{a} is allowable stress.

5 0
3 years ago
A rigid 10-L vessel initially contains a mixture of liquid and vapor water at 100 °C, with a quality factor of 0.123. The mixtur
masya89 [10]

Answer:

Q_{in} = 46.454\,kJ

Explanation:

The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.

Q_{in} = U_{2} - U_{1}

Properties of water at initial and final state are:

State 1 - (Liquid-Vapor Mixture)

P = 101.42\,kPa

T = 100\,^{\textdegree}C

\nu = 0.2066\,\frac{m^{3}}{kg}

u = 675.761\,\frac{kJ}{kg}

x = 0.123

State 2 - (Liquid-Vapor Mixture)

P = 476.16\,kPa

T = 150\,^{\textdegree}C

\nu = 0.2066\,\frac{m^{3}}{kg}

u = 1643.545\,\frac{kJ}{kg}

x = 0.525

The mass stored in the vessel is:

m = \frac{V}{\nu}

m = \frac{10\times 10^{-3}\,m^{3}}{0.2066\,\frac{m^{3}}{kg} }

m = 0.048\,kg

The heat transfer require to the process is:

Q_{in} = m\cdot (u_{2}-u_{1})

Q_{in} = (0.048\,kg)\cdot (1643.545\,\frac{kJ}{kg} - 675.761\,\frac{kJ}{kg} )

Q_{in} = 46.454\,kJ

3 0
2 years ago
Print reading for industry unit 9 review questions
RideAnS [48]
Hhhhhhhhhhhhhhh. High
4 0
3 years ago
Read 2 more answers
A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m
natima [27]

Answer:

a. 0.4544 N

b. 5.112 \times 10^{-5 M}

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O

NaOH\ Mass = Normality \times equivalent\ weight \times\ volume

= 0.3200 \times 40 g \times 21.30 mL \times  1L/1000mL

= 0.27264 g

NaOH\ mass = \frac{mass}{molecular\ weight}

= \frac{0.27264\ g}{40g/mol}

= 0.006816 mol

Now

Moles of H_2SO_4 needed  is

= \frac{0.006816}{2}

= 0.003408 mol

Mass\ of\ H_2SO_4 = moles \times molecular\ weight

= 0.003408\ mol \times 98g/mol

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

= \frac{acid\ mass}{equivalent\ weight \times volume}

= \frac{0.333984 g}{49 \times 0.015}

= 0.4544 N

b. And, the acid solution molarity is

= \frac{moles}{Volume}

= \frac{0.003408 mol}{15\ mL \times  1L/1000\ mL}

= 0.00005112

=5.112 \times 10^{-5 M}

We simply applied the above formulas

4 0
3 years ago
Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s
ioda

Answer: r = 0.8081; s = -0.07071

Explanation:

A = (150i + 270j) mm

B = (300i - 450j) mm

C = (-100i - 250j) mm

R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

270r - 450s - 250 = 0

150r + 300s = 100

270r - 450s = 250

solving simultaneously,

r = 0.8081 and s = -0.07071

QED!

5 0
3 years ago
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